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Stoichiometryby Todd Bennett

Section I

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Unit Conversiona) Dimensional Analysisb) Density Determinationc) Typical Conversions

Elemental Analysisa) Mass Percentb) Empirical Formulasc) Molecular Formulasd) Combustion Analysis

Solution Concentrationa) Units and Terminology

i. Molarityii. Molalityiii. Mass Percent (in $olution)iv. Density

b) Dilutionc) Beer's Law

Balancing Keactionsa) Standard Balancingb) Limiting Keagents

Keaction Typesa) Common Reactionsb) Oxidation States

Test.Taking Tipsa) Oeneral Adviceb) Mathematical Tricks

i. Addition and Subtractionii. Averagingiii. Multiplicationlv. L,lvtsron

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SpeciaLiztng in MCAT Preparation

Stoichiometry Section GoalsKnow how to convert one kind of concentration unit into another-o?

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..*S Un<lerstand dilution and its effect on concentration of a solute.w W Dilution involves a reduction in the concentration of a solute in solution by the addition of solvent

( to the mixture. The addition of solvent therefore dilutes the concentration, but does not change themoles of solute. The equation that you must recall is based on the constant number of solute moles:Mittltid'Vir,itial = Mfinal'Vfittal.

standard reactions from chemis

The concentration of a solution can be measured in terms of molarity, molality, and density. Youmust know the definitions of each unit and how they differ from one another. Atthough the testdoes not feature a great deal of math, you should have an idea of how to convert between units.

Understand the difference between empirical and molecular formulas.Know the difference between the molecular formula (actual ratio of atoms in a molecule) and theemoirical formula (simplest whole number ratio of the atoms in a molecule). Be familiar with theexpleriments and inforriation needed to determine both of the formulas.

Know the effect of standard conditions.Standard conditions are defined as 1 atm. and 298 K for thermodynamics, but STP (standardtemperature and pressure) is defined as 1 atm. and273 5. Yily..ulculations of gas volume use theideal gas assump^tion that at STP, one mole of gas occupies 22.4htets.

The most commonlv recurrjng reactions in general chemistry that you are expected to know includecombustion, single'replace*e"nt, double diiplacement, and protori transfer, to name iust a few. Youmust recognize t-hese reactions and have a basic understanding of them.

Recognize the limiting reagent in a reaction and know its effect on the reaction.The limiting reagent dictates the amount of product that can be formed. and consequently the percentyield for a Teu.tlott. Using only starting va'lues and the stoichiometric equation, you inust be ableio determine which reacta"nt is ihe limiting reagent in the reaction.

Ilnderstand the stoichiometric ratios in combustion reactions.In the combustion of both hvdrocarbons and carbohydrates, there is a consistent relation betweenthe number of oxygen moleiules on the reactant side, and the number of water and carbon dioxidemolecules that fo'rtr on the product side. Know each reaction so that you may easily balance thecoefficients.

General Chemistry Stoichiometry Introduction

The perfect spot to start any review of general chemistry is the basics, whichtraditionally include stoichiometry and chemical equations. The mostfundamental perspective of a chemical reaction, where bonds are broken so thatnew bonds can be formed, is at the atomic and molecular levels. Due to theminute size of atoms, we can never actually view a chemical reaction (so statesHeisenberg's uncertainty principle). We must therefore rely upon developingmodels that can account for changes in all of the atoms and molecules involvedin a chemical reaction or physical process. At the molecular level, we considermolecules. At the macroscopic level, we consider moles. Stoichiometry allowsus to convert one into the other and to shift between these two perspectives. Thenumber of molecules is converted into the number of moles using Avogadro'snumber (6.022 x 1023). The concept of a mole is based upon the amount ofcarbon-12 that is contained in exactly 12.0 grams of carbon, a quantitydetermined by knowing the volume of a 12.0-9 carbon sample, the type ofmolecular packing in it, and the dimensions of the carbon atom. This task ofquantifying atoms in a mole is similar to guessing the number of peas that arecontained in an aquarium. It is important that you utilize the mole concept tounderstand, and later to balance and manipulate, chemical equations.

In the stoichiometry section, we focus on those skills needed to solve ratioquestions. Stoichiometry is most commonly thought of as the mathematicalportion of general chemistry. The MCAT, however, has relatively fewcalculations. It is a conceptual test, emphasizing logical thought process ratherthan calculations. For some of you, this is great news. But before celebrating toomuch, consider where the mathematical aspects of general chemistry fit into a

conceptual exam. The MCAT does involve some math, but it is not toocomplicated. Math-related calculations required for MCAT questions involvemaking approximations, determining ratios, setting up calculations, andestimating the effect of errors on results. The initial problems presented in thissection involve slightly more calculations than you should expect to see on theMCAT. Some of them may look familiar to you from your general chemistrycourses and should stimulate your recall. As the section proceeds, less emphasisis placed on calculating and estimating, and more emphasis is placed on the artof quickly determining ratios and approximating values.

The focus of the stoichiometry section is problern-solving, with special attentionto the idiosyncrasies of each type of problem. Definitions of important terms arepresented with sample questions and their solutions. Answer solutions discusstest strategy and the information needed to obtain the correct answer. Eachproblem in the stoichiometry section represents what we might call the "book-keeping" of reactions in general chemistry, and it offers an ideal opportunity tobegin work on fast math skills as well. As you do each of the questions, learn thedefinitions and develop an approach that works well for you. You may want toconsider multiple pathways to arrive at the correct answer. It is important thatvou be able to solve questions in several different ways and to get into themindset of the test writers. As you read a passage, think about the questions thatcould be asked about it. If a passage gives values for various masses andvolumes, there will probably be a question about density. If it gives values formoles and solution volume, there will probably be questions on concentrationand dilution. Use your intuition and common sense as much as you can, andmake every effort to develop your test-taking logic.

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General Chemistry Stoichiometry Unit Conversion c

:$$ilffiiii$$ uHDimensional AnalysisDimensional analysis is a mathematical conversion from one set of units intoanother. It involves multiplying a given value by a conversion factor or a seriesof conversion factors until the value is finally expressed in the desired units.Converting from one set of units to another is a critical skill needed to eliminateincorrect answer choices in the physical sciences section. Be systematic whenconverting between units. The standard measurements that can be expressed ina variety of units are distance (1 m = 1.094yd,2.54 crn = 1.00 in, and 1.609 km =1.00 mile), mass (1.00 kg = 2.205Ib and 453.6 g = 1.00 lb), volume (3.79 L = 1.00

gal and 1.00 L = 1.06 qt), and time (3600 s = 1.00 hr). Always convert units as

they appear in a problem into the units indicated in the answer choices (the so-called "target units").

Example 1.1Sprinters can run 100 meters in just under 10 seconds. At what average speed inmiles per hour must a runner travel to cover 100 meters in 10.0 seconds?

A. 3.7 miles/hourB. 11.2 miles/hourC. 22.4 miles/hourD. 36.0 miles/hour

SolutionThe first task is to determine the given units and the target units. From there,convert the given units into the target units. We are given 100 meters in 10

seconds, but the answer choices are expressed in miles per hour. Use the correctconversion factors, as follows:

Given . 100 m * ? x? x? = miles10.0 s hour

Conversion of distance: 100 m* miles -miles10.0s m s

Conversion o1 1i*". 100 mx s - h10.0 s hr hour

Overall:100m* 1km * lmile *3600s= 100x3600 miles10.0 s 1000 m 1.609 km L.00 hr 10 x 1000x 1.609 hour

100 x 3600 = 3600 = 36 miles10 x 1000x 1.609 10 x 10x 1.609 L.609 hour

36 < 36 < 36, where 36 = 1g and 36 = 36. So L8 < 36 < 36

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A frequent task in chemistry is the conversion between various types oftemperature units, volume units, pressure units, and concentration units.Chemists, like most scientists, employ the MKS system, so the conversion fromconventional units less commonly used in science to MKS units is routine.Example 1.2 demonstrates the interconversion between the conventionalFahrenheit unit and the scientific Celsius unit of temperature.

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General Chemistry Stoichiometry Unit Conversion

Example 1.2At what temperature is the numerical value the same, whether the units are inCelsius or Fahrenheit?

A. 32"B. 0'c. -40'D. -273"

SolutionThe formulas for conversion between Celsius and Fahrenheit are as follows:

T"p= 9 T.c+32 .'. T"6=5 1t"r-sz;59Answering the question requires setting T'F = T"C.

T"p= 9 T"c + 32becomes: T=2T +32,soT= 1.8 T +3255

T = 1.8 T + 32= -0.8 T = 32 :. T = -40'

Density DeterminationThe density of a material or solution is the mass of the sample divided by thevolume of the sample. Density is a measured quantity, determinedexperimentally. Understand the techniques used to measure density. The termspecific graaity refers to the density of a material relative to the density of water,and may be used in a question in lieu of density. For our purposes, specificgravity means the same thing as density, but it has no expressed units.Determining density is a typical example of dimensional analysis.

Example L.3Exactly 10.07 mL of an unknown non-volatile liquid is poured into an empty25.41-gram open flask. The combined mass of the unknown non-volatile liquidand the flask is 34.12 grarns. \A/hat is the density of the unknown liquid?

o. 34.13 8L0.07 mL

C. 1.0.07 g

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SolutionThe density of the liquid is found by dividing the mass of the liquid by thevolume of the liquid. This results in units of grams per milliliter, whicheliminates choice D. The volume of the liquid is 10.07 mL, so 10.07 should be inthe denominator. This eliminates choice C. The mass of the liquid is thedifference between the final mass of the flask and liquid combined, and the massof the flask (34.12 - 25.41), which is equal to 8.71 grams. This means that thenumerator should be 8.71. The correct answer is choice B. Incidentally, thequestion did not state the reason for using a non-volatile liquid. The liquid mustbe non-volatile, to prevent any loss due to evaporation from the open flask. Inthe event the liquid evaporates away, then the mass you determine is too small,due to the loss of vapor molecules.

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General Chemistry Stoichiometry Unit Conversion

Density questions may on occasion involve a more intricate conversion of units.For any density question, keep in mind that the target units are mass solutiondivided by volume solution. The mass percent of a solute is mass solute dividedby mass solution. The product of the density and mass percent is mass solutedivided by volume solution. Converting the mass of solute into moles of soluteyields molarity. Example 1.4 shows this.

Example 1.4\A/hat is the molarity of a3"h NaCl solution with a density of 1,.05 grams/ml?

A. 0.497 M NaClB. 0.504 M NaClC. 0.539 M NaClD. 0.724M NaCl

SolutionThe first step is to determine the units you are looking for, which in this exampleis moles solute per liter solution. You must find both moles solute and literssolution. The density of the solution is 1.05 grams/ml which means that one

liter of the solution weighs 1050 g. Three percent (3%) of the solution is sodiumchloride, so the mass of sodium chloride is 0.03 x 1050 g. This is the same as 37o

of 1000 g+3% of 50 g, which is 30 g + 1.5 g = 31.5 g of NaCI per liter solution.The grams of sodium chloride are converted to moles by dividing by themolecular weight of NaCl (58.6 grams/mole). The unit factor method is shownbelow:

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= 1.05 x 1000x 3 moles NaCl = 3.15 x lOmoles NaCl100 x 58.5 L solution 58.6 L solution

On the MCAT, you will not have time to solve for values precisely, so you mustmake an approximation. Select the answer that is closest to that approximation.

31.5 > Q =1, so the value is greater than 0.500 M58.5 60 2

31.5 < 33 = 11 =5L, so the value is less than 0.550 M58.6 60 20 100

The value falls between 0.500 M and 0.550 M, so choices A and D are eliminated.Next you must choose between 0.504 M and 0.539 M. The value is not close

enough to 0.504 M, so you should choose C, and be a wise student! Wise

students are a good thing. Some questions on the MCAT may presentmathematical set-ups, without solving for an exact number.

The physical sciences section of the exam incorporates physics and generalchemistry, so from the beginning of your review, make a conscious effort toconsider physics when working on general chemistry and to consider general

chemistry when working on physics. Determining density is a problem conunonto both disciplines. Example 1.5 shows an approach to the concept of densitythat is more typical of what is found in a physics problem'

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General Chemistry Stoichiometry Unit Conversion

Example 1.5What can be concluded about the density of a metal object which, when placed ina beaker of water at room temperature, sinks to the bottom?

A. The density of the metal is less than the density of either water or ice.B. The density of the metal is less than the density of water, but greater than the

density of ice.C. The density of the metal is greater than the density of water, but less than the

density of ice.D. The density of the metal is greater than the density of either water or ice.

SolutionWhen an object floats in a liquid medium, its density is less than that of themedium surrounding it. The fact that it floats means the buoyant force pushingupward against it (pmedium.Vobject.g) is greater than gravitation force pushingdownward (weight = mg = pobject.Vobject.g). Thus, an object floats whenPmedium > Pobject. Because the metal object sinks in water, it must be denserthan water. Ice floats in water, meaning that ice is less dense than water and thusless dense than the metal object. The density of the metal must be greater thanthe density of either water or ice. The correct answer is therefore choice D.

Typical ConversionsIn chemistry, conversions between products and reactants are common, so themole concept is frequently employed. The mole concept is pertinent in theinterconversion between moles and mass, using either atomic mass (forelements) or molecular mass (for compounds). These calculations involve usingthe unit factor method (also known as dimensional analysis.)

Example 1.5F{ow many moles of NaHCO3 are contained in 33.6 grams NaHCO3?

A. 0.20 moles NaHCO3B. 0.40 moles NaHCO3C. 0.50 moles NaHCO3D. 0.60 moles NaHCO3

SolutionThe first step in determining the number of moles is to determine the molecularnass of NaHCO3. The mass is 23 + 1. + 12 + 48 = 84 grams. The number of molesot NaHCO3 is found by dividing 33.6 by 84, which is less than 0.50. ThiseLi.srinates choices C and D. The number is greater than 0.25 (21 over 84) andnus greater than 0.20, so choice A is eliminated. The only value left is choice B,

-"{0 moles.

3er-ond determining the moles from.grams for the same compound are questions-,q-here the moles of products are determined from the grams of reactants. These:;estions require converting from grams of a given substance to moles of the;:';en substance, and then expressing the quantity of a final substance in terms of:ones, grams, or liters. By balancing the reaction, the mass of a selected product:at is formed in the reaction can be calculated based on the mass of a selected

=a,ctant (which must be the limiting reagent). Examples 1.7 and 1.8 involveretermining moles, mass, and volume from the given values.

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General Chemistry Stoichiometry Unit Conversion Ge

Example 1.7Based on the following reaction, how many grams of water would form from0.33 moles CaH16O reacting with an unlimited amount of oxygen gas?

CaH16O(g) +6O2G)

-

aCO2G) +5H2O(g)A. 18.00 gramsB. 24.00 gramsC. 30.00 gramsD. 36.00 grams

Solutionwith an excess of oxygen, the limiting reagent in this reaction is C4H16o. Theamount of water formed is determined by the 0.33 moles of CaH1Oo reactant.Using the balanced equation, the ratio of H2O to moles CaHlgO is 5 : 1, so I.667moles of water are formed. At 18 grams per mole, this means that fewer than 36grams but more than27 grams are formed. This makes choice C the best answer.

0.33 molesC+Hroo " -

5 *,olujl_ro_ "

tt qt31 =l x 5* 18 gH2o = 30 gHzo1 mole CalItOO 1 mole H2O 3 "

Example 1.8

How many liters of Coz(g) result from the complete decomposition of 10.0 gramsof CaCO3(s) to carbon dioxide and calcium oxide at STP?

CaCO3(s) ---+

CaO(s) + COz(g)

A. 1.12 litersB. 2.24litersC. 3.36litersD. 4.48liters

SolutionYou are asked to determine the amount of product from a known quantity ofreactant. The first step rn problems of this type is to make sure the reaction isbalanced. In this case, it is already balanced. The mole ratio of the twocompounds is 1 : 1. The required conversion involves changing from massreactant, to moles reactant, to moles product, and finally to volume product.This is one variation of unit conversion via mole ratio calculation. In addition,there is th" "g-m - m - g" conversion and the "v -m -m - g" conversion. youneed three steps to go from grams reactant to the target (liters product). Unitsare important here. The units for the mass of reactant is grams. You need tomultiply mass by moles and dir.'ide by grams. This is the same as dividingby theMW. The second step is to read the mole ratio from the balanced equation. L:rthis reaction, the mole ratio is 1 : 1 (the units of both numerator and denominatorare moles). The third and final step is to convert from moles product into litersproduct (i.e., multiplr' br' liters and divide by moles.) This is done bymultiplying by the molar volume of the product gas, which at STP (standardtemperature and pressure) is 22.-1 Liters.

10 gramsCaCO3 x , -1*ot" a1a9? ,. 1 mole CQ *22.4liters COz

100 gramsCaCO3 l mole CaCO3 l mole CO2

= 1o x Z2.-l = 2.24liters Co2100

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I General Chemistry Stoichiometry Elemental Analysis

Elemental analysis determines the atomic composition of an unknown molecule.It is based on the idea that all molecules of the same substance combine atoms ofthat substance in the same way. Lr other words, water always has two hydrogenatoms and one oxygen atom. Because of this feature of structural uniformity, it ispossible to determine the atomic composition of any molecule. The fundamentalprocess of elemental analysis involves oxidizing an unknown completely andcollecting the products. The amount of each element that was present in theunknown compound can be determined from the amount of oxidized product.These mass values can be converted to mass percent and mole ratio values. Inthe determination of the empirical formula, the mass percent is converted to arelative mass value and then a relative mole value. The mass percent of anelement within a compound must be determined prior to determining theempirical formula for an unknown compound. An empirical formula, you mayrecall, is the simplest whole number ratio of the atoms in a molecule.

Mass Percent (Percent Composition by mass)The mass percent of a particular element within a compound is found bydividing the mass of that element by the mass of the compound and thenconverting the fraction to a percentage. This is shown in Equation L.1.

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Mass percent can never exceed 100% for any component element. Determiningthe mass percent of an element from the molecular formula is a straightforwardtask, although the math may be challenging. Mass percent questions can beasked in a concepfual or mathematical manner. Mass percent is independent ofthe total mass of the sample of compound.

Table 1.1 shows the relative masses of oxygen and carbon from different samplesof carbon dioxide. This demonstrates the law of multiple proportions. Atomscombine in a fixed ratio in terms of mass and moles. Note that the outcome is thesame in all four trials measuring the ratio of oxygen gas that reacts with a knownmass of carbon. The experiment involves oxidizing a known amount of carbonand collecting the product gas. The mass of this product gas is determined, andthe mass of oxygen is assumed to be the difference between the initial and finalweighed masses of the carbon sample.

Mass Carbon Mass Orygen Mass Or/Mass C

L.33 g 3.53 g 3'53 1. nn =2.6s/ I.JJ

1.07 g 2.87 g 2'8' / r.0, =2.68

i.11 g 296 g''nu / r-rt=2'67

1.27 g 3.39 g 3'39 / 't.27 =2'67

Table 1.1

The mass ratio of oxygen to carbon in the four trials avelages out to be2.67 :1',which is roughly 8 : 3. This means that for the oxidation product of carbon, theratio of oxygen to carbon is 8 grams to 3 grams, equivalent to 2 moles to 1 mole.

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General Chemistry Stoichiometry Elemental Analysis jExample 1.9How much calcium metal combines with one gram of oxygen to form CaO?

A. 1.00 g CaB. 1.25 gCaC. 7.67 gCaD. 2.50 g Ca

SolutionFrom the molecular formula, the mole ratio of calcium to oxygen is 1 : 1. Theatomic mass of calcium (Ca) is 40.08, while the atomic mass of oxygen (O) is16.00. The mass ratio for the compound is 40.08 to 16.00, which reduces to 2.505 :

1, which rounds to 2.50 to 1. This means that 2.50 grams of calcium combine with1.00 grams of oxygen to form CaO. Choice D is best.

Example 1.10What is the mass ratio of iron to oxygen in Fe2O3?

A. 1.08 g Fe to 1.00 g OB. 1.63 g Fe to 1.00 g OC. 2.33 gFe to L.00 g OD. 3.49 g Fe. to 1.00 g O

SolutionFrom the molecular formula, the mole ratio of iron to oxygen is 2 : 3. The atomicmass of iron (Fe) is 55.85, while the atomic mass of oxygen (O) is 16.00. The massratio of iron to oxygen for the compound is 2(55.85) to 3(16.00), which equals771,.7 :48.0. This ratio is approximately equal to 116 :50, or 232:100, whichreduces to 2.32: 1. Both numbers must be increased proportionally to keep theratio the same. Choice C is a ratio of 2.33 to 1, which is the closest of the choices.This means that 2.33 grams of iron combine with 1.00 grams of oxygen to formFe2O3. Choice C is best. You should note that iron and oxygen can combine tomake other compounds (with different molecular formulas). One of thesecompounds is FeO, with a mass ratio of 55.85 to 16.00, which reduces to a ratio of3.49 : 1.00. The mass ratio (and mole ratio) of an oxide can be used to identify aspecific compound. This process is known as combustion analysis.

Examples 1,.9 and 1.10 demonstrate how mass percent questions can bemathematical. Mass percent questions can also be asked in a conceptual manner,where the relative mass percentage of a specific element is compared for severalcompounds. Examples 1..1.1., 1..1.2, and 1.13 demonstrate some different forms ofthis type of question, starting with typical examples and graduating to moreabstract ways of asking for mass percent.

Example 1.11VVhat is the mass percent of oxygen in carbon dioxide?A. 27.3%B. 57.7%c. 62.5%D. 72.7%

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General Chemistry Stoichiometry Elemental Analysis

SolutionThe mass of carbon in CO2 is 12 grams, and the mass of oxygen in CO2 is 32grams. The total mass of CO2 is 44 grams, so the mass percent of oxygen is theratio of 32to 44. This ratio reduces to 8 over 11.

MassoercentO= 328o x 100% -32 *'J.00"h =3 x 100%' 44 gCOz 44 11

Quick Calculation Technique:

Quick calculations require knowing the values of selected fractions. One-eleventh is equal to 0.091; therefore, eight-elevenths is equal to 8(0.091) = 0.728.This method gets an exact value and is very fast, if you know how to do it.

8 = 8x I = 8x 0.091 = 0.728 = 72.8"h11 11

Narrowing-Down-Choices Technique:On a multiple-choice exam, you can eliminate answers by narrowing down therange into which the answer fits. 8 over 11 is less than 9 over 12,but greater than

7 over 10. A range has been established between 9 and !. 9 ou"t 12 is 75"/o,12 10

and 7 over 10 is70"h, so the correct answer falls betweenT}% and75"/'.

-9 >-8 > T,where 9-=75"/o and,f-=70o/o.So:75"/.>L> 70%72 1L 10 12 10 11

C:roice D is the best answer.

Example 1.12lVhat is the mass percent of nitrogen in NHaNO3?

L. 28%B. 35%c. 42%D. 50%

SolutionThe total mass of the nitrogen in the compound is 28 g/moIe, because there aretrvo nitrogen atoms in the compound at1"4 g/mole each. The mass of thecompound is 28 + 4 + 48 = 80 g/mole. You must divide 28 by 80 quickly. The

couunon denominator of both is 4. Reducin gby 4yields a fraction of Z.20

Masspercen,N =,, ?89T-- x 100% =D* 100"h = J-x fi}%80 gNHaNO3 80 20

Quick Calculation Technique:

Quick calculations may involve getting a denominator to some easy-to-usenurnber, such as 10, 100, or 1000. It is easy to convert a fraction into decimals orpercents when the denominator is either 10, 100, or 1000. For this question, a

denominator of 100 works well. To convert 20 to 1"00, we must multiply by 5.

Multiply both numerator and d.enominatorby 5, to change the fraction -7 ' b E.The percentage is 35%, so choice B is correct.

20 100

28=7 =7xS =35 =0.35=35%80 20 20 x5 100

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General Chemistry Stoichiometry Elemental Analysis Gr

Narrowing-Down-Choices Technique:Narrowing down the range into which the answer fits can be applied to anymultiple-choice math question. 28 over 80 reduces to 7 over 20. The value of 7over 20 is greater than 7 over 21., but less than 8 over 20. A range has been

established between Z and !. The value of 7 over 21 is 33.3"h, and the value of21. 208 over 20 is 40"h, so the correct answer falls between33.3% and 40"/o. Only choiceB fits in this range, so choice B must be the correct answer.

7 <J- < 8,where J*=33.3"h and! =40"/o. So33.3% <f-<+0"/,21. 20 20 21 20 20

Example 1.13\A/hich of the following samples yields the MOST moles of sodium cation?

A. 1.0 gNaClB. 1.0 g NaBrC. 1.0 g NaNO3D. 1.0 g Na2CO3

SolutionThis question is a subtle way of asking, "Which salt has the greatest mass percentof sodium?" All choices are 1.0 g of compound, so the most moles of sodium arefound in the compound with the greatest mass percentage of sodium. Choices A,B, and C have the same number of sodium atoms in the compound (one), so theyeach have the same numerator in the mass percent formula. The best choice ofthose three salts is the lightest compound (resulting in the smallest denominatorwhen calculating mass percent). The lightest compound of the three choices A,B, and C is the salt with the lightest anion, which is choice A, NaCI. Now thequestion involves comparing the mass percent of sodium in NaCl to the masspercent of sodium in Na2CO3, choice D.

Mass percent Na in NaCl = -1u o q-*i- x r00"/. = =.?3

x 700'/,58.5 g NaCl 58.5

Mass percent Na inNa 2Cos = - 1u 9i *:-

x 100% - 46 x 100% =D xrll"/o106 gNa2CO3 106 53

23 > 23 , so choice D is the best answer.53 58.5

Empirical FormulaAn empirical formula for molecules uses the smallest whole number ratio of theatoms in a compound. It is the formula that gives the relative numerical valuesfor each element in the molecule in such a way that the numbers in the ratiocannot be reduced without involving fractions. An empirical formula may ormay not be the actual formula of the molecule. It is calculated from the masspercentage of each element within a compound. You may recall from yourgeneral chemistry courses that we start by assuming a 100-gram sampie, so thatthe percentages can be changed easily into mass figures. From this point, it is amatter of converting from mass into moles, using the atomic masses for eachelement. The empirical formula is a whole number ratio of these mole values.Empirical formulas must include whole number quantities as subscripts.

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General Chemistry Stoichiometry Elemental Analysis

Example 1.14\4/hich of the following is an empirical formula?

A,. C2H6B. CeHsC. CaHroD. C6H6

Solution-\n empirical formula for a molecule is defined as the formula in which the:onstituent atoms are in their smallest possible whole number ratio. Choice A,CtH6, can be reduced to C1H3 (normally written as CH3), so choice A is not an

enpirical formula. Choice C, C4H1g, can be reduced to C2H5, so choice C is not

-. empirical formula. Choice D, C5H6, can be reduced to C1H1 (normally.'.-iitten as CH), so choice D is not an empirical formula. This eliminates all of the

-:.oices except choice B, C3Hg. The ratio of 3 : 8 cannot be reduced any further,

': C3Hg is an empirical formula, making choice B the correct answer. In the case

-: C3Hg, the empirical formula and the molecular formula are the same, because---.e compound is completely saturated with hydrogens' CeHtO has too many- -,'Jrogens and is not a possible formula. Organic chemistry rules can help to-: . = nme on formula questions.

E"r,a-mple 1.15,',1a: is the empirical formula for a compound that is 72"/o C,i r ::.rcrsed solely of carbon, hydrogen, and oxygen?

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:,: lution: . : =:.pir-ica1 formula calculations, assume a 100-gram sample of the compound... -.'.--am sample in this case would contain 72 grams C, 12 grams H, and 16

::::^, ,J. The 16 grams of oxygen are determined from the difference between'i*: :-ai: of carbon plus hydrogen and the 100 grams of sample. Next, you mustr :: ::-l fie grams of each element into the corresponding moles of each element.'- r r - ::,:.r glams to moles, divide the mass of the element by its atomic mass' In:". ::*ie 12 grams of C is equivalent to 6 moles of C, 12 grams of H is equivalent: -- :.l-es of H, and 16 grams of O is equivalent to l mole of O. This is a

r l:::-*::1-'- easv example, because the ratios turn out to be whole numbers. Inrl;r: -,-,'i,€r€ thev don't come out whole, you must divide the mole quantity of.i :i :-::r-er1t in the compound by the lowest mole quantity for any of the, ritr :r i-i l-{orvever, for this example the best answer is choice B. Drawn below,F i --:"::--i- ia""'out of empirical formula calculations. It is often easier just to plugr, ri:.-i -r i r ar. equation such as this, because you don't have to show your work- ::: l-lC-\T. Do the questions as quickly and carefully as you can/

.II1 r *:;r-:::"i organization in your path to a solution.

fl percentage hydrogen O percentage oxygen = CZ2}I]O1f. = COHTZO I-ola. -ass hydt.gen molarmass oxygen 12 "I 16

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General Chemistry Stoichiometry Elemental Analysis

Example 1.16What is the empirical formula of an oxide that is 60% oxygen by weight?

A. SOB. SOzC. SzOsD. SO3

SolutionHere again you should assume a 100-gram sample. A 100-gram sample would

have 60 grams of oxygen and 40 grams of sulfur. The moles of O = 60, and the16

moles of S = n-. 60 is more than double 40, so there are more than two oxygen32 16 32

atoms per sulfur atom. This eliminates choices A and B. By reducing 4'9- to2!," zz 1.6

we find that the ratio of oxygen to sulfur is 3 : 1. This makes choice D the bestanswer.

$ percentage sulfurO percentage oxygen = S40060 = SZq060 = S103molar mass sulfur molar mass oxygen 32 16 1'6 1'6

The test emphasizes ratios, so the more numerical intuition you develop, thebetter. To make problem-solving less mathematical, focus on eliminating choicesby comparirtg relative ratios. An alternative way to ask an empirical formulaquestion with reduced math is shown in Example 1.1,7.

Example 1.17If a molecule is composed of only two elements (X and Y), and if X and Ycombine in equal mass quantities, and if Y is less than twice as heavy as X, whichof the following molecular formulas is NOT possible?

A. XYB. XYzC. XsYzD. X3Y

SolutionIf Y were exactly twice as heavy as X, then equal masses of X and Y would resultin exactly twice as many moles of X as Y, a2:1 ratio of X to Y. Because Y is less

than twice as heavy as X, there are fewer than twice as many moles of X as Y.

Thus, the ratio of X : Y must be 2 : l or smaller. This limiting ratio is true of allthe answers except choice D. The wording of this question allows for thepossibility that the molecular mass of X is equal to or greater than Y.

Molecular FormulaThe molecular formula is the actual mole ratio of the elements within thecompound. The molecular formula is found by multiplying the empiricalformula by the whole number ratio (including 1, in some cases) of the molecularmass to the empirical mass. Therefore, conversion from the empirical formula tothe molecular formula requires knowing the molecular mass of the compound. Ifthe molecular mass is double the empirical mass, then all of the elements in the

empirical formula are doubled to get the molecular formula.

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General Chemistry Stoichiometry Elemental Analysis

Example 1..L8

What is the molecular formula for a compound that is 82.76% C, has a molecularmass of 58.1 grams per mole, and is composed solely of carbon and hydrogen?

A. CH2B. CzHsC. CsHsD. C4H1s

SolutionChoice B is eliminated immediately, because a hydrocarbon cannot have an oddnumber of hydrogens. For neutral C2H5 to exist, it would have to be a freeradical. Continuing to use organic chemistry logic, choice A is not physicallypossible. One carbon requires four bonds to hydrogen atoms to form a stablemolecule (methane). C1H2 would be a carbene (:CH2), which is not stable due toits lack of an octet. To decide between choices C and D, you must first find theempirical formula, and then convert that into the molecular formula.

( percentage carbon H percentage hydrogen - C82.76Ht7.24 =Ce .ggHtZ.Z+molar mass carbon molar mass hydrogen 72 1

empirical formula: C#+# = CtHtTs = C1H2.5 = C2Hs

An empirical formula of C2H5 when multiplied by a whole number cannot yieldC3Hg, so choice C is eliminated. This leaves only choice D. The correct ratio ofmolecular mass to empirical mass confirms that choice D is the best answer. Themolecular formula is found using the molecular mass of 58.1 grams per mole.The empirical mass is 2(12) + 5 = 29. This value is only half of the molecularmass, so the formula must be doubled to yield C+HfO. This question could havebeen solved in seconds by seeing that only choice D has a molecular mass of 58.

Example L.L9

-{n unknown stable gas is composed of 13.10% H, 52.23"h C, and the remainderO. A 0.10-mole sample weighs 4.61 grams. \Alhat is the molecular formula fordre compound?

A. C2H6OB. C3H3OC. C3HeOD. CaHsO

Solution-{ stable compor.rnd made of carbon, hydrogen, and oxygen cannot have an oddrumber of hydrogens, so choice C is eliminated. Neutral C3H9O would have to:e a free radical. The remaining choices obey the octet rule. The molecular mass

= -16.1, so choice A is the correct answer. That is the method you should use on a

=tultiple-choice exam. Now let's confirm that by using the molecular formula.iirst you must assume a 100-gram sample, resulting in 13.10 g}{,52.23 g C, and34.67 g O. Next, the numbers are converted into moles, and then the values are

=duced to a whole number ratio. The calculation of the empirical formula is

=hown below, where a formula arrangement is used to help keep track of the;alues,

Cq?23fiEJq 094"62- = Ca.aHtg.t oz: =C+.+Huto?z =C2H'6o112 7 1.6 2.2 2.2 2.2

The empirical mass is 2(12) + 6 + 1,6 = 45. This value is equal to the molecular:lass, so the empirical formula is the molecular formula.

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General Chemistry Stoichiometry Elemental Analysis G

Combustion Analysis (an Experimental Procedure)Combustion analysis entails determining the mass percent of each componentelement in an unknown compound. It is accomplished by oxidizing theunknown with excess oxygen (to ensure complete combustion), followed by theseparation and collection of all of the oxidized products. This is an experimentalprocedure, which makes it a likely topic on a conceptual exam such as theMCAT. When a hydrocarbon is oxidized, carbon dioxide and water are formed.Carbon dioxide and water can be separated using various methods. One methodinvolves passing the CO2 gas and H2O vapor across a hydroscopic salt of knownmass. The hydroscopic salt absorbs the water, and thus increases in mass.

The hydroscopic salt must not react with carbon dioxide. A good choice for thehydroscopic salt is either calcium chloride or magnesium sulfate (both of whichyou should have used as drying agents in your organic chemistry lab). Once thewater is absorbed, the remaining gas is passed across a sample of KOH of knownmass. KOH is a base that reacts with acidic carbon dioxide to form potassiumbicarbonate (KHCQ). The potassium hydroxide salt absorbs the carbon dioxide,and thus increases in mass. Knowing the masses of CO2 and H2O, we can

determine the masses of carbon and hydrogen by multiplying the mass percentof each element by the mass of its respective oxide product that was collected.Upon dividing these numbers by the mass of the original sample, the mass

percent of hydrogen and carbon in the original sample is determined.

The carbon dioxide and water can be separated and collected using a differentmethod than passing the gases over salts that bind the products. By lowering thetemperature, carbon dioxide and water can be converted to their solid states.Because solids do not flow, they can be collected easily. The math is the same,

once the quantity of each product has been established.

Figure 1-1 shows a typical apparatus used in a combustion analysis. The oxygentank serves to provide excess oxygen to the system constantly. The pressurevalve is a one-way valve designed so that oxygen can flow into the samplechamber, but no gases can flow back. The resistor in the base of the samplechamber provides heat to initiate the oxidation. The tube to the right of thesarnple chamber is connected to a vacuum, to generate a low pressure. Once thereaction is complete, Valve #3 is closed so that no gases are lost to theenvironment, and Valve #1 is opened. Gases flow into the region above themagnesium sulfate. Magnesium sulfate absorbs moisture from the gases. After atime, Valve #2 is opened so that gases flow into the region above sodiumhydroxide. Sodium hydroxide absorbs carbon dioxide gas. Oxygen gas is thenused to flush any remaining gas in the sample chamber through the system'

Pressurevalve Valve #1 Valve #2 Valve #3

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General Chemistry Stoichiometry Solution Concentration

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#ffiilffffirfiffixiiffi$ffi$ffsr#Units and TerminolgySolutions are mixfures formed by the addition of a solute to a solvent. A solutionmay contain several different solutes. The amount of solute is measured relativeto the amount of solvent, which results in a certain concentration for the solution.Concentration units include molarity (moles solute per liter of solution), molality(moles solute per kilogram of solvent), mass percent (mass solute per mass ofsolution), and density (mass solution per volume solution). The concentration ofa solute can be changed by changing the amount of solvent. Addition of solventto solution is referred to as dilution and results in a lower concentration by anymeasurement. Paramount to solving problems involving concentrations anddilution is an understanding of the different units'

Molarity\{olarity (M) is the concentration of a fluid solution defined as the moles of asolute per volume of solution, where the volume is measured in liters (L). ToCetermine the molarity of a solution, the moles of solute are divided by the litersof solution.

Erample 1.20"r\hat is the molarity of 500.0 mL of solution containing 20.0 grams of CaCO3(s)?

-{. 0.15 M CaCO3(aq)B. 0.20 M CaCO3(aq)C. 0.33 M CaCO3(aq)D. 0.40 M CaCO3(aq)

Solution\tolarity is defined as moles of solute per liter of solution. In this question, youru.rst convert from grams CaCO3 into moles CaCO3 by dividingby the molecular:iass of CaCO3, and then dividing this value by the liters of solution:

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100 Srams//mole

-.10 moles CaCOg =0.20 M CaCO3 =0.40 M CaC03 = 0.40 M CaCO3, choice D-r.50 L solution 0.50 1

laese questions can be trickier if the units are milligrams, milliliters, or:jlimolar. The question uses similar math, but there are more oppoltunities to:::ke an error. A common error to avoid is the "factor of a thousand" error. To:e.ome more conscious of possible trick questions, ask yourself: "If I were-u-nting this test question, what would I ask?" If you consider questions from the:.*t rr.riter's point of view, the tricks become more apparent.

\folality}-{rlality (m) is the concentration of a fluid solution defined as the moles of a.-.-ute per kilogram of solvent. The molality of a solution does not change with:.nperature, so it is often used to determine a change in the solution's:::lperature when the change depends on concentration. Notable examples of::s include boiling-point elevation and freezing-point depression. To determine:,e nnolality of a solution, the moles of solute are divided by the kilograms of the::1.;ent.

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General Chemistry Stoichiometry Solution Concentration G

Example L.21\A/hat is the molality of a solution made by adding 7.46 gKCl to 25a g of water?

A. 0.20 m KCI(aq)B. 0.33 m KCI(aq)C. 0.40 m KCI(aq)D. 0.50 m KCI(aq)

SolutionMolality is defined as moles of solute per kilogram of solvent. In this question,you must convert from grams KCI into moles KCl by dividing by the molecularmass of KCl, and then dividing this value by the kilograms of solvent:

7.46 grarnsKCl= 0.10 moles KCI

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0.10 moles KCI -0.10 m KCI =0.40 m KCI = 0.40 m KCl, choice C0.25 kg H2O 0.25 L

Mass Percent (in Solution)Mass percent is the concentration of a fluid solution defined as the mass of soluteper mass of solution multiplied by one hundred percent. The mass percent of asolution remains constant as temperature changes. To determine the masspercent of a solution, the mass of solute is divided by ihe mass of the solution(where both are measured in grams). Mass percent is a unitless value, becausemass is divided by mass.

Example 1.22What is the mass percent of a 1.0 m NaCl(aq) solution?

A. 5.53% NaCl by massB. 5.85% NaCl by massC. 6.22%NaCIby massD. 9.50% NaCl by mass

SolutionMass percent is defined as grams of solute per grams of solution. In thisquestion, you must convert from moles NaCl into grams NaCl by multiplying bythe molecular mass of NaCl, and then dividing this value by the total mass ofsolution. The total mass of solution is the sum of the mass of solute and the massof solvent. It is easy to forget to consider the mass of solute in the total mass,which leads to the incorrect answer choice B.

1.0 moles NaCl(aq) x 58.5 8/-ole = 58.5 g NaCl

Total mass solution = 58.5 g NaCl + 1000 g H2O = 1058.5 g solution

Mass % NaCl = 58'5 g NaCl , where 58.5 < 58.5 = 5.85%

1058.5 g solution 1058.5 1000

Only choice A is less than 5.85%, so choice A is the best answer. Sometimesquestions that would normally require a calculator for a precise answer can bedetermined well enough without one to answer a multiple-choice test question.

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General Chemistry Stoichiometry Sotution Concentration

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DensityDensity (p) is the concentration of a fluid solution defined as the mass of solutionper volume of solution. The density of a solution varies with temperature. Thedensity of a solution is uniform throughout, so a small sample of the solution has

the same density as the entire solution. To determine the density of a solution,the mass of a sample of the solution is divided by the volume of the sample(usually measured in milliliters). Examples 1.3 and 1.5 addressed the topic ofdensity.

DilutionDilution involves the addition of solvent to a solution, thus resulting in anincrease in the volume of the solution and a decrease in the concentration of thesolute in solution. Equation 1.2 below describes simple dilution where a solventjs added to solution. Determining the concentration when two solutions are

mixed requires more work than simple dilution.

Mir,itiul'Vinitiil = Mfinal'Vfinut (1.2)

I*then working with ditution questions, be aware of a common twist that theititers can employ. Their question may ask for aolume added tathet than askingior the final total volume. Percent dilution may also be discussed. Multiple:ontainers are used in standard dilution procedure, so rinsing ensures that the;.rn-rcentration of solution on the walls of the new containers are equilibrated withi,e contents they will hold. You may recall filting a volumetric pipette with ao:lution in general chemistry lab, then draining the pipette before filling it with::e sample to be transferred. This is done to ensure that any residual liquid in:te pipette has the same concentration as the solution being transferred and that

=.i- rvater in the pipette is rinsed away.

Lrample 1.23r'*:"at is the molarity of a solution made by mixing 200 mL pure water with 100

:10.75 M KCI(aq)?

-q- 0.25 M KCI(aq)ts 0.50 M KCI(aq)C 1.50 M KCI(aq)D. 1.25 M KCI(aq)

i,ol,ution],.-;;L15e water has been added to the solution, the concentration must decrease,

': jloices C and D are eliminated. Solving this question involves using Equation

- - :o determine the effect of dilution on the molarity. The initial molarityl,{-.rut) is 0.75 M, the initial volume (Vmiti"t) is 100 mL, and the final volume-'-_:r-xj is 300 mL. The question requires solving for the final molaritl (Mfi"ail.

Mitiud'Vir',itial = Mfinal'Vfinal "' 0'75 M'1-00 mL = M1i1n1'300 mL

0.75 M.100 mL=- =300 mL

lk inat molarity is 0.25 M, so choice A is the best answer. Because the moiarityl* je;reased by a factor of three, the dilution process in this example is referredu as a threefold dilution. That is, when two parts solvent are added to one part;:r.::.on. the volume is tripled and the dilution is threefold. This terminologym,* ; be unJamiliar at first, but in a short time it should make sense'

l.{:.rl = Mittitiul'Vinitiut .'. Mfinal

Vfinalo.?s M(r) = 0.25 M

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General Qhemistry Stoichiometry Solution Concentration e

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Example L.24How many milliliters of water are needed to dilute 80 mL 5.00 M KNO3(aq) to1.00 M?

160 mL H2O320 mL H2O400 mL H2O480 mL H2O

SolutionAgain, Equation 1.2 should be employed to calculate the change in concentrationof a solution after dilution from the addition of solvent. First we must solve forthe final volume. You are provided with values for M;r.,i1i61, Mfinat, and Vini1ix1,so you can manipulate the equation to solve for V1ir.,21.

Vfinal -Minitial'vinitial =5.00Mx8OmL = 5x g0mL = 400mlMfinal 1.00 M

The question asks how much water is added, not the final volume. The volumeadded is the difference between the initial volume (80 mL) and the final volume(a00 mL). The difference between the two values is 320 mL, so 320 mL of watermust be added to 80 mL of 5.00 M KNO3(aq) to dilute it from 5.00 M KNO3(aq) to1.00 M KNO3(aq). Choice B is the best answer.

Example 1.25Which dilution converts 6.00 M HCl(aq) to 0.30 M HC(aq)?A. 11 parts water to 1 part 6.00 M HCI(aq)B. 19 parts water to 1 part 6.00 M HCl(aq)C. 20 parts water to L part 6.00 M HCl(aq)D. 27 parts water to 1. part 6.00 M HCI(aq)

SolutionHydrochloric acid goes from 6.00 M to 0.30 M, which is a twenty fold dilution.This means that the final volume is twenty (20) times the initial volume. whendealing with answer choices that present the dilution in terms of parts, the ratiois volume of solvent added to volume of original solution. For the final volumeto be twenty times greater than the initial volume, nineteen parts must be added.

Vfinal =Mittitiul =r -Yfinat =9.00 y =20 :.vfinal = 20 (vini1i21)

Vinitiul Mfinal Vir,itiul 0.30 M

Vadded = Vfinal - Vi.,itial = 20 Vi61i21 - Vi.,itiul = 19 (V6i1i31)

The ratio of the volume added to the volume of solution initially present is 19 : 1,so the best answer is choice B. A part can be any set volume. A20 :1 dilutionwould result in a final volume that is 21 times the initial volume, so the finalconcentration would be less than 0.30 M.

A solution can be diluted by adding solvent or another solution to it. Theaddition of pure solvent is known as a simple dilution. Mixing two solutions ismore complicated than a simple dilution where pure solvent is added, becausesolute from both initial solutions must be considered. The final concentration liessomewhere between the two initial concentration values before mixing. The finalconcentration is a weighted average of the initial concentrations.

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I General Chemistry Stoichiometry Solution Concentration

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Example 1.26What is the final concentration of Cl- ions after mixing equal volumes of 0.20 MKCI(aq) with 0.40 M CaCl2(aq)?

A. 0.20 M Cl-(aq)B. 0.30 M Cl-(aq)C. 0.40 M Cl-(aq)D. 0.50 M Cl-(aq)

SolutionThe salt KCI yields one chloride ion when it dissociates in water, so the chlorideconcentration is 0.20 M. The salt CaCl2 yields two chloride ions when itdissociates in water, so the chloride concentration is 0.80 M. The finalconcentration equals the total Cl- ions from both solutions divided by the newtotal volume. Because equal volumes are mixed, the final concentration will bean average of the two initial concentrations.

0.20 M Cl- + 0.80 MCI- = 1.00 M Cf = 0.50 MCI-22

I{ the volumes are not equal, then a weighted average yields the finalconcentration. For this example, choice D is the best answer.

Example L.27lVhat is the concentration of K+ ions in solution after 25.0 mL of 0.L0 MK2SOa(aq) is added to 50.0 mL of 0.40 M KOH(aq)?

A. 0.25 M K+(aq)B. 0.30 M K+(uq)C. 0.33 M K+(aq)D. 0.50 M K+(aq)

SolutionThe salt KOH yields one potassium ion when it dissociates in water, so the K+concentration is 0.40 M. The salt K2SOa yields two potassium ions when itdiq,s6sislss in water, so the K+ concentration is 0.20 M. The final concentrationequals the total K+ ions from both solutions divided by the new total volume.tseca,tse unequal volumes are mixed, the final concentration is a weighteda\rerage of the two initial concentrations. The mixture involves 25 mL 0.20 M K+n'ith 50 mL 0.40 M K+, so the final concentration must fall between 0.20 M and0.40 M. This eliminates choice D. If the two volumes were equal, the finalconcentration would be 0.30 M K+, the average of the two concentrations' Butbecause there is more of the more concentrated solution, the final concentration is

greater than 0.30 M, so choices A and B are eliminated. Only choice C remains.

25mLx0.20MK+ *50ml-x0.40MK+ =1 (0.20MKl *L 10.+oMK+)75rnL 75mL 3 3

=Z +.8- Vtf* =1MK* = 0.33 MK+,choiceC333

Beerts LawWhen electromagnetic radiation is passed through a solution, the solute mayabsorb some of the light. The light absorbed is in a specific wavelength range/and the intensity of the absorbance varies with the concentration of solute. Ageneric absorbance spectrum for a hypothetical solute is shown in Figure 1.-2.

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General Chemistry Stoichiometry Solution Concentration (fr

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Because the absorbance of lighi varies with concentration, absorbance can be

used to determine the concentration of a solute. This is the essence of Beer's law.Beer's law is expressed in Equation 1.3, where e is a constant for the solute atl,rrru* (the wavelength of greatest absorbance), C is the solute concentration (C =[Solute]), and I is the width of the cuvette (length of the pathway through whichthe light passes).

Absorbance = €Cl (1.3)

The key feature of this equation is its expression of the principle that absorbanceis proportional to concentration. By knowing the absorbance for solutions ofknown concentration, the concentration of an unknown solution can bedetermined by comparing its absorbance value to the known values.

Example 1.28For 100 mL of a solution with an absorbance of 0.511, what amount of water mustbe added to reduce the absorbance to 0.100?

A. 389 mL H2OB. 411 mL H2OC. 488 mL H2OD. 511mL H2O

SolutionFor this question, a hybrid of Equations 1.2 and 1.3 should be employed todetermine the volume that must be added to dilute the solution. Because

absorbance is directly proportional to concentration, Equation 1.2 can be re-written as follows:

Absinili6l.Vinitial = Abs661'Vii1a1

First, we must solve for the final volume. You are provided with values forAbsi1i1ix1, Abs1i1n1, and Vi61ia1, so you can solve for V6nn1.

Vfinal - Absinilinl'Vit iUut - 0.511 x 100 mL - 5.11 x 100 mL - (1 1 mT

Abs1i61 0.100 1

The question asks for how much water is added, which is the difference between

the initial volume (100 mL) and the final volume (511 mL). The differencebetween the two values is 41,1, mL; therefore, 4L1 mL of water must be added.The best answer is choice B.

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n General Chemistry Stoichiometry Balancing Reactions

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Standard BalancingLet us briefly address the process of balancing chemical reactions. Reactions arelvritten from reactants to products. Because of the law of conservation of matter,the number of atoms must be identical on each side of the reaction. The twosides of the reaction are separated by an arrow drawn from left to right.

C5H12(l)+Oz(g) € COz(g)+H2O(g)

There are carbon atoms, hydrogen atoms, and oxygen atoms on both sides of thereaction. To balance the reaction, keep track of the atoms on each side of thereaction. Start with the compound whose atoms are least present in the reaction(carbon and hydrogen are present in only two compounds each, so we start withCSHTZ). Starting with one CSHIZ, the atoms must be balanced step by step:

1C5H12(l) + Oz(S) ---l> COz(g) + H2O(g)

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1C5H12[) + Oz(g) -->

5 CO2(g) + H2O(g)

5C 5C72F{ ?H?O 10+?O

Balance hydrogen atoms by multiply H2O by six:

1 C5H12(l) + Oz(g) ---r> 5 CO2(g) + 6 H2O(g)

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Balance oxygen atoms by multiply 02 by eight:

1 C5H12[) +8O2(g) --+

5CO2G)+6H2O(g)

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Example 1.29iVhat are the correct coefficients needed to balance the following reaction?

: Co(OH)3(s) + H2SOa(ae) ------+ Co2(SOa)3(aq) + H2O(l)

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SolutionBalancing equations requires that you keep track of each atom. In this case,because of cobalt, the ratio of Co(OH)3(s) to Co2(SOa)s(aq) must be 2 : 1, whicheLiminates choices B and D. The correct answer is found using water. Two molesof Co(OH)3(s) and three moles of H2SOa(aq) have a total of tn 12 H atoms and 1,8

O atoms, making choice A the best answer.

Copyright @ by The Berkeley Review Exclusive MCAT Preparation23

General Chemistry Stoichiometry Balancing Reactions Gt

:Balanced equations can be used to determine the amount of a product from agiven amount of reactant. As we saw in Example 1.2, balanced equations can beused to determine how much product is formed from a given mass of a reactant.We refer to these questions as gram-to-mole-to-mole-to-gram conversions, wherethe overall conversion process is from grams reactant to grams product.

Example 1.30How many grams of water are formed when 25.0 grams of pentane (CSHrz)reacts with oxygen?

A. 18.8 g H2OB. 25.0 g H2OC. 37.5 g H2OD. 75.0 B F{zO

SolutionStep 1: Convert grams reactant to moles reactant by dividing by molecular massof the reactant:

grams reactant " mole reactant

= moles reactantgrams reactant

Step 2: Convert moles reactant to moles product using the coefficient ratio fromthe balanced reaction:

moles reactant t moles Product = moles product

moles reactantstep 3: Convert moles product to grams product by multiplying by molecularmass of the product:

. grams oroductmotes product X *- = gfZrfitS product

moles product

The overall conversion is as shown below:

25.0 gC5H1z ,.T4S#1Z* 6 mgle-I{z-o * 18 g-Hzo =27.s BHzo- 72 gCSHTZ L mole CSFITZ l mole H2O

The ratio of water to pentane comes from the balanced oxidation reaction. Theproduct of 6 x 18 is 108, which is greater than 72. This means that the original25.0 grams is multiplied by a number greater than 1, which in tum means thatthe final number is greater than 25.0 g. This eliminates choices A and B. 108 over72 is less than two, so the final value is less than 50.0 grams, so choice D iseliminated. The only answer that remains is choice C,37.5 g.

Limiting ReagentsDetermining the limiting reagent in a reaction requires comparing the number ofmoles of each of the reactants. The limiting reagent is the reactant that isexhausted first, not necessarily the reactant with the lowest number of moles.when the limiting reagent is completely consumed, the reaction stops, regardlessof the amount of the other reactant. To determine the limiting reagent, theamount of all reactants and the mole ratio of the reactants must be known. If theratio of the moles of Reactant A to Reactant B is greater than the ratio of ReactantA to Reactant B from the balanced equation, then Reactant B is the limitingreagent. If the ratio of the moles of Reactant A to Reactant B is less than the ratioof Reactant A to Reactant B from the balanced equation, then Reactant A is thelimiting reagent.

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Copyright @by The Berkeley Review 24 The Berkeley Review

m ittrre

General Chemistry Stoichiometry Balancing Reactions

Example 1.3LAssuming that the following reaction between oxygen and hydrogen goes tocompletion, which statement is true if 10.0 grams of hydrogen are mixed with64.0 grams of oxygen?

2HzG) + Oz(g) --+ 2 H2O(t) {' ,,u

A. The limiting reagent is oxygen.B. 74.0 grams of water will form.C. 3.0 moles of hydrogen will be left over following the reactionD. 68.0 grams of water will form.

SolutionLr limiting reagent reactions, you must decide which reactant is depleted first.Limiting reagent questions often look like ordinary stoichiometric questions. Therule is simple: If they give you quantities for all reactants, it is probably a limitingreagent problem. In this question, you are given quantities for both hydrogenand oxygen. 10 grams of H is equal to 5 moles of }{2, and 64grams of oxygen is 2rtoles of 02 (remember your diatomic elements!) From the balanced equation,ir-e learn that twice as many moles of hydrogen as oxygen are needed. Thenumber of moles indicates there is a 5 :2 ratio of hydrogen to oxygen, which isgreater than a 2 : 1 reaction ratio, so oxygen is depleted first. The correct choice isanswer A. The question gives you opposing choices in A and B. One of these:-,vo choices must be true. The correct choice is A.

Exarnple 1".32l\-hat is the limiting reagent when 22.0 grams C3Hg are mixed with 48.0 grams

'l?C3H3(l) + Oz(s) € COz(g) + H2O(g)

-{.. Oxygen is the limiting reagent.ts. Propane is the limiting reagent.C. Water is the limiting reagent.D. There is no limiting reagent.

Solution},is question is more difficult than the previous question, but you are still:eciding which reactant is depleted first. Because the limiting reagent is a

=actant, choice C (a product) is eliminated. To solve the question, stick to this

'::nple rule: Compare the actual ratio of the two reactants to the balanced*:'.ration ratio of the two reactants. In this question, you are given unequal mass:-uantities of C3Hg and 02 and a mole ratio that is not l. : 1. Good luck.i-:rnember, the first step is to balance the reaction.

1 C3H3[) +5 O2(g)

--> 3CO2(g) +4H2O(g)

- - grams of C3Hg is 22 moles of C3H6, which is 0.50 moles CSHg. 48 grams of 0244

. 49 moles of 02 (remember your diatomic elements!), which is L.50 moles of.C,2.t2

::.rm the balanced equation, we see that we need 5 moles of 02 for L mole of*,.1{3. The number of moles indicates there is a 1.50 : 0.50 ratio of 02 to C3H8,i,-:.ich is less than 5 : 1. This means that oxygen (O2) is depleted first. Oxygenr- is the limiting reagent, so choice A is the best answer.

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General Qhemistry Stoichiometry Reaction Types

Common ReactionsThere are some reaction types that are standard reactions in inorganic chemistry'

Included among the common reaction types are the following five: 1)

precipitation ,"uition, (also known as double-displacement reactions), 2) acid-

tase reactions (also known as neutralization reactions), 3) composition reactions,

4) decomposition reactions, and 5) oxidation-reduction reactions (electron-

transfer reactions). Oxidation-reduction reactions can be categorized as either

single replacement or combustion. Each reaction type will be addressed in more

aetiit in later sections, so let us consider each type of reaction in minimal detail

in this section.

Precipitation ReactionsA reaction that involves two aqueous salts being added together to form

spectator ions and a solid salt precipitate that drops out of solution is known as a

precipitation reaction. It may iho 6e referred to as a double-displacement reaction,

although that term is not as useful in describing the chemistry' Drawn below is a

sample precipitation reaction:

Na2CrO4(aq) + Sr(NO3)2(aq) + 2NaNO3(aq) + SrCrO4(s)

Aqueous Salt Aqueous Salt Ions Precipitate

Precipitation reactions can be recognized by the solid salt 9"-q" product side_of

the equation. Recognition of the type of reaction is useful for predicting_the

product. Recogniziig a precipitate is highly bene{icial in identifying a d91ble-

displacement ieaction. -

The following solubitity rules can be helpful inidentifying the likelihood of a precipitate's forrning:

1. Most salts containing alkali metal cations (Li+, Na+, K+, Cs+,

Rb+) and ammonium (NH+*) are water-soluble'

2. Most nitrate (NOa-) salts are water-soluble'

3. Most salts containing halide anions (CI-, Br-, I-) are water-soluble(with heavy metal eiceptions such as Ag+ and Pbz+)'

4. Most salts containing sulfate anions (SO+2-) are water-soluble(with exceptions such-as BaZ+,Pb2-,lH82*, and Caz+)'

5. Most hydroxide anion (oH-) salts are only slightly water-soluble.

KOH ind NaOH are substantially soluble, while Ca(OH)2,

Sr(OH)2, and Ba(OH)2 are fairly soluble in water'

6. Most carbonate anion (COg2-), chromate anion (CrO42-)'

phosphate anion (po43-),'and sulfide anion (t-) salts are only

slightly water-soluble.

Acid-Base ReactionsA reaction between an acid (a proton donor) and a base (a proton acceptor) forms

a neutral salt and water. Foi now, recognize that proton donors (acids) must

have a proton on an acid (H-Acid). Acids to recognize are-{!1, HBr' HI' HNO3'

H2SO4-,and NH4+. Bases to recognize are NaOH, KOH, LiOH, and CaCO3'

HCIOa(aq) + LiOH(aq) + LiClOa(aq) + H2O(l)

Acid Base Salt Water

Acid-base reactions can be recognized by the formation of a salt and water on the

prod.uct side of the equation. Aqn"ont acid-base reactions can be identified by

ihe transfer of an H fiom the acid to the hydroxide of the base on the reactant

side of the equation.

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s General Chemistry Stoichiometry Reaction Types

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Composition ReactionsA composition reaction involves the combining of reactants to form a product.The number of reactants exceeds the number of products in a compositionreaction. Entropy decreases and more bonds are formed than are broken incomposition reactions.

PCl3(s) + Cl2(g) + PCl5(g)2 Reactants 1 Product

Composition reactions may fall into other reaction categories as well. In thesanrple reaction, when PCl3 reacts with Cl2,PClg is oxidized and Cl2 is reduced.

Decomposition Reactions-{. 'C.ecomposition reaction is the opposite of a composition reaction. It involves:eactants decomposing to form multiple products. The number of reactants is:es-s than the number of products in a decomposition reaction. Entropy increaseslnd more bonds are broken than are formed in decomposition reactions.

CaSO3(g) ----+ SOz(g) + CaO(s)L Reactant 2 Products

-;ie composition reactions, decomposition reactions can also fall into other::action categories as well.

,Crildation-Reduction Reactions"u:: oxidation-reduction reaction involves the transfer of electrons from one atomlr: another. Loss of electrons is defined as oxidation, while gain of electrons is:e:-raed as reduction (LeoGer). The atom (or compound) losing electrons is:n:.:rsing reduction, so it is referred to as the reductant (reducing agent), while thet:om (or compound) gaining electrons is causing oxidation, so it is referred to as:"e oxidant (oxidizing agent). The oxidation states must change in an oxidation-:eCuction reaction. A sample reaction (below) shows how magnesium is losing:-ectrons (thus being oxidized and having an increase in oxidation state) to-omine (which is being reduced and having a decrease in oxidation state):

Mg(r) + Br2[) ------> MgBr2(s)Reductant Oxidant Salt

Combustion Reactionsilombustion reactions are a special case of oxidation-reduction reactions, wherene oxidizing agent is oxygen gas, and the products are oxides. Typical exampleslf combustion reactions include the oxidation of organic compounds, such as:r-fuocarbons and carbohydrates, into carbon dioxide and water.

1C3H3(aq) +5O2(s) € 3CO2(s) +4 H2O(l)Hydrocarbon Oxygen Carbon dioxide Water

Combustion reactions of both hydrocarbons and monosaccharides balance in a:redictable manner, as shown below:

Hvdrocarbon Combustion

C*H, +,. * l)

o2(g) -------> x Co2(g) *I-uzo(g)

Monosaccharide Combustion

C1H21O1 + xO2(g) + xCO2(g) + xH2O(g)

Copyright @by The Berkeley Review 27 Exclusive MCAT Preparation:w

General Chemistry Stoichiometry Reaction Types

Example 1.33The following reaction is an example of which type of reaction?

MgCl2(aq) + AgNO3(aq)

--

Mg(NO3)2(aq) + AgCl(s)

A. Cation-crossoverB. Oxidation-reductionC. NeutralizationD. Double-displacement

SolutionOf the generic reactions with which you are familiar, there are typical features tonote. In this example, you have two salts undergoing an exchange reaction toyield a precipitate. This makes it a precipitation reaction. No proton wastransferred (eliminating neutralization), and no oxidation states changed(eliminating oxidation-reduction). Cation-crossover is a fabricated name, sochoice A is eliminated. Choice D, double-displacement, is another name for aprecipitation reaction. In double-displacement reactions, you have two saltsundergoing a reaction where they exchange counterions, and one of the newcombinations forms a precipitate. This is shown in a generic fashion below:

MX(aq) + NY(aq) --------+ MY(aq) + NX(s)

Oxidation StatesAssigning an oxidation state to an atom is a matter of distributing electronswithin a bond based on which atom is more electronegative. The oxidation stateof an atom can be determined by assigning it a value of positive one (+1) forevery bond it forms with a more electronegative atom and assigning it a value ofnegative one (-1) for every bond it forms with a less electronegative atom. Theoxidation state is a sum of all these bonding values. In general chemistry, it isoften easiest to say oxygen is -2 (except in molecular oxygen and peroxides),hydrogen is +1 (except in molecular hydrogen and hydrides), and halides are -1(except when they are a central atom in an oxyacid). The sum of the oxidationstates of the elements in the compound must equal the overall charge, so theoxidation state of any remaining atom can be determined by finding thedifference between its charge and the sum of the known oxidation states.

Example L.34What is the oxidation state of manganese in KMnO4?

A.B.c.D.

SolutionTo simplify this example, we will consider that O = -2, arrd alkali metals = +1. Inthis case, K = *1, and there are 4 Os valued at -2 each, for a net oxidation state of-8. Summing oxygen and potassium yields a total of -7. This means that for themolecule to be neutral, the Mn (the only atom remaining) must cancel out that -7

by being +7. In other words, the sum of the oxidation states equals themolecule's formal charge (zero). So in this case, the oxidation state of manganeseis +7. The correct €rnswer is choice D.

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Copyright @by The Berkeley Review 2A The Berkeley Review

General Chemistry Stoichiometry Test-Taking Tips

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I eneral Advice-:: stoichiometry section of this review course is best learned by trial and error: - .-', €x?rnples (i.e., practice with many problems.) For the most part, to be

----:essful in stoichiometry requires being fast at math and being able to see

-:-::Lediately what a question is asking for. These are skills that are acquired and- ,: :lecessarily memorized. Keep in mind that on a multiple-choice exam, the::-: has been done for you, so all you need to do is approximate the answer.'- =:e is no universal shortcut that works in every situation, but finding a range

:::n which only one answer choice fits is a good approach to most questions.

-.::lion will also prove useful on the MCAT. Traditional testing at major-:.','ersities genera\ rewards memorization over intuitive skills, but preparing

: -:e MCAT forces you to hone your analytical and intuitive skills as you recall-:::.:n facts from memory. You should try to emphasize this thought process

'':.',' and regularly throughout your review studies.

- ::.e examples above, several topics and styles of questions were presented.;,=: - :e I'ou move on to the practice questions in the passages, make sure that you

:=rstand the basic principle of each topic and the math typically used to-.::,'.'cr these questions. Math tricks may prove helpful, even for the conceptual

. ,-...ions in stoichiometry. Keep in mind that you are not graded for showing- :: rvork on the MCAT, so don't solve every problem to the last decimal place.

': ?.--,.-ze each question only well enough to eliminate three wrong answers. Be:::ise and efficient rn your problem-solving, not exhaustive. Generally, the- =

-:rons ask you to decide which fraction (or ratio) is larger. This can be done,.---.'by converting all the fractions to values over the same denominator, and

"-:',g for an answer choice that falls within a range. Keep it simple.

'i:.:hematical Tips and Shortcuts- --e the MCAT does not require elaborate calculations, you still must be able to, .. -"'''ith ratios and percentages. Do not use a calculator when practicing for the- -T. The following strategies are useful ways to calculate a value quickly and: ::ir approximation without tables or a calculator:

. : lition and Subtraction:: -:,-ng numbers and adding common terms is a useful way to make addition- : subtraction easier. To split a number, consider how you would round it, and

.' .: split it into the rounded number and the difference between the original and' -,.led numbers. 193 rounds up to 200, so 193 canbe thought of as 200 - 7. 826-- js down to 800, so think of 826 as 800 + 26. Adding and subtracting

' ; - -rr€S Linking like terms. Thus, adding 193 to 826 canbe thought of as:

793 +826=200- 7+ 800 +26= 200 + 800 +26-7 = 1000 +19 =1019- , ::action is done in a similar way. 826 - 193 can be thought of as:

826 - 193 = 800 + 26 - 200 + 7 = 800 - 200 +26 + 7 = 600 + 33 = 633

-.. approach may seem awkward at first, but it is effective and easy when::-:g and/or subtracting several numbers at once. For instance, consider: ::g 213 to 681., then subtracting 411.

l-3 + 681 - 477 =200 + 13 + 700 - 19 - 400- 11 = 200 +7A0 -400 +13 - 79 - 77

=500-17=483

- . :' rght @ by The Berkeley Review 29 Exclusive MCAT Preparation

General Chemistry Stoichiometry Test-Taking Tips

Averaging TermsAveraging terms involves estimating a mean value, and then keeping a runningtally of the differences between the actual values and the estimated average. Tofind the average difference, the running tally is divided by the number of valuesbeing averaged. For instance, the average of 25, 33,21",28, and 30 can be thoughtof as being around 28 (the median value), so the actual average is 28 + /- theaverage difference:

The total difference is -3 + 5 -7 +0 +2 = -3

\vVhen the total difference is divided by 5, it yields an average difference of - 0.6

The average of the five values is thus 28 - 0.6 = 27.4

MultiplicationMultiplication can also be made easier by splitting numbers as you would roundthem. For instance, 97 is 100 - 3. Only one number need be split inmultiplication. Thus, multiplying 97 by 121. can be thought of as:

97 x121. =(100 -3)x121, =(100 x121)- (3x121) =12,100-363

12,100 - 363 = 1,1,,700 + 400 - 363 = 7L,700 + 37 = 1L,737

DivisionIf you memorize the following set of fraction-to-decimal conversions, thenproblems involving division will be far easier:

L = 0.200,L = 0.L66,L = 0.I4g,L = O.IZS,I = 0.111,-1* = 0.091,*1- = 0.083567891.772

Memorizing these decimal values can be useful in several ways. For instance, thedecimal equivalent of the fraction 18 / 66 can be found in the following manner:

18 = 3 = 3x! = 3x (0.091) = 0.27366 1,1. 11

Knowing these decimal values is also useful for estimating in decimal termsfractions that are just less than 1. For instance, the decimal equivalent of thefraction 11./12 can be found in the following manner:

77 =12-1 =72 __L- = 1_ 1 - 1_ 0.0g3 = 0.91712 12 12 12 72

These decimal values are also useful in deciding what to multiply a denominatorby to convert it to some number close to 10, 100, or 1000. For instance, thenumerator and denominator of the fraction 47 / 742 should be multiplie d by 7 ,

because 0.143 = 1,/7 so7 x1.43 is nearly 1000 (actually, it's 1001). The decimalequivalent of the fraction 47 /1.42 can be found in the following manner:

47 =7x47 -g2g -329 + ahttle =0.32g +alittle742 7 x142 994 1000

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StoichiometryPassages

l2 Passages

IOO Questions

Suggested Stoichiometry Passage Schedule:After reading this section and attending lecture: Passages I - III & VI - VIIIQrade passages immediately after completion and log your mistakes.

Following Task I: Passages IV V & IX, (2O questions in 26 minutes)Time yourself accurately, grade your answers, and review mistakes.

Review: Passages X - XII & Questions 87 - IOOFocus on reviewing the concepts. Do not worry about timing.

I. Density ExPeriment

II. CombustionAnalYsis

III. Empirical Formula Determination

IV. Molar Volume of a Cas

V. Elemental AnalYsis

VI. Dilution DxPeriment

VII. Solution Concentrations and Dilution

VIII. Beer's Plot and Light Absorption

IX. Beer's Law DxPeriment

X. Reaction lYPes

XI. Calcium-Containing Bases

XII. Industrial Chemicals

Questions Not Based on a Descriptive Passage

Stoichiometry Scoring Scale

Raw Score MCAT Score

84 - 100 15 - 15

66-83 lo- l2

47 -65 7 -9

34-46 4-6L-33 l-5

(r -7)

(8 - 15)

(14 - 20)

(2r - 26)

(27 - 33)

(34 - 40)

(4r - 47)

(48 - 54)

(55 - 6l)

(62 - 68)

(6e - 78)

(7e - 86)

(87 - loo)

oassage | (Questions 1 - 7)

A student fills a 50-mL graduated cylinder exactly-.,fway with water, adds a previously weighed sample of an

- .l,nown solid, and records the new water level indicated by-: markings on the side of the graduated cylinder. After'' : rrding the volume, she removes the unknown solid and.r:s water to the cylinder to raise the volume back to:-.:rsely 25 mL, replacing any water that may have adhered

::e solid. This procedure is repeated for a total of five- - r,:.rwn solids, and it is discovered that each of the solids

- ,:-: to the bottom of the graduated cylinder. Table 1 shows- ' l"la for all five trials.

Lnknown Mass Volume Reading

I 9.63 s 31.42 mL2 12.38 s 31 19 mL3 14.85 s 29.95 mL4 8.22 s 28.00 mL5 5.64 s 26.41mL

Table I

- ,::ond experiment is conducted with liquids, using a

--.- ,,olumetric cylinder (one that holds exactly 10.00- ,,..rrion) that weighs 42.61 grams when empty. In:::::.:e trials, unknown liquids are poured into the

'::: :'.:ctly to the 10.00-mL mark on the cylinder each

-.- : :r're combined mass of the cylinder and the liquid is, ,: - . .ble 2 shows the results of the second experiment.

: lr-noq'n Mass of Cylinder with Liquid

51.33 e

58.72 s,

53.21 s.

49.03 s

Table 2

,-.: ali of the unknown liquids are immiscible- ,: :-ssolve) in water, how many of the unknown, :.: :loat on water?

t-"t\

-.:;.: unknown solids is the DENSEST?

- : l:re Berkeley Review@ 33 GO ON TO THE NEXT PAGE

7 3. What would be the volume of a 20.0-gram picce of' unknown Solid #1?

A. 13.3 mLB. 15.0 mLC. 25.0 mLD. 30.0 mL

4. How many of the unknown solids can float on Liquid#7?

A. 0B. Ic.2D.3

5 . Which of the following sequences does NOT accurately

reflect the relative densities of the unknown liquids?

A. Liquid #7 > Liquid # 6 > Liquid #8

--B: Liquid #8 > Liquid # 6 > Liquid #9" --c": Liquid #7 > Liquid # 8 > Liquid #9

?-{ Liquid #7 > Liquid # 8 > Liquid #6

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6 , How would the results in Experiment 2 differ from the

actual results, if a heavier graduated cylinder had been

used?

A. Both the mass of the cylinder with the liquid and

the density of the liquid would increase.

B. The mass of the cylinder with the liquid wouldincrease, while the density of the liquid woulddecrease.

p. The mass of the cylinder with the liquid would/' increase, while the density of the liquid wouldremain the same.

,E'. The mass of the cylinder with the liquid woulddecrease, while the density of the liquid wouldincrease.

7 . Which of the following is NOT a unit of density?

,-s,K. ='mLB. !i

cm-ksc. -L

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Passage ll (Questions B - 13)

Elemental analysis is often a preliminary study instructural analysis. A sample compound is placed into achamber with a positive pressure of oxygen gas flowing in.The chamber has an ignition coil that is heated by a cunent.As the reaction proceeds, the pressure in the chamber buildsup. After a short time, Valve #l is opened to allow theproduct gas mixture from the reaction to flow into anevacuated tube containing some magnesium sulfate, whichabsorbs water vapor from the product gas mixture. ThenYalve #2 is opened, allowing the gas to flow into a secondevacuated tube containing some sodium hydroxide, whichabsorbs carbon dioxide from the product gas mixture.. Theapparatus is shown in Figure l. The oxygen tank providesoxygen in excess throughout the process. Valve #3 isconnected to a line that can either evacuate the system orsupply nitrogen to the system.

Valve #1 Valve #2 Valve #3

a(s) NaOH(s)

Figure 1

Four different samples are analyzed. The sample mass ofeach unknown substance is approximately two grams. TableI shows the sample mass placed into the reaction chamber,and the initial and final masses of magnesium sulfate and:odium hydroxide in the side tubes.

Unknown Sample Mass MgSO4 Tube NaOH Tube

I 2.011 gInit: 40.00 g

Fin: 41.21 g

Init: 30.00 g

Finl. 32.94 E

tr 1.995 gInir: 40.01 g

Fin: 41.26 gInit: 30.00 g

Fin: 34.39 s

m 2.003 g Init: 40.00 g

Fin: 42.00 e

Init: 30.00 g

Fin: 34.89 s

IV 2.001 gInit: 40.00 g

Fin: 41.75 eInit: 30.00 g

Fin: 35.99 e

Table I

8. The relative mass percent of carbon in the fourcompounds is BEST described by which of thefollowing?

A. I>II>III>IVB. I>III>II>IVC. IV>II>III>ID. IV>III>II>I

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9. The properties of sodium hydroxide shouldbeing:

A. hydrophobic and semi-reactive with CO2.

.h-l"Vdrophobic and highly reactive with CO2.(-E. hydroscopic and semi-reactive with CO2.

D. hydroscopic and highly reactive with CO2.

10. Why is the oxygen tank attached to a pressure valve?

A. It absorbs excess oxygen.B. It helps cool the reaction chamber.C. It ensures that oxygen gas is in excess.D. It ensures that oxygen is limiting.

11. Which of the four unknown compounds islikely to contain oxygen?

A. CompoundlB. CompoundtrC. Compound ItrD. Compound IV

12. Why are the U-tubes containing the two saltsin the order that they are?

A. To ensure that water is absorbed before theinteract with the NaOH chamber

B. To ensure that carbon dioxide is absorbed begases interact with the NaOH chamber

C. To enhance the reaction between water anddioxide

D. To absorb any excess oxygen gas before itwith NaOH

13. What solid is being formed in the second tubeproduct gas mixture interacts with the salt?

A. Magnesium bicarbonateB. Magnesium carbonateC. Sodium bicarbonateD. Sodiumcarbonate

J includ rassage lll (Questions 14 - 20)

Exactly 10.0 grams of an unknown organic compound is-rred into a flask. The compound is then exposed to excess

" '.'.sen gas to oxidize it to CO2 gas and H2O gas. The' -lized vapor flows through a tube filled first with copper

.le to ensure complete oxidation. The vapor then flows': ugh 100.00 grams of powdered anhydrous sodium sulfate,-,:h binds water vapor to form 112.16 grams of hydrated. . The vapor continues to flow through 100.00 grams of' , :ered anhydrous sodium hydroxide, which binds carbon

-Je vapor to form 123.19 grams of bicarbonate salt.alve?

The unknown compound contains only oxygen, carbon,' - -r'drogen. The mass percent of carbon in the compound

,':ermined to be greater than 50%. In a subsequentr.:-ment, the compound is found to have a molecular mass-- -

'. here between 70 and 80 grams per mole. When the. ;ontaining the unknown compound is left uncapped,

. .:ents slowly evaporate.

- ie information from the combustion reaction can be- .::ed into mass percent for both carbon and hydrogen.LEAS' : rltiplying the grams of CO2 times the mass of one- . arom and dividing by the mass of carbon dioxide, the

. .: the carbon in the original sample can be determined.: -;SS of hydrogen in the original sample can be found in'-:_ar manner. These two mathematical procedures. :. , rhe grams of product molecules into the grams of' ::rm. The final numbers are the grams of carbon and' .-3n. respectively, in the unknown compound. To

-. re the mass percent, the mass of the atom is divided'; rt?SS of the sample. The mass percent of oxygen in--, towll compound is determined by difference.

|]rang3

le gasi

:tbre rL

I carb;lr

t reaif

, -: information from the mass percents of the: .-nt atoms can be used to determine the empirical- . iormula of the lowest coefficients) for the unknown: -.:d. To determine the molecular formula from the- -.- formula, the compound's molecular weight must be' For compounds containing only carbon, hydrogen,

.-en, the molecular formula must always have an even',: rl hydrogens. Molecular formulas with an odd' ,: :i carbons and oxygens, however, are possible.

i r.0 grams of the unknown compound described in-': -:assage were oxidized, what would be observed?.. . The moles of CO2 would double, while the

percentage of carbon in the sampie would remain:he same.

! , The moles of CO2 would double, and thefercentage of carbon in the sample would alsotouble.

The moles of CO2 would remain the same, and the:ercentage of carbon in the sample would also.:main the same.

i The moles of CO2 would remain the same, while::'te percentage of carbon in the sample would:oub1e.

Iter lk

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15. The percentage of carbon by mass in the unknowncompound can be calculated as:

A. 23.79 x 12 x 10 x l00Vo44

B. 23.19 x4Lx10x 100Vot2

C.23.79xf2xJ_xtoovo44 10

D. Z3.jg *44* f_x t00%o12 10

16. What can be said abour rhe boiling point (b.p.) andmelting point (m.p.) of the unknown compound relativeto ambient temperature (T)?

A. m.p. > T2, and b.p. > T"B. *.p. > T2, and b.p. < TaC. m.p. < T2, and b.p. < TuD. m.p. < Ta, and b.p. > Tu

17. Which of the following formulas CANNOT be amolecular formula?

A. C2H4OB. C2H5OC. C3H6OD. CaH3O2

18. How many moles of water are formed from thecombustion of 10.0 grams of the unknown compound?

A. 0.31 moles H2O(i)B. 0.69 moles H2O(l)C. 1.10 moles H2O(l)D. 1.38 moles H2O(t)

19. What is the empirical formula for the unknowncompound?

A. C3H6O2B. CaHsO2C. CaH16OD. C3H19O

2 0. What is the mass percent of carbon in C5H2O2?

A. 26.4Vo

B. 51.1Vo

C. 60.8VoD. 68.2Vo

Passage lV (Questions 21 - 26)

A researcher completely oxidizes exactly 1.00 grams ofan unknown liquid hydrocarbon in a containment vessel toyield carbon dioxide and water vapor. The two gases thusformed are collected and analyzed for quantity. The watervapor is collected by passing the gas through a tubecontaining anhydrous calcium chloride. The carbon dioxidegas is collected by passing the remaining gas through a tube

containing anhydrous sodium hydroxide. The mass of the

carbon dioxide gas thus collected is 3.045 grams at STP.

The carbon dioxide gas is regenerated upon heating the

sodium carbonate and this gas is found to occupy a volume of1.55 liters at STP. The experimental apparatus is shown inFigure 1.

co,. H,o. and -Jq Jt-t fg*"*"!rr5r"","-ffiCaClz(s) NaOH(s)

Figure 1

In a second experiment, the researcher places a 5.0-mLaliquot of the unknown liquid into a capped 1.00-liter flask.The cap has a tiny hole in the top, and the empty flask withcap weighs exactly 120.00 grams. The compound is heated

until it reaches a gentle boil. The vapor escapes through the

tiny pore in the cap. The liquid continues boiling at 31"C,

until none of it remains visible in the flask. The heat source

is removed from the flask, and the contents are allowed tocool back to ambient temperature. As the flask cools, the

vapor in it condenses into a small pool of liquid at the base

of the flask.

The flask and cap are then massed with the condensed

liquid present. The entire system is found to have a mass ofexactly 122.32 grams. This means that the mass of the

liquid is 2.32 grams. It is assumed that at the moment when

the heat source was removed, the flask was completely filledwith vapor from the liquid and that all of the air originally inthe flask was displaced. Table 1 lists the molar volume foran ideal gas at selected temperatures.

Temperature (K) Molar Volume

2"t3 22.4r L288 23.64 L298 24.46 L304 24.96 L3t3 25.69 L323 26.5r L

Table

Copyright O by The Berkeley Review@ 36

D.

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21. What is the mass percent of carbon in CO2 gas?

A. 25.0Vo

B. 21.3Vo

c. 3l.4voD. 35.0Vo

22. How can the molecular weight of this unknown liqube determined?

^ 22.41 grams

2.32 moleD 24.96 grams

2.32 mole

C . (2.32 x 22.41) grams

mole

D. (2.32 x24'961Eramsmole

2 3 . If the mass percent of carbon in the unknown compouis found to be 82.9Vo, what is the empirical formulathe unknown hydrocarbon?

A. CH2B. C2H5C. CH3D. C2H7

2 4. Using the data from the first experiment, how can

mass percent of carbon in the unknown compounddetermined?

6. 1.55 x 12.011 x t00Vo22.41 x 1.00

g, 1.55 x22.41 x l2.0ll x l1OVo1.00

g. 22.41 x l2.Oll x t1OVo1.55 x 1.00

p, 22.41x1.00 xI11Vo1.55 x 12.011

2 5. How many moles of CO2 gas were formed in the

experiment?

1.55 -o1",96,22.41

1.55 -o1", gg,23.6422.41

^o6" gg,

1.55

1.00 moles CO222.41 x 1.55

r. srryftDm

rL-

n,

0-

A.

C.

B.

n liquii

)mpoun;rmula o.

i can thlt

round hr.

nthefu

l: . If in the second experiment the organic vapor had notfully displaced all of the air from the flask by the timethe heat was removed from the flask, how would theresults have been affected?

\ . The mass of unknown liquid collected would be toogreat, so the calculated molecular mass would betoo high.

B . The mass of unknown liquid collected would be toosmall, so the calculated molecular mass would betoo high.

C , The mass of unknown liquid collected would be toogreat, so the calculated molecular mass would betoo low.

D . The mass of unknown liquid collected would be toosmall, so the calculated molecular mass would betoo low.

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Passage V (Questions 27 - 33)

The empirical formula for a compound can be determinedusing the technique of elemental analysis. For hydrocarbonsand carbohydrates, the process involves trapping andremoving water vapor and carbon dioxide gas and thenquantifying the amounts collected. The products can betrapped in many different ways. In this experiment, thetrapping of the gases is accomplished by passing the productgas through a series of low-temperature gas traps. Bylowering the temperature, the gas can be converted intosolids, which cannot flow and thus are easily collected at thebases of their respective temperature traps. The system isevacuated completely and then flushed with nitrogen gas toremove any remaining air. The vapor from the reactionvessel is then allowed to enter one trap at a time. The gas

remains isolated in the region ofeach trap for a short interval,to allow any gases to form a solid or freeze into a liquid. Theapparatus is shown in Figure 1.

Vacuum+

Reactionvessel

Gastrap I

Gastrap II

Gastrap III

Oil bubbler

Figure IIt is important to allow the excess oxygen gas to flow

out of the system. To accomplish this, the line is fitted witha one-way oil bubbler. The oil bubbler maintains the closedsystem by not allowing air to flow into the system, whileallowing the pressure to equilibrate with the environmentthrough venting.

27 . At what temperature should the first trap be held inorder to collect water vapor?

A. 25"C (standard temperature)B. 0"C (melting point of ice)C. -33"C (boiling point ofFreon refrigerant)D. -196"C (boiling point ofliquid nitrogen)

28. The temperatures of the successive traps (i.e., Trap I,Trap II, and Trap III) should be set in what manner?

A. The temperatures should gradually increase, so thateach gas is selectively removed one trap at a time.

B. The temperatures should gradually increase, so thateach gas can be trapped into all three traps,allowing one to determine the moles by difference.

C. The temperatures should gradually decrease, so thateach gas is selectively removed one trap at a time.

D. The temperatures should gradually decrease, so thateach gas can be trapped into all three traps,allowing one to determine the moles by difference.

29. What additional piece of information is necessary todetermine the molecular formula for the experimentalcompound?

A . The volume of CO2 collectedB. The volume of the water collectedC . The volume of the hydrocarbon before the reaction

was carried outD . The molecular mass of the hydrocarbon

3 0. If an unknown compound were combusted in thepresence of excess oxygen, what by-product of thecombustion would be collected to determine thepercentage of sulfur within that compound?

A. SO2B. CS2C. H2SD. Ss

31. The mass percent of oxygen within a compound cannotbe determined directly using elemental analysis. Whichof the following is NOT an explanation for this?

A. Oxygen gas does not exist in the solid phase at anytemperature.

B . When a carbohydrate is oxidized, the oxygen of theunknown carbohydrate can be found in both waterand carbon dioxide.

C . The procedure requires adding excess oxygen, so the

oxygen atoms from the carbohydrate cannot bedistinguished from the oxygen reactant.

D. Oxygen in the carbohydrate, being fully reduced,does not react with oxygen gas.

3 2. Why is the bubbler filled with mineral oil?

A. The oil traps out any unreacted organic vapor.B. The oil can transfer heat to warm the gas rapidly.C. The oil prevents back-flow of gas from the outside

environment.D. The oil filters out any liquid products from the

reaction.

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33. Which of the following is NOT associated withincreasing mass percent of carbon in a hydrocarbon?

I. An increase in the mass of carbon per gram ofcompound

tr. An increase in the mass of water formedoxidation of one gram of the compound

trI. An increase in the mass of hydrogen per gramthe compound

A. II onlyB. Itr onlyC. IandtronlyD. II and Itr only

ith ar ::ssage Vl (Questions 34 - 40)r?

of the f ilution reduces the concentration of a solute by adding

: solvent to the solution. The addition of solvent.-:ises the volume of solution while having no effect onupot

--.:les of solute. Molarity is defined as moles solute per

i - : rlution, so the denominator is increased by the addition:am o' .'.ent. while the numerator is unaffected. To determine

: , .:Jentration, use:

M;Vi = \'t1Y,

Equation 1

- =:: Mi is the initial molarity, V1 is the initial volume,I . the final molarity, and V1 is the final volume.

- ..rtion can be described by the relative concentration of- ..r1 and final solutions. For instance, a fifty percent

. = involves a reduction of the molarity by fifty percent.

' ruld result from doubling the volume of the solution,:,:1by mixing one part solvent with one part solution.

-. .olution is diluted as a solvent is added to it in a

"-;iric flask, until the desired volume is reached. To

--- --omplete transfer of the solute, the original flask is' : rsed with the new solvent, and then the contents are

. .; , rto the volumetric flask. The laboratory instructions. .--old dilution are:

, , a volumetric pipette with a sample of solution': r a beaker and then discard the solution. Repeat this

:::-'edure two additional times to equilibrate the

, ..:entration of the solution on the walls of the pipette.1 the concentration ofthe solution in the beaker.

- . :t-s the treated volumetric pipette, transfer ten

- .iliters of solution to a 100-mL volumetric flask.

: ..e pure water through the pipette and into the

-imetric flask until the flask is roughly eighty:: rent full.

::. ihe pipette aside, and continue to add water to the'-,i: until the base of the meniscus is flush with the

' -nL line on the volumetric flask.

- '--.:h of the following solutions has the GREATEST- ..rity'?

\, . j7c by mass KBr in water:" ':Vc by mass KCI in water- . jVc by mass NaBr in waterl '-c bv mass NaCl in water

- rr-rrlvert 300 milliliters of 0.150 M solution to a

-:ion with a concentration of 0.0075 M, how much. :: must be added?

., , 6,00 liters:l . -<,70 liters- :,30 litersI : 10 liters

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36. What is the final Cl- concentration after you mix 50.00

mL 0.25 M HCI with 25.00 mL 0.50 M NaOH?

A. 0.33 MB. 0.25 Mc. 0.17 MD. 0.15 M

3 7. Why in step III is water passed through the volumetricpipette?

A . To ensure complete transfer of solutionB . To measure the volume of the water added

C. To cool the volumetric pipetteD . To warm the water prior to mixing

38. Which of the following mixtures results in a 10-folddilution?

A. 9 parts solvent with 1 part solutionB. 10 parts solvent with I part solutionC. 10Vo solvent with90Vo solutionD . 91% solvent with 9Vo solution

3 9. Addition of water to an aqueous salt solution would do

all of the following EXCEPT:

A. lower the molality.B. lower the molarity.C. increase the density.D. increase the mass percent ofsolvent.

40. Which of the following would MOST dilute 0.10 MLiCl(aq)?

A. The addition of 100 mL HzO(l) to 25 mL 0.10 MLiCl(aq)

B. The addition of 200 mL H2O(l) to 60 mL 0.10 MLiCl(aq)

C. The addition of 50 mL H2O(l) to 15 mL 0.10 MLiCl(aq)

D. The addition of 150 mL H2O(l) to 50 mL 0.10 MLiCl(aq)

Passage Vll (Questions 41 - 47)

There are many ways in which the concentration of a

solution can be expressed, including:

Molarity: The concentration of a solution as measured inmoles solute per liter solution.

N/r - moles solutelit"rs sotution

Molality: The concentration of a solution as measured inmoles solute per kilogram solvent.

moles solutekilogram tof

""r,t

Percent solution: The percent of solute in a solution bymass or moles.

7o Solution by mass - mass solute x l00Vomass solution

7o Solution by moles = moles solute x IOOVototal moles in solution

Density: The mass of the solution divided by the volumeof the solution.

^ - mass solutionn - uolur. rolr'rl'ion

The concentration of a solution can be expressed in anyof these units, which can be converted into one another as

long as the molecular mass of the solute and solvent areknown. For instance, when the percent solution by mass ismultiplied by the density, the result is mass of solute pervolume of solution. When the mass of solute is convertedinto the moles of solute (which requires knowing themolecular mass), the molarity can be determined. Thepercent solution by mass can be converted into molality bysubtracting the mass of solute from the mass of solution tofind the mass of solvent. That determines the denominator.To get the numerator, the mass of solute is converted intomoles solute, and solving for the molality becomes a simpledivision problem.

Adding solvent to a solution dilutes the solution andthus reduces the concentration of the solute in the solution.Addition of solvent to the solution decreases all of the abovemeasurements of concentration, with the exception of thedensity. The density change of a solution depends on therelative density of the solvent and solution.

4 1. Adding water to an aqueousconcentration alwavs decreasesEXCEPT:

A. density.B. molarity.C. molality.D . mass percent of the solute.

solution of knownall of the following

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42. When 1.0 grams of a salt are dissolved into 100 mLwater, the volume of the solution is greater than 1

mL but less than 101 mL. What can be said aboutconcentration of the solution?

A. The molality of the solution is greater thanmolarity of the solution; the density of the solis greater than that of pure water.

B. The molarity of the solution is greater thanmolality of the solution; the density of the solis greater than that of pure water.

C . The molality of the solution is greater thanmolarity of the solution; the density of the solutiis less than that of pure water.

D. The molarity of the solution is greater thanmolality of the solution; the density of the so

is less than that of pure water.

4 3. An organic compound with a density that is less1.00 g/ml- is added to an organic liquid, also wdensity that is less than 1.00 g/ml.. What can beabout the concentration of the solution?

A. The molality of the solution is greater thanmolarity of the solution; the density of the solis greater than that ofpure organic liquid.

B. The molarity of the solution is greater thanmolality of the solution; the density of the so

is greater than that ofpure organic liquid.The molality of the solution is greater thanmolarity of the solution; the relative densitithe solution and the organic liquid cannotdetermined without more information.

D. The molarity of the solution is greater thanmolality of the solution; the relative densitithe solution and the organic liquid cannotdetermined without more information.

44. Given that a solute is denser than the solventwhich it dissolves, what is TRUE of the concenmeasurements of different solutions made up sothe two components?

A. The solution with the greatest density alsogreatest molarity and molality.

B. The solution with the greatest density alsogreatest molarity, but the molality is theall of the solutions.

C . The solution with the greatest density alsolowest molarity and molality.

D. The solution with the greatest density alsolowest molarity, but the molality is the sa

all of the solutions.

45. F{oqm

I _lrl n

-{. Ifr. :C.:m.5

T,m mw,-@ilmm"J

w&rJ[- U

l- il.

c- il.

D-[

ffiim,rffiC.

mL o-

an 10fout thr

ran the

olutior

ran the

olutior

ran tht

;olutior

ran the

;olutior

:ss tha:with r

be saii

han thsoluticrr

han tbt

solutiolt

han th;ities o;

lnot br(

han thr

sities m

nnot nr

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:ntrati'nc

,o1e1r' d

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r has

!ame

r5. How many milliliters of pure water must be added to100 mL 0.25 M KBr to dilute it to 0.10 M?

.{. 100

B. 150

c. 250D. 500

To achieve the same chloride ion concentration as 1.0

:rams NaCl(s) dissolved into 100 mL solution, hownany grams of MgCl2(s) must be added to enough'.', ater to make 100 mL of solution?

\. 1.0 x 1 x 58'4 gMgCl2(s)2 94.9

B. 1.0 x2x 58'4 g MgCl2(s)94.9

C, 1.0 x1 x94'9 gMgCl2(s)2 58.4 "

D. 1.0 x 2 x94'9 g MgCl2ts)-58.4 "

l:r'en two compounds, Compound A and Compound B,.,:d the fact that B has a higher molecular mass than A,: -t A is denser than B, which of the following mixtures.'. ruld have the greatest mole fraction of A?

{. The mixture of 1.0 grams Compound A with 1.0

grams Compound B3. The mixture of 1.0 moles Compound A with 1.0

moles Compound B

D,

The mixture of 1.0 mL Compound A with 1.0 mLCompound BThe mixture of 1.0 x 1023 molecules Compound A

',vith 1.0 x 1023 molecules Compound B

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Passage Vlll (Questions 48 - 54)

The absorbance of visible light by colored aqueoussolutions is directly proportional to the concentration ofsolute in the solution. Based on this fact, the concentrationof a solution can be determined by monitoring the absorbance

at one wavelength of light. For best results, the detectorshould be focused on the wavelength of highest absorbance(known as l".u*;. The relationship between absorbance and

solute concentration is expressed as

Absorbance = e[C]l

Equation L

where E is the molar absorbtivity constant of the solute, [C]is the concentration of solute, and 1 is the path length of the

light passing in through the cuvette.

A student measures the absorbance for a series of standard

solutions. Once enough data points are collected, the

molarity of another solution using the same solute in an

unknown concentration is analyzed by comparing itsproperties with the experimental data. The molarabsorbtivity constant and cuvette path length remain constant

throughout all the frials, so any difference in absorbance

between the unknown and the reference compounds can be

attributed to differences in solute concentration. Figure 1 is agraph ofthe student's data collected for the standard solutions.

Concentration (in molaritY)

Figure 1

Table I lists the same data summarized graphically above.

Molarity Absorbance

0.10 0.093o.20 0.1 88

0.30 0.2180.40 0.3630.50 0.4560.60 0.5600.70 0.636

Table 1

4 8. If the concentration of a solute were doubled, what

would happen to the absorbance of the solution?

A . It would increase by a factor of four.B. It would double.C . It would be cut in half.D . It would decrease by a factor of four.

4 9. How couldexplained?

the plateau of the following curve be

Solute added to solution

A. The solvation catalyst in solution has becomesaturated.

B. The reverse reaction is favored at higher soluteconcentration.

C. As more solute is added to the solution, the solute

that is already dissolved begins to repel the solvent.D . A maximum solute concentration has been reached,

because no more solute molecules can dissolve intosolution.

50. Adding 50 mL of pure water to a 10.0-mL sample ofaqueous salt solution with an absorbance of 0.518would yield a new absorbance of:

A. 0.518.B. 0.104.c. 0.086.D. 0.259.

e()X

+i

C)O

!a.o

5 L. The concentration can be

the following equations?

A. tcl =Abse.l

B. tcl = t'lAbs

C' tcl-Abs'le

D' tcl - Abs'e

I

found according to which of

5 2. According to the data from the experiment, what is the

concentration of an unknown solution, if it has an

absorbance of 0.242?

A. 0.197 MB. 0.240 Mc. 0.258 MD. 0.289 M

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53. What are the units of e?

A. M.cmB. M.cm-lC. cm.M-lp. 1

M.cm

54. Which of the following relationships may be TRUE?

L As the molarity increases, the absorbance inc

II. If Compound X has a lower molar absorbticonstant (e) than Compound Y, then to haveabsorbance readings for separate solutions ofand Y(aq), the concentration of Y must be gthan the concentration of X.

III. Absorbance = e[C]l at all ], where absorbancelight can occur.

A. I onlyB. II onlyC . I and III onlyD . tr and Itr only

UE?

reases.

'btiviqe equal

f X(aqgreatel

rnce clj

rassage lX (Questions 55 - 61)

-\n experiment to ascertain the effects of solution" ::ntration on the absorbance of visible light studies

-:rons of varying concentration for three different. -:.runds. The solutions are analyzed in cuvettes using a- -\-IS spectrometer set at fixed values specific for each

-:rund. The goal is to maximize the absorbance, so the, - ;.ength of maximum absorbance is used for each

-:'rund. For Compound M, the spectrometer was set at:n: for Compound Q, the spectrometer was set at 413.:ld for Compound T, the spectrometer was set at 691

Table 1 shows concentration and corresponding. ": r: ance for each solution.

Trial Contents Absorbance

0.10 M Compound M 0.362

iI 0.10 M Comoound O 0.299

Itr 0.10 M Compound T 0.511

n' 0.06 M Compound M 0.211

0.06 M Comoound O 0.1 80

\l 0.06 M Compound T 0.307\TI 0.03 M Compound M 0.109

\]tr 0.03 M Compound Q 0.090

x 0.03 M Comoound T 0.r53

Table 1

-:e absorbance of each solution was compared to the: :nce of a sample of distilled water, which remained in

- ; ln the spectrometer for the duration of the study., --se all three compounds have absorbance bands in the-.: range, they are all observed to have a distinct color:1 .nev are exposed to white light. The absorbed color is- ::plementary color of the observed color. The visible-:l:n ranges from a wavelength of 400 nm to 700 nm.

'i-hy is the spectrophotometer set at 561 nm for the:ral involving Compound M?

A , 561 nm is the average wavelength of absorbance forthe complementary color of what is absorbed.

B , 561 nm is the average wavelength of absorbance forthe color that is absorbed.

C . 561 nm is the wavelength of maximum absorbancefor the complementary color of what is absorbed.

D . 561 nm is the wavelength of maximum absorbancefor the color that is absorbed.

How can a compound's molar absorbtivity constant be: L,tained, if absorbance varies with cuvette length?

{. e = Abs.[Compound].1

R c- Abs

[Compound].1

r.- c - [Compoundl'lAbs

D. e=Abs1

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Concentration

43

5 7. Which solution has the GREATEST concentration?

A. Compound M solution, with an absorbance 0.400

B. Compound Q solution, with an absorbance 0.250

C. Compound T solution, with an absorbance 0.500

D. 0.10 M KCl(aq)

5 8. Which graph accurately shows absorbance as a functionof concentration for Compounds M, Q, and T?

5 9. What can be expected for other solutions?

I. 0.05 M Compound M has an absorbance of A =0.181 at l, = 561 nm.

II. A solution of Compound Q with an absorbance ofA = 0.225 at ?t, = 413 nm has a concentration of0.075 M.

m. 0.11 M Compound T has an absorbance of A =0.611 at l, = 710 nm.

A. I onlyB. I and II only

C . I and III onlyD. II and III only

6 0. To form a solution of Compound T with an absorbanceof 0.250 atlv = 697 nm, what must be done?

A. Mix 10.0 mL 0.10 M T with 5.0 mL H2O

B. Mix 20.0 mL 0.10 M T with 20.0 mL H2O

C. Mix 19.0 mL 0.10 M T with 20.0 mLH2OD. Mix 10.0 mL 0.10 M T with 9.0 mL H2O

61. What are the observed colors for each solution?

A. M: green; Q: violet; T: red

B. M: red; Q: violet; T: green

C. M: green; Q: yellow; T: red

D. M: red; Q: yellow; T: green

Concentration

Concentration Concentration

Passage X (Questions 62 - 68)

Stoichiometric reactions can be classified into sixcategories:

Combination Reaction: Occurs with the combinationofreactants to form one product.

PBr3(l) + Br2O '+ PBr5(s)

Decomposition Reaction: Occurs with thedecomposition of one reactant to form two products.

CaCO3(s) + CaO(s) + COzG)

Single-Replacement Reaction: Occurs with theexchange of either the cations or the anions in a salt, but notboth. A single-replacement reaction is also referred to as ano xidation- re ductio n r eac tion.

3 Mg(l) +2 ScBr:(g) -* 2 Sc(s) + 3 MgBr2(s)

Metathesis Reaction: Occurs when two cationsexchange their anions. At least one precipitate falls out ofsolution. A metathesis reaction is also referred to as adoub le -dis plac eme nt r e action.

AgNO3(aq) + KCI(aq)

-> KNO3(aq) + AgCl(s)

Combustion Reaction: Occurs with the addition ofoxygen to a reactant to form oxidized products, (usuallycarbon dioxide and water, when dealing with hydrocarbons andcarbohydrates).

I C3Hs(g) + 5 O2(g) "+ 3 COzG) + 4H2O(r)

Neutralization Reaction: Occurs with the reaction of anacid with a base to form water and a salt.

HNO3(aq)+KOH(s) + KNO3(aq)+H2O(l)

Each reaction is unique from a stoichiometricperspective. When a solid is formed by the reaction of ions(in the metathesis example above), it is referred to as aprecipitate. Gases can also be formed from reactions. Theneutralization of sodium bicarbonate (NaHCO:) yields carbondioxide gas. It is possible to categorize all inorganicchemistry reactions by reaction type.

6 2. How should the following reaction be classified?

Ba(NO3)2(aq) + K2SO4(aq) +

2 KNO3(aq) + BaSOa(s)

A. Combination reactionB. DecompositionreactionC . Single-replacement reactionD. Metathesis reaction

Copyright @ by The Berkeley Review@ 44 GO ON TO THE NEXT P

6 3. A precipitate is LEAST likely to occur as a result ofA. combustion reaction.B. decomposition reaction.C . single-replacement reaction.D. metathesis reaction.

64. What is the precipitate formed when aqueous sodiodide reacts with aqueous calcium nitrate?

A. CaIB. NaNO3C. CaI2D. Na2NO3

6 5. What is the gas formed when magnesiumneated with hydrobromic acid?

A. Hydrogen gas (H)B. Hydrogen gas (H2)C. Carbon dioxide gas (CO2)D. Magnesium bromide gas (MgBr2)

6 6. What type of reaction is LEAST likely to formdioxide gas?

A. Combustion reactionB. Metathesis reactionC. Decomposition reactionD. Neutralization reaction

6 7. How should the following reaction be classified?

CaBrz(l) + Cl2(g) + CaCl2(s) + Br2(l)

A. Combination reactionB. DecompositionreactionC . Single-replacement reactionD. Metathesis reaction

68. What type of reaction is MOST likely to hanegative value for the change in entropy (AS)?

A. Combination reactionB. DecompositionreactionC . Single-replacement reactionD. Combustion reaction

're'urs$,age Xl (Questions 69 - 78)

- '';ium carbonate is found in many everyday products,r rrLr

'- ., narble, chalk, and antacids. Commercially, it can be

| : :rid as either calcium oxide or calcium carbonate.r,,. , -:r oxide converts to calcium carbonate upon exposure

r -r":i- n dioxide under high pressure. At room temperature,r,,. -:, ;arbonate is relatively insoluble in water. Theri "- ,-." of calcium carbonate increases as the pH of the

luilir -u . -! solution decreases, because calcium carbonate isrlrii, , R.eaction I can be combined with Reaction II tor rr' :* :alcium oxide into calcium carbonate in water.

CaO(s) + HzO(l) + Ca(OH)z(aq)

Reaction I

-: DH)2(aq) + CO2@) -+ CaCO:(s) + HzO(l)

Reaction II

, : OH)2 and CaCO3 both readily form a relativelym -:-e white solid precipitate in water. Because of this. ':.;bility, the products of both Reaction I and Reaction

- :: :;.sv to isolate from solution. Fortunately, calciumr"', *-. r -;e is more soluble in water than is calcium carbonate,,ur" ":. :trlows for the selective precipitation of calciumrlr- . - r:e in Reaction II. Because of this, industrial processes

,,' ,,,.ating calcium carbonate are primarily water-based., , -:r carbonate can also be formed according to the

'r ;',.-,S equilibrium reaction:

CaO(s) + CO2(g): CaCO3(s)

Reaction III

,il I Ja OH)2 is considered to be which of the following?

\ , An amphoteric saltB. A non-metal hydroxideC . An Arrhenius acidD , An Arrhenius base

- 'i''hich of the following molecules are NOT held::gether by ionic bonds?

r. cot2-II. COz

I. CaO

A. I onlyB, tronlyC . III onlyD , I and II only

- ::.. right @ by The Berkeley Review@ 45 GO ON TO THE NEXT PAGE

7 1. The mass percent of calcium is LEAST in which of the

following molecules?

A. CaO(s)B. Ca(OH)z(s)C. CaCO3(s)D. CaCl2(g)

7 2. If 10.00 grams of Ca(OH)z(s) produces 5.00 grams ofCaCO:(s), what is the percent yield for the reaction?

A. 31Vo

B. 68Vo

C. 14Vo

D. l00Vo

7 3. What is the mass percent of calcium in CaO?

(Ca = 40 g/mole O = 16 g/mole)

A. 28.6Vo

B. 50.0vo

c. 66.1Vo

D. 7l.4%o

7 4 . 28.0 grams of CaO(s) when reacted with 10.00 grams ofH2O(1) would yield which of the following?

(Ca= 40 g/mole O = 16 g/mole H = 1 g/mole)

A. 37.00 grams Ca(OH)2 with leftover waterB. 37.50 grams Ca(OH)2 with leftover CaO

C. Exactly 38.00 grams Ca(OH)2 with no leftoverD. Exactly 42.60 grams Ca(OH)2 with no leftover

75. Which of the following is required to neutralize 5.00mL of 0.20 M CaCO3(aq)?

A. 10.0 mL 0.10 N NaOH(aq)B. 10.0 mL 0.10 N HNO:(aq)C. 10.0 mL 0.30 N HrPO+(aq)

D. 10.0 mL 0.20 N HCI(aq)

7 6. What is the final concentration of Ca(OH)z(aq) after

50.0 mL of pure water are added to 5.0 mL of 0.50 MCa(OH)2(aq)?

A. 0.055 M Ca(OH)z(aq)B. 0.050 M Ca(OH)z(aq)C. 0.046 M Ca(OH)z(aq)D. 0.025 M Ca(OH)2(aq)

7 7 . If 20.0 mL of 0.20 M CaCl2(aq) were mixed with 30.0mL of 0.30 M CaCO3(aq), what would the final Ca2+concentration be?

A. 0.233 M Ca2+1aq;B. 0.250 M Ca2+1aq;C. 0.260 M Ca2+1aq;D. 0.267 M Ca2+1aq;

78. If 50.0 grams of CaCO3(s) are completely neutralizedwith HCl, how many liters of COZ(g) form at STp,knowing that one CaCO3 yields one CO2?

A. 4.48LB. 10.00 Lc. 1t.20LD. 13.56 L

Copyright @ by The Berkeley Review@ 46 GO ON TO THE NEXT PA

Passage Xll (Questions 79 - 86)

Every year, a substantial amount of the countless tons ofchemicals produced worldwide is used in the manufacture offertilizers and plastics. Described below, according to acommon element each one contains, are some of the typicalchemical fertilizers used in America.

Potassium:

Salts containing potassium are referred to as potash.most common forms are K2SO4, 2 MgSOa.K2SO4, andKCl. The amount of potassium per gram of salt is importanto know when determining the quantity of fertilizer neededa job. The compound richest in potassium (by mass per<is potassium oxide, K2O. The potassium conteni ofpotash is expressed as a fraction of the potassium ipotassium oxide (K2O).

Nitrogen:

Salts containing nitrogen are very useful as fertilizers.most common forms are NH4NO3, (NHa)2S04, uNHaH2POa. A very common organic fertilizer containinitrogen is urea, H2NCONH2, which is made from amand carbon dioxide. Ammonium sulfate is also made fammonia by combining Reaction I with Reaction II:

2 NH3(aq) + COzG) + H2O(l) ----+ (NHa)2CO3(aq)

Reaction I

(NHa)2CO3(aq) + CaSO4(aO

-->(NHa)2SOa(aq) + CaCO3(s)

Reaction II

Phosphorus:

f0" fr]tmn

A.E.c.n"

rmJt-:-c-D.

ilhrydil-

l-

c-

L-

Salts containing phosphorus are also very usefulfertilizers. The most common form is Ca(H2pOaCalcium bisdihydrogenphosphate is produced by Reaction

2 Ca5(POa)3F(t + 7 H2SOa(aq)

-.;3 Ca(H2POa)2@q) + 7 CaSO4(aq) + 2 HF(e)

Reaction III

Fluoroapetite (Ca5(POa)3F) is added as the limitreagent in Reaction III. This is done to maximize theyield of phosphorus in the reaction. Industrially, theyield of a reaction and the mass percent of its productcritical in terms of profit margin for the fertilizer producer.

79. Which of the following compounds has tGREATEST amount of potassium per gram?

I{(

I

IU

,[

A. K2OB. K2SO4C. KCID. KNO3

ilI

$ l"t. Which of the following compounds has the LOWESTmass percent of nitrogen?

^{. H2NCONH2B. NHaNO3C. (NHa)2SOaD. NHaH2POa

\\hat is the mass percent of nitrogen in urea?

A..23.3VoNB.3l.9VoNC . 46.6V0 ND. 66.'7VoN

frhy is Ca5(POa)3F the limiting reagent in the;1 nthesis of Ca(H2POa)2?

-\. It prevents leftover Ca5(POa)3F from beingrvasted, so that P-containing compounds are

conserved.ts " It allows leftover Ca5@Oa)3F to be wasted, so that

P-containing compounds are conserved.C. It prevents leftover Ca5(POa)3F from being

rvasted, so that P-containing compounds are notconserved.

D . It allows leftover Ca5(POa)3F to be wasted, so that

P-containing compounds are not conserved.

L i l,:r.0 grams of (NH+)zCO: are used in Reaction II to:':';in 10.0 grams of (NH+)ZSO+, then what is the

rtr:ent yield for the reaction?

s . Less than 50Vo

M . Greater than 50Vo, but less than 7 5Vo

ll . Greater than75%, but less than l00%o

D ., Greater than loj%o

T:e rrolarity of potassium is GREATEST for which oftlrc :lliowing solutions?

4., -0.0 g KCt in enough water to form 100 mL of>olution

tons ofrture ofrg to

l.t4, a

ded

erceofium

rS.

4,tarru

):(ac f

)03(sr

:ful,o4

lon

il. 10.0 g K2SO4 in enough

solution1: 10.0 g K2CO3 in enough

roiutionDt. 111.0 g KNO3 in enough

solution

water to form 100 mL of

water to form 100 mL of

water to form 100 mL of

",,,, nnnr:git O by The Berkeley Review@ 47 GO ON TO THE NEXT PAGE

85. From which of the following reactions is it EASIESTto isolate the desired product?

A . A reaction yielding the product as a precipitateB. A reaction yielding the product as a liquidC . A reaction yielding the product as a gas

D. A reaction yielding the product as an aqueous solute

86. Which of the following relationships must be TRUE?

I. pKal(H3pOa) is less than PKa(HF)

tr. pKa2(H3pOa) is less than PKa2(HzSO+)

m. pKa(HF) is less than PKal(H2SOa)

A. I onlyB. II onlyC. IandtronlyD. IIandItronly

Questions 87 - 100 are NOT basedon a descriptive passage

8 7. If 25 grams of oxygen are combined with 20 grams ofpropane gas, which of the following statements wouldbe TRUE after the compound is ignited?

A. 45.0 grams of carbon dioxide forms.B. 38.0 grams of water vapor forms.C. Oxygen is the limiting reagent.D. Propane is the limiting reagent.

88. The hemoglobin in red blood corpuscles of mostmammals contains approximately 0.33V0 iron byweight. If osmotic pressure measurements show thatthe molecular weight is 68,000 for hemoglobin, howmany iron atoms must be present in each molecule ofhemoglobin?

A. 1

8.4c.224D. 400

89. 9.00 grams of a sugar are burned in a containmentvessel, and all the COz is collected. The volumeoccupied by the CO2 at STP is 6.72 liters. If the

molecular weight of the sugar is 180 g/mole, what isthe ratio of 02 to CO2 in the balanced equation?

9 0. What is the molecular formula for an unknown gas withthe empirical formula C2H3O, if 1.00 grams of the

unknown gas occupies 26OmL at STP?

A. C2H3OB. CaH6O2C. C6H9O3D. CsHl2Oa

91. What volume of O2G) is produced from 1.0 g BaOupon its decomposition to Ba(s) and OZ(g) at STP?

A. 0.074 LB. 0.100 Lc. 0.148 LD. 0.166 L

A. 3:3B. 3:6C. 6:3D. 6:6

Copyright @ by The Berkeley Review@ 4A GO ON TO THE NEXT P

92. For the following reaction, calculate the massMg2P2O7(s) that is formed from the decomposition2.0 grams MgNHaPOa(s).

MgNHaPOa(s) -+ Mg2PzOz(s) + 2 NH3(g) +

A. 0.8 grams

B. 1.7 grams

C. 2.2 grams

D. 2.8 grams

9 3. A compound containing 50Vo by weight of Element(atomic weight = 40) and SOVo by weight of Element(atomic weight = 80) is one in which:

A. the molecular formula isXZorZX.B. the simplest formula isXZorZX.C . the simplest formula is XZ2 or Z2X.D . the simplest formula is X2Z or ZX2.

94, 11.89 grams of hot iron are exposed to a

stream of pure oxygen for ten minutes. At the end

this time, the completely oxidized sample weighs 1

grams. The empirical formula for the compoundformed is MOST accurately written as:

A. Fe3O2.

B. FeO.C. Fe2O3.D. FeO3.

95. A stable compound consisting of 53.4Vo C, ll.0%o

and the remainder O has a molecular weight ofgrams/mole. The molecular formula for the

is:

A. C5H6O2.B. C3H6O3.C. C4H26O.D. C4H16O2.

96. In reducing CrO42-(aq) to Cr2O3(s), how

oxidation state of chromium change?

A. From +6 to +3B. From +4 to +3

C. From +3 to +4D. From +3 to +6

does

fl

nd16.

1t

)f

-\n unknown metal is found to combine with oxygen ina 2 : 3 ratio in the molecular formula. The metal oxiders approximately 53Vo metal by mass, and the remainderrs oxygen. What is the MOST probable identity of thenetal?

.\. CalciumB, IronC. ChromiumD. Aluminum

thich of the following organic compounds has the

;REATEST mass percent of carbon?

\. Acetic acid (CH3CO2H)B. Ethanol (CH:CHzOH)C. Methyl acetate (CH3CO2CH3)D. Glucose (C6H12O6)

l! -,i hen one gram of each of the following organic

- nrpounds is burned (oxidized), which one yields the:REATEST amount of carbon dioxide (by mass or-":ies)?

\ , Acetic acid (CH3CO2H)B. Ethanol (CHTCHzOH)tl , Methyl acetate (CH:COZCH:)D, Glucose (COH1ZOO)

l,r'-hat is the mass percentage of chlorine rn\1g(ClOa)2?

.\. 7t.0voB. 40.6Vo

C . 31.8Vo

D. 15.9Vo

;;:i,ght @ by The Berkeley Review@ 49 YOU ARE DONE.

1.8 2.D 3.A 4. B 5.A 6.C1.D 8D 9.B 10.C 11.D 12.A

13. C 14. A 15. C 16. D 17. B 18. B19. C 20. B 21. B 22. D 23. B 24. A25. A 26. D 27. C 28. C 29. D 30. A31. A 32. C 33. D 34. D 35. B 36. C37. A 38. A 39. C 40. A 4r. A 42. A43. C 44. A 45. B 46. C 47. C 48. B49. D 50. C 51. A 52. C 53. D 54. C55. D 56. B 57. A 58. D 59. B 60. c61. D 62. D 63. A 64. C 65. C 66. B67. C 68. A 69. D 70. D 1r. D 12. A73. D 14. A 15. D 16. C 11. C 18. C79. A 80. D 81. C 82. A 83. B 84. C85. A 86. A 87. C 88. B 89. D 90. B91. A 92. B 93. D 94. C 95. D 96. B97. D 98. B 99. B 100. C

Stoichiometry Passage Answers

1' Choice B is correct. In order for a liquid to float on water, the liquid must be both immiscible in water and lessdense than water. All of the liquids are assumed to be immisciLle in water, accord.ing to the question, so theonly stipulation that remains to be considered is the density of each liquid. The densiiy of watir is defined as1'00 grams per milliliter. We know that the mass of the empty volumetric cylinder is 42.6r grams and itscapacity is 10'00 mL. For the density o1,u"y liquid to be less thin 1.00, a 10.00-mL sample of the liquid musthave a mass of l1s; t\n 1o.00 grams. This *"un, that when 10.00 mL of a liquid is added to the volumetriccylinder, the liquid and cylinder must have a combined mass of less than 52.61 grams. If the combined mass isless than that, then the density of the solid must be less than 1.00. According to tubl" 1, unknown Liquid #6(with a combined mass with the volumetric cylinder of 51.33 grams) and unkriown Liquid #9 (with a combinedmass with the volumetric cylinder of 49.03 grams) are the or,!, t*o' lrr,known liquids with combined. masses ofless than 52.67 gtams. Only Liquid #6 and Liquid #9 can float on water. The best answer is therefore choice B.

2' Choice D is correct. This question requires evaluating the density for unknown Solids #2, #3, #4, and #5.Density is defined -as

mass per volume. In this case, the'volume of ihe solid is obtained by taking the volumereading from the chart for each unknown ald subtracting 25.00 mL for the volume of the water iready in thecylinder' This method of measurement is known as thei'volume by displacement technique.', The followingtable shows the values of mass, volume, and density obtained for each solid:

Wl

[uM!@

ds

Unlcrown Mass (e) Volume (mL)

31.19-25.00=6.79

29.95-25.00=4.95

28.00-25.00=3.00

26.47-25.00=1.41

Densitv

72.38g =2.00

g6.19mL mL1'4.859

=3.00 g

4.95mL mL8.229 _rrn g

3.00 mL mL5.649

=4.00 g

2 12.38

3 14.85

4 8.22

5 5.64

cffiil,@

0oddcL@ll

ffiftu[r@tifi

rdmrffi&c

@

1.41mL mLThe unknown solid with the greatest density is unknown Solid #5. The correct answer is thus choice D. Thenumbers could have been compared to one another to obtain the relative values.

3' Choice A is correct.- The density of unknown Solid #1 must be determined first. It is found by dividing its massby its volume (found by difference):

p = 9.639 - 9.639 = j.5o g

' 37.42 - 25.00 mL 6.42mL mLBecause the density t: 1q0 grams per milliliter, a20.0-gtam piece of the solid must have a volume of 13.3 mL,because 20'0 grams divided by 13.3 mL is equat to r.s-o gru*, per milliliter. The correct answer is choice A.Choices C and D should have been eliminated immediailly, because when the volume of an object is greater

:lii,T: Tjl:j$ t:l:"',1,t or.the object is less jl1" 1 09: soud #1 sinks when placed into waier, indi"catingthat its density is greater than that of water, which is 1.00 grams per milliliter. Distinguishing between

choice A and choice B requires looking more closely at the ratios.

4' Choice B is correct. In order for a solid to float on unknown Liquid #7, it must have a density less than thatliquid. The density for unknown Liquid #7 is:

, _ 58.72-42.619 _ 1.6.11g = L611 g

10.00 mL 10.00 mL mLonly Solid #1, with a density of only 1.50 grams per milliliter, has a density less than 1.611 gramsmilliliter. The best answer is choice B. The teGtirre densities of Liquid #7 andsoiia +r are shown below:

pliquido, = #*. = #k,psoria+r = ##,= ##, - #k, ##Copyright O by The Berkeley Review@ 50 Section I Detaited

Choice A is correct. The following table shows how to solve for the densities of the four liquids:

uI less i

Unlmown Mass (s)

6 51.33 - 42.61- = 8.72

7 58.72 - 42.67 = L6.77

8 53.2L - 42.67 = 10.60

9 49.03 - 42.61. = 6.42

Volume (mL)

10.00

10.00

10.00

10.00

Densitv ( I ).-mr-8.729

=o.gr2 E

10.00 mL mL1.6.71.9

=L67I g

10.00 mL mL1o.6og

=1.060 I10.00 mL mL6A29

=0.642 E

o the:d as

d itsmustretricrss isrd #6,ined

es ofB.

r #5.iumen therving

I mn*ce A.:ea

atinq

r. tl:d

10.00mL mL

":om the calculations, the relative densities in descending order are: Liquid #7 > Liquid #8 > Liquid #6 >

-rquid #9. The answer choice that does not follow this pattem is choice A: Liquid #7 > Liquid #6 > Liquid #8.-lhis question could also have been solved by comparing the masses in the chart for the cylinder and liquid:-.nbined. All of the liquids have the same volume (10.00 mL) and were in the same cylinder, so the same mass

-o subtracted from each in determining the liquid's mass. This would have saved much time.

[hoice C is correct. If the mass of the cylinder were heavier than 42.61, grams, then the reading for the mass of:-.e liquid and cylinder combined would be greater than it was in Experiment 2. However, the mass of liquid in: ired volume is the same, so its density does not change. No matter what container is chosen to hold it, the:=nsity of a liquid is an invariant property of that liquid. This is best reflected in answer choice C.

Chroice D is correct. Density is defined as a measure of mass per volume. The units for density should therefore:=lect a mass unit divided by a volume unit. In answer choice A, the mass is measured in grams and the volume--. reasured in milliliters, which makes choice A acceptable. In answer choice B, the mass is measured in, -:-rces and the volume is measured in centimeters cubed, which makes choice B acceptable. You should recall--:t a milliliter is a centimeter cubed. In answer choice C, the mass is measured in kilograms and the volume is:=asured in liters, which makes choice C acceptable. In answer choice D, the weight (and not necessarily the* '.s) is measured in pounds and is divided by an area (dimension squared) and not by volume. Choice D is thus: :..easure of pressure, not density. The correct answer is choice D.

-hoice D is correct. Given that all of the samples were of nearly equal mass (between 1,.995 g and 2.011 g), the:.::est mass percent of carbon is in the compound that has the greatest mass of carbon. The compound with the;:=":est mass of carbon produces the greatest mass of carbon dioxide gas. This means that the easiest way to

. -' e this question is to compare the amount of carbon dioxide collected for each sample, as listed in column 4:-= \aOH tube column) in Table 1. From the data in the NaOH tube column, the greatest mass of CO2 is" *=;ted from Compound IV (35.99 - 30.00 = 5.99), eliminating choices A and B. Because a greater mass of CO2 is

r - .,::.rc€d by oxidizing Compound III than by oxidizing Compound II, choice D is the best answer.

ll: ::ce B is correct. The role of the sodium hydroxide salt is to bind CO2, not to bind H2O. This means that ther,.: =:rould be both hydrophobic (non-water-binding) and reactive with carbon dioxide, making choice B the

" "x r rrlsw€r. The term "hydroscopic" refers to a compound with a high affinity for binding water.

-,: : irce C is correct. The pressure valve is designed to allow oxygen gas to flow from the tank into the system' -:: a threshold pressure is maintained. The oxygen partial pressure can be controlled and maintained at a::- .evel. The oxygen tank is left open with a positive pressure of oxygen to ensure that oxygen gas is: --:.ual1y flowing into the system, so choice A is eliminated. Nothing was mentioned about the temperature of

-" : :',,\'gen gas, so choice B is eliminated. Oxygen gas is always present, so it is in excess and is not a limiting'', r i:r:t. This eliminates choice D and makes choice C the correct answer.

l.- : ice D is correct. The greatest mass percent of carbon is found in the compound with the smallest number of' ;:r atoms in its formula. Thus, the compound least likely to contain oxygen is the one that produces the most

" - :pon combustion. That is Compound IV. In additionr pure carbon when oxidized yields 7.33 g CO2.- : ::.-r-rnd IV yields 6.00 grams CO2, so it is close enough to pure carbon to assume no oxygen is in it, choice D.

ri * ri: 3 by The Berkeley Review@ 5l Section I Detailed Explanations

12. Choice A is correct. When the first valve is opened, the gas is exposed to magnesium sulfate, which binds water,but not carbon dioxide. This means that water is bound first, leaving an atirosphere of excess oxygen gas andcarbon dioxide. This is important, because the sodium hydroxide compound can bittd both carbon iioxide andwater. Using magnesium sulfate first to remove the water vapor ensures that all water is removed. from the gaswhen the second valve is opened, exposing the gas to the sodirm hydroxide salt. Any increase in the mass oJmagnesium sulfate is due to the binding of water. Any increase in the mass of sodium hydroxide is due to thebinding of carbon dioxide. If the gases were first exposed to sodium hydroxide, the increase in mass would be dueto both water and carbon dioxide. Choice A is the best answer. Excesi oxygen leaves the system as a free gas.

Choice C is correct. The solid formed in the second tube results from the reaction of sodium hydroxide (NaOHwith carbon dioxide (COil, so it must be a sodium salt. This eliminates choice C. The following reactiorconfirms that the best answer of the given choices is sodium bicarbonate, choice C:

NaOH(s) + CO2(g) _-+ NaHCO3(s)

Choice A is correct. If 20.0 grams of the unknown were oxidized, instead of 10.0 grams, then the amountcarbon dioxide and water formed as products would double. This eliminates choices L and D. The mass percrof carbon should remain the same, because the mass of carbon dioxide formed and the mass of the compound bodoubled. The best answer is choice A. The mass percent of carbon is constant, because the formula is cbnstant.

Choice C is correct. To determine the mass percent of carbon in the unknown compound, the grams of carboncarbon dioxide (COz) are divided by the total number of grams of original compound. Tie mass of carbdioxide formed is found by subtracting 100.00 grams of original sodium hydroxide from the 1zg.7g gramsbicarbonate salt. The difference is the mass of carbon dioxide that binds the salt.

!8.. (llt:tr

fldi

13.

c

cffieI,ffi

74-

1,5.

23.79 gramsCo2 x 72gC

,r"*"t-#qThis makes choice C the best answer.

x100% =23.79x]2xLxfi)%44 10

16. Choice D is correct. In the first sentence of the passage, we read that the unknown organic compound is porinto a flask. The term "poured" implies that the compound flows, which makes it a-fluid. ThL fact thai itbe poured into a flask means that it is flowing down, which defines it more specifically as a liquid.compound is a liquid at ambient temperature (room temperature). This means that the melting point itthan ambient temperature, because at room temperature ii has already melted into a liquid. rhe"boiling poiis greater than ambient temperature, because at room temperature it has not yet boiled- into a gur. eeJrrsereadily evaporates, the boiling point may be close to ambient temperature; bui because it is a liquid at ambitemperature, the boiling point must be greater than ambient temperature. The best answer is choice D.

Choice B is correct. As stated at the end of the passage, the molecular formula for a compound with jthydrogen, oxygen, and carbon cannlt have an odd number of hydrogens. An odd number of hydrogens resultsan odd number of bonding electrons (electrons present in bonds). Considering that there are two electrons Ibond, an odd number of bonding electrons results in only half of a bond somewhere in the compound (whlequates to a single electron, or free radical.) A half-bond is not stable (possible to isolate physically), so choB,-with five hydrogens in the formula, is not possible. Knowledge from organic chemistry .un prorru usefulsolving general chemistry questions. You must incorporate informition from many sources to excel at this

Choice B is correct. Thgre are 1'2.76 grams of water produced from the oxidation of 10.0 grams of the 1nkno*This can be determined by subtracting 100.00 grams for the mass of the anhydrous sodium sulfate from the firmass of 772.76 grams for the hydrated sodium sulfate salt. The moles of waier formed are greater than 0.5 nbecause 12.16 grams divided by 18 is greater than 9 grams divided by 18. The moles of waler formed is less1.0 moles, because 72.1'6 grams divided by 18 is less than 18 grams divided by 18. The best answer is therechoice B.

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Choice C is correct. The question asks for the empirical formula, but because information is given about themolecular mass, you may wish to consider the molecular formula first, then eliminate choices based on thatinformation. According the passage, the molecular mass has a value between 70 and 80 grams per mole, sochoice B (MW = 48 + 8 + 32 = 88) and choice D (MW = 96 + 10 + 16 = 122) are eliminated. The only choices with amass in the range of 70 grams to 80 grams are choice A (which has a mass of 74 grams per mole) and choice C

(which also has a mass of 74 grams per mole). Choice A is 36 percent carbon, while choice C is 4& percent' 74 74'carbon. The mass percent of carbon is greater than fifty percent, so choice A is eliminated, leaving choice C as

the best answer. The exact numerical value for the mass percent of carbon can be approximated as follows:

23.79 gramsCo2 x 72gC- MgCOz

10J grrt* ""1*"*"x 100% = 23.79 x L x J- x 100"/'

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Choice C is both the molecular formula and the empirical formula.

Choice B is correct. The mass percent of carbon in C5H12O2 is found by dividing the mass of five carbon atoms(60 amu) by the molecular mass of C5H12O2 $O + 12 + 32 = 104 amu). The value determined when 60 is dividedby 704 is less than60"/o, because it is less than 60 divided by 100. The best choice for a value close to, but lessthan, 60% is choice 8,57.7"/o. You really should choose answer choice B, if you know what's best for you.

mass percent of carbon - 5 x 12 = 60 < 60 = 60ohffi+12+32 704 100

:: Choice B is correct. The mass percent of carbon in carbon dioxide is the mass of carbon (12.011) divided by the

massof carbondioxide (44.009)x100%. Thefraction12 reducesto 3 =3*a-=3x0.091 =0.273x1,00%=27.3"/o.44 11 11To make math easier, you should memorize the following set of fraction-to-decimal conversions. Ademonstration of how these fractions are useful is found on page 30.

1= 0.333,L= 0.250,L=0.200,L= 0.1.66,L= 0.143,1 = 0.125,13456789 = 0.100,! = 0.091,! = 0.091,! = 0.08310 11 72

This trick works well, once learned. It may seem awkward at first, but try it. The correct answer is choice B.

:" Choice D is correct. The second experiment was conducted to determine the molecular mass of the unknownLiquid. At 304 K, the boiling point of the liquid (and thus, the temperature of the gas), the volume of gas isexactly 1".00 liters, assuming that the flask is completely filled with vapor from the liquid. Molar volume atthis temperature is 24.96literc per mole. At 304 K, the gas has a density of 2.32 grams per liter. \Alhen thisvalue is multiplied by molar volume at 304 K the liters cancel out and the remaining units are grams per mole.The molecular mass is 2.32 g/L x 24.96 L/mole = 2.32 x 24.96 g/mole = 58 g/mole. The best answer is choice D.You should always consider units when looking at math questions. The question asks for a molecular weight,rvhich has units of grams per mole. The numbers in the answer choices arc 2.32 (which is in grams per liter),22.41 (which is in liters per mole), and 24.96 (which is in liters per mole.) To get the target units, grams perliter are multiplied by liters per mole, choices C or D. Choice D has the correct molar volume of the unknownliquid at 304 K.

:i: Choice B is correct. You know the molecular mass of the unknown compound (58 grams per mole) and the masspercent (82.9%), so you can make an intuitive determination of the molecular formula. The mass of carbon in thecompound must equal either 12, 24, 36, or 48 (multiples of 12). Given that 82.9% of 58 is closest to 48 of themultiples of. 12, it can be assumed that the molecule has four carbons. This leaves ten hydrogens (to round outthe 10 grams in 58 grams per mole not accounted for by the four carbons); so the molecular formula is C4H19,rvhich has an empirical formula of C2H5. AII of this intuition should lead you to pick B.

if you want to solve for the empirical formula exactly, here is the math using the compact formula, where a100-gram sample is assumed. Do lengthy math calculations only if they are absolutely necessary.

Cl2gFltz.t = C6.gHl7] = C6.eH17.1 = C1H2.5 = C2H51.2 1 6.9 5.9

5.' Section I Detailed Explanations

24. Choice A is correct. The mass percent of carbon in the unknown compound must be less than 100%. Thisautomaticallr' elimrnates choices B, C, and D. If you didn't notice thatlact, then the problem can be solvedknort'ing that the mass of carbon in the CO2 is equal to the mass of carbon in the unknown

"omporrnd. The mass of

carbon rn CO2 n.ill be the moles of CO2 0.55/22.41) multiplied by the mass of carbon (12:011). When this isdir ided b'r- the mass of the sample (1.00), the result is answer choice A.

Choice A is correct. The key fact in the passage is that the 3.045 grams of carbon dioxide gas formed occupies1.55 L at STP. The question asks for moles of carbon dioxide, so the liters must be converted to moles. The.-?"Lut:i9tt is given in the chart as 22.4liters per mole. To convert to moles, either the 3.045 CO2 grams must bedivided by the molecular mass of 44 grams per mole, or the 1.55 liters CO2 must be divided Uy in" 22.41litersper mole. The second option is answer choice A, so pick that. If you wish to get an exact value for the moles, youmust get the denominator close to 100. This can be done by scaling the numerator and the denominator (i.e.,adding a value to each that is equivalent to multiplying both by the same value).

1.55 x 4

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This is fairly precise, but unnecessary on the MCAT. The addition of 0.775 to the numerator and addition of 1.1..2to the denominator accomplish the goal of making the denominator close to 100 while scaling proportionally.

Choice D is correct. If the organic vapor had not displaced all of the air in the flask at the time the heat wasremoved, then the flask would not have been filled with pure organic vapor at that time (as was assumed).The actual amount of organic vapor in the flask would be less thin the aisumed value (10b%), so the actualamount of liquid collected would be too low. If the measured mass of the liquid is too small, ttren the value inthe numerator is too small for the calculation of the molecular mass, so the caiculated molecular mass is too lowas well. Pick D to feel that happy sensation of correctness.

27. Choice C is correct. Water must be frozen out of solution, so the temperature of the trap should be less than thefreezing point of water (0'C). This eliminates choices A and B. Th; temperature of the trap cannot be lowerthan the sublimation point of carbon dioxide, however; otherwise, the car6on dioxide gur -ontd sotidify in thetrap along with the water vapor. The trap's temperature must be greater than the sub-limation point oicarbondioxid,e, but that temperature is neither given in the passage nor is it common knowledge, unless yo, happen toown the Jeopardy@ Chemistry game. To play it safe, choose a temperature just less i-han the boiling pornt ofwater. The value closest to 0"C without exceeding 0'C is -33"C. The best answer is therefore choice C.

Choice C is correct. The successive traps should be aligned in a manner to isolate each gas separately. Tmoles are not determined by difference, so choices B and D are eliminated. The trick here is determining trelative temperature sequence. If the traps were set first to isolate carbon dioxide by deposition (convertinfrom gas into solid), then water vapor would freeze out along with the carbon dioxide gas, and thus thcompounds would not be separated. By freezing water vapor first, the carbon dioxide

"u. puss to a later tr

where it can in turn be isolated free of water. Because the freezing point of water is O"C and the sublimatipoint of carbon dioxide is -78"C, the temperatures of successive traps should gradually decrease. Choose C.

29. Choice D is correct. To determine the molecular formula from the empirical formula, the molecular mass ofcompound must be known. This fact makes choice D the best answer. The volume of the products and redctanlis dependent on the mole ratio in the reaction, but not exclusively on the molecular formula. The volume machange depending on other conditions, such as temperature. This eliminates choices A, B, and C. pick DDetermination of the molecular formula from the empirical formula involves comparing the empirical massthe molecular mass. If the empirical mass is equal to the molecular mass, then the empirical ?ormula is I

molecular formula. Choose D for the satisfaction of knowing you got this correct.

Choice A is correct. Combustion analysis involves oxidizing compounds and collecting their oxide gases. I

gas can be oxidized like carbon (given that carbon and sulfur are roughly equivalent in electronegutirrity;.means that sulfur oxide can be collected and analyzed like the oxide of carbon. The oxide oisulfur thatformed is uncertain, but the only answer choice showing oxidized sulfur is choice A, where sulfur carriespositive four (+4) oxidation state. Sulfur can also be oxidized into sulfur trioxide (SOS), but that is not listeda choice.

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Choice A is correct. If oxygen gas were trapped out from the gas mixture, there would be no way to knowwhether the oxygen gas collected was from the original hydrocarbon, or from the excess oxygen gas needed toensure the complete combustion. This makes choice C valid. It is true that oxygen from the carbohydrate isfound in both the carbon dioxide product and the water product; thus, the oxygen from the compound cannot beisolated. This makes choice B valid.

Oxygen in the carbohydrate is fully reduced (has a negative two (-2) oxidation state when bonded to carbon andhydrogen). When looking at complete combustion reactions, remember that oxygen atoms from the compound donot react with oxygen gas at all. This makes choice D valid. Oxygen gas has a sublimation point around -180"C(so oxygen gas undergoes deposition when the temperature drops below -180"C); thus, oxygen gas can be

collected out of the air using a liquid nitrogen trap. Nitrogen liquid has a boiling point of -796'C, so thetemperature of the trap must be -196'C. The best choice is therefore choice A, because oxygen can exist as a

solid. The correct answer was a double negative.

Choice C is correct. As stated in the passage, the bubbler vents any pressure buildup within the system whilekeeping the system closed. The bubbler is a one-way valve that allows gas to effuse from the system, but does

not let gas infuse from the environment. The oil in the bubbler is intended to interact only with the gas, so nounreacted organic vapor should dissolve into the oil. Choice A is a valid possibility in the reaction (a truestatement), but it is not relevant to the goals of the experiment. Answer choices like choice A are difficult toeliminate sometimes. The oil in the bubbler is not serving as an oil bath, so choice B is eliminated. The bubblerkeeps the system closed by preventing infusion of gaseous compounds from the outside. The best answer is choice

C. Liquids cannot flow into the oil in the bubbler, so choice D is eliminated.

Choice D is correct. An increase in the mass percent of carbon is defined as an increase in the amount of carbonper gram of the compound, so statement I is valid.

The amount of water formed depends on the number of hydrogens in the compound, which depends on the mass

percent of hydrogen. As the mass percent of carbon within a hydrocarbon increases, the mass percent ofhydrogen in the hydrocarbon decreases. This reduces the amount of water formed from one gram of compound, so

it is not true that the mass of water formed increases upon oxidation. The decrease in mass percent of hydrogenreduces the mass of hydrogen formed per gram of compound. This means that neither statements II nor III isassociated with an increasing mass percent of carbon within a hydrocarbon. Choice D is the best answer.

Choice D is correct. The four salts listed as choices are all present in the same mass quantity in solution (4"/o oftheir respective solution.) All of the solutions have 96o/o water solvent, so the salt with the greatest number ofmoles has the highest molarity. Given that all of the salts have an equal mass, the greatest number of molesbelongs to the salt with the lowest molecular weight. Sodium (Na) is lighter than potassium (K), and chlorine(Cl) is lighter than bromine (Br), so NaCl is the lightest salt of the choices. Pick D, and feel fresh.

Choice B is correct. The trick here lies in the wording: "...how much water must be added?" This questionrequires that you use the dilution equation, Mi.,itiulVi.,itial = MfinalVfinat, where M is molarity and V is volume.

0.15 M x 300 mL = 0.0075 M x V6nu1 .'. Vfinal =0.15M x 300mL 15

= 15 x 4 x 100mL

0.0075 M 0.75

= 6000 mL = 6.0 L

This value is V;inn1, not the answer to the question (volume of water added)! To finish with 6.0 liters ofsolution from an original volume of 0.3 liters, 5.7 liters must be added. Choice B is the correct answer.

Choice C is correct. As far as chloride ion concentration is concerned, it doesn't matter whether you add purewater or NaOH(aq) to the HCI(aq) solution, because Cl- is just a spectator ion in the acid-base reaction. NeitherNaOH(aq) nor pure water increases the moles of chloride anion in the solution. Solve the question by using thedilution equation, Mi^itiulVi.ritial = MfinalVfinal, where M is molarity and V is volume:

0.25M x 50mL = Mfinal x75mL.'. Mfittul =0.25M x 50mL - 50 x 0.25M = x 0.25M = 0.167M

75mL 75

The final molarity is 0.1,67 M, which makes choice C,0.77M, the best answer.

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38.

37. Choice A is correct. Pure water is passed through the volumetric pipette to flush any residue out of thevolumetric pipette. Not all of the contents of the pipette may come free because of the adhesion of water to the

glass, so rinsing ensures that the residual solution in the pipette is forced from the pipette into the newsolution. The rinsing is not quantitative, so it does not measure a volume. This eliminates choice B. Notemperature is mentioned, so the water used to rinse the pipette is not responsible for heating or coolinganything. This eliminates choices C and D. The best answer is choice A.

Choice A is correct. A ten-fold dilution is defined as a dilution that results in a final concentration of a solutionthat is ten percent of its original concentration. For this to occur, the final volume of the solution must be ten

times the initial volume. The final volume is the sum of the initial volume and the added volume, so tfollowing math can be applied:

Vfinal = Vititiul + Vadded and Vfinal = 10Vini1in1, so 10Vitli1iul = Vinitial + Vadded

gVi.,itial = Vadded, so 1 part solution is mixed with 9 parts solvent.

Ten parts solvent mixed with one part solution yields a final volume that is eleven times larger; thus, the ficoncentration would be one-eleventh, which is less than ten percent. This eliminates choice B. Choice C isreverse of a ten-fold dilution. Choice C is a dilution to ninety percent of the original concentration. Choice Dthe same as choice B. Pick choice A to feel that jovial tingle of correctness.

39. Choice C is correct. The addition of water to an aqueous salt solution increases the mass of the solvent witchanging the moles of solute, so the molality decreases (given that the denominator increases.) This machoice A valid. Addition of water to an aqueous salt solution increases the volume of the solution wichanging the moles of solute, so the molarity decreases (given that the denominator increases.) This machoice B valid. The salt solution is denser than the pure water, so the addition of water (a less dense solutito the salt solution decreases the density of the solution, making choice C invalid. Addition of water toaqueous salt solution increases the mass of the solvent without changing the mass of solute, so the massof solvent increases. This makes choice D valid. The correct answer is choice C.

Choice A is correct. The greatest dilution results from the greatest relative addition of solvent. The greadilution involves the greatest ratio of solvent added to initial solution present, and it does not depend ontotal volume of the initial or final solution. The ratio of solvent added to the initial volume of solutionchoice A is 100 : 25, which reduces to 4 : 1. The ratio of solvent added to the initial volume of solution in choB is 200 : 60, which reduces to 10 : 3. The ratio of solvent added to the initial volume of solution in choice C is: 1"5, which reduces to 10 : 3. The ratio of solvent added to the initial volume of solution in choice D is 150 :

which reduces to 3 : 1. The greatest ratio of solvent added to initial volume of solution is found in choice A,best answer.

Choice A is correct. if the density of the solution is less than the density of water, then the addition of wto the solution may actually increase the density of the solution. The molarity and the molality of the solualways decrease as solvent is added, because the denominator in both molarity and molality increase wthe numerator remains the same. As solvent is added to the solution, the mass percent of solute in soludecreases, because the mass of solute in solution remains constant while the mass of solution is increasing.the density does not always decrease upon the addition of water to an aqueous solution; thus, the best answerchoice A. ln rare cases where the aqueous solution has a density of less than 1.00, the density increases

water is added.

Choice A is correct. Calculating the molarity and the molality of the solution involves the moles of salt innumerator. The numerator is the same in both the molarity and molality, so the difference between thevalues depends solely on the denominator. The mass of water is 0.100 kg, while the volume of solutiongreater than 0.100 liters. This means that the denominator in the molarity calculation is greater thandenominator in the molality calculation. The larger the denominator, the smaller the value. This meansthe molality is greater than the molarity, which eliminates choices B and D. The density of pure water is 1

grams per mL. The density of the aqueous salt solution is the mass (L01 grams due to salt and water) dividedthe volume. The volume of the solution is less than 101 mL, so the density of the solution is greater than 1

(given that the numerator is 101 and the denominator is less than 101). The density of the solution is gthan the density of water, so the best answer is choice A.

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Choice C is correct. Because the density of the solvent is less than 1.0 grams per mL, the volume in litersexceeds the mass in kilograms for the solvent. When comparing molality and mblarity, both values have thesame numerator' You are looking for the value with the smaller denominator. The mass of solvent is less thanthe volume of solution, so molality is greater than molarity. This eliminates choices B and D. The density isthe mass of the solution divided by the volume of the solution. The density values are less than 1, but are notknown specifically. In addition, there is no information about the mixing process. The density cannot bedetermined without information about the density of the components. The beifanswer is choice C.

Choice A is correct. All of the solutions are made up of one solute and solvent, but in different ratios. The soluteis denser than the solvent, so it can be concluded that the solution with the greatest density has the highestconcentration of solute. As solute concentration is reduced, the density is also reduced. Increas-ing the amount ofsolute relative to solvent increases both molality and molarity. The best answer is choice A.

Choice B is correct. This questions requires the use of the dilution equation, Mi.,itiulVi.ritial = MfinalVfinal,n'here M is the molarity and V is the volume. Plugging the given values into the equation yields:

0.25Mx100mL=0.10Mxv6i1a1 .'.Vfi.ut-0'25Mx100mL-0!25x100mL=2.5x100ml=250mL0.10 M 0.10

The final volume (Vmat) is 250 mL, but Vfinal is not tlne answer. The question asks for the volume of wateradded' To start with 100 mL of solution and finish with 250 mL of solution, 150 mL of water must be added.

Choice C is correct. The units of molarity are moles solute per liter solution. Because both solutions have thesame final volume (100 mL), the question reduces to asking how many grams of MgCl2 yield the same number ofrtoles of Cl- as 1.0 grams NaCl. There are two chlorides per magnesium chloride molecule, so only half thenumber of moles of MgCt2 as moles of NaCl is required. This accounts for the factor of one-half in the;alculation. The grams of NaCl must be converted into moles of NaCl, which in turn are converted into grams of\lgcl2 after accounting for the mole ratio of the two salts This calculation requires knowing the molecular mass:or both salts. Because the magnesium chloride is heavier than sodium chloride, intuition tells us that more*rams of the magnesium chloride are needed to form a mole quantity equal to that for the sodium chloride.ilhis means that in the calculation, the molecular mass ratio term should be greater than 1, making choice Che best answer. The unit factor method solution is as follows:

l.0gramsNaCl x lmoleNaCl r lMgClz r 94'9gramsMgClz = 1.0 x t x94.g qramsMec6

58.4 grams NaCl 2 NaCl l mole MgCl2 Z 58.44 o

lhe final answer is less than 1.0 grams MgCl2.

Choice C is correct. The mole fraction of a compound in a solution is found by dividing the moles of the given-ompound by the total moles of the solution. Choice B has a mole fraction of 0.50 for both Compound A andCompound B, because there is one mole of each compound, and the total number of moles is two. In an equal-rass mixfure (one gram each), the greater number of moles is present in the compound with the lower molecular

=ass. Compound A is lighter than Compound B; therefore, in equal masses of Compound A and Compound B,--:,ere are more moles of Compound A than Compound B. The mole fraction of Compound A is therefore greater--:nn 0.50 in choice A. This eliminates choice B. For an equal volume solution (in choice C there is one mL each),--:e relative masses can be determined by the densities. Because Compound A is denser than Compound B, equal'.'olumes of Compound A and Compound B result in the mass of Compound A being greater than the mass ofiompound B. The mass percent of Compound A is greater than 50%, as it is in answer choice A; thus, choice C:esults in an even larger mole fraction of Compound A than in the equal mass solution of choice A. The equal-rolecules solution (choice D) has the same mole fraction as the equal-mole solution (0.50 for each). Because::oice B and choice D are the same answer, they are both eliminated. The greatest mole fraction of Compound-:";s found in the equal-volume solution, choice C.

Choice B is correct. According to the absorbance equation (and the data and the graph), when the concentration-: I solute is doubled, the absorbance of the solution containing it doubles. This occurs because twice as many=-olecules are present to absorb light. The best answer is choice B. Choices C and D should have been:-rminated immediately, because the absorbance increases as the concentration of solute increases. Choice A is'*-:minated, because the absorbance does not increase as the square of an increase in molarity.

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49. Choice D is correct. At low concentrations, the relationship between concentration and absorbance is linear, as

predicted from the absorbance equation. At higher concentrations, deviation begins to occur, until theabsorbance reaches a maximum. There is no reaction occurring, so choice B can be eliminated. No solute woulddissolve into the solvent if the two repelled one another, so choice C can also be eliminated. Choice A mavsound tempting, if you are blindly choosing an amount on the basis of words you recall from biochemistry. Thesaturation of the solvation catalyst sounds like a good answer (especially considering that Michaelis-Mentenkinetics shows a similar graph when saturation of some sort occurs). The problem here is that "saturation oisolvation catalyst" means nothing, as there is no solvation catalyst. Also, catalysts affect the rate (not theconcentration) of reactants, products, solutes, or other components of a solution. Choice A should be elimjnated"Once water (the solvent) has interacted with as much solute as it can, no more solute can dissolve into solutionThe molarity reaches a maximum, so the absorbance reaches a maximum. The graph would be a line thatends at the saturated concentration. The best answer is choice D.

50. Choice C is correct. The addition of 50 mL of water to 10 mL of solution results in a dilution to one-sixth oforiginal concentration value. The absorbance of light should also be reduced by a factor of one-sixth. One-sixof 0.518 is a value less than 0.100, so the only possible answer choice is choice C.

51. Choice A is correct. Dividing the equation Absorbance = s[C]l by e and I isolates the value of soluconcentration (tC]). The result is that choice A is the best answer. This question may have seemed too simple tyou. On occasion, there are some simple questions about the MCAT, so don't try to find tricks that aren't there.

Choice C is correct. From the data in Table 1, we know that the absorbance of a solution with a concentration0.20 M is 0.188, while the absorbance of a solution with a concentration of 0.30 M is 0.278. If the absorbancethe unknown solution is 0.242, then the molarity of the solution must lie between 0.20 M and 0.30 M.average of 0.20 M and 0.30 M is 0.25 M, and the average of 0.188 and 0.278 is 0.233, so the absorbance for a 0M solution should be roughly 0.233. Because 0.242 is slightly greater than 0.233, an absorbance value of 0.242associated with a concentration value that is slightly greater than 0.250 M. The best answer is choice C.

53.

54. Choice C is correct. As solute concentration (molarity) increases, the absorbance increases. This ma

Choice D is correct. Isolating the value for e in the absorbance equation yields absorbance divided(concentration.pathlength). Absorbance is unitless, while concentration is in molarity, and pathlength issome fraction of meters (like centimeters). The units of e therefore must involve inverse molarity timescentimeters. The best answer is choice D.

statement I valid. At this point, the answer choices narrow to either choice A or choice C. If you are strapfor time, just look at statement III. Beer's law applies at all wavelengths (I*u* is chosen because it isgreatest value, and thus is the easiest wavelength at which to obtain an accurate measure of absorbBecause statement III is valid, choice C is the best answer. The absorbance depends on both the mabsorbtivity constant and solute concentration, so if Compound X has a lower constant than Compound Y,the solution with Compound X must be more concentrated than the solution with Compound Y, in order to haan equal absorbance. This makes statement II invalid. The best answer is choice C.

55. Choice D is correct. The spectrophotometer measures the absorbance of light, so the wavelength setcorresponds to an absorbed color, not a complementary color. This eliminates choices A and C, both of wrefer to the reflected (or complementary) color. To obtain the most accurate value, the spectrophotometer isat the wavelength corresponding to maximum absorbance. This makes choice D the best answer. Choice B

coincidentally correct in some cases, where the absorbance band is symmetric. This does not make choicebetter answer than choice D, but it does raise an important point: Remember that the test writers reward r

for choosing the best answer, not just a correct answer.

Choice B is correct. The question states that the absorbance varies with cuvette length. From the passage.know that absorbance varies with the concentration of solute in the solution. It makes sense thatabsorbance depends on the molar absorbtivity constant (e), so the Beer's law relationship of Abs. = e.[C].1 cardeduced. Dividing both sides of ihe equation by [Compound].I yields choice B. From the term ''

absorbtivity," it can be inferred that absorbance is in the numerator and molarity in the denominator, r.r-1

makes choice B the only possible answer. Units may not always be the best route to the correct answer.

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58Copl'right O by The Berkeley Review@ Section I Detailed Explanat

theruld:na)'Thenter,,n oii theated.tiont jus:

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Choice A is correct. This question requires converting absorbance to concentration. -{ 0.10 \1 soluion oi

Compound M has an absorb-an ce of 0.362, so if the absorbance is 0.400, the molarity is greater than 0.10 \1. This

eliminates choice D. Compound e with an absorbance of 0.250 is less that 0.10 M (where the absorbance is

).299), so choice B is eliminated. Compound T with an absorbance of 0.500 is less that 0.10 M (r.t'here the

absorbance is 0.511), meaning choice C is eliminated, leaving choice A as the best answer.

Choice D is correct. From the data in Table 1, we see that Compound T has the greatest absorbance of the three

:ompounds, so choices A and B are eliminated. The absorbance values for Compound M at a concentration equal

:r that of the other compounds are closer to the values of Compound Q than Compound T. That means that the

:etter graph is choice D. This is a question where it easy to make a mistake by choosing answer choice C

,., rthoul examining the other choices more closely. A brief survey of former students showed that roughly one-

::ird of you will make a careless mistake on this question and choose C.

Choice B is correct. Compound M has an absorbance of 0.217 at56L nm when the concentration is 0.06 M.

--:erefore, when the concentration is reduced to 0.05 M (a value that is aboutt7"h less), the absorbance should

,= :educed to a value that is five-sixths of 0.277, which is roughly 0.181. Statement I is valid. A solution of

- -:i,pound e with an absorbance of 0.225 at 41.3 nm must have a concentration between 0.06 M (where the

,-.-.rbur,." is 0.1g0) and 0.10 M (where the absorbance is 0.299). Because 0.225 is a little closer to 0.180 than

-:i. the concentration should be a little closer to 0.C6 M than 0.10 M, which makes statement II valid, and

: -:(es choice B the only possible choice. Statement III is invalid, because at 697 nm, 0.10 M Compound T has an

::.-,:bance of 0.511. fhi, -uun, that at the same wavelength, with a molarity of 0.1'1 M, the absorbance is

, : *: 0.56+. Given that the wavelength is not 697 nm (the wavelength where maximum absorbance is

:.=:r-ed), the absorbance is actually lJss. This makes statement IIi invalid, agreeing with the selection of- ,:e B. pick B to gain the prestige, i onor, and pride that goes with knowing you got another question right'

- - :,-ce C is correct. Compound T has an absorbance of 0.511 when it has a concentration of 0'10 M. An absorbance

-rl is just less than half of the 0.511 value, so the concentration must be just less than 0.05 M (half of 0'10

he question is then, "Which solution results in a concentration of Compound T that is just less than

:l.l: The starting solutions are all 0.10 M Compound T; so to achieve a solution that is just under 0'05 M, the

:*- . -:.: of water adied should be just a tiny amount greater than the amount of Compound T solution. This is"- : r-.'€l tvith choice C. The absorbance for each choice is listed below:

Choice A, 70 / ts (0.511) ,1 / z (0.511) > 0.25; therefore, choice A has too high an absorbance'

Choice Bt 20 / so (0.511) =1/z(0.511) > 0.25; therefore, choice B has too high an absorbance.

Choice ct 79 / gs (0.511) .1 / z(0.511) t.19 / zs (0.511) may equal 0.25; therefore, choice C is good.

Jhoice D, 10 / tg (0.511) ,7 / z (0.511) > 0.25; therefore, choice D has too high an absorbance.

. _ *= of ,9 / gg(0.5i1) should be close enough to 0.250 to convince you to choose C.

. : :: D is correct. Although colors were not reviewed in this section, it is a topic that frequently is discussed' : ,,:n the absorbarr.u oT Hght. One of the goals of these review passages is to expose you to this kind of

:- . .= i material. In each ,uJtion, there are sJme questions on topics that have yet to be discussed' Such is

,:,: :..re. The color observed for the solution is ihe complementary color of what is absorbed. The visible

, .'-:: . from roughly 200+ nm to 400 nm. Red light is the least energetic of vi-sible light, so red light has a

. , : , -,-,,,elength of"approximately 700 nm. Violet lignt t the most energetic of visible light, so violet light

, , - : rr:-,r' l.ivelengih of approximatety 400 nm. Compound Q absorbs tight with a wavelength of maximum, - : :-- :: i,.,u") of Zf g ,-rrr,.

-Thi, corresponds to violei light, so the color observed for the solution is yellow

:-:-ementary color of violet). This eliminates choices A and B. Compound T absorbs light with a

- : r l--r of maximum absorbance (l-u*) of 697 nm. This corresponds to red light, so the color observed for, -:.,.. rs green (the complimerrtary color of red). This eliminites choice C, and makes choice D thebest

r , 'r,: : should be able to answer this with the background knowledge that 400 nm is at the violet end of

L r = s;1ectrum, 700 nm is at the red end of the visible spectrum, and a working knowledge of-^rrrs. Compound M is irrelevant in solving this queition. Appearing red (reflecting red light)' : ,:.li LUIL

nr :-:: -een light (the complementary coior of red) *as absorbed. Green light is in the middle of the. ::,::::.irn, so It could have a warrelength of 561 nm. There is no reason for you to know the exact l" values

, - . - -: but you must be able to deduce the best answer to this question. Absorbed light, reflected light,

i,r,,- ^ . - : r i r^r-,- col,ors, the color wheel, and the visible spectrum will all be addressed in the next chapter'

59tana ,: - Berkeley Review@ Section I Detailed ExPlanations

62. Choice D is correct. It is not a combination reaction, because no compounds combined to form a single compou

so choice A is eliminated. It is not a decomposition reaction, because no comPound broke apart to form multi

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compounds, so choice B is eliminated. It is not a single-replacement reaction, because no metal compound w

formed or consumed, so choice C is eliminated. Becairse the barium cation (Ba2t) and the potassium cation (K

have exchanged their respective anions in this reaction, the reaction is classified as a metathesis reac

(known also is a double-displacement reaction). The best answer is choice D. By picking choice D, you can

iatisfied knowing you picked a correct answer. Go ahead, feel good!

Choice A is correct. A precipitate is a solid that forms from solution and deposits on the bottom of the flask

a reaction proceeds. The reaction least likely to form a precipitate is the reaction that is least likely to form

solid product. In choice A, the combustion reaction forms water liquid and carbon dioxide gas, but does not fc

a solid of any kind. For this reason, choice A is the best choice. In a decomposition reaction, a solid can

formed, as in the example offered in the passage. In the single-replacement and metathesis reactions, salts

formed as products, so precipitates are both likely and probable'

Choice C is correct. The reaction of aqueous sodium iodide (Nal(aq)) with aqueous calcium nitr

(Ca(NOg)z(aq)) is a double-displacement (metathesis) reaction, so the precipitate formed in the reaction m

be one oJ tt "

t*o possible product salts. Calcium is a dication (carries a +2 charge), so choice A is elimina

because the calcium salt is not neutral when bonded to only one iodide anion. Sodium and nitrate are thighly soluble in water, so neither sodium nor nitrate is found in the precipitates. Both choice B and choice

are therefore eliminated. The best choice is answer C. The reaction is given below:

2 Nal(aq) + 1 Ca(NO3)2(aq) .+ 2NaNO3(aq) + 1 CaI2(s)

Choice C is correct. The reaction of aqueous hydrobromic acid (HBr) with magnesium carbonate (MgCO3) i

neutralization reaction. It is stated in the purrug" that carbonates, when neutralized, yield carbon diox

gas. The best choice is answer C. The reaction is drawn below:

2HBr(aq) + 1 MgCO3(s) + L H2CO3(aq) + 1MgBr2(s)

1H2CO3(aq) -+ 1H2O(aq) + I COzG)

2 HBr(aq) + 1MgCO3(s) -+ 1H2O(aq) + 1 CO2(g) + 1MgBr2(s)

64.

65.

66.

67.

you should recall that baking soda (sodium bicarbonate) effervesces when an acid is added. The bubbles

form are carbon dioxide guri Uydtogen gas cannot exist as H (molecular hydrogen is a diatomic gas),

magnesium bromide is an"ionic titt 1*li"tt-is definitely not a gas at room temperature), meaning that choi

and D should be discarded immediately. To eliminate choice B, you had to recognize that magnt

oxidation-reduction reaction'

carbonate is a base, and not a reducing aglnt. _Magnesium metal is rich in electrons, so it would make a s

reducing agent, but not magnesium .utiot (Mg2*). The addition of HBr to Mg(s) would yield H2(g) by way

Choice B is correct. Carbon dioxide gas is a common product in combustion reactions, so choice A is elimi

The example reaction for decomposiiion shows calcium carbonate (CaCO3) decomposing to form:4ttlTan4 carbon dioxide. Choice C is thus eliminated. The same reaction is possible with calcium sulfite (CaS(

so sulfur dioxide can be formed in the decomposition reaction of calcium sulfite. The passage states that

neutralization of sodium bicarbonate (NaHCO3) forms carbon dioxide gas; therefore, choice D is elimi

The only choice left is answer choice B, the metathesis reaction, where cations are interchanged, but no r

dioxide gas is formed. If the anion begins as carbonate, it finishes as carbonate, and not carbon dioxide'

Choice C is correct. In the reaction, the chlorine gas is reduced to chloride anion and exchanged for the b

anion, which is oxidized into bromine liquid. 'The

product is not the result of coupling (combining)

reactants, so it is not a combination reactiin. This eliminates choice A. The products are not the result <

breakdown (decomposition) of a reactant, so it is not a decomposition reaction. That eliminates choice B'

anion in the salt is exchanged, while the cation is not. This describes a single-replacement reaction, m

choice C is valid. Because the cation remains the same, the reaction cannot be a metathesis reaction,

eliminates choice D. The best answer is choice C'

Copyright @ by The Berkeley Review@ 6() Section I Detailed Ex

ilrnd,dplewas(K*):tionL feel

i58. Choice A is correct. A negative value for AS results when the reaction gains order (or loses randomness). Ir, ucombination reaction, the number of molecules decreases from reactant to product, which is an increase in order.As such, a combination reaction is likely to have a negative value for AS. This makes choice A the best answer.In a decomposition reaction, the number of molecules increases from reactant to product, which is an increase indisorder. A decomposition reaction is likely to have a positive value for AS. This eliminates choice B.

In a single-replacement reaction, the number of molecules remains the same on each side of the reaction. Bothsides of the equation have a precipitate and a solute, which leads to no conclusion about the change in order forthe system. A single-replacement reaction can have either a positive or negative value for AS. Although asingle-replacement reaction may have a negative AS, choice C is not as good an answer as choice A. In acombustion reaction, there is an increased number of molecules from reactants to products. The products aregases, so there is a large positive value for AS. This eliminates choice D.

Choice D is correct. Ca(OH)2 yields a hydroxide anion (OH-) when added to water, which makes Ca(OH)2 anArrhenius base. Ca(OH)2 cannot donate a proton, so it is not an Arrhenius acid. This eliminates choice C.Choice A is also eliminated, because it cannot be an amphoteric species, unless it can act as both an acid and abase. Calcium is a metal, so Ca(OH)2 is a metal hydroxide, not a non-metal hydroxide. Choice B iseliminated. The correct answer is choice D.

Choice D is correct. Although COtz- is charged and thus an ion, the anion itself is held together by covalentbonds (three o-bonds and a resonating r-bond). Carbon dioxide is held together by covalent bonds between thecentral carbon and the two adjoining oxygens atoms. Only CaO (III) has an ionic bond, and it is considered to bepartially ionic. Calcium carries a +2 charge, and oxygen carries a -2 charge. This makes D the correct choice.The structures and bonds are shown below:

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Choice D is correct. All of the compounds contain only one calcium per molecule, so the smallest mass percent ofcalcium is found in the compound with the greatest molecular mass. The mass percent is found by dividing themass of calcium by the mass of the compound. The molecular mass for CaO is 56 grams per mole; for Ca(OH)2 itis 74 grams per mole; for CaCO3 it is 100 grams per mole; and for CaCl2 it is 111 grams per mole. The heaviestcompound is CaCl2 (choice D), meaning choice D has the greatest denominator and therefore the smallest masspercent of calcium. This makes choice D the best answer.

Choice A is correct. To calculate the percent yield, the actual yield (in moles) is divided by the theoreticalvield (in moles). The actual yield in moles is 5 9/100 g.moles-r, while the theoretical yield in moles is 10 g/74B.moles-l. The percent yield is found as follows:

6o

percentyield =*ffl91 x 100% = fr|fr- x1N./, = #, x 100% = ffi x 100% = 37"h

,4 g-,"I"tThe percent yield is 37"/o, which makes choice A the correct answer.

Choice D is correct. The mass percent of calcium in calcium oxide is the mass of calcium (40 grams per mole)divided by the mass of calcium oxide (56 grams per mole) and multiplied by 100%. 40 / 56 x I00"/" is greater than50%, which eliminates choices A and B, and is greater than 66.7"/o, which eliminates choice C. Only choice Dis greater than66.7"h, so choice D must be the correct answer.

Rangemethod: 408 x 100% > 40 x 100% =Z x 700% = 66.7%-56g603

Calculationmethod: 4!3 =5 = 5 xl= 5 x 0.143 = 0.775 = 77.5ohffig77

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I n:-'right @ by The Berkeley Review@ 6l Section I Detailed Explanations

74. Choice A is correct. Because matter can be neither created nor destroyed, the total mass of reactants must equalthe total mass of the products. There are 38.0 grams of reactants, so choice D is eliminated for having morethan 38.0 grams of products. The question from here becomes a limiting reagent problem. We must decidewhether there is leftover water (is CaO the limiting reagent?), leftover calcium oxide (is water the limitingreagent?), or an exact mixture (there is no leftover and thus no limiting reagent.) The moles of CaO equal2S/56(which is one-half) and the moles of H2O equal 10/18 (which is greater than one-half.) Because the calciumoxide reacts with the water in a one-to-one ratio, there is leftover water. This means CaO is the limitireagent in the reaction, so the reaction does result in leftover H2O. The only choice with leftover water is A.

75. Choice D is correct. The N in the answer choices stands for normslitq, which is the molarity of the protothat can be generated by the acid. Normality is the molarity of an acid times the equivalents of hydroniper molecule. CaCO3(aq) reacts according to the equations shown below. A 2 : 1- ratio is needed to reachneutralization.

CaCO3(aq) + 2 H+(aq) ------> H2CO3(aq) + Ca2+1aq; H2CO3(aq) -* H2O(1) + CO2(g)

5 mL x 0.2 moles/liter = 1 mmole CaCO3, so 2 mmoles of a strong acid (which yield H+) are needed toequivalence. Choice A is a strong base, so eliminate it. Choice B yields only 1 mmole of H+, and choice C yie3 mmole of FI+, so eliminate both choices B and C. Choice D yields the 2 mmole of H+ that are needed. Theanswer is choice D. For a monoprotic acid, normality and molarity are the same. However, for a diproticthe normality is double the molariiy; and for a triprotic acid, the normality is triple the molarity.

Choice C is correct. This is a case where Equation 1.2 in the text, Mi'riliulVinitial = MfinalVfinal, rrust be used.

0.500Mx5.0mL=Mfinatx55.0mL.'.Mfir.,ut=5/55x0.50.'.Mii'ul=1/ttx0.50<t/tO*0.50=0.050

The final molarity is therefore less than 0.05 M. Without pounding numbers, you should see that choice Cthe best choice (the choice just less than 0.050 M). Choice D is too far below 0.050 M to be reasonable. Dosolve math questions exactly, but find a range into which only one answer choice fits. In this case, the rangefrom just less than 0.050 (maybe 0.040 or so) to 0.050. OnIy choice C fits into that range.

Choice C is correct. The final concentration can be determined by taking a weighted average ofconcentrations for the two initial solutions before they are mixed. A quick observation you should apply toquestion is that if the mixture were a fifty-fifty mixture by volume (the mixture of equal volume solutithen the final concentration would lie exactly between (be the mean value of) the two values. Inparticular question, the mean value would be 0.250 M. Because there is an excess of the higher-concentrasolution (30 mL at 0.30 M, with only 20 mL at 0.20 M), the answer should be closer to 0.30 M than 0.20 M (it greater than 0.250 M). This eliminates choices A and B. The exact value can be solved for as follows:

(20 mL x 0.20 M) + (30 mL x 0.30 M)

76.

(mffiM

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77.

Total moles =Total volume 50mL

= 4+950

- 13 = 0.26, choice C.50

78. Choice C is correct. 50 grams CaCO3(s) is equal to 0.50 moles CaCO3(s), because the molecular mass of Cis 100 grams per mole. Because there is only one carbon in CaCO3(s), 0.50 moles CaCO3(s) yield 0.50 molesCO2. 0.50 moles of CO2 at STP occupies a volume of: 0.50 moles x 22.4liters/mol" =1 / 2x22.4 = 11,.2L. Pick C.

79- Choice A is correct. The greatest amount of potassium per gram of compound is found in the compound withgreatest mass percent of potassium. The greatest mass percent of potassium is found in K2O, as was mentithe passage. This means that the best choice is answer A. The mass percents for the four salts are as follows:

2x39.7 2x39.'l

utu

(lmilrildllWftl

ffituMWnfu1ftum&m'urfuildAilii

t@@sMs4Wm-

K2():(2x39.1) + 76

661. 39.1 =39.L39.1 + 35.5 74.6

=78.2 _39.794.2 47.7

K2SOa: 78.2 =39.1174.2 87.1(2x39.1) +32+$x16)

KNO3: 39'I - 39.739.7 + 74 + (3 x 16) 101.1

The relative mass percents when comparing the four values are' 39.1 > 39.1 > 39.1 > 39.L

The greatest mass percent of the choices is found with K2o (choice A). 47 '1' 74'6 97 '1 101'1

Copyright @ by The Berkeley Review@ 62 Section I Detailed Explana

)rel-

ni-

Choice D is correct. The compounds in choices A, B, and C all have two nitrogen atoms per compound, while thecompound in choice D has only one nitrogen atom in the compound. The lowest *u5 p"r""niof nitrogen is inchoice D, because the four compounds are roughly comparableln molecular mass, but they have differeri-urr",of nitrogen. The mass percents for the four choices are shown below:

A. H2NCoNH2 + 28gN * 2a - -7 < L609 H2NCONH2 60 15 2

B. NHaNO3 =

D. NH4HzPOa .+

28-60

(NHa)zSo+ = 28gN

132g (NHa)2SOa

B0g NHaNO3

14g N1159 NHaH2PO4

28gN _?a _80

7_20

35%

7_JJ

C.

1E

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- 14 < 14"/.115

21*"/"

>29132

28

80> 28 (= 14)

230 1ls

Choice C is correct. The mass percent of nitrogen in urea is just less than 50%:

Mass percent urea =28gN

609 H2NCONH2lhe only answer choice that is just less than 50% is choice C.

Choice A is correct. According to the passage, the phosphorus content of the compound is critical, so no:nosphorus should be wasted in unreacted reactant. This can be prevented by adding the reactants in a ratio::-at avoids the presence of any leftover phosphorus-containing reactant. The goal is to conserve the:iosphorus-containing compound, which makes choice A the best answer.

Choice B is correct. The percent yield for a reaction is defined as the actual quantity of product isolated from-:-e reaction mixture, divided by the theoretical quantity of product that should form, as determined from the-:miting reagent of the reaction. According to the balanced equation, for every one mole of (NHa)2CO3 that:eacts, one mole of (NHa)2SO4 forms. This means that 10 grams of (NHa)2CO3 (which is 1,0/96 moles\Ha)2CO3) should produce 1,0 /96 moles of (NHa)2SO4, which is (10 /96 x 732) grams (NHa)2SOa. Dividing::-e mass of (NHa)2SO4 obtained (10 grams) by the theoretical mass of (NHa)2SOa that should have formed-3 75 grams) determines the percent yield. The calculation of the yield is shown below:

Theoretical*urr, 10 x13Z=1320 = 330 - 110 > 104 +6 = 133 =13.759696248884Rangeforthepercentyield:50%<5 =10 <10'00 <3 = \0 =75"/o7 74 73.75 4 73.33

-:.e value for the percent yield lies somewhere between 50% and 75"/", which makes choice B the best answer.

Choice C is correct. All of the choices have the same volume (100 mL) and the same mass of salt, so the greatest::larity of potassium is present in the solution with the greatest number of moles of potassium in solution.i -ence, the greatest molarity results from using the compound with the greatest *uss pet""nt of potassium,;:jch corresponds to the salt with the lowest molecular mass. Choice D iJ ehminated, because NOj is heavier

:-;n Cl, so KCI has a lower molecular mass than KNO3, and thus a greater mass percent of potassium than in:'.\O3. Choice B is eliminated, because SO4 is heavier than CO3, so K2CO3 has i lower molecular mass thanr '-SO4, and thus a greater mass percent of potassium than K2SOa. The question boils down to determining the:=-afive mass percents of potassium in KCI and K2CO3. The molecular mass of KCI is74.5 grams per mole, and*-: molecular mass of K2CO3 is 138.2 grams per mole. Potassium carbonate (K2CO3) yields two pbtassium ions:'.: molecule, so to have the same mass percent of potassium as KCl, its molecular mass would have to be double:*'a: of KCI (which equals 149 grams per mole.) Because K2CO3 is less than twice as massive as KCl, the:=-alive denominator is smaller when calculating the mass percent of K2CO3 than when calculating the mass:':,:cent of KCl. The greatest mass percent of potassium, and thus the most grams of potassium in a 10.0-gram;r-l sample, is found in K2CO3. This means that the highest concentration of potissium is present in the:. -ieous K2CO3 solution, so choose answer choice C for best results.

8_7 -160152

,,: :-::.t O by The Berkeley Review@ 63 Section I Detaited Dxplanations

85.

86.

Choice A is correct. It is easiest to isolate a solid (precipitate) from solution, because the solid can be filteredfrom solution rather easily. Equally, a liquid and gas can flow, while a solid cannot, so removing liquid and gasto leave solid behind is easy. This is the crux of many lab techniques. A liquid must be distilled away fromsolution, a solute must be either distilled, extracted, or removed by chromatography, and a gas must becollected in a gas-trap free of air. The best answer is thus choice A.

Choice A is correct. This is another case where the question focuses on a topic we have yet to discuss. There area few things you should recall. First, the lower the pKu value is, the stronger the acid is. Second, when the pHof the solution exceeds the pKu of an acid, that site exists predominantly in the deprotonated state; StatementIII can immediately be eliminated, because the first proton of sulfuric acid is strong, meaning it has a very low(negative) pKu value associated with it. From the products of the last equation in the passage, it can be seen

that H2POa-,5C.42-, and HF all simultaneously exist in solution. Sulfuric acid has lost both of its protons,phosphoric acid has lost its first proton, and hydrofluoric acid has yet to lose a proton. The first proton ofphosphoric acid is lost more readily than the proton of HF, which means that pKnl of H3PO4 is lower thanthe pKu value of HF. Statement I is thus true. Because the second proton of sulfuric acid has been lost, whilethe second proton of phosphoric acid has not been lost, the second proton of sulfuric acid is more acidic than the

second proton of phosphoric acid. The result numerically is that pKu2 of H2SOa is less than pKn2 of H3PO4.Statement II is thus false. Only statement I is true, so the best answer is choice A.

Choice C is correct. This is a limiting reagent question. The balanced equation for the reaction is as follows:

1C3H3(g) +5 O2(g) + 3CO2(g) +4H2O(g)

There are 25/32 moles O2(g) and 20/44 moles CSHS(g) as reactants. According to the mole ratio frombalanced equation, the moles of O2(g) must be five times that of CSHA(g). The moles of O2(g) needed to rewith 20 / 44 moles of C3Hg(g) is 1,00 / 44. The value of 25 / 32 is less than 1, so there is less than one mole of 02present. The value of 700 / 44 is greater than 2, so more than two moles of O2(g) are needed. This meansoxygen (OZ(S)) is depleted before CSHS(g), making O2(g) the limiting reagent and choice C the best answer.

88. Choice B is correct. If you know this answer from your biology classes, trust your knowledge and don't wtime doing the calculation. You should know that hemoglobin contains four iron atoms. That is the pointthis question. The math is time-consuming, so save time wherever you can. If you didn't know that facthemoglobin, then doing the math was necessary. The mass of iron in hemoglobin is the mass percent of itimes the totalmass of hemoglobin. 0.33% x 68,000 =.0033x68,000 = 3.3x 68=204+20.4=224.4 gramsFe.number of irons is therefore 224.4/55.8, which makes 4 (choice B) the best answer choice.

89. Choice D is correct. The first method which probably comes to your mind is the method learned fromchemistry classes. In this method, the 9 grams of carbohydrate are converted to 0.050 moles of carbohydratedividing by the molecular mass of 180. The 6.72Liters of carbon dioxide at STP are then divided by 22.41iper mole to yield 0.30 moles CO2. The ratio of CO2 to 02 in the combustion of a monosaccharide is 1 :

Therefore, 0.30 moles of oxygen reacted as well, and thus the mole ratio of carbohydrate to 02 to CO2 is0.30 : 0.30, which equals L :6: 6. This is a long-winded, but valid, solution. A shorter method is as follows:

In the combustion of a monosaccharide, the mole ratio of oxygen reactant to carbon dioxide to water is always1 : L. A monosaccharide has the molecular formula C1H21O., so it has a molecular mass of 30n (12n + 2n + 1

grams per mole. For a molecular mass of L80 grams per mole, the value of n is 6, so the formula formonosaccharide is C6H12O6. The mole ratio of O2to CO2 is therefore 6 : 6. Pick D and feel relaxed.

90. Choice B is correct. Choices A and C are eliminated, because they cannot be a molecular formula.compound with just carbon, oxygen, and hydrogen, the number of hydrogens must be an even number. Withodd number of hydrogens, the bonding does not work out. To decide between choices B and D, you mustthe molecular mass of the compound, by dividing the mass by the volume and then multiplying this valuethe molar volume:

1 gram * 22.4 literc _ 22.4 grams - 22.4 grams = 4 x 22.4{ams = g9.6grams

0.26 liters L mole 0.26 moles 0.25 moles mole mole

Only CaH6O2 has a molecular mass just under 90 g/rnole, so choice B is the best answer.

)

I

87.

Copyright @ by The Berkeley Review@ 64 Section I Detailed

Choice A is correct. This is a question testing your knowledge of the irnit factor method. Barium oxide (BaO)can form barium metal and oxygen gas (O2) upon decomposition. These are the only two products possible fromthe decomposition of barium oxide. Hence, for every two moles of BaO, one mole of 02 forms. To solve thisproblem, the 1.0 grams of BaO must be converted to moles, and the moles of BaO in the 1.O-gram sample must bemultiplied by one-half. This gives the moles of 02 formed. Multiplying the moles of O2by 2Z.4literc per mole(the volume for one mole of gas at standard temperature and pressure), gives liters of oxygen gas produced bydecomposition:

1.0 grams BaO x 1 mole BaO ,. 1*ll"=O2, * 22.4Iiters - 22.4 = 22.4 < 1

153 grams BaO 2 moles BaO 1 mole 153 x 2 306 10

Only choice A is less than 0.1 liters, so pick choice A and make yourself one point wiser.

Choice B is correct. To determine the grams of product, the conversion must go through moles. This question is a

standard grams reactant-to-moles reactant-to-moles product-to-grams product conversion.

2.0 grams MgNHaPOa , TM9ZPZOZ

13Zs8ran.sl 2 MgNHaPOa/mole

22 222.6 s1137.3

The mass of Mg2P2O7(s) is greater than 1, but less than 2. The only answer that falls between 1 and 2 is answer,hoice B.

Choice D is correct. We know nothing of the molecular mass, so any formula we determine is the empirical:ormula (simplest formula). This eliminates choice A. Because the elemental mass of. Z is twice the elemental:rass of X, a compound with equal mass quantities of X and Z contains twice as many moles of X as moles of Z.llre mole ratio of X to Z is therefore 2 : 1; thus, the empirical formula is either X2Z1 or 21X2. Pick choice D to:eei correct.

Choice C is correct. The math here is beyond the MCAT level, but the point of such a question is to encourage.'iru to use intuition and simplify the question. You can always eliminate some of the answer choices, and at-east narrow your guess from random choice to a fifty-fifty chance of success. This question is simplified by,.rowing the possible oxidation states of iron. From your experience with hemoglobin, you know that iron has a:ridation state of either +2 or +3, and that oxygen has an oxidation state of -2 when it is coupled with a metal.

' oxygen has an oxidation state of -2, then iron cannot have an oxidation state of +2 or +3 in either Fe3O2 or-eO3. Choices A and D are eliminated based on outside knowledge. This problem can now be solved using your:rathematical intuition. In the iron oxide sample, there are 11.89 grams of iron and 5.10 grams of oxygen (found:-,'subtracting 11.89 from 16.99). The molecular weight of iron (Fe) is 55.85 grams per mole, and the molecular,';eight of oxygen (O) is 16.0 grams per mole. An empirical formula is based on mole ratios, so the elemental::.asses must be converted into mole quantities. 5.1 over 16 is just less than one-third, so there are roughly 0.30:,.rles of oxygen in the iron oxide. 11.89 over 55.85 is just over one-fifth, so there is roughly 0.20 moles of iron in::.e iron oxide. There are more moles of oxygen than iron, which eliminates choice B (FeO has equal moles of,:rn and oxygen) and leaves choice C as the correct answer by default. Minimize doing math by using your::ckground knowledge and logic.

Fe11.se 05.10 - Fer*O1 - Fe0.2* OO.SS- = Fe2O355.85 16.0 5 3

tliir Choice D is correct. It is not possible to have 26 hydrogens and only 4 carbons in a stable compound. You may

=tall from organic chemistry that the maximum number of hydrogens in a hydrocarbon or carbohydrate is 2n +I -,r-here n is the number of carbons in the compound. This eliminates choice C. Perhaps it is easiest to solve-:r: question by first determining the number of carbons in the molecule. The mass percent of carbon in the:-"riecule is 53.4"/o. When this percentage is multiplied by 90 grams/mole, it tells us that just over half the mass

-. carbon. Carbon has a mass of 12 grams per mole, so the value must be a multiple of L2. The mass due to carbon*. 48 grams, which is the mass of four carbon atoms. The molecular formula therefore contains four carbons,:-:ninating choices A and B. The only choice left is answer D, which does in fact have roughly 11% hydrogen.a-ran, the goal in preparing for your exam is to be able to solve these questions quickly and using as much::-ution as possible. You are not rewarded only for being thorough; you are also rewarded for being fast. Take

-: dcal short cuts whenever they present themselves. Do not blindly repeat the calculation techniques you-:::ned in your general chemistry courses.

,, r: ,::it @ by The Berkeley Review@ 65 Section I Detailed Explanations

,222.6.eYgzl_zo_-r = ?-: ?rr9 = HsMszpzel mole Mg2P2O7 737.3 x 2 1.37.3 -

iul

16'

si

;."

96. Choice A is correct. Because this is a reduction reaction, the oxidation state must decrease, which eliminachoices C and D. By assuming oxygen has a -2 oxidation state, it is possible to reduce this problem to simalgebra. In the teactant, the sum of the oxidation states for the four oxygen atoms is -8. hr order for the ovecharge on CrO42-(aq) to be -2, the chromium must have an oxidation staie of +6. The sum of +6 and -8 is the icharge of -2. In the product Cr2O3(s), each chromium must be +3 to cancel out the three oxygen atoms (ateach) to make the overall molecule neutral. For this reason, choice A is the best answer.

97. Choice D is correct. The metal combines in a 2 :3 ratio with oxygen to form an oxide that is 53% metal byAssuming a 100-gram sample implies that there are 53 grams of metal and 47 grams of oxygen in the 100-grsample of the metal oxide. This means that the ratio of (53 divided by the molecular mass of themetal) to (47 divided by the molecular mass of oxygen/ which is 16) is equal to 2 : 3. This can be solvedhand or by intuition. Rather than solve the math exactly, it is a good idea to plug the values for allanswer choices into the setup, and get a rough estimate:

53 g metal . 47 go -2:i +Given that4zis roughly 3, 53 g metal

mustbe about2MW"r"."l *81 , 76 MWmqhl

/moleThe molecular masses of the four answer choices are 40.0, 55.9,52.0, and 273 respectively. 47 dividedby 16

slightly less than 3, so 53 divided by the molecular mass of the metal must be slightly less than 2.eliminates the first three choices (40.0, 55.9, and 52.0), leavin g 27 .0 (aluminum) as the best answer. Theanswer is choice D. You could also have eliminated choice A from knowing that calcium cannot achieve aoxidation state (or charge), so calcium cannot combine with oxygen ina2:3 ratio.

98. Choice B is correct. The mass percent of carbon is the mass of carbon in each compound, divided by themass of each compound, multiplied by L00%. For each compound, the mass percent can be determined as

Acetic agicL: 24 x lCf,,h60

Ethanol: b x 100% Methyl acetate: 35 x fiO%46 74Glucose: 72 x

180

24-n_L 24,?3_]. %,37_7 72-90_76060246462747421801802

Of the choices, only choice B (ethanol) has a fraction greater than one-half. This implies that the greamass percent of carbon is found in ethanol. The best answer is therefore choice B.

99. Choice B is correct. The MCAT test writers can ask about the same concept in many ways. It is more imthat you walk away from this questi.on knowing the concept than it is getting the question correct. Thecarbon dioxide results from the compound with the greatest mass percent of carbon, as long as there aremasses of each sample present initially. There was one gram of each compound initially, before oxidatiorLour only concern is with the mass percent of carbon. The greatest mass percent of carbon is found in eththe greatest amount of carbon dioxide results from the oxidation of ethanol. The best answer is choice B.

100. Choice C is correct. Because there are two Cl in the Mg(ClOa)2 molecule, the mass percent of chlorine ismass of two Cl over the mass of Mg(ClOa)2.

Mass chlorine =Masscompound 243+(2x35.5)+(8x16)

7L > 7L > 71 ,whereTl =L=13-3%,and fI -1-25%.Sog3.g%> 71 >ZS%213 223.3 284 213 3 284 4 223.3

The mass percent is a little less than 33.3"/', so choices A and B are eliminated. The value is just a littlethan 33.3%, because the denominator (223.3) is just a little greater than 213. The best answer is the choiceis slightly less than 33.3o/o, which is 31..8"h, choice C. Choice D can be eliminated, because it is less than

?vi55 =71=71243 + 71, + 128 223.3

66Copyright @ by The Berkeley Review@ Section I Detailed Dx

Section IIAtomicTheory

by Todd Bennett

- tr3s

Filter

Prism

Atomic Structurea) Subatomic Particlesb) Isotopesc) Average Atomic Massd) Classical E,xperiments and Machinery

i. Thomson E,xperimentii. Mass Spectroscopyiii. Millikan Oil Drop Experimentiv. Rutherford ExperimentHeisen berg's UncertainLy PrincipleAtomic ModelHydrogen trnergy Levels

Electronic Structurea) Electronic Theoryb) Effective Nuclear Chargec) Electronic Spin Pairingd) Spin Pairing and Magnetisme) Electronic Density and Orbitalsf) Collective Orbital View o[ Energy Levelsgl Electronic_Configurationh) Quantum Numbers

Periodic Trendsa) The Periodic Tableb) Ceneral Elemental Periodic Trends

i. Atomic Radiusii. lonization Energyiii. Electron AttiniLyiv. trlectronegativily

c) Periodic Families (Groups)

Light Absorption and Emissiona) Excitation and Relaxationb) Atomic Spectrum of Hydrogenc) Electromagetic Spectrumd) Visible Spectrum and Colors

iii. Emitted Coloriv. Reflected Color and the Color Wheel

e) Fluorescencef) PhotoelectricE,ffect

Nuclear Chemistryal Nuclear Particlesb) Nuclear Decay and Capturec) Half-Life

e)

0g)

Photographicplate

PEBITruEYl)n.n-v.r.E-'w'

Specidrrztng in MCAT Preparation

Atomic Stnrcture Section Goals

ov Be familiar with the location, mass, and charge of sub-atomic particles.

av

An atom, from the chemist's perspective, is composed of a nucleus made up of protons and neutrons,surrounded by orbiting elections. You must know the Bohr model and the properties of each particlewith regard to its charge, mass, and effect on the atom. For instance, additiohal electrons result inthe formation of an anibn, while a decrease in electrons results in the formation of a cation.

Be familiar with energy levels and energy transitions.The electrons orbit in distinct, quantized levels. There is a base (lowest-energy) level, and excitedstates above that to which electrons jump when energy is added to the atomil system. Energy isabsorbed when an electron is elevated from the ground state to an excited state. Ehergy, in the frirmof a photon, is released when an electron drops"to the ground state from an excited siite.

Understand electronic configurations.Electrons fill the orbitals according to a defined sequence, described by the Aufbau principle. Thekeys facts to recall about the prindiple are that ther-e are two electrons-per orbital, and there is ones-orbital per level, three p-orbitals per level, five d-orbitals per level, and seven f-orbitals per level.The filling order has exceptions in the transition metals, lanthanides, and actinides. There are

exceptions that allow for half-filled shell stability, as is seen with chromium.

Understand qgantum numbers.Quantum numbers are used to define the orientation and location of an electron. There are fourquantum numbers: n, Im1, and ms. You must be familiar with the quantum numbers and how todetermine them for a given electron within an atom.

Be familiar with periodic trends.Knowing the periodic trends for the main-group elements is essential. The trends that you must befamiliar with ihclude ionization energy, atomic radius, electron affinity, and electronegativity (althoughelectronegativity of the noble gases need not be committed to memory). [t is important that you befamiliar with eath basic trend and the reason for that trend. Be able tb explain ahy deviations fromstandard periodic behavior. t r

Understand isotopes, average atomic mass, and isotopic labeling.Isotopes are atoms with the same number of protons, but a different number of neutrons in theirnuclei. For some elements, there are several isotopes. The elemental mass (average atomic mass)found in the periodic table is a weighted average of all of the isotopes of that given element. [sotopesare chemically equivalent, so they are hard to detect without using a mass spectrometer. lsotopes,because of th'eir ihemical similaiity, are substituted for one another in som-e reactions to servb as1abe1s.

Understand nuctear decav and half-life associated with first-order decay.It is essential that you understand the breakdown and buildup of nuclei through the gain and lossof nuclear particles. It is also important to be able to determine the concentralion oFa species bycombining'the initial concentratibn with half-life information. The half-life is the period of tim-erequired for half of a given sample to decay to some product. Most examples will involve determiningthe concentration at a given time for a reaction (or process) that follows fjrst-order decay kinetics-.

@?

"v

'3

General Chemistry Atomic Theory Introduction

l:rm the work of Thomson, Millikan, Rutherford, Bohr, Pauling, and others, we::','e a modern view of the atom and its fundamental structure. The core of the:::rL by this view is a nucleus composed of protons and neutrons held together:', an unbelievably strong force. The details of the nucleus are not well..- Jerstood, but using a simplistic model, the majority of the mass and all of the: :.:live charge of the atom is found at the core. The nucleus is surrounded by::::ling electrons that stay in distinct orbits, no two of which are exactly the;.:,e (Hund's rule). Coulomb's law explains the mutual attraction of the orbiting:-n::rons and the nucleus. Electron energy levels are based on Coulomb's law,:-:'ough quantum mechanics is invoked to explain the overall behavior of the::::ing electron. The closer an electron is on average to the nucleus, the more:l:'.lv it is held, and the greater the energy required to remove that electron.:::r the atom. The space in which an electron is believed to orbit is referred to:; zn orbital Each orbital is distinct from all other orbitals, and electrons have the:::-rn of spinning clockwise or counterclockwise as they occupy the orbital.-l---,- two electrons may occupy each orbital at the same time, and they must"-:'"'e opposite spins to do so. These distinct orbitals are quantized energy levels

'':.re the electrons are said to reside. This fundamental idea is the basis of all.::::ic theory.

r: ::lic behavior and electronic configurations are explained by the concept ofr.':"::onic bookkeeping. The ease or difficulty of gaining, removing, or sharingi: :iectron are determined by the location of the electron. An electronic. ::::quration is an account of all of the electrons in an atom. Quantum numbers, -. specific for each individual electron. Questions about electronic.:::guration and quantum numbers on the MCAT should be some of the

,-::ler questions you see, so be sure to get them correct, as they are worth thei:. amount of points as the more difficult topics. Understanding the

1 :'E:nents and applications associated with the absorption and emission of- r-: LS critical to performing well on the MCAT. Absorption spectroscoPy,.:*rs. drld fluorescent tubes are jusi a few examples of devices thatutilize light

'i" i: ','ou are expected to understand for the MCAT.

I : -.iiiv important as understanding the structure of an atom and the energetics:. orbiting electrons is seeing the effect the structure has on the reactivity of

:1 : ::om. There are distinct trends in atomic radius, ionization energy, electron*::.-,::n-, and electronegativity that can be traced back to the filling of electrons in'I'E rif,rr. Knowing the reasoning behind the periodic trends is more importanti';,- nemorizing the direction of each trend and notable exceptions to the trend.

-'= -ast of the topics that fits under the heading of atomic structure is nuclear'::,*::-rstry. This section should be one of the easier sections, as scientists do not.:.i: :e:stand the topic in enough detail to expect you to have a deep.*::=:standing of it. Nuclear chemistry is made easy by knowing the particles*- - re processes of decay and capture, and your ability to do algebra. Know the,il::,:itions of isotopes, nuclear decay, nuclear capture, nuclear particles, and-* ---rre. This chapter starts with an elementary look at subatomic particles andl:i:-es with an overview of the atom, the energy states of its particles, and:

":*o:tions between energy levels.

69 Exclusive MCAT Preparation

General Chemistry Atomic Theory Atomic Structure

Subatomic ParticlesOur most fundamental view of matter is that atoms are composed of three sub-atomic particles. These particles are the proton, neutrory and electron. Each sub-atomic particle is unique in its properties and position. A proton and neutron arecomparable in mass, while an electron is roughly 1/ ;3OO of the mass of a proton.The proton and electron carry the same magnitude of charge, but with oppositesign. Physical properties associated with each particle are listed in Table 2.1

Particle Mass (kg) Mass (amu) Charge (C) Charge (e)

Electron 9.11 x 10-31 5.49 xt0-4 -1.602* 10-19 -1

Neutron I.67 x 10-27 1..0087 0 0

Proton I.67 x t0-27 I.0073 I.602 x 1O-19 +1

Table 2.1

The nucleus is made up of neutrons and protons. Mass and charge are the twomeasurements repeatedly used to describe subatomic particles and whole atoms.By convention, the mass of an atom is said to be due only to the protons andneutrons, because the electrons are essentially massless. An atom normallycarries a neutral charge, unless it has lost or gained electrons. In a neutral atom,the number of protons equals the number of electrons. A convenient shorthandnotation is used to describe every element. There is the mass number (A) used todescribe the number of protons and neutrons in an atom. There is also theatomic number (Z) used to describe the number of protons in an atom (and thusthe number of electrons in a neutral atom). The notation used to represent eachatom is shown in Figure 2-1.

Symbol The A number tells us that the combinednumber of neutrons and protons is L95.TheZnumber tells us that the number ofprotons is 78. Combining these factstells us that there are 'J,17 neutrons and

platinum-195 78 protons in this atom.

Figure 2-1

IsotopesIsotopes are atoms of the same element that contain a different numberneutrons but the same number of protons within their nuclei. Isotopes, hthe same number of protons, are chemically similar but have different amasses. Isotopes react the same way chemically and thus can be distinguonly by mass separation techniques (such as mass spectroscopy). Isotopesoften used as markers in chemical labeling experiments and as tags inmagnetic spectroscopy studies. An isotope can be traced from the source (

it is added) to the endpoint in a physiological system, a biochemical pathway,a reaction mechanism. Typical examples of isotopes include 1H (standahydrogen), 2H (deuterium), and 3H ltrltrum;. Common isotopes used instudies include deuterium, tritium, carbon-13, carbon-14, phosphorus-32,iodine-121. These isotopes can be monitored either by the radioactive decayemit or by nuclear magnetic resonance imaging (known as NMR spectroscopy).

number

\195

78

.r'number

lemental

,/t

Copyright @ by The Berkeley Review 70 The Berkeley

ure General Chemistry Atomic Theory Atomic Structure

ub-ub-are

:on.

site

lVO

ms.

mdily)m/mdltothelUS

lch

ofrngnicredareearereor

rrdingndle)-

).

EH

:d;.

f:tS

rd

Example 2.1:low do the subatomic particles n261'l and27 Aldiffer?

-\. The two are isotopes with a different number of neutrons.B. The two are isotopes with a different number of protons.C. The two are ions with a different number of electrons.D, The two are ions with a different number of protons.

SolutionSecause the symbol is Al, the atomic number is always 13, implying that there

-e 13 protons. This eliminates choices B and D. In either case there is no charge,. r there are thirteen electrons present and the species is not an ion. This

=iminates choice C and leaves chbice A as the "orreit

answer. In26Al, there are,l neutrons (because 13 and L3 sum to 26), while in 27 Al, therc are 14 neutronsrecause 1,4 and 13 sum to 27). In26Al, there are 13 neutrons, 13 protons, and 13

-,ectrons, while in 27 Al, therc are L4 neutrons, 13 protons, and 13 electrons. The.,','o differ by one neutron, making them isotopes. Choice A is good.

Example 2.2',hat symbol represents the neutral atom with fifteen electrons and sixteen

,.-utrOnS?

+ i4Ga

E. i8P

:. ?4p

D ?;S

: llution:=;ause there are fifteen electrons, there are also fifteen protons in the neutralr::n. This makes the atomic number 15, which is associated with the element

- -.:sphorus, and eliminates choices A and D. The mass is 31 (15 protons and 16

-.-:trons), so the symbol is ?1P, choice C.

::.ample 2.3:rch of the following is an isotope of element 35 containing 44 neutrons?

{", 31Br-

3 79gt

: SlBr

l ;9Kr

5,r'np1i61

1,.: atomic number of bromine is 35, regardless of which isotope is being: :served. Because the element in this question is bromine, choice D is eliminated

-::rediately. When the 35 protons of bromine are coupled with 44 neutrons, the::.s of the bromine isotope is 79 amu. The answer which shows a mass of 79

i:.-r is choice B. Choice A is an anion, which is formed upon the addition of an

'- -=:iron.

- - n-right @ by The Berkeley Review 7l Exclusive MCAT Preparation

General Chemistry Atomic Theory Atornic Structure

Average Atomic MassThe average atomic mass of an element is a weighted average of the masses of allof the isotopes, an ave_rage that takes into account abundance. The reference forall isotopic masses is rzC, which^is assigned a mass of 12.000 amu. All isotopicmasses are measured relative to rzC.

Example 2.4IrVhen a sample of magnesium is subjected to mass spectroscoplr, it is found thatthere are three detectable isotopes. What is the average atomic mass ofmagnesiqm, given that the relative isotopic abundance is796/o2nMg,70"h25Mg,and11"/"26vtgz

A. 23.71.8 / ^ol"B. 24.318 f oroLC. 25.838 /

^o1"D. 26.928 / ^oteSolutionThe math associated with this question is time-consuming, so before diving intoit, review the answers to see what can be eliminated. The average mass has tofall within the range between the lightest and heaviest isotopes. In other words,the average is in the middle somewhere. This eliminates choices A and D. Themost abundant isotope is24Mg, so the average atomic mass should be close to 24grams per mole. Because the heavier isotopes are 21"/" the total mass of thesample, the average atomic mass is a little more than 0.21 grams above 24. Thebest answer is choice B. The mathematical solution is shown below, but inpreparation for this exam, use rigorous math only to confirm infuitively obviousanswers.

Average atomic mass = 79%(24) + 10%(25) + 11"/.(26)

= (0.79 x24) + (0.10 x 25) + (0.11x26)

= (0.79 x 24) + (0.10 x Qa + \) + (0.11 x (24 + 2)

= (0.79 x24) + (0.10 x 2a) + (0.10 x 1) + (0.11 x24) + (0.11 x 2)

= (0.79 x 24) + (0.10 x 2\ + (0.11 x 24) + (0.10 x 1) + (0.11 x 2)

= (1.00 x24) + (0.10 x 1) + (0.11 x2)=24 + 0.10 +0.22=24.32

The math is shown in an intuitive, step-wise fashion that does not require a

calculator. The number used as the center point, 24, was chosen knowing thatthe average value was around 24.

Classical Experiments and MachineryClassical experiments in chemistry are studies that determined the fundamenfeatures of matter. These experiments are important, because they definenature of matter and the smallest units of matter. Of interest when lookingmatter are charge, mass, location, and composition. There are three classiexperiments: the Thomson experiment (used to determine the sign of cthe Millikan oil drop experiment (used to determine the magnitude ofand the Rutherford experiment (used to determine the location ofparticles). The experiments as listed here are modified from their originalto emphasize the rationale, rather than the procedure. Of equal signifitoday is the mass spectrometer (used to determine the charge to mass ratio forparticle), which was developed to support the classical experiments.

Copyright @by The Berkeley Review 72 The Berkeley

General Chemistry Atomic Theory Atomic Structure

Thomson ExperimentThe Thomson experiment determined the existence of opposite charges in anaiom andthat charge is.a fixed quantity. Thomson deflected a stream oicharged:articles (electrons) using an eitet.rai electric field (the plates of a capacitlr).lecause the stream of particles bent in a uniform rasruon, Thomson concluded-:'at there was a consistent charge-to-mass ratio for the particles.a -tcedure and Apparatus

-- the Thomson experiment, a beam of electrons was generated travering left to:-iht, as shown in Figure 2-2. Thomson observed ihut *h"r, he appiied an.-:ctric field (a positively,charged plate on one side and a negatively charged:-iie on the other) to an electron beam, he could deflect it by in exact amount,:=rending on the strength of the field (charge on the plates) ind the mass of the.-::tron. Reversing the plates of the externufri"la gets the opposite deflection.

Figure 2-2

, ,":s and Conclusions:" ::r-ise the direction of the deflection changed when the orientation of the field- :: red, Thomson concluded that there must be two types of charge that oppose- . :nother. Because the arc of the deflection was consiant, Thomson corrci rd"d

:' .: =lectrons have a fixed charge-to-mass ratio, measured tobe 1]6 x1,08 C/ g.

.:,.:::d Thomson Experiment (in MCAT testing style)--:: is presented is an adaptation of the rho-son experiment, with changes

,, ;" j on current knowledge. The simplified apparatus has un accelerating fiild" : shown), which accelerates particles to the left, a doubre filter to ulsrrr": :r:rn linear, perpendicular trajectory, and an electric field perpendicular to. ','.ctor of entry. Charged particles are deflected according to their sign of' :--:e. Four particles are considered. Their pathways are shown ln Figure I-3.

++++++++++++++++Particle acceleratedtrom left to right

Figure 2-3

II

il

Double *r"rll

No applied electric field

Electric field on

: , ::ght @ by The Berkeley Review lc Dxclusive IUCAT Preparation

General Chemistry Atomic Theory Atomic Structure

Pathway I is attributed to an electron beam, while Pathways II, III, and fV are notidentified. The MCAT test writers are more interested in ascertaining yourintuitive understanding of problems than your superficial knowledge of facts.Questions are likely to introduce obscure particles and test you on theirproperties, which must be derived from the experiment. Consider the followingquestion based on the data in Figure 2-3.

Example 2.5Pathway II is taken by which of the following?

A. A positronB. A neutrinoC. A gamma rayD. A muon

SolutionThe particle following Pathway II must be negatively charged, based on thedirection of its deflection. It deflects in the same direction as the electron.Although some of the particles in the answer choices may seem unfamiliar, thequestion still can be answered. Based on the name, you should deduce that apositron is positively charged. This eliminates choice A. A positron is an anti-electron, in that it has the same mass as an electron, but the opposite charge.Based on the name, you should deduce that a neutrino is neutrally charged. Thiseliminates choice B. A neutrino is essentially massless and carries no charge.They are difficult to detect. Detection of neutrinos is done through collision andscintillation. Choice C is eliminated, because a gamma ray is a photon. A photondoes not bend in an electric field. Photons may be refracted, but that requires achange in medium. The only choice left is a muon, choice D. In all likelihoodyou do not know what a muon is. This question is not testing your knowledge ofparticles; it is testing your reasoning abilities. A muon traces a different pathwaythan the electron, because it is about 200 times as massive as an electron.

Copyright @by The Berkeley Review 74 The Berkeley

General Chemistry Atomic Theory Atomic Structure

\[ass Spectroscopy-- irass spectrometer is designed to measure the charge-to-mass ratio for a-:rrged particle. This is accomplished by sending a particle into a perpendicular:::rretic field and observing the degree to which it curves. The degree of arcing:::jus of curvature) for a particle can vary with mass, initial velocity, magnitude.: iarge, and the strength of the magnetic field. As momentum increases (either:.:is or initial velocity), the particle deflects less, so the radius of curvaturei- ::eases. As the charge magnitude increases, the force causing deflectionr:::eases, so the particle deflects more, causing the radius of curvature to:,:":ease. By comparing the curvature for an atomic or molecular ion to a known*:iard, the mass of the unknown ion can be determined. The mass,:"::rometer is used in general chemistry to determine isotopic abundance.;;,:.:trl that isotopes are the same element with a different number of neutrons. In:r":ic chemistry, the mass spectrometer is used to determine molecular mass

"::: fragmentation behavior to help elucidate the structure of an unknown:,:.:ound. Figure 2-4 shows a basic schematic design for a mass spectrometer.:= that mass spectrometers may have a velocity selector situated earlier in the

: i:: ,\-ay than the deflecting region.

xxx

XX 12) r"l

Figure 2-4

"' i;:ir€ and Apparatusl'r',r :-.a-rs spectrometer embodies a simple concept. Force depends on mass, so

''| *:r ll1 equal force is applied to different masses, they accelerate at different'r *s llhe mass spectrometer takes advantage of this by accelerating charged: r*- -'r'q in motion using a magnetic field. The procedure is as follows:

O -{n element or molecule is ionized using high-energy electronrmpact. The ionized particle is accelerated from the cathode plate.The electric field strength is adjustable, so the velocity of the particlecan be set to any desired value.

A The particle passes through the double filter to ensure a uniformperpendicular beam.

'6 The particle is deflected in a counterclockwise, radial fashion by theaerpendicular magnetic field oriented into the page.

S The radius is measured using a detector, where the detector simplydetects collisions. The mass-to-charge ratio is calculated from theradius of the arc.

r''::;ht @ by The Berkeley Review ID Exclusive MCAT Preparation

General Chemistry Atomic Theory Atomic Structure (

Example 2.6\A/hich of the changes to the system would increase the radius of curvature in themass spectrometer?

A. Using a doubly ionized rather than singly ionized elementB. Using an isotope with fewer neutronsC. Increasing the accelerating voltageD. Increasing the magnetic field

SolutionThe trick here is translating what the answer choices mean in terms of thephysics of the apparatus. Using a doubly ionized element (+2 cation) rather thana singly ionized element (+1 cation) results in a greater q value, so a greaterdegree of deflection is observed. This reduces the radius of curvature (r), sochoice A is invalid. Using an isotope with fewer neutrons results in a reducedvalue for mass, so a greater degree of deflection is observed. This reduces theradius of curvature (r), so choice B is invalid. Increasing the accelerating voltageresults in a greater velocity, so a lessened degree of deflection is observed. Thisincreases the radius of curvature (r), so choice C is the best answer. Increasingthe magnetic field results in a greater force, so a greater degree of deflection isobserved. This reduces the radius of curvature (r), so choice D is invalid. Thisquestion can be answered from a conceptual perspective or based on therelationship of variables described in a formula, with equal success. Whetheryou use equations or intuition is a matter of personal preference and timing.

The mass spectrometer can be used to determine the charge-to-mass ratio for theelectron and the proton. The mass of a neutron is obtained by looking at themass difference between known isotopes. For instance, the mass differencebetween 1H* (a proton) and 2H+ (deuteiium ion) is the mass of one neutron. Themass spectrometer can also be used to determine the isotopic abundance for thecomponent atoms of each element. Common isotopes that should be memorizedinclude: 12C lthe most abundant isotope of carbon), 13C (used in carbon NMR).14C (used in carbon dating, because it undergoes decay), iH

1ttr" most abund.antisotope of hydrogen),2H (deuterium, used in proton NMR solvents), 3H (tritium,used in radio{abeling experiments), 235t (used in nuclear fission), ut't4 238g 16.most abundant isotope of uranium).

Millikan Oil Drop ExperimentThe Millikan oil drop experiment is a difficult experiment to perform. Its aim L.

to suspend a charged oil drop in an electric field. To do this, an electron must beadded to the oil drop or the oil drop must be ionized by impact, before it isplaced into the electric field. We wiil consider adding the electron here. Thiscannot be accomplished easily, given that the oil drop is neutral and has noaffinity for the negatively charged electron. Falling oil drops pass through a

beam of electrons where some oil drops are penetrated at random by an electron.Ideally, the electron penetrates the core of the oil drop and comes to rest, due tothe viscosity of the oil. The suspension of an electron in the oil drop produces a

charged oil drop. Enough charged oil drops continue to fall, one of whi&eventually passes through a pore in the upper plate of a capacitor. The twcrplates of the capacitor are separated, but both lie within the walls of a glasscylinder. Figure 2-5 is a basic schematic of the apparatus used in the Millikan oildrop experiment.

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General Chemistry Atomic Theory Atomic Structure

oil drop*o .

€

llqrI

i'.*I

If suspended, q

mg.'. O='E

e

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Figure 2-5

. -:cedure and Apparatus

-:. ihe Millikan experiment, a fine mist of oil droplets is allowed to fall through a

:r-L\- por€ in the upper capacitor plate into a region where a uniform electric field

=r:sts. The oil droplets fall because of gravity, so Millikan set out to apply a force-:.at could stop the droplets from falling. If the droplets are suspended, then the-: ice applied equals the gravitational force (a known quantity thai is dependent,:- the mass of the falling object). The applied force is formed by charging the oil::rplets by exposure to either an electron beam or an x-ray beam (only one of the:eams is applied in different trials of the experiment). An electric field is applied: - suspend the charged droplets. The uncharged oil drops fall unaffected by the=:id. Despite the difficulty of this experiment, Millikan obtained enough valid:n:a so that an average measurement was put forth and accepted by the scientific: -,nmunity at large.

O The falling oil drop gains a charge by engulfing an electron as it falls.On rare occasions, the falling oil drop both gains an electron andpasses through the pore in the upper plate. The electric fieldstrength is adjustable, so that the oil droplet can be suspended.

O If the particle is suspended (or falting at a constant velocity), then thenet force is zero, so mg = - qE. Because we know g/ we can set theelectric field strength (thus we know E); and because we candetermine the average mass of an oil drop (we know an average m),we can solve for q, the charge of the electron.

; : ::tlts and Conclusionsl:-. charge of an electron has a fixed numerical value that is the same for all

=-=cirons. The value for this fundamental unit of charge is 1'6 x 10-19 C. The

-:.erge of a proton is found to have the same magnitude, but opposite sign of the.-i:tron. When a proton is combined with an electron, there is no net charge.l]:e Millikan oil drop experiment was difficult to carry out, so even after many::.s, there were only two significant figures in the final number.

lLe Thomson experiment, the Millikan oil drop experiment, and the use of the: lss spectrometer generally address questions about the subatomic particles in:=:irs of what they are. Tlne where they are questions are answered by interactions' -:h incident light and particle beams. The Rutherford experiment is the most

-.:=r-iJicant location experiment we will consider.

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Rutherford ExperimentThe Rutherford experiment determined that atoms have dense nuclei with nearlyall of the atomic mass centtally concentrated, and that metals have uniformlyspaced atoms in their microscopic composition. The Rutherford experimentrelies upon a technique that can be employed to find a ball in the bushes in thedark. "what does finding a ball in the bushes at night have to do with theMCAT?" you may ask. well, when a ball is lost in the bushes, it is easier to findits location by looking for its shadow than by hunting for the ball itself. Byshining light through the bushes and observing the shadow cast against thebackground of a wall, a circular shadow can give hints as to the location of thespherical ball. Depth in the bushes can be obtained by first moving the flashlightcloser to the bushes, then moving it farther away. The size of the shadow varieswith position of the light source, so the relative position of the ball to the lightsource is determined from shadow dimensions. In essence, the ball can belocated and its dimensions can be found without ever seeing the ball. This sameprinciple was used by Rutherford to find subatomic particles. A subatomicparticle is smaller than a ball, so a light source of significantly shorter wavelengthmust be employed. In this example, we use x-rays as the light source. In theanalogy, the bushes had to be thin for the light to pass through them, so a thinstrip of gold foil is used for the study. Rather than looking for a shadow on thewall, the experiment uses photographic paper to collect the x-rays that passthrough. Figure 2-6 shows the basic design of the experiment.

Lead reaction vessel

X-ray beamPhotographic plate

orLuminescing screen

X-ray sourceLead filter

Figure 2-6

Procedure and ApparatusAn incident beam (x-rays, alpha particles or electrons may be used) is focusedand aimed at a thin slice of gold metal, thin enough that the beam is able topenetrate and pass through the gold foil. Gold is chosen, because it has a largenucleus. If alpha particles are used, a luminescent screen is placed around thegold foil to detect where the particles pass through the foil and strike theluminescent screen (which glows when struck by an alpha particle). If x-rays areused, a photographic plate is placed around the gold foil to detect where thephotons pass through the foil and strike the film. In the alpha-particle version ofthe experiment, some particles are deflected by the gold sample, resulting inparts of the luminescing screen never illuminating.

Results and ConclusionsBecause the incident beam mostly passes straight through the sample, withdeflection being observed in only a few cases, it is concluded that the atom ismade up predominantly of empty space. The mass associated with the atomoccupies very little space and is not spread uniformly through the material.Atoms are composed of a nucleus holding the mass (the protons and neutrons.)This dense nucleus carries all of the mass besides that of the electrons. But giventhat the electrons are of such low mass, it is not possible to discern theirwhereabouts from the Rutherford experiment.

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General Chemistry Atomic Theory Atomic Sh:ucturee

The Rutherford experiment disproved the diffuse particle model (referred to as

the plum pudding model) In the version of this experiment using the x-ray beam,

the photographic screen displayed the output shown inFigure2-7.vvrtre

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lark spots represent areas where no x-rays struck the film. In essence, the spots

::e the shadows of the sub-atomic particles. Based on the distribution of the dark

':ots, it is concluded that the mass of gold is not evenly distributed in the gold, ,,1, but is in fact found in concentric, dense nuclei. Because the nuclei (spots) are

='.'enly spaced, the gold atoms must be arranged in a lattice structure. This is:€:ause the shadow pattern on the film mimics the distribution of particles in the

:..lerial. Gold is chosen because it is malleable and it has a massive nucleus,

::s it diffracts x-rays more readily than lighter elements, like aluminum. This is, :,1,- x-rays in medical imaging show bones and teeth (rich in calcium nuclei)::-:.er than tissue (rich in the light carbon, oxygen, and nitrogen nuclei). In order"-. ztalyze blood using x-rays, a heavy salt must be added to the solution, often::i:'rnr iodide.

I.l anple 2.7::t hypothesis did the Rutherford experiment support?

L -{toms combine in definite proportions'i -\toms contain subatomic particles.

- l:otons and electrons carry opposite charges.l, Sriids are made of atoms with a dense nucleus and vast empty space

:etlveen nuclei.

: : -ubion,--::e A is Dalton's law of definite proportions. Choice B is particle theory.,-' : r:e C is the Thomson experiment, conducted by observing the deflection of

-.: :Eam in a cathode ray tube when an external electric field is applied. Thei --* e:ford experiment involves the bombardment of a thin piece of metal foil

: t:ther high-energy photons (x-rays) or a beam of electrons (Rutherford did. ._: ::1 sepaiate experiments.) The photons pass through and strike an

. -- -:.ating screen, which forms a shadow pattern indicative of the material's- ::. Voit of the beam passes through, with a minimal amount being either

',, "::ed or diffracted. The conclusion is that the material is essentially empty

,;::-= ',r-ith a few dense nuclei scattered throughout the material. The spots on:".. :.::een are uniformly spaced, so the nuclei must also be evenly spaced withinl'" : .-.:nent. This makes choice D the best answer.

-", .-:ation of the electrons is not determined by Rutherford experiment.. .:. : : :, *r s are too small and moving too fast to locate precisely. This is described'- ---='-senberg's uncertainty principle.

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tonlriaL.

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General Chemistry Atomic Theory 61qmic Structure

Heisenberg's Uncertainty PrincipleThe Heisenberg uncertainty principle quantifies the idea that it is not possiblesimultaneously to identify a particle's position (where something is) and velocity(how fast and where it's going). Equation 2.1 is the mathematical version of ttreHeisenberg uncertainty principle, where Ax is the uncertainty in position andA(mv) is the uncertainty in momentum.

Ax.A(mv) ,h / nn (2.1])

The basic premise here is that you can know either where something is, or howfast it is going, but not both at the same time. Think of using a camera to focuson a moving ball. If your aperture is small and the shutter speed is fast, then thepicture of the ball shows you where it is, but you don't know where and how fastit's going. If your aperture is large and the shutter speed is slow, then the pictureof the ball is a streak that shows you where and how fast it's going, but you don'tknow exactly where the ball is. Because we cannot locate an electron's preciseposition, we settle for a view where the electron is observed over time. Thisresults in orbitals as a model for the orbiting electron.

Atomic ModelThe Bohr model presents a simplified picture that explains the quantization oflight and the reproducibility of spectra. The basic premise is that electronsoccupy specific circular orbits about the nucleus, and thus the electrons havespecific energy levels (associated with each orbit). Electrons can exist only inspecified orbits (electronic shells), so each energy level of an atom is quantized.Figure 2-8 shows this:

n=4n=3

tt=2Lq-t

r:_r

Lz-t

n=1Electronic energy levels Electronic shells

Figure 2-8

The energy levels are spaced accord.ing to the energetics of transition between thelevels. More energy is required to carry out transitions when the electron isnearest to the nucleus. The electrons are situated in various energy levels(known more accurately as orbitals). These are quantized states that electronsoccupy. Principle energy levels are numbered 1 to -, where n = 1 is the lowestelectronic energy level. Energy must be absorbed by the atom for an electron toelevate to a higher energy level. This is referred to as both excitation of. anelectron and absorption of energy. Conversely, energy is emitted when anelectron drops from a higher energy level (excited state) to the lowest energylevel (ground state). A good analogy to electrons climbing energy levels is a

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General Chemistry Atomic Theory Atomic Structure

:ocket ship escaping Earth's gravitational pull. it takes less energy to increaseliqtance as the rocket ship pulls farther away from the Earth. Likewise, it takes

-ess energy for an electron to increase energy levels as the electrons increase to::gher energy levels (farther from the nucleus). For instance, it requires more.lergy for an electron to go from the n = 2 level to the n = 3 level, than for the.,ectron to go from the n = 3 level to the n = 4 level in any atom. This is why the::'rergy levels for electron states are drawn closer and closer as the principle::rantum number (n) increases in Figure 2-8. This picture becomes a little more

- rmplicated if the rotational energy levels associated with an atom are combined;";ith the electronic energy levels. At the level of understanding needed for

"_:.swering MCAT questions, we ignore the rotational energy levels when we

-:"ok at electronic energy levels. We shall consider only the principal energy level,",'hen considering electrons. Equation 2.2 is used to determine the energy of an

=-ectron in its principal energy level.

',2h2: = energy (principal energy level) m = mass of an electron (9.11 * 10-31 \A)r = nuclear charge e = the charge of an electron (1'.6. x 10-rv C)

: = the electronic energy level h = Plank's ionstant (6.63 x 19-34 J'sec)

-:.e mass of an electron, m, the charge of an electron, e, and Planck's constant, h,::e all constants, so when Equation 2.2 is considered as a proportionality, it:€,:omes Equation 2.3.

zn2mzlea

o2EnL

*

(2.2\

(2.3)

l.

Lrample 2.8.-_::ording to Figure 2-8, how would the photon from an n = 4 to n = 2 transition

-,:lpare to the photon from drr = 2 to n = 1 transition?

\. Then = 4 ton = 2 transition is twice as energetic asthen = 2ton = 1

transition.E. The n = 4 to n = 2 transition is more than twice as energetic as then = 2 ton =

1 transition.The n = 2 ton = L transition is twice as energetic as the n = 4lransition.The n = 2 to n = 1 transition is more than twice as energetic as the n? transition.

S nlutionli.e distance between the n = 4 and n = 2 levels is less than the distance between

= 2 and n = 1, so transition energy is greater from the n = 2 level to the n = 1

.=.,'e1. The photon released from an n = 2 level to n = 1 level transition has more::.ergy than the photon released from an n = 4 level to n = 2level transition. This:-:n-inates choices A and B. The transition energy from the n = 2level to the n =- -evel is more than twice the transition energy from n = - to n = 2, so it is:=initely more than twice the energy of the n = 4 level to n = 2 level transition.

-:roice D is the best answer. This energy difference is shown in Figure 2-8. On-*-,e MCAT, you should assume that the diagrams are drawn to scale, unless

: -:.erwise noted in the question or passage.

ton=2

=4ton=

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General Chemistry Atomic Theory Atomic Structure

The energy required to excite an electron from a lower level (orbital) to a higherlevel is often in the ultraviolet (UV) to visible range of electromagnetic radiat"ion,so the energy given off as the electron drops back down to the lower level (orbit)is emitted as light energy. This is the basic principle behind spectroscopy.Lower energy levels have less absolute energy and thus are more stible states inwhich an electron can exist. The smaller the gap between energy levels, the lessenergy that is given off, and therefore the longer the wavelengin of hght that isemitted. One formula is important for understanding the relationship betweenthe speed of light, the frequency and wavelength of light, and light energy. Theenergy of a photon and its wavelength of light are inversely propo*ional.Equation 2.4 sums this up, where E is the energy of the photon, v is frequency, cis speed of the wave, and l, is wavelength.

Hydrogen Energy LevelsHydrogen is the simplest atom to study, because it has only one electron and oneproton. Much of our atomic theory is extrapolated from what we know abouthydrogen. Because energy is quantized and the energy of the electron dependson features of the hydrogen atom, the energy levels can be calculated. Equation2.5 represents the energy of the different levels of hydrogen.

E- (2.s)

Energy levels are defined as being negative relative to a free electron. If theelectron is in the 11 = @ energy level, then E = 0, and the electron is free from anucleus. Considering that photons are absorbed and emitted when electronschange energy levels, the more useful application of the energy equation involvestransition energy. Equations 2.6 and2.7 show the relationship between transitionenergy and the corresponding wavelength of the photon involved.

(2.4)E - hv =hc)"

-2.178xrr"(#)

AE = E6ru1 - Einitiul ... AE = -2.77gx10-18/--L - 1 \L.LIV LW (;il il;Jr -hc

AE

The ionization energy of hydrogen from its ground state (from the n = L level) is1312 kl/mole. Because of the squaring of the principle energy level, theionization of an electron in hydrogen from the n = 2 level is one-fourth of thatvalue (328 kJlmole). The transition energy from the n = 1 level to n = 2level isthe difference between the two values,984 kjlmole.

(2.6)

(2.7)

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General Chemistry Atomic Theory Electronic Structure

EIe fffiHi'iffiG- #ffiHI-lectronic Theory:, atoms, the electrons orbit in distinct shells. Not all shells can hold the same:.-::rber of electrons. Shells farther from the nucleus have a greater radius, and-:us a greater capacity to hold electrons. Equation 2.8 gives the maximum.,::.1pancy of electrons in a shell, where n is the principle quanfum number.

Number of electrons in shell = 2 (n)2 (2.8)

: -:.re 2-9 shows electrons of the lithium atom in their respective shells. The firsti- .-l holds two electrons, so the third electron must occupy the second shell. The

'-:': shell is the core shell, while the outermost shell is the aalence shell-

valence ,h"ll 2"0 energY level--+(electron occupancy up to 8) /

Core shell 1st energY level(electron occupancy of 2)

Nucleus: made up of bothneutrons and protons

Figure 2-9

i:ective Nuclear Charge (Nuclear Attraction)l:irLing electrons are held in their orbits by an attractive electrostatic force to the- -;leus. In addition to nuclear attraction, electrons are also repelled by other:-::hons. The net force is responsible for holding the valence electrons in place.l: t net charge exerted upon the valence electrons is referred to as the effectiae

' :.".. -:y charge. The effective nuclear charge accounts for attraction to the nucleus,--: ision from core electrons, and minimal repulsion by other valence electrons.' :en approximating the effective nuclear charge (Zefil, the nuclear charge isr::ed to the core electron charge (a negative term). Figure 2-10 shows effective- - -iear charge increasing while moving left to right across the periodic table.

i-.ucleus: +3; core electrons -2. 7 _,1.. Leff - TL

nucleus: +4;core electrons -2.7 ,-=+)" "ett - '-

Figure 2-10

:.en we move from left to right in the periodic table, the nucleus of each,-::eeding atom adds a proton and the valence shell adds an electron' The effect:: -i.e extra valence electron is not as significant as the effect of the additional'::,:on. As a result, the effective nuclear charge increases as the periodic table is

i'::::ned from left to right.

@ @

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General Chemistry Atomic Theory Dlectronic Structure

Each electron in an element travels a unique pathway, dictated by severalprinciples. Electrons are difficult to track, so we have theories to explainelectronic behavior, but they are just models to explain observed behavior.According to the Heisenberg uncertainty principle we cannot see an electron, butwe can study its pathway over time. Charged particles in motion createmagnetic fields, so by studying the magnetic field generated by a movingelectron, it is possible to learn about the pathway and position of the movingelectron. This is why two of the four quantum numbers associated with anelectron refer to magnetism that results from a moving electron. We shall blendthe many ideas about the electron that have evolved over time, starting with themost simplistic model, the Bohr model.

Electron Spin PairingElectrons fill orbitals in a pre-determined sequence, filling evenly into orbitals ofequal energy with like spin (all orbitals get a single electron, said to be "spin up"),before placing a second electron with opposite spin into each orbital. Thephysical reality is that electrons may spin either clockwise or counterclockwiseabout their axis. spinning charged particles generate magnetic moments, so thetwo opposite spins produce opposite magnetic fields. The magnetic fieldsgenerated by electrons revolving about their axis are referred to as either spin up(implying that the spin produces a magnetic field vector oriented upward) orspin down. By convention, electrons are said to fill orbitals spin up first, beforefilling spin down. Figure 2-11 shows the electron filling of lithium-7 andberyllium-9, where arrows represent electrons, and the orientation of the arrowimplies spin.

1s

it is paramagnetic.

1s

it is diamagnetic.

Figure 2-L1

The shells represent energy levels an electron can occupy, while orbitalsrepresent the region in which the electron is likely to be found. An s-orbital hasspherical electron density. The difference between the 1s and 2s orbitals lies intheir dimensions. The 1s has a smaller radius and has no nodal shells (regionswhere the electron has zero probability of existing.)

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General Chemistry Atomic Theory Electronic Strrctrrelrre

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Spin Pairing and MagnetismAparamagnetic species is defined as an atom or molecule that contains at least oner.rnpaired electron. In organic chemistry, paramagnetic compounds are referredlo as radicals. An unpaired electron is an electron that has no second electronspin paired with it. By convention, the first electron into an orbital is said toenter in a spin up fashion, thus an unpaired electron is a spin up electron in an

orbital that has no spin down electron. Because the electron is unpaired, it issusceptible to magnetic fields. If an external magnetic field is applied to a

paramagnetic species, the electron spins align with the field. This induces a

nagnetic moment into the compound, thus making it magnetic. This is to say

:hat paramagnetic species can have magnetism induced into them.

.\ diamagnetic species is defined as an atom or molecule that contains no unpairedelectrons. Al1 electrons in the atom or molecule are spin-paired, meaning thater.ery electron that is spin up will have a spin down electron sharing its orbital.Bv convention, the first electron into an orbital is said to enter in a spin upiashion, so the second electron is a spin ilown electron. Because all of theelectrons are spin-paired, diamagnetic compounds are not susceptible to:ragnetic fields. If a magnetic field is applied to a diamagnetic species, half of theelectron spins align with the field, forcing the other half to align against the field.\o magnetic moment is induced into the compound. This is to say that,:-iamagnetic species cannot have magnetism induced into them.

Electron Density and Orbitals-\tomic orbitals are three-dimensional pictorial representations of the region-,,.'here an electron is likely to be found. Because we observe electrons over time,-,",'e look at where the electron usually is, and draw a probability map of the:lectron distribution. It's like look at a spinning fan. You cannot see each bladea= they turn, but you can see over time the area where they spin' Figute 2-\2:epresents the electron density of an electron in an s-orbital over time, the orbital:epresentation, and the probability map based on distance from the nucleus.

' ':i.1 '. l"r '." '

Electrondensity map

Orbitalrepresentation

Figure 2-12

lhe electron density map shows that electrons are found most often near the:.ucleus. This is also represented by the graph of the probability of finding an:rectron as a function of its distance from the nucleus. The orbital representation

-. tvpically used by chemists to depict the s-orbital. The size of the sphere varies,,-ith the electron density map, depending on the atom. The shape of an orbital is

:efined by the distribution of electrons about the nucleus. Ninety-five percent of-:.e time, the electron can be found within the boundaries of the orbital. We will-:,ok at the s-, p-, and d-orbitals in substantial detail, while f-orbitals will be

:onsidered, but in minimal detail. Most common elements do not have electrons,':cupying the f-orbitals.

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General Chemistry Atomic Theory Dlectronic Structure

S-orbitalsS-orbitals result from spherical distribution of the electrons about the nucleus.Figure 2-13 shows three different s-orbitals, where the principle quantumnumber represents the energy level and the average distance from the nucleus.

2s

Figure 2-13

P-orbitalsP-orbitals result from barbell-like distribution of the electrons about the nucleus.Figure 2-14 shows the three different p-orbitals, each oriented about a differentaxis. Electrons are not found at the nucleus in p-orbitals. Absence of electrondensity at any point is referred to as a node. P-orbitals have one node at thenucleus that is part of a nodal plane between the two lobes.

P* P, Py

Figure 2-14

D-orbitalsD-orbitals result from double barbell-like distribution of the electrons about thenucleus. Figure 2-15 shows the five different d-orbitals, each oriented differently.D-orbitals have two nodal planes, and electrons are not found at the nucleus.

@1s

d"" dry

Irl dxz, d*r, and drr, lobes lie

dr"

between the axes

z

In d*z _ rz and drz,lobes lie on the axis

d"z -rz

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Figure 2-15

The Berkeley Review

General Chemistry Atomic Theory Dlectronic Structure

I-orbitalsl-:rbitals result from triple barbell-like distribution of the electrons about the. ::leus. There are the seven different f-orbitals, each oriented about a different:-ane or axis. Electrons are not found at the nucleus. F-orbitals have three nodal: -anes. Little chemistry is carried out with the f-orbitals, so they are unconunon.

Ll.cllective Orbital View of Energy Levelsl':bitals result from probability calculations, where energetics is considered.l:rierent orbitals are associated with different energies. Conceptually, we use,::itals to show the energy and most frequent location of an electron. Figure 2-

-: .hows orbitals with relative size emphasized, from lowest energy levels to the--iher energy levels. Levels are spaced according to energetics. Arrows:.:resent electrons and their spin orientation. Because there are twelve electrons:: r\vn, the element represented is magnesium.

w,*# ro-ffi ,o*ffi 3d.'rm r.rE

3p*C>@ t rC@ Bp,

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The upward single-headedarrow represents an electron inthe 1s orbital with its magneticspin orientation up.

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3w

_ll_,r"

8

The downward single-headedarrow represents an electron in

/ I the ls orbital with its magnetic

I i spin orientation down.

f@Figure 2-16

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Electrons within an element fill the energy levels starting from the lowest energy.This means that the electrons within an element follow a specific filling order.There are some rules to consider when looking at electronic configurations.

Pauli's exclusion principle: No two electrons can have the same set of quantumnumbers (n, l,rry,rnr).

Hund's rule: Electtons completely fill lower energy levels before starting to fillhigher energy levels. kr a degenerate set of orbitals, electrons singly occupy eachorbital before a second electron pairs up within the same orbital. Figure 2-17demonstrates Hund's rule.

1tNot allowed, because electrons fill Allowed, because electrons have filleddifferent orbitals before pairing up. each orbital singly without pairing.

Figwe2-17

Aufbau principle: Electrons are added one by one to the shells, starting with thelowest energy level, and then into sequentially increasing energy levels. Thenumbers in Figure 2-18 represent the sequence of addition for the electrons.

,,J_fI l"

"^ol Illu.ol I

Figure 2-18

Electronic ConfigurationElectronic configurations are shorthand notation for the electrons present in anatom and their energy levels. Electrons fill according to a set pattern, one that isderived from the Aufbau principle chart shown in Figure 2-19. By drawing thetable and then sequentially following the arrows, the orbital filling sequence isgenerated. Couple this information with the orbital occupancy, and electronicconfigurations are seen to be systematic. For instance, the first line shows thatthe 1s level fills first, to an occupancy of two electrons. The next arrow crossesthrough the 2s level, so the 2s orbital is filled next. From here the third arrowshows that the 2p level followed then by the 3s level are filled. It continues downthe chart. The first break from numerical sequencing comes when the 4s level isfilled before the 3d level, despite the fact that the perimeter of the 3d level iscloser to the nucleus than the perimeter of the 4s orbital. The reason for theapparent discrepancy is that the energy of the level is based on an averageposition of the electron, not the extreme position. Ionizing electrons are notremoved from the atom in reverse order, however. Outer shell electrons arealways removed first when forming cations. Figure 2.8 shows only the first fivearrows, but the pattern continues. You should also be able to deduce theelectronic configurations for neutral atoms, cations, anions, excited states, andany exceptions to the rules.

AJ

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General Chemistry Atomic Theory Electronic Structure

4f

5f 59

2:

3:

4:

5:

Fills the 1s orbital to: 1s2

Fills the 2s orbital to; 2s2

Fills the 2p then 3s orbitals to: 2p63s2

Fills the 3p then 4s orbitals to: 3p64s2

Fills the 3d,,4p then 5s orbitals to: 3d104p65s2

.c".

E

rl

D.

Figure 2-19

- rEure 2-19 shows that the electrons fill according the orbitals listed by the-quential arrows. It works by following the arrows sequentially. Arrow 1 goes-iough 1s, so the Ls orbital is filled first. Arrow 2 goes through 2s, so the 2s:rrital is filled next. Arrow 3 goes first through 2p, then through 3s, so the 2p:r:ital is filled after the 2s orbital, followed by the filling of the 3s orbital. The:rocess is repeated arrow after arrow until all of the electrons have been::counted for. Althougli. g-, h-, and j-orbitals exist in theory, the periodic table:-rntains no elements that have electrons in either g-, h-, or j-orbitals.

fxample 2.9ll"e electronic configuration for manganese is which of the following?

rs2zs2zp6 g"2916 96.7

7s22s22p6 gt2g16 965

7s2 2s2 2p6 9"2 g16 4"2 g 65

7s2 2s2 2p6 9"2 g16 4"2 36,7

5,u-lutionl'{anganese (Mn) is element number 25, so a neutral manganese atom must::ntain 25 electrons. This eliminates choice B (only 23 electrons) and choice D::rtaining 27 electrons). Because the 4s orbital is filled before the 3d orbital,: ':,ice A is eliminated. This leaves only choice C.

srample 2.10

':. element in which column of the periodic table is diamagnetic?

\ Column 1 (alkali metals):r Column 2 (alkaline earth metals)

: Column 7 (halogens)

iqlution-- ;iamagnetic compound has all of its electrons spin-paired. This means thatr,t:e must be an even number of electrons in the element. Based on the evenr ':lber constraint, choices A and D are eliminated. Column 6 elements (the:,ilcogens) have a valence electronic configuration of ns2np4, which results in*,', : p-orbitals having only one electron each. This means that chalcogens are; '::magnetic, eliminating choice C. This means that the alkaline earth metals in::r--unn 2 are diamagnetic, with a valance electronic configuration of ns2. Thel---r;a,line earth metals and the noble gases are diamagnetic. The best answer is: ::ce B.

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General Chemistry Atomic Theory Electronic Structure (

Electronic configurations may include abbreviations based on filled core shells.A filled core is represented by the, noble gas that contains those same electrons.For example, aluririnum (Al) is k22s22p61s23p1, which is equivalent to drawingit as [Ne]3s2gp1. Thir shorthand is typical, In addition, you must be aware ofsome common exceptions to the Aufbau principle. Half-filled d-shell and filledd-shell stability results when a single electron is elevated from a lower energylevel that is paired (usually the s-orbital) to yield even distribution of electrons inthe d-level. Half-filled d-shell stability is seen with chromium, molybdenum,and tungsten. Filled d-shell stability is seen with copper, silver, gold, and somesay platinum. Figure 2-20 shows the electronic configurations for chromium andcopper, exceptions to the Aufbau filling order.

Half-filled d-shell stability in chromium: [Ar]4s13d5 rather than [Ar]4s23d4

Filled d-shell stability in copper: [Ar]4s1gd10 rather than [Ar]4s23d9

Figure 2-20

Elements in the same column of the periodic table have similar valence shells andelectronic configurations, with the notable difference being the shell number. Forinstance, Wa is iNe]gs1 and potassium is [Ar]4s1. This means that alkali metalsare s1 metals, and exhibit similar chemical behavior, given their commontendency to lose one electron. Blocks in the periodic table are named after thelast electron in the electronic configuration. Alkali metals fall into the s-block byvirtue of their last electron in an s-orbital. So far, we have viewed ground stateelectronic configurations. Ground state electronic configurations occur when theelectrons occupy the orbitals in the exact predicted order, starting from leastenergetic and filling orbitals that are progressively of higher energy. An excitedstate electronic configuration occurs when any electron absorbs energy andmoves to a higher energy level than it normally occupies in the ground state.The absorption and emission of energy, usually in the form of a photon, is

associated with the excitation and relaxation of an electron, as it moves betweenthe ground and excited states.

Example 2.11

\A/hich electronic configuration represents an excited state?

A. Fi 7s22s22D6B. 19. 1r22r2jo3C. He: 1s2

D. Li:1.s22p1

SolutionAn excited state electronic configuration does not follow energetic sequence. Ar.excited state has at least one electron in an energy level higher than what i:drawn as standard for the ground state. Be sure not to confuse an ion (eithe:cation or anion) with an excited state. A cation is an atom that has a deficit of al

least one electron and thus carries a positive charge. An anion is an atom thaihas an excess of at least one electron and thus carries a negative charge. In thl'question, choice A is a fluorine anion (it contains an extra electron), and choices E

and C are normal. For Li, it should have 1s22s1 as a ground state. The electron -configuration given in the answer choice has the last electron in a 2p-orbitarwhich is of higher energy than the ground state 2s. This makes choice D th.correct answer, because it is an excited state.

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Copyright O by The Berkeley Review 90 The Berkeley Revier ffi,[lL'''::,.{

General Chemistry Atomic Theory Electronic Structure

Quantum NumbersQuantum numbers are a set of four numbers that uniquely describe an electron;r-ithin an atom. Four factors describe an electron: the shell, the orbital, therrientation of the orbital, and the alignment of the magnetic field resulting fromie precessing electron. These terms translate into the quantum numbers. A:undamental rule with quantum numbers is that no two electrons can have thesame set of quantum numbers (this is the Pauli exclusion principle). Quantum:.umbers are used to describe the motion and location of each electron in an

=Lement. Quantum numbers describe the shape of an electron's cloud, theromentum of the electron, the orientation of the electron density, and the::tation of the electron about its axis. There are four quantum numbers used to:escribe an electron; n, l, rr.1, and ms. There are also rules that must be followed

'','ren assigning quantum numbers to an element. Each number has its specific:.ridelines, which often depend on the other quantum numbers. Table 2.2 lists':e rules for assigning quantum numbers.

Table2.2

:: -e 2.3 shows the correlation of quantum numbers to electrons within an atom.

Quantum # Rules

nPrinciple (n): Describes the shell (average radius of theelectron from the nucleus and its energy level) in which theelectron resides. It can be any integer greater than zero.

IAngular Momentum (l): Describes the orbital (shape of theelectron cloud formed by the orbiting electron) in which theelectron resides. It must be less than the value of n. It can bea positive value or zero.

m1Magnetic (m): Describes the orientation of the orbital about a

plane or axis. It can be any value in the range from negative I

to positive l, including zero.

msSpin (mr): Describes the rotation (counterclockwise orclockwise) of the electron about its axis. It can be eitherpositive or negative one-half (spin up or spin down).

I m1 ms orbital totalelectrons

Description

0 0 +1) 1s 2 tst tsJ

0 0 a1)

2s 2 zsI zsl

1 +1, 0 t1)

2p 6 2p*I 2puI 2prl 2p*!Zpul2prl

:i0I

0 +12

3s 2 gs'l gsJ

1 +1,0 +1) 3p 6 3p*'l 3put 3pr'l 3p*J 3puJ 3prJ

-0 0 +12

4s 2 +st +sJ

2 01-.) +1, a12

3d 103dxyt 3dxzI 3dyzI 3dx2_yzI 3dz2I3d*.rJ gdxz! 3dvzl 3dx2 - v2I gdz2!

Table 2.3

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General Chemistry Atomic Theory Electronic Structure (

When assigning quantum numbers to an electron, you must keep in mind that"no two electrons orbiting the nucleus of the same element have the same set ofquantum numbers." To assign quantum numbers, you must first describe theelectron in words. Consider an electron in the third shell. There are eighteenelectrons held in the third shell, so we need to be more specific. An electron inthe third shell can be in either an s-orbital, a p-orbital, or a d-orbital. For sake ofargument, let's consider a p-orbital. There are three p-orbitals, each capable ofholding two electrons, so there are six electrons that can be found in the 3p level.We need to be more specific. Each p-orbital has a different orientation in space(p* along the x-axis, pu along the y-axis, and p" along the z-axis). For sake ofargument, let's considei the p-orbital aligned on the x-axis. This is the p*-orbital.Two electrons can be found within a 3p*-orbital, one with a magnetic spinmoment upward and the other with a magnetic spin moment downward. Bydescribing the electron as spin up, the electron is unique. There is only oneelectron that can be spin up within a 3p*-orbital. It took four terms to narrow itdown to a unique electron, hence there are four quantum numbers. Let'sconsider that same electron:

A 3p*-spin up electron has the following description in words and thereforethese corresponding quanfum numbers:

Third shell therefore, the principle quantum number (n) is 3

P-orbital therefore, the angular momentum quantum number (/) is 1

X-axis orientation therefore, the magnetic quantum number (m) is -1

Spin up therefore, the magnetic spin quantum number (ms) is + 12

The quanfum numbers for the electron are: n - 3, I = 1, m1- -L, and ms = + 1. The2

skill you must (re)develop is getting the four numbers quickly. Determining nand I is relatively easy. The n-value is the shell number, so it is pretty much a

given. The l-value is the orbital. Orbitals increase from s to p to d to f, and so on,and the l-values increase from 0 to 1 to 2 to 3 and so on. An l-value of 0corresponds to an s-orbital, an l-value of 1 corresponds to a p-orbital, and so

forth. The challenging part is finding the m7 and ms values. In all likelihood,you have no idea why x-axis orientation leads to the conclusion that ml = -1.

That's just the way they do it. Just as x comes before y and z in the alphabet, -1

comes before 0 and +1 numerically. To get the m7 and ms values, you must drarn'

out the energy levels. The assignment of m7 values for the p-orbitals is shown inFig:re2-21,.

Pzml= +1

m

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Chemistry convention tells us to number the different orbitals from -l to +'sequentially. The middle orbital always has an m7 value of 0. This is true for a1l

orbitals. The assignment of m7 values for the d-orbitals is shown tnFigwe2-22.

dxyml= -2

d*,ml=-1

dytml=0

Figure 2-22

d*z -rzm7-+L

d12

ml= +2.

Copyright @by The Berkeley Review 92 The Berkeley Revieu

General Chemistry Atomic Theory Electronic Structure

Flectrons are filled into orbitals one at a time from left to right, with spin upqoing in first, followed by spin down once each degenerate orbital has one:iectron. Filling spin up first is a convention and does not represent the physicalreality of electrons. When an electron is spin up in an orbital, it has a spin:uantum number (mr) of +1 12.

tNhen an electronis spin down in an orbital, it:.as ms = -1 /2. For the 3p* spin up electron, the filling is shown in Figure 2-23.

Pxml= -L

Pzml= +'1,

Pyml=0

Figure 2-23

lLe electron falls into the first p-orbital, so m7 is -1. The electron is spin up, so ms.i. -- /2-

, : be able to apply these quantum numbers, keep in mind that each electron--iin an element has a unique set of quantum numbers. An electron can be

:=-cribed in terms of words (such as an electron in the second energy level in a: --:bital with x-orientation and spin up is a 2p" t) or in terms of numbers (n = 2,

= 1. r1l = -1, rns = +1 /2). Quantum numbers follow arbitrary guidelines. For:-i:ance, the electron in a p-orbital oriented on the z-axis in the third shell with: -:. up has the quantum numbers n = 3 (for the third shell), I = 1 (because the: rirofl is in a p-orbital), trrt= +l(for the z-axis), and me = +1 / 2(for spin up).

:r-ample 2.12'* --I the following are true of an electron EXCEPT:

'- tiecttons in a lower energy level can absorb energy and elevate to a higher:r'rergy.

: :ractly the same amount of energy is emitted when an electron relaxes:etrveen the same states.

." ::.ere are many energy absorptions possible, but they are always of a

: :ecisely known energy.- ..r. electron in the n = 1 energy level can be found at an infinite number of

: stances from the nucleus.

: : . "tf,ion-: :r.odel of atoms is a positive concentric nucleus surrounded by orbiting: . r r,fls. These electrons may occupy only specific orbits, which have distinct' :i:€s and pathways. According to this description, electrons in a lower-.::,'level can absorb energy and elevate to a higher energy, so choice A is. -: and thus eliminated. According to this description, exactly the samer: . -:.t of energy is emitted when an electron relaxes between the same states, so

:= B is valid and thus eliminated. According to this description, there are-.---, possible energy absorptions possible, but they are always of an exact" ,::I,'. so choice C is valid and thus eliminated. According to this description,,;' =-:::ron in the n = L energy level is found at only one distance from the

. - = '-i, not at an infinite number of distances. This makes choice D invalid, and

''" -.:: --i.€ best answer.

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General Chemistry Atomic Theory Periodic Trends

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PeriodicTrendS''''.''..'::..::'::'::''The Periodic TableThe periodic table is organized according to valence electrons. Figure 2-24 showsthe shell of the periodic table. The blocks are named after the last electron thatfills each respective atom in that section of the periodic chart. For instance, anelement in the D-block has its last electron (a valence electron) in a d-orbital.Understanding the blocks helps to understand the periodic trends. The S-blockcomprises only metals. The D-block houses the transition metals, where periodicbehavior is not necessarily an obvious trend. The P-block includes metals andmetalloids in its left side and non-metals in its right side.

ttllltttttttttttrttltttttt

- - + - - + - - + - - + - - + - - S:btoak - - + - - + - - + - - + - - + - -llltltltltttll!ltltltltltt

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General Elemental Periodic TrendsPeriodic trends refer to any chemical behavior that can be matched to a trendwithin the periodic table. All chemical properties depend on the valenceelectrons, so periodic trends ultimately grow out of valence electron trends. Anr-feature of an atom that affects how tightly a valence electron is held contributesto periodic trends. The two major factors are the effective nuclear charge and thevalence shell, both of which support periodic trends. They are listed below:

O As you move from left to right across a period in the periodic table, theeffective nuclear charge increases.

@ As you descend a family in the periodic table, the valence shell increases/ sc

the distance of the valence electron from the nucleus of the atom increases.

The effective nuclear charge (2"1) is the net charge exerted upon the outermos:electrons (valence electrons). This value is empirically determined and takes intiaccount attraction due to the protons, shielding due to the neutrons, anirepulsion due to the core electrons. It is generally approximated as the proto:charge minus the electron repulsion. The effective nuclear charge affects horr'tightly the electrons are held, which affects the ionization energy, the electro:affinity, and the atomic radius. The effective nuclear charge increases across :row in the periodic table. Although we generally approximate the effectir':nuclear charge, it can be derived from the ionization energy. This procedur:assumes that ionization energy is purely related to the effective nuclear chargeand fails to account for shell stability, particularly filled-octet stability.

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Copyright @ by The Berkeley Review 94 The Berkeley Revien

General Chemistry Atomic Theory Periodic Tfends

Exarnple 2.13r\tLich of the following atoms has the GREATEST effective nuclear charge?

-A.. CarbonB. FluorineC. SodiumD. Sulfur

Solution:lfective nuclear charge increases from left to right in the periodic table, so the.lement in the column that is farthest to the right has the greatest effective:,uclear charge. Fluorine is to the right of carbon within the same period, so it:,as a greater effective nuclear charge (2s16). Choice A is eliminated. Sodium is inre first column of the periodic table, so it has the smallest effective nuclear:rarge. Choice C is eliminated. The correct answer is choice B.

: =riodic trends depend on both the effective nuclear charge (affecting the

'::ength with which valence electrons are held) and the valence shell (affecting-.:.e distance between electrons and the nucleus.) Periodic trends as we movei"-nl left to right across a row of the periodic table (period) are attributed to:.;reasing effective nuclear charge. Periodic trends as we move up through a::-urrm of the periodic table (family) are attributed to decreasing valence shells.l}le net result of these two effects is represented by the bold arrow shown in::rure 2-24. As we move along the pathway of the bold arrow, the following

=:reral atomic trends are observed:

@ The atomic size decreases (the radius of the atom is defined as the distancerrom the center of the nucleus to the exterior of the valence electron cloud).

'g The ionization energy increases (the energy required to remove theoutermost electron from the atom).

,0 The electron affinity increases (the energetics associated with an atomgaining an electron).

{D The electronegativity increases (the tendency to share an electron withaaother atom within a bond).

iuanple 2.14',::ch sequence accurately lists increasing ionization energy of the atoms?

4- Br>F>Cl>Te3 O>S>P>Brl- Br<FcCl<TeI O<S<P<Br

i'c,lution1'e appearance of fluorine in the middle of a trend should get your attention as a,';-'--B orSWer, eliminating choices A and C. Oxygen is directly above sulfur inrt periodic table, so oxygen has a greater ionization energy than sulfur. The::ect answer is choice B.

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@neral Chemistry Atomic Theory Periodic Trends

None of the trends is uniform or perfect. The effective nuclear charge does notuniformly increase when we scan across a period. There are deviations in sometrends. One of the more common deviations is seen with electron affinity andionization energy/ due to half-filled stability and filled-shell stability. Forinstance, nitrogen has a greater ionization energy than oxygen, because uponionization, nitrogen loses its half-filled p-shell. On the contrary, oxygen gainshalf-filled stability upon being ionized. The test writers may not prey on theseexceptions, but ihey certainly can emphasize the conceptual aspects by looking atthe factors that affect periodicity. For instance, rather than ask about atomicradius, they may ask about ionic radius.

As a general rule, cations are smaller than neutral atoms, because the loss ofelectrons allows the atom to compact more tightly, given the diminishedrepulsion associated with the missing electrons. As a general rule, anions arelarger than neutral atoms, because the gain of electrons causes the atom toexpand, given the enhanced repulsion associated with the additional electrons.Valance electrons account for the size of anions, neutral atoms, and cations.Extra electrons repel and thus increase the atomic size of an aniory while a loss ofelectrons results in less repulsion and a smaller radius for a cation.

Example 2.15\zVhen strontium (Sr) becomes an ion, what is observed?

A. It forms a +L cation that is smaller than Sr.

B. It forms a +1 cation that is larger than Sr.

C. It forms a +2 cation that is smaller than Sr.D. It forms a +2 cation that is larger than Sr,

SolutionChoice C is correct. Strontium (Sr) is found in the second column of the peritable. Alkaline earth metals lose two electrons to gain octet stability. Asstrontium carries a +2 charge, so choices A and B are eliminated. Cationssmaller than neutral species, because there are fewer electrons and thusrepulsion. This makes choice C the best answer.

Example 2.15Which of the following ions is the LARGEST?

A. CTB. Na+c. K+D. Br-

SolutionWithin a period, anions are larger than cations, so chloride (Cl-) is largersodium cation, and bromide (Br-) is larger than potassium cation. This elimichoices B and C. Because Br is lower in the periodic table than Cl, Br is largera neutral atom than Cl. This same trend holds true, if both Br and Cl pick upsame number of electrons. In this case, both bromide and chloride picked upelectron each, so bromide, with its electrons in a higher valence shell, isthan chloride. The answer is choice D.

Copyright O by The Berkeley Review 96 The Berkeley

General Chemistry Atomic Theory Periodic Trends:nds

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Atomic Radiuslhe atomic radius is the distance from the center of the nucleus to the edge of the'.'alence cloud of electrons. However, most of the empirical measurements of::omic radii that exist are not from electron density maps, but instead areietermined by dividing bond distances between like atoms in half. Because of:''"'erlapping electron clouds, this method does not generate a true atomic radius,:ut rather a covalent bonding radius. However, it does lead to internally:-rnsistent values. Atomic radii are measured in units of picometers. The radius:: an atom decreases as a family in the periodic table is ascended, because the:::.mber of electronic shells decreases. The radius of an atom decreases as a:eriod in the periodic table is scanned from left to right, because the effective:--rclear charge increases. The trend is fairly uniform from left to right, with no::<Linct exceptions due to half-filled stability.

.,rure 2-25 lists the atomic radii of the first twenty elements. As a general::servation, within a period, atomic radius decreases as the atomic number:areases.

0.7

0.6

0.5

0.4

0.3

0.2

0.1

7 B 710 1.1 12 13 14

Atomic Number

Figure 2-25

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General Chemistry Atomic Theory Periodic Trends

The sudden increase in size from He to Li, Ne to Na, and Ar to K is attributed tothe expanded valance shell associated with the additional electron. For instance,the electronic configuration of Ar is k22s22p69s23p6, while K isk22s22p6gs2g?64s1. The fourth shell (n = 4 quantum level) has a larger radius,so the potassium atom is larger than the arSon atom. Table 2-25 terminates at

element 20, because from element 2'J. to 30 the radius stays roughly equal, since

the electrons are being added to the third quantum level (3d orbitals). This does

not affect the radius of the electron shell drastically. The transition metals have

very similar atomic radii, although they are not exactly equal'

The trend is consistent through the elements listed, with the exception of heliumand hydrogen. The larger atomic radius of helium when compared to hydrogengoes against the discussed trend in effective nuclear charge. The best explanationfor this deviation involves both the shielding effect of the two neutrons in the

helium nucleus and the electron repulsion experienced by electrons in the firstquantum level where they are closer together than in any other quantum shell.

In other words, the electrons in the n = 1 level repel one another more thanelectrons in the tt = 2 level, because they have the smallest interelectronicdistance. This repulsion forces the electrons away from one another, resulting ina greater area being occupied by the orbiting electrons. Keep in mind that the

electrons, not protons ot neutrons, define the radius of an atom. The atomic radiiof atoms may be used to predict the bond length within molecules. The smallerthe atomic radius of the atom, the shorter the bond it forms when sharingelectrons with another atom. Shorter bonds are stronger bonds, so there exists a

correlation between an element's location in the periodic table and the strength ofthe bonds that element can form.

Example 2.17Which of the following elements has the LARGEST atomic radius?

A.OB.FC. NeD. Na

SolutionThis is just a simple case of reading from the periodic table' The element in the

lowest and furthest left position is sodium, Na, so choice D is the best answer.

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General Chemistry Atomic Theory Periodic Ttends

Ionization EnergyIonization is the process of losing an electron from the valence shell. \Atrhen anatom is ionized, it becomes a cation. The energy required to remove the outer-most electron from the valence shell is known as the ionization energy. A genericreaction for ionization is shown below, where E represents any element.

E(g) + E+(g)+e-

The energy required to carry out ionization depends on the attraction of theelectron to the nucleus, its distance from the nucleus, and the stability of itselectronic configuration. Because several factors influence ionization energy, it isioo difficult to calculate, and thus it is generally evaluated in a qualitative sense.

figure 2-26 lists the ionization energies of the first twenty elements in the:eriodic table. Within a row in the periodic table, ionization energy increases as

re atomic number increases. This is a general trend, but with some exceptions.

2500

2400

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1100

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1 m * ;[l O by The Berkeley Review 99

15 t6 t7 18 t9 20

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General Chemistry Atomic Theory Periodic Trends

Ionization energy for an element generally increases as you move left to right inthe periodic table. Notable exceptions occur when there is half-filled stability ofthe energy level and when there is an s2-shell. Ionization energy for an elementincreases as you ascend a column in the periodic table, with the element higherup the column having the greater ionization energy. This is because as thenumber of electronic shells decreases, the proximity of an electron to the nucleusincreases, and thus the attraction to the nucleus increases. The sudden decreasein ionization energy from He to Li, Ne to Na, and Ar to K is attributed to theexpanded valance shell (and thus reduced attraction) associated with theadbitional electron. For instance, the electronic configuration of Ne is 1s22s22p6,

while for Na it is 1s22s22p63s1. The third shell (n = 3 quantum level) has a largerradius, so the sodium atom can more easily lose an electron than the neon atom(with its outermost electron being more attracted by the nucleus). As withatomic radius, from element 21 to element 30 the ionization energy remainsroughly equal, because the electrons are being removed from the same 4s-orbital.The effective nuclear charge on the 4s-electrons does not change drastically. Theexceptions in the transition metals are also due to half-filled and filled d-shellstability.

Ionization energy may be used to predict the oxidation and reduction potentialsof an atom. The easier it is to ionize an atom, the easier it is to oxidize that atomby one electron. This leads to a larger (more positive) value for the oxidationpotential. A low ionization energy for an atom correlates to a smaller (or morenegative) value for the reduction potential of the cation that is formed.

Example 2.18\Alhy is the ionization energy of beryllium greater than the ionization energy oilithium?

A. Be has a larger principal quantum number than Li.B. Li has a greater density than Be.

C. Be has a larger effective nuclear charge than Li.D. Li has a bigger proton count than Be.

SolutionBoth beryllium and lithium have their last electron (the electron lost uponionization) in a 2s-orbital. This eliminates choice A, because the principlequantum number (valence shell) is the same for both. Lithium is less massive

and larger than beryllium, so it is less dense. This eliminates choice B. Lithiu:rhas three protons, while beryllium has four, so choice D is a false statement,Only choice C remains. The difference between lithium and beryllium lies in the

effective nuclear charge. The beryllium nucleus has four protons, while the

lithium nucleus has only three protons. The greater number of protons increases

the attractive pull on the electron. Because the pull is greater, the effectir-enuclear charge is greater. The greater the effective nuclear charge, the greater the

ionization energyi therefore beryllium has a greater ionization energy tharlithium, because it has a greater effective nuclear charge than lithium.

The MCAT test writers design questions to encompass logical analysis. To test a

concept and generate the illusion of unfamiliarity, they can make subtle change=

to a question. The second ionization energy can be tested in a questior:.Consider the periodic trends for elements, but remember that when you evalualeions, reading from the periodic table does not always give the right answer.

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Copyright @by The Berkeley Review 100 The Berkeley Revieu

General Chemistry Atomic Theory Periodic Ttends

Example 2.19:{ow does the ionization energy of sodium compare to the ionization energy ofnagnesium?

1.

B.

Na has a greater ionization energy, because it has the greaternuclear charge.Mg has a greater ionization energy, because it has the greaternuclear charge.

C. Na has a greater ionization energy, because it has the greater electronic shell.D. Mg has a greater ionization energy, because it has the greater electronic shell.

S olution.:.:s is just a simple case of reading from the periodic table. Both magnesium:--.1 sodium are in the same period (third), so they have the same valence shell.':-s eliminates choices C and D. The element that is furthest to the right in the:=::.odic table is magnesium, so magnesium has the greater ionization energy.-:.oice B is the best answer, because Na and Mg are in the same row of the:'=::odic table, where effective nuclear charge is the reasoning behind periodic: -::erences.

lnmple 2.20-,:-,.'- does the second ionization energy::zation energy of magnesium?

of sodium compare to the second

I-. \a has a greater second ionization energy, because it has the greater effective:,uclear charge.

effective

effective

\1g has a greater second ionization energy, because.if ective nuclear charge.\a has a greater second ionization energy, because

it has the greater

it has the smaller.lecfronic shell.

- \lg has a greater second ionization energy, because it has the smaller:lecfronic shell.

:.:,ution-". second ionization energy is the energy associated with losing the second, :-::orr, which takes the element from +L to +2. For sodium, an octet is obtained" .:sLng the first electron, thus the second electron lost drastically destabilizes." = =lectron cloud. This makes the second ionization energy very high. For:: =.esium, an octet is obtained by losing the first and second electrons, thus the,,'-,:,d electron lost stabilizes the electron cloud. This makes the second: -ation energy very low for magnesium, eliminating choices B and D. Because

' . ::e talking about shell stability, the best (albeit not perfect) answer is choice

Na+ -------+ Na2+

622"22O6 1"21s22ps

Mg* ->

Mg2*

1"22"22'.69s1 1s22s22p0*'" . .-ectronic configurations show that sodium loses octet stability upon its, .: i ionization, while magnesium gains octet stability upon its second"-.ion.

..king about the second ionization energy, the concept rather than the:::ization of periodic trends is tested.

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General Chemistry Atomic Theory Periodic Tfends

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Electron AffinityElectron affinity measures the tendency of an element to gain an electron. It is ameasurement of the energy absorbed or released when an electron is added intothe valence shell. Electron affinity can be either negative or positive, meaningthat gaining an electron can be either exothermic oiendotherrnic. The generiireaction for electron affinity is shown below, where E represents any elemfnt:

E(g) + e- --+

E-(s)

Figure 2-27lists the electron affinity for the first twenty elements in the periodictable. Within a row in the periodic table, electron affinity correlates with atomicnumber, but there are some extreme spikes in the trend.

-100

-725

-150

-175

-200

-225

-250

-275

7 8910ltL213Atomic Number

Figure 2-27

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General Chemistry Atomic Theory Periodic lfends

The biggest deviations are attributed to the stability associated with a filled s-shell. The graph may seem confusing at first, because the lowest numbers have

'Jre greatest electron affinity. The numbers listed are energy released upon:aining an electron, so a negative number refers to an element with a highrlectron affinity. The sudden increase in electron affinity (energy released upon

=aining an electron) from Be to B, Mg to Al, and Ca to Ga is attributed to the

-rstability of one electron in the p-level. For instance, upol gainilS 1n electron,:re electionic configuration of Mg goes from 1s22s22p63s2 to 1's22s22p63s23p1,

.'hich creates a new energy level, and is unfavorable. Upon gain{g a-n electron,

--.e electronic configuration of Na goes fromLs22s2zp6ZJ to 1's22s22P63s2, whichi-Js the s-shell and generates stability. From element 21 to element 30, electron::f,nity is erratic, because the d-shell stabitity is changing. No trend for electron

-:inity is evident in the transition metals.

-:ke ionization energy, the energy associated with electron affinity depends onre attraction of an electron to the nucleus, its distance from the nucleus, and the-:ability of its electronic configuration. Because several factors influence electron.--irLity, the trend across a period is erratic. In general, an element releases more::.ergy upon gaining an electron as you move left to right in the periodic table.-r:astic exceptions occur when there is half-filled stability of the energy level and

:-en there is an s2-shell. In general, an element also releases more energy upon:,iing an electron as you ascend a column in the periodic table. This is because,, the number of electronic shells decreases, the new electron is closer to the- ::leus, and thus the attraction to the nucleus increases.

-:r.ample 2.21-:.t electron affinity of an element is MOST similar to which of the following: : -:erties?

l" Electronegativityl lonization energy: Cxidation potentiall, Reduction potential

: : Xution--

= electron affinity for an element measures the energy associated with the gain: re electron. Choices B and C are out, because both of them deal with losing

.: =-ectron. Electronegativity is not the best choice, because it deals with ther" r:ng of electrons in a bond, not the gaining of an electron. The best answer is

:.r" : -;e D, reduction potential, because reduction is the gain of an electron. An, s:..€flt with a high reduction potential has a high electron affinity.

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General Chemistry Atomic Theory Periodic Trends:

1. 2 3 4 5 6 7 I 910 11 12 1,3 14 15 16 17 1.8 19

ElectronegativityElectronegativity is a measure of an atom's tendency to gain and retain anelectron from a neighboring atom within a bond. It is formally defined as theability of an atom to attract towards itself the electrons in a chemical bond.Electronegativity is related to both ionization energy and electron affinity. Thisis to say that the electronegativity of an atom depends on both the electronaffinity and ionization energy of that atom. Linus Pauling generated a method tomeasure electronegativity, and created a scale, referred to as the Pauling scale.

Electronegativity is measured on a relative scale, with the values measured fromthe electron distribution within a bond. The standards are 0.9 for sodium and 4.0

for fluorine, and all other values are based on dipole moments associated withbonds to these atoms.

The electronegativity of an atom increases as the periodic table is ascended,because as the number of electronic shells decreases, causing the attraction to thenucleus to increase. The electronegativity of an atom increases as the periodictable is scanned from left to right, because the effective nuclear charge increases.The trend in electronegativity is very clean, showing no exceptions. Figure 2-28reflects these trends.

Atomic Number

Figure 2-28

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General Chemistry Atomic Theory Periodic Tlends

\Alhen the electronegativity values of the two atoms within a bond are close, thebond is covalent. When the electronegativity difference exceeds 2.0, then thebond is ionic. Figurc 2-29 represents the electron clouds and shows the relativeelectronegativity of the atoms involved in an ionic bond and a covalent bond.

Unequal sharing(AEneg =2.1, :. the bond is ionic)

Relatively equal sharing(AEneg = 1.0, .'. the bond is covalent)

Figare 2-29

Example2.22:lectronegativity difference between bonded atoms is BEST determined by:

.\. measuring the bond length.B, measuring the dipole moment.C. calculating the difference in electron affinity between the two elements.D. calculating the difference in ionization energy between the two elements.

Solutioniecause electronegativity measures the tendency to share an electron, and the

' pole moment represents the degree of sharing between two atoms in a bond,

-.e best answer is choice B. Electronegativity is related to both electron affinity.:-C ionization energy, so electronegativity can be estimated knowing both,:-ization energy and electron affinity, but not just one of them. Choices C andI are eliminated, because you need both to approximate the electronegativity::-'ierence. Choice A is eliminated, because bond length dictates bond strength,: -t not necessarily the relative electronegativity. This question may seem'^allenging, because the terms are interconnected.

:'tample 2.23l-.e lrend in electronegativity increases with which of the following?

r. Ionization energyI -{tomic radius- -\tomic numberl, Number of valence electrons

::lution-- . question is close to verbatim in reproducing a question from a recent MCAT: :3. Electronegativity follows a cleanly predictable trend, so choice A is, -::unated, because ionization energy follows an erratic trend. Electronegativityr- :reases as atomic radius decreases, so choice B is eliminated. Increasing atomic'-::.ber sounds tempting; but when a new shell is formed, electronegativity:.-- -:s, while atomic number increases. This can be seen in going from fluorine to";-:. to sodium. This eliminates choice C. As the number of valence electrons

-" :::ases, we are moving from left to right across a period of the periodic table.

-- ::easing valence electrons does not affect shells. From left to right in a period,.":::onegativity increases. This makes choice D the best answer.

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General Chemistry Atomic Theory Periodic ltends

Periodic Families (Groups)Columns in the periodic table (families) contain elements that have the samevalence electron count and thus show similar chemical reactivity. They arereferred to as groups of elements. some of the groups you should know aie thealkali metals (first column of H through Fr), the alkaline earth metals (secondcolumn of Be through Ra), the chalcogens (sixth column of o through po), thehalogens (the seventh column of F through At), and the noble gases (last columnof He through Rn). The last column of elements is observed to be generally non-reactive. They are the inert gases. we shall address the general properties ofeach group, without emphasizing trivia. The MCAT test-writers often misleadyou with trivial information in the passage. It is up to you to recognize the trivia.

Alkali Metals (Group I)Alkali earth metals are in the first column of the periodic table. Included arelithium, sodium, potassium, rubidium, cesium, francium, and to some extenthydrogen. Hydrogen can act as both a halogen and an alkali metal. Thecouunon feature is that their valence shell is ns1, where n is any integer greaterthan one. As neutral elements, they are strong reducing agents, because theyreadily lose an electron to become a +1 cation with a filled octet. They are someof the strongest reducing agents (most favorably oxidized). They react with anycompound or element that has a high electron affinity. Their reactivity increasesas you descend the column, mostly because it is easier to lose an s-electron from ashell that is further out (greater n quantum number). Their cation form is verysoluble in water with almost any anion.

All alkali metals react favorably with water to form the metal hydroxide andhydrogen gas. Reaction 2.1 is the generic reaction:

2M(s) + 2H2O(g) ---+

2MOH(aq) + Hz(g)

Reaction 2.1

The oxides they form are variable with the metal. Lithium forms an oxide (MzO),sodium forms a peroxide (MzOz), and potassium, rubidium, and cesium fsuperoxides (MOz). Reactions 2.2, 2.3, 2.4, 2.5, and 2.6 show the oxidatireactions of the alkali metals:

4 Li(s) + Q(g) -----+ 2Li2O(s)

Reaction 2.2

2 Na(s) + Oz(g) ----+Reaction 2.3

K(s) + Oz(g) +Reaction 2.4

Rb(s) + Oz(g) +Reaction 2.5

Cs(s) + Oz(g) +Reaction 2.6

Na2O2(s)

KO2(s)

RbO2(s)

CsO2(s)

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General Chemistry Atomic Theory Periodic lfends

Besides reacting with oxygen, alkali metals are oxidized by halogens, nitrogen,nd hydrogen. Reactions 2.7,2.8, and2.9 are a sampling of these reactions.

6l,i(s) + Nz(g) € 2Li3N(s)

Reaction 2.7

2 Na(s) + HzG) -->

2 NaH(s)

Reaction 2.8

2Cs(s) + Br2[) € 2CsBr(s)

Reaction 2.9

Lxample 2.24',l:rat happens to sodium metal when it is added to water?

{.. It is oxidized to yield sodium hydroxide, which is insoluble in water.B It is oxidized to yield sodium hydroxide, which is soluble in water.:. It is reduced to yield sodium hydride, which is insoluble in water.D" It is reduced to yield sodium hydride, which is soluble in water.

r'olution--: shown generically in Reaction 2.1, a metal hydroxide is formed upon addition-: an alkali metal to water. Because sodium is going from neutral to +1 when-,sng an electron to oxygen, sodium is oxidized. This eliminates choices C and

- You may recall from your acid-base chemistry experience that sodium- . :roxide is a strong base, and it readily dissociates in water. This makes NaOH.:",. soluble in water. The best answer is choice B.

4,&aline Earth Metals (Group II)*;aline earth metals are metals from the second column of the periodic table.*-:,uded are beryllium, magnesium, calcium, strontium, and barium. Most:,::-.-llium complexes are covalent in nature. The common feature is that their:-'nce shell is'.rr2, where n is any integer greater than one. As neutral elements,

:':,.- are strong reducing agents, because they readily lose two electrons to: tr": rme a +2 cation with a filled octet. However, they are not as reactive as alkali: =:als. Like alkali metals, they are strong reducing agents. They too react withL:',- compound or element that has a high electron affinity. Their reactivityi: ::eases as you descend the column, because the first and second ionization,:=:gies both decrease. Their cation form is not as soluble in water as are the.-" rli metals, primarily due to their +2 charge and smaller radius.

--.ratrine earth metals, except beryllium, react favorably with water to form a

::::a1 hydroxide and hydrogen gas. Reaction2.l0 is the generic reaction:

M(s) + 2 H2O(g) ------'> M(oH)2(aq) + HzG)

Reaction 2.10

-:: alkaline earth metals all form oxides (MO) when oxidized by oxygen gas.-:= generic reaction is shown in Reaction 2.11.

2M(s)+Oz(g)+2MO(s)Reaction 2.1L

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General Chemistry Atomic Theory Periodic T?ends

Besides reacting with oxygen, alkaline earth metals can also be oxidized byhalogens, nitrogen, and hydrogen. Reactions 2.I2, 2.I9,2.I4, and 2.1.5 arc arandom sampling of these reactions:

Mg(s) + 2 HCI(aq) ----+ MgCl2(aq) + Hz(g)

Reaction 2.L2

3 Sr(s) + 2 N2(g) ---'€- 2 Sr3N2(s)

Reaction 2.13

Be(s)+Hz(g)+ BeH2(s)

Reaction 2.L4

Ba(s) + I2(g) -+ BaI2(s)

Reaction 2.15

Chalcogens (Group VI)Chalcogens are metalloids and non-metals from the sixth column of the periodictable. Included are oxygen, sulfur, selenium, tellurium, and polonium. Oxygen,sulfur, and selenium are non-metals, while tellurium and polonium aremetalloids. The common feature is that their valence shell is ns2np4. They formseveral covalent molecules with non-metals. As neutral elements, they areoxidizing agents, because they gain two electrons to become a -2 anion with a

filled octet. However, their reactivity decreases as you descend the column,because the first and second electron affinities are not as great. They are ofteninsoluble, although it varies with their counterion (cation). Oxygen exists as adiatomic molecule (O), sulfur and selenium exist as octatomic molecules (sg andSe6), and tellurium and polonium exist in vast molecular matrices.

Halogens (Group VII)Halogens are non-metals from the seventh column of the periodic table.Included are fluorine, chlorine, bromine, iodine, and astatine. The commonfeature is that their valence shell is ns2np5. They form covalent molecules withnon-metals and ionic compounds with metals. As neutral elements, they arestrong oxidizing agents, because they readily gain an electron to become a -1anion with a filled octet. However, their reactivity decreases as you descend thecolumn, because the electron affinity is not as great. They are often soluble"although it varies with their counterion (cation). They all exist as a diatomicmolecules (X2), although little is known of astatine due to its radioactivity.

Noble Gases (Group VIII)Noble gases are non-metals from the eighth (and last) column of the periodictable. Included are helium, neon, argon, krypton, xenon, and radon. Thecommon feature is that their valence shell is complete at ns2np6. For the moslpart, they form no bonds and exist as monatomic atoms. Thanks to the work ofNeil Bartlett of U.C. Berkeley, xenon and krypton are known to form compoun&with halogens. The compounds with fluorine show more electron densitnaround fluorine, implying that fluorine is more electronegative than either xenor"or krypton.

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General Chemistry Atomic Theory Light Absorption and Dmission

l..i$hfi ii1ffi il#ufi#ftndfi r.i'ffi diliibmissionExcitation and RelaxationIt requires energy to ionize an atom, because one of its electrons is removed froma stable environment and placed into a less stable environment. However, whena quantum of energy less than what is required to ionize the element is absorbedbv the element, then an electron can be excited to a higher energy state (known asan excited state). Different forms of energy may be absorbed to excite an electron-rp to a higher energy level. For instance, heat energy can be absorbed to excite

:eleased as a photon. This is seen with fireworks. Some different forms of.nergy include light energy, thermal energy (usually vibrational kinetic energy),nechanical energy (usually translational kinetic energy), and electrical energy. It-. important to consider different energy forms, because there are many things:apable of absorbing one form of energy and emitting a different form.

:.!:sorption (Excitation): The gain of energy by an element or molecule, resulting-:r the excitation of an electron from a lower energy state (often the ground state): r a higher energy state (an excited state). The form of energy absorbed can vary.:.,tission (Relaxation): The loss of energy by an element or molecule, resulting in--:.e relaxation of an electron from a higher energy state (which must be an excited

':ate) to a lower energy state (which can be the ground state or another excited

':ate, but of lower energy). The form of energy emitted can vary.

a3sorption of energy is defined as any process in which a photon is absorbed by: compound to excite an electron in the compound to an elevated electronic::.ergy level (excited state). This means that the electron goes to a higher energy,:ite. The period of time that the electron remains in this elevated energy level is:=ferred to as the lifetime of the excited state. A compound can store energy by:-aintaining a high population of electrons in elevated energy states. When the.-=ctron relaxes back down to the ground state, energy is emitted in the form of a::.oton. Absorption and emission are therefore opposite processes. Figure 2-30:ro\vs the absorption and emission processes for a theoretical pair of energy.-,'els. Because the energy levels are equal, the energy of the photon absorbed for

- ,:itation is equal in energy to the photon released upon relaxation.

:rcited state Excited state

r,r'absorbedANNA,>

Electron excitesup to a higherenergy level

Electron relaxesdown to a lowerenergy level

hv emitted..,AAAAA

I

I

I,:ound state

{bsorption (Excitation)

Ground state

Emission (Relaxation)

Figure 2-30

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General Chemistry Atomic Theory Light Absorption and Dmission

q qir generic example, the energy of the photon absorbed is equar to the energyof the photon emitted. In actuality, there is more than one singular energv levelfor the ground state and the excited state due to the coupling of electricafenergylevels and rotational energy levels associated with the atom. Transitions involvea change in electronic state as well as a change in the rotational energy of theatom. hr molecules, there are vibrational energy levels to consider, in addition tothe rotational and electronic energy levels. The result is that the ground state andexcited state exists as a band of energy levels, not a single level. This means thatmultiple possibilities exist for the energy of transition. Rather than a single lineabsorption or emission spectrum being formed, a range is formed. This is whyFigure 1-2is abroad peak and not a sharp line. Figure 2-31 shows different typesof emission and absorption spectra where the color range is shown. Absorptionspectra show all light except what was absorbed, which appears as a black line,due to the absence of light. They are black lines in a rainbow. Emission spectrashow only the emitted light. They are colored lines against a dark background(due to only selected frequencies being emitted). They are stripes of color.

Wavelength

Wavelength

Figure 2-31

The lifetime of an excited state is often in the picosecond to nanosecond range, ssthe interval of combined excitation and relaxation is very fast. The conversicmbetween energy forms is exploited in many devices. Incandescent bulbsthermal energy resulting from the resistance associated with electrical flow andemit light energy (although some energy is dissipated in the form ofthrough conduction and convection). The operation of tube lighting involves thiconversion process. Plate charges build up at each end of the tube, creatingelectric field. A cation in the gas tube is accelerated by the electric force. ascollides with other gas molecules, its kinetic energy is transferred to themolecule, which absorbs the energy to excite an electron. Upon relaxing,energy is released as light. In reality, the tubes must run on alternating(AC), so that the cation never reaches a plate. This is why tube lights, suchfluorescent tubes, are actually high-frequency strobe lights.

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General Chemistry Atomic Theory Light Absorption and Emission

Atomic Spectrum of HydrogenHydrogen is the most studied element in terms of its absorption of light.separate samples of hydrogen gas exhibit exactly the same emission indabsorption spectra. This repeatability of a spectrum is attributed to thequantization of the energy levels. what is meant by "quantized energy levels" isthat the absorption and emission of light by hydrogen occurs in distinctjncrements. Because hydrogen has distinct energy levels, it also has distinctlransitions between energy levels, so that photons of the same frequency areabsorbed by all hydrogen molecules. However, with the large numbei of energyievels, there are several transitions that are possible. For hydrog"n, gtonpr -of

lransitions are named for the energy level to which they relax. For instance, alltransitions down to the n = 1 level fall into the Llrman series and emit phoions in-\e ultraviolet region of the EM spectrum. The energy levels and transition series:or hydrogen are shown in Figure 2-32.

n=2

aE6-? =-he-Le-z

AE". =bhs-z

AE,^= hclt".+-z

68, " --he-Lg-z

n=1

Figure 2-32

l--: series are named for electronic transitions within hydrogen alone. only the::-rner series emits photons in the visible range. when a sample of hydrogeni---",-S, it is because electrons are relaxing down to the n = 2level, which is stilian',::ted state. Figure 2-33 shows the line spectrum associated with the Balmer

-J--.'E- -t>.

Balmer series emission spectrum

l"slLl.z Lz-z Lo_z \s_z \+_z

Wavelength.----.>

Figure 2-33

lll Exclusive MCAT Preparation

General Chemistry Atomic Theory Light Absorption and Emission

The Balmer series emits photons in the visible range. In the Balmer series, 7q_2is656 nm and 4.6-2is 410 nm, so they fall in the visible range of the spectrum. Thisalso means that x4-2 and 1,5-2 fall in the visible range. The Faschen seriesinvolves photons in the infrared region of the EM spectrum. The Brackett seriesinvolves photons in the low infrared and microwave regions.

Example 2.25Given that 1"3-2 is 656 nm, which of the following is NOT true?

A. All Balmer series transitions are less than 800 nm.B. All Lyman series transitions are less than 400 nm.C. The lowest energy Paschen transition is greater than 800 nm.D. The highest energy Brackett transition is less than 800 nm.

SolutionThis requires looking closely at Figure 2-32. Transitions to the n=2 level fall intothe Balmer series. The lowest energy transition in the Balmer series is from n=3to n=2 and is given as 656nm. All other transitions in the Balmer series are moreenergetic, so these transitions emit photons with a wavelength less than 656nm.Choice A is a valid statement. Transitions to the n=1 level fall into the Lymanseries. The lowest energy transition in the Lyman series is from n=2 to n=1, andit is of significantly higher energy than the n=- to n=2 transition. This meansthat all photons in the Lyman series are of higher energy and lower frequencythan the transitions in the Balmer series. According to Figure 2-33, the Balmerseries emits photons near 400nm, so transitions in the Lyman series emit photonswith a wavelength less than 400nm. Choice B is a valid statement. Transitions tothe n=3 level fall into the Paschen series. The lowest energy transition in thePaschen series is from n=4 to n=3, and it is of significantly lower energ'y than then=3 to n=2 transition. This means that the lowest energy photon in the paschenseries is of lower energy and higher frequency than the transitions in the Balmerseries. The Balmer series emits a photon at 656nm, so the lowest energytransition in the Paschen series emits a photon with a wavelength greater than656nm, and according to Figure 2-32, higher than 800nm. Choice C is a validstatement. Transitions to the n=4 level fall into the Brackett series. The highestenergy transition in the Brackett series is from n=* to n=4, and it is of muchlower energy than the n=3 to n=2 transition. This means that the highest energ'yphoton in the Brackett series is of lower energy and higher frequency than thetransitions in the Balmer series. The Balmer series emits a photon at 656nm, sothe lowest energy transition in the Brackett series emits a photon with awavelength greater than 656nm, and according to Figure 2-32, higher than800nm. Choice D is an invalid statement, and thus is the correct answer choice.

Electromagnetic SpectrumPhotons have associated with them a distinctive frequency, energy amount, andwavelength. The relative energetics of various radiation must be known. If youever have difficulty recalling the electromagnetic radiation spectrum, think aboutthe fact that you wear sunglasses to protect your eyes from ultraviolet radiation,not visible radiation. This means that ultraviolet radiation is of higher energythan visible light. Also, lead shields are used to filter out x-rays, so they must beof even higher energy than ultraviolet photons. Conversely, we are fairly safefrom radio waves considering the fact that millions of radio waves pass throughus every second. Common sense can be applied to recall the relative spectrum.Table Z.Alists the conunon EM ranges in terms of frequency and wavelength.

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General Chemistry Atomic Theory Light Absorption and Dmission

EM Radiation Wavelength (m) Frequency (Hz) Common Usage

AM to3 - to4 105 to 106Cheaper radio

communication

FVI t-702 107 to to8Expensive radiocommunication

\'{icrowave 10-4 - 1 168 1o 1612Satellite, cell phones,radar, heating water

lnJrared 10'6 - 1o-3 1611 1o 1614Line-of-sight com,molecular ID, heat

|isible 4 xLo-7 -7 x1o-7 4.3 x 1.014 - 7.5 x1014Vision, fiber opticcommunication

L-llraviolet rc-8 - rc-7 1615 1o 1616Bond-breaking,

exciting electrons

\-ray 19-12 - 1g-8 1616 1o 1620Nucleus detection,core e- ionization

giiTIITt? 16-15 - 161-10 1918 1o 1923Nuclear excitation,world destruction

Table 2.4

- : lrmon practical uses of EM radiation are something that should be learned,:':*ause of the tradition of test writers asking about such devices. It should be::-l:r:Lon knowledge that radio waves and low-energy microwaves are used in,:::.nunications. Microwave communications include satellite transmissions,:::Le television, cellular phone networks, and airport landing systems.l-::orvaves are also used for cooking and heating things that contain water,

:,::ause the frequency required to rotate a water molecule is found in the:r::owave region. Remote control units (also known as line-of-sight-::rnunicators) often use infrared signals to send information. The frequency of:. E\{ radiation used in these common tools and appliances to which we are:-:.ctly exposed cannot be near !700 cm-1,3000 cm-1, oi SSOO cm-1, because those,:= =e bond-stretching frequencies associated with molecular bonds common to

'.* :limal life. Ultraviolet light is used to break bonds, and x-rays are used for-: : grng atomic structure (including bones and teeth).

,' 'oible Spectrum and Colors-. ,= ','isible spectrum runs from red (around 700 nm), orange, yellow, green, blue,:::':gh violet (around 400 nm). The relative wavelength, frequency, and energy: ::.otons in the visible spectrum of light (by decreasing wavelength, increasing

::::-;ency, and increasing energy) is given in Figure 2-34.

Red-.':.-627nm 627 lo594nm 594to561 nm 56L1o477nm 477to438nm 438to400nm

)

Figure 2-34 +?-*-

= order of the colors of the visible spectrum is often recalled using ther: r:ronic ROY G. BiV, an acronym for the spectrum of visible light in sequence: :.:reasing energy. It is also important to know that the visible range of light is

::::. approximately 700 nm for red light to approximately 400 nm for violet light.- -::,,-iolet spectroscopy ranges from 20 nm to 400 nm.

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General Chemistry Atomic Theory Light Absorption and Emission I

, .--, '

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Example 2.26For some substance X, AE1 yields yellow light, while AE3 yields blue light. \Alhatmust be true of the photon associated with AE2?

A. It is probably orange.B. It has a wavelength of 700 nm.C. It is probably green.D. It has a wavelength of 400 nm.

SolutionAE2 falls between AE1 and AE3, so the photon emitted must be between yellowand blue light in the visible spectrum. According to the spectrum mnemonic(ROYGBiV)/ green light falls between yellow light and blue light. This eliminateschoice A and makes choice C the best answer. A wavelength of 200 nmcorresponds to red light, and a wavelength of 400 nm corresponds to violet light,eliminating both choice B and choice D.

Color is a phenomenon associated with vision. Light in the 400 nm to 200 nmrange can be detected by the eye and processed in the brain. Color is the result oflight within this range being absorbed by the cones of the eye. Rods also detectlight (in the green range), but this is for the purpose of intensity analysis ofdetails in the image. Consider the eye to function like a visible lightspectrophotometer, analyzing stimuli for intensity and wavelength. There arethree kinds of cones, each one responsible for detecting different frequencvranges of light. Color, in a photon sense, can result from one of two phenomena.Color is perceived from either the emission of light at a specific frequency or thereflection of light at a specific frequency. Hence, there are two types of color toconsider: emitted and reflected.

Emitted ColorColors is emitted from a source that radiates visible photons. Atoms emit lightenergy when an electron relaxes from an excited state to a lower level. Figure 2-35 illustrates the concept of emitted color.

Figure 2-35

An electron relaxes from an excited state and releases a photon of somewavelength. All of the photons in Figure 2-35 are of essentially the samewavelength. The color that is observed is the color emitted. Simply put, witLemitted color, "What you get is what you see." Emitted colors can be seen atnight, because their source is emitting light. A good test to determine whether a

color is an emitted color is to ask whether it can be seen in the dark. If theanswer is yes, then it is an emitted color. Examples of emitted light sourcesinclude: neon light tubes, television screens, fireworks, and glow-in-the-dark ink.

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General Chemistry Atomic Theory Light Absorption and Emission

Reflected Color and the Color WheelReflected colors are produced when incident light (we'll consider it to be white1ight, for the sake of simplicity) strikes a surface, and the surface absorbs certaintrequencies, thus reflecting only a portion of the incident light to the eye. Thereflected radiation appears as a color that is a combination of the reflectedphotons. The color we observe is the complementary color of the frequency thathad the highest intensity of absorption. This is where the color wheel can help.Complementary colors are on opposite sides in the color wheel. Given that white-ight is a combination of all colors, it is a combination of all complementary pairs.This is a over-simplification, because the primary colors of light are differentnom the primary pigment colors, but we shall use the two interchangably to aidn answering chemistry questions. Complementary pairs include red and green,:lue and orange, and yellow and violet. A color wheel is shown in Figure 2-36.

The color wheel is used to determine complementarycolors. Red and green are opposing on the color wheel,so they are complementary colors. Moving around thewheel, violet is the complementary color of yellowand blue is the complementary color of orange.

Figure 2-35

-hJorophyll appears green in the presence of white light. The conclusion is thatSorophyll contains a pigment that has a maximum intensity absorbance of red:Eht, the complementary color of green. One type of chlorophyll has anrrsorbance maximum at740 nm, while another has an absorbance maximum at:,50 nm. Both of these values correspond to red light. Figure 2-37 illustrates the::ncept of emitted color as it relates to chlorophyll.

Incident Reflected color(appears green)white light

(absorbs red) I,out = 700tnm

Figwe2-37

Let'lected colors can be seen only in the presence of an extemal light source. If a-:,lor cannot be seen in the dark, then it is a reflected color. Examples of reflected

-f,ht sources include paint pigments, pen inks, and fabric dyes.

(absorbs red)

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General Chemistry Atomic Theory Light Absorption and Emission

Example2.27\A/hich of the following statements is NOT true?

A. A yellow paint pigment absorbs light with a Imax of 411 nm.B. A green light bulb emits light with a l.-u* of 522nm.C. A blue shirt has a dye that absorbs light with a fmax of 611 nm.D. A television screen glowing violet emits light with & Imax of 573 nm.

SolutionThis question entails comparing reflected and emitted colors. The first thing youshould ask yourself is whether the color can be seen at night. If the answer is yes,

then it is an emitted color and what you see is the color involved in the electronictransition (emission). If the answer is no, then it is a reflected color and what yousee is the complementary color of the light involved in the electronic transition(absorption). Choice A is a paint pigment, which cannot be seen at night. Theyellow color is a reflected color, so according to the color wheel, violet light has

been absorbed. Violet light has a wavelength near 400 nm, so choice A is valid.Choice B is a light bulb, which can be seen at night. The green color is an emittedcolor, so green light has been emitted. Green light has a wavelength between 561

nm and 477 nm, so choice B is valid. Choice C is a fabric dye, which cannot be

seen at night. The blue color is a reflected color, so according to the color wheel,orange light has been absorbed. Orange light has a wavelength between 627 nmand 594 nm, so choice C is valid. Choice D is a radiating screen, which can be

seen at night. The violet color is an emitted color, so violet light has beenemitted. Violet light has a wavelength near 400 nm, so choice D is invalid, ancthus the best answer to this question.

The absorption of energy followed by the emission of a photon of light is a usef'tiprinciple that can be applied to many forms of spectroscopy, including infraredspectroscopy, in organic chemistry to identify bond types and functional groups.It is also involved in UV-visible spectroscopy, where a compound is subjected tcincident light at a specific frequency, and the relative intensity of transmittedlight is measured. Conclusions about the concentration and the kind o:compound itself can be inferred from the results. Classical experiments tha:involve measuring quantities of components in a gaseous mixture often focus or.

analysis of the emission and absorbance spectra of the gases. This is at the hear:of atomic absorption spectroscopy, used to analyze planetary gases.

The MCAT is not printed in color, so questions involving color will preser,lnumerical data for wavelengths or graphs of absorbance spectrum. Figure 2-35

shows an absorbance spectrum and its analysis.

Abs =

l,*u* is the wavelength at the greatestabsorbance. A I^o" at 520nm means thatyellow light is absorbed, so the compoundappears violet to the human eye.

Wavelength (in nanometers)

Figure 2-38

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The Berkeley Relieu

General Chemistry Atomic Theory Light Absorption and Emission

Fluorescence .Fiuorescence most simply understood is the conversion of ultraviolet light intovisible light. A material with a semi-stable excited state can absorb an ultravioletohoton, relax a slight amount giving off infrared photons, and then relaxcompletely back to its original state. The visible photon emitted is of slightly lessenergy than the ultraviolet photon absorbed, because of the loss of energy when:he infrared photon was lost. The next energy range down from ultraviolet is,.'iolet light in the visible range. This is why fluorescing materials appear purple.The term "black light" refers to an ultraviolet light (it emits light outside of the-.-isible range) that when shined upon fluorescing material appears purple. You:ray have seen this in organic chemistry lab when you analyzed tlc plates. This:oncept does not need to be interpreted at any higher level, so this rougherplanation is adequate. Fluorescence will be addressed in a passage.

Photoelectric Effect-he photoelectric effect is exactly what the name implies. An incident photon:1uses the release of an electron. It is a fancy way of saying that a compound can

-: ionized with a photon, as long as the photon has energy greater than the.:r:zation limit of the material. The energy holding the electron to the surface of::te solid material is referred to as the binding energy. Excess energy (energy in:,.cess of the binding energy) becomes kinetic energy for the electron that is

=::ritted. The photoelectric effect is really quite simple, but confusion sometimes,:'qes from the different terminology used in chemistry and physics. Don't be: - rled; just say no to physics (just kidding, but I'm a chemistry nerd and just had:- say thet). Equation 2.9 is a physics equation for the photoelectric effect and::uation 2.10 is a chemistry equation for the photoelectric effect. Both say the: ii:'e thing, but use different symbols to do so.

hv =O (2.e)

(2.10)

L-: photoelectric effect is the principle at work in solar panels. Photons of all:-::erent wavelengths and energy are emitted by the Sun. This energy can be. , -lected at the Earth's surface. Some solar photons are sufficiently energetic to. ::i ar electron when they strike a metal surface. To do so, the photon must be: rn energy high enough to remove the electron from the valence shell of the

:.:a1. The loss of an electron from the metal creates electrical flow (electricity),:::h can be converted to either potential energy (stored in an electrochemical: or mechanical energy (used to power a device).

-..-. :he energy of the incident photon increases, the kinetic energy of the: r:rarged electron increases. Figure 2-39 shows an apparatus devised to: : asure the photoelectric effect, while Figure 2-40 is a graph of the electron's*-:ic energy as a function of the frequency of the incident photon.

+ l*rr22

hv = B.E. + kinetic energy of emitted electron

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General Chemistry Atomic Theory Light Absorption and Emission

Incident beam ofDetector of KE

Battery

Figure 2-39

Electric circuit

Figure 2-40

The threshold frequency corresponds to the photon equal in energy to theionizing energy. Any photon with a frequency less than the threshold irequencydoes not have enough energy to ionize the material, so no electron is ejected.

Example 2.28which material probably has the LowEST threshold frequency for thephotoelectric effect?

A. BoronB. CarbonC. SodiumD. Sulfur

SolutionThe threshold frequency corresponds to the binding energy of the material (B.E- hvo). Each atom holds onto its valence electrons with a different bindenergy, therefore each atom requires a different amount of energy to undionization. The lowest threshold frequency for the photoeiectric ecorresponds to the material that is easiest to ionize. This question is askingthe material with the lowest binding energy. of the choices, only one is a rrxso only one has a low binding energy. The best answer is choice c, sodium.answer corresponds with lowest ionization energy as well.

hv=hvo+KE"-

vo: threshold frequency

Frequency of incident photon (Hz)

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General Chemistry Atomic Theory Nuclear Chemistry

Nffin$ ::::ffieffi$ffiNuclear ParticlesSimply put, nuclear chemistry is the chemistry that the nucleus of an atom canr.rndergo. Instead of referring to it as reactiaity, it is referred to as instability of thenucleus. The science community knows little of nuclear behavior at this point inhistory, but there are theories. From our perspective, however, all we will.oncern ourselves with is nuclear decay and nuclear capture. The process ofparticle loss from the nucleus that results in a different nucleus is referred to asnuclear decay. In physics, it may also be referred to as fission The process ofparticle gain by the nucleus which results in a different nucleus is referred to as

ruclear capture. In physics, it may also be referred to as fusion. As a generalru1e, particles with mass less than 55 amu undergo fusion, while particles withmass greater than 56 amu undergo fission. Particles that we consider to be lost or:ained are alpha particles (helium nucleus), beta particles (an electron), positronsa positively charged particle with the mass of an electron), neutrons, and

:.eutrinos (an uncharged particle with the mass of an electron). The nucleus also:.as ground states and excited states associated with it (just as the orbiting:-ectrons have energy levels), so nuclei can also undergo photon absorption and

=:nission. The high-energy photon is a gamma ray. Table 2.5 lists some of the:rlrunon nuclear processes. You should treat these reactions as exercises in=-nple algebra. The total mass and number of protons should be equal on both--les of the reaction. This is because mass and charge are conserved (along with-romenfum) in each process.

Table 2.5

'* -iclear Decay and Capture- :edvy element undergoes a decay process to increase its nuclear stability. A,::.: element undergoes a capture process when struck by a high-energy particle" ':rcrease its nuclear stability. There are a few points you should note for these:: . :esses. With B-decay, the electron may or may not be added to the electrical, "=,' of the atom. If it is, the atom remains neutral. If the electron escapes, the

Process Reaction Tracking Notes

,:-Decay tSBx -+ $u +r[fii The mass works: 720 = 4 + 176.The proton number works: 50 = 2 + 48.

-Decay tBBx -+ le +rllz The mass works: 120 = 0 + 120.The proton number works: 50 = -1 + 51.

. --Decay t38x -+ ?" * t?$q The mass works: 120 = 0 + 120.The proton number works: 50 = 1 + 49.

* m1ss10n 138x" -+ hv + 1fr$x Mass and proton number do not change,The nucleus relaxes to emit a photon.

-r-Capture t!$x * $u -+1!$xThe mass works: 120 + 4 = \24.The proton number works: 50 =2 + 48.

:-Capture t38x *_roe +1lgeThe mass works: 720 + 0 = 120.The proton number works: 50 + -1 = 49.

-Capture t3$x * le -+1!lzThe mass works: t20 + 0 = 120.The proton number works: 50 + 1 = 51.

.-rbsorption t38x * rrv -+ 138X. Mass and proton number do not change.The nucleus absorbs a photon to excite.

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General Chemistry Atomic Theory Nuclear Chemistry

compound becomes a cation. usually the particle escapes, and an ion is formed,although we typically ignore this fact in nuclear chemistry. The representation ofthe beta particle as having a -1 for its Z term is not accurate, but ii works for thealgebra of the equation. Beta capture and positron decay both result in the sameelement being formed. Likewise, beta decay and positron capture both result inthe same element being formed. The positron behaves just as a beta particledoes, only it carries a positive charge and is an anti-electron. This makeJit anti-matter, but for our purpose that doesn't matter.

The reactions in Table 2.5 are simplified. We have ignored conservation ofenergy and conservation of momentum. To balance these more accurately, themass of the electron (beta particle) and positron (anti-electron) cannot be treatedas zero. A neutrino is required in some cases to balance the equation. Solvingdecay and capture questions in the manner that these examples are presentedwill work fine on test questions you may see. The questions you may see willinvolve determining the particle that was lost in a process, or determining thefinal elemental product after a nuclear reaction has taken place. It is sort of oddthat if you reverse the first two letters in nuclear chemistry the phrase becomesunclear chemistry. Given the science community's level of understanding at thepresent time, unclear is an accurate depiction of the subject.

Example 2.29which of the following elements results from two consecutive alpha decays of2I0 1.1z

4. 210p,

B. 2069i

C. 202y1

9. 202p;

SolutionAlpha particles are helium nuclei (alpha-helium sounds like alpha-helix, soremember the phrase "alpha helium"). An alpha particle has a mass of 4 amu,thus losing two alpha particles results in the loss of 8 amu. The element thatremains after two consecutive alpha decays has four fewer protons than theoriginal element (from 85 to 81) and has a mass eight less than the originalelement (from 210 to 202 amu). At is element number 85, so the final elementmust have atomic number 81. Consulting a periodic table shows that element 81i" 11. 20211is the best answer, so pick C.

Example 2.30Electron capture by a nucleus results in which of the following?

A. An increase in atomic number by one, and a mass increase of one amu.B. An increase in atomic number by one, and no change in atomic mass.C. A decrease in atomic number by one, and a mass decrease of one amu.D. A decrease in atomic number by one, and no change in atomic mass.

SolutionElectron capture involves the gain of an electron by the nucleus. An electrona -L charge and no mass. This means that the element should experiencedecrease in its positive charge by one, and no change in its mass. This isdescribed in choice D.

Copyright O by The Berkeley Review 120 The Berkeley

General Chemistry Atomic Theory Nuclear Chemistry

Example 2.31Which of the following processes would NOT result in the formation of tritium?

-{. Alpha decay of 7Li

B. Positron emission by 3He

C. Beta decay of 4He

D. Gamma emission from 3H

Solution-\n alpha particle is a helium nucleus (4H"), so when 7Li loses a helium nucleus,--he mass decreases by four to three and the number of protons decreases by twoto one. The final product is tritium 13H;. Choice A is valid. A positron is a mass-iess positively charged particle that when lost, converts a proton into a neutron.The mass remains the same, but the 3H" .onrr"rt into tritium (3H). Choice B isvalid. Beta decay involves the loss of an electron from the nucleus. This convertsa neutron into a proton. The mass does not change and thus remains at four.The number of protons increases by one to three. The product is 4Li, which isNOT tritium. The best answer is choice C. Gamma emission does not change theparticle, thus choice D is valid.

Example 2.32L:i the conversion from 165go in157Gd,, at least two alpha particles were emitted.What else was emitted?

A. A third alpha particleB. One neutronC. One beta particleD. One positron

SolutionIn converting from 1659e to 157Gd, the mass decreases by eight, which equatesto the mass of two alpha particles. This means that any other particles emittedare massless. This eliminates choices A and B. The atomic number has decreasedbv three. Loss of two alpha particles decreases the atomic number by four, so theother particle emitted must increase the atomic number by one. Loss of apositron decreases the atomic number by one, so choice D is eliminated. A betaparticle must also have been emitted in addition to the two alpha particles.Choice C is correct.

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General Chemistry Atomic Theory Nuclear Chemistry

Half-LifeThe half-life of a sample of material is the period of time required for 50% of the

concentration of material to decay to a different (possibly stabler) form. It is most

corunon to see half-lives associated with first-order decay processes. The graphs

shown in Figures 2-4'1, and 2-42 show the concentration of a component over time

for first-order decay (exponential decay) and zero-order decay (linear decay).

The graphs shown in Figures 2-43 and 2-44 ate the derivative graphs of Figures

2-41 md2-42,whichrepresent the change in rate as a function of change in time.

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Zero-Order Decar-(linear decay)

Half-life decreases

Figure 2-46

Time+Figure 2-43

In Figures 2-45 and.2-46, the graphs are marked al50% and 25'h of the initia1

concJnftation values. These are the various concentrations after consecutive haiJ-

lives. It takes one half-life to cut the concentration in half, and a second half-liJe

to cut that concentration in half again, which results in one-quarter of the originai

concentration. Figure 2-45 shows half-life as a function of time for a first-order

decay. Figure 2-46 shows half-life as a function of time for a zero-order decay.

txl100o/o

50%

25%

txl100%

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First-order decay(exponential decay)

Zero-order decay(linear decay)

First-order decay(exponential decay)

Zero-order decay(linear decay)

First-Order Decay(exponential decay)Half-life is constant

Copyright @ by The Berkeley Review

Figure 2-45

122 The Berkeley Revieu

General Chemistry Atomic Theory Nuclear Chemistry

Figure 2-45 demonstrates that for a first-order process, the hatf-life is constant,regardless of concentration. As the concentration decreases, the duration of thehalflife remains the same. Figure 2-46 demonstrates that for a zero-orderprocess/ the half-life is directly proportional to the concentration of thecompound. As the concentration decreases, so does the duration of the halflife.The second half{ife is half as long as the first half-life, because the concentrationchange is half as large. As a point of interest, for a second-order process, the half-rije is indirectly proportional to the concentration of the compound. As theconcentration decreases, the length of the hatf-tife increases.

lihese graphs should look familiar from biology and physics. In biology,bacterial growth is exponential, and enzyme kinetics can be either exponential orlinear. In physics, displacement, velocity and acceleration follow similar:atterns. The point is that you must understand what these graphs tell youabout some values increasing or decreasing as a function of time. Half{ifeanalysis is most commonly evaluated using the upper graphs.

ftr the MCAT, you can expect to have to solve some half-life questions. As far as:lalf-life problems are concerned, there are two methods by which they are

=olved. One method is to use Equation 2.11, the first-order decay equation:

Ct = Co e-kt (2.LL)

,---here C1 is the concentration at a given time, Co is the initial concentration, k isre rate constant for the decay process, and t is the elapsed time. This works but:equires the use of natural logs, which can be time-consuming.

lhe second method involves the sequential summing of half-lives required to:each a specific concentration. The test traditionally uses numbers that work well;-;ith this method. It is the most time-efficient method. When a problem asksion'much time passes until a certain percentage of the original quantity remains,:: is easiest to figure how many half-lives are required to reach that percentagend then convert the quantity of half-lives to total time. When using this:rethod, use reasonable approximations. For instance, if the halfJife is 1L0 years,:en 335 years is three half-lives plus a little bit. There is 50% remaining after the:rst half-life ,25"/o remaining after the second half-life, and 12.5"h remaining after:re third half-life. Thus, there is just under 72.5oh rcmaining after 335 years.

Example 2.33l-{orv many half-tves are required to decornpose93.75"h of some material?

A*2ts.3c-4D.5

5olution)ecomposing (decaying) 93.75% of the initial material results in 6.25%:emaining. The quick and easy method is to cut the percentage in half:;rntinually until 6.25 is reached.

100% -+ 50% -+ 25h -+ 12.5% -+ 6.25%

-: requires a total of four half-lives to reach a point where 6.25% remains, which..:-. a decomposition of 93.75% of the material. The best answer is choice C.

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General Chemistry Atomic Theory Nuclear Chemistry

Example 2.34If the initial concentration of a toxic material is 1500 parts Per million and the

half-life is 4.5 days, how long witl it take for the level to reach a concentration of100 ppm?

A. 13.6days \Set -' -l{oB. 17.8 daysC. 1.8.2 daysD. 22.3 days

Solutionwe must sequentially halve 1500 until a number close to L00 is reached.

1500ppm -+ 750Ppm -+ 375ppm -+ 187-5Ppm + 93.75ppm

After four half-lives, there is a little less than the target value of L00 ppmremaining. This means that just under four half-lives are needed to reach 100

ppm. Four half-lives are eighteen days, so the best answer is slightly less than 18

days. The best answer is choice B'

Sequential halving of the concentration in first-order decay questions is far easier

than using Equation 2.1L or other rate equations. The test-writers may give you

the appropriite equations needed to solve the question in longhand, but this

methbd should be your method of choice. Just because the test writers provideyou with a piece of information, does not mean that you need to use it.

This technique can also be applied to exponential growth problems, as seen inbiology with population genetics that obey exponential growth behavior. Withgto*ih, yon iotttitnally double the concentration rather than cut it in half. For

instance, if a bacterial population doubles in L2 seconds, then in one minute itincreases 32 times (25).

X -+ 2X -+ 4X -+ 8X -+ 16X -+ 32X

Example 2.35If the half-life of a particular atom is 20.4 minutes and the initial concentration bL200 parts per million, what will the concentration be after exactly one hour?

\r*n

,x1 sazppm \ zc o -) (' oo -! i oo _-' \ ! \ "'rB4-gstpp^ , 't- 31,,C./155ppm

2fr't+sppmSolutionOne hour is not exactly equal to three half-lives (three half-lives = t hour and

minutes). After three half-lives, we have the following:

1200ppm -+ 600Ppm -+ 300ppm + 150ppm

We haven't quite used a full three half-lives, so the concentration should

have decreased all the way down to L50 ppm. This means that the cor

answer should be just over 150 ppm (150+ ppm). The only ernswer choice that

slightly greater than 150 ppm is L55 ppm, choice C.

Copyright @ by The BerkeleY Review 124 The BerkeleY

I. Classical Experiments

II. Isotopic Abundance and Average Atomic Mass

III. Bohr Model of Hydrogen

ru. Paramagnetism and Liquid Crystal Displays

V. Migration through a Membrane

VI. Ionization Dnergy

VII. Transition Metal Trends

VIII. Lasers

IX. Paint Pigments

X. Fluorescence and Phosphorescence

XI. Flame Test

XII. Glyphosate

XIII. Technetium Decay

XIV. Cold Fusion

Questions Not Based on a Descriptive Passage

Atomic Structure Scoring Scale

Kaw Score MCAT Score

84 - 100 t5-1566-85 lo-1247 -65 7 -9

34-46 4-6t-55 1-3

(l - 8)

(e - 14)

(15 - 2t\(22 - 28)

(2e - 55)

(36 - 42)

(43 - 4e)

(5O - 56)

(57 - 65)

(64 - 6e)

(7O - 76)

(77 - Br)

(82 - 88)

(8e - e5)

(e6 - lOO)

Passage I (Questions 1 - B)

Two of the more famous classical experiments in:stablishing the properties of subatomic particles are theThomson experiment and the Millikan oil drop experiment.The Thomson experiment determined that the mass-to-charge:atio for an electron is 5.686 x 10-12 kg/C. The Thomson:rperiment involved the application of a perpendicular:lectric field to a beam of electrons produced by a cathode ray:ube. The Thomson apparatus is shown in Figure 1.

Upper cathode plate

| .. Lower anode plateI t'w'

Electron beam

Figure L

The double filter ensures that the electron beam is in a,.rsle direction. The electron beam, when exposed to the:.:ctric field generated by the two plates, is bent towards the:': sitively charged plate. The curvature is constant as long as

:e field strength is constant. The bending of the electron:t.im is attributed to both the attractive force of the.:positely charged plate and the repulsive force of the like:. .rrged plate. A magnetic field may also be used to bend the: i.-tron beam. When using a magnetic field to deflect the: :--trons, the radius of the arc of the electron beam can be:::ployed to determine the mass-to-charge ratio using::.ration 1.

m=Brqv

Equation 1

-,: Equation 1, the mass of the electron is m, the charge ofr: electron is q, the strength of the magnetic field is B, the*:,:rus of the arc of the electron pathway is r, and the velocityl ne electron is v.

In the Millikan oil drop experiment, after ten years ofr^::a collection, it was determined that the charge of an: ::Jtrorl is 1.6 x 10-19 C. The experiment involved chargingr -.il drop by bombarding it with an ionizing electron beam.i,:-.;121 oil drops passed through the ionizing beam, but only, ':-* gained a charge. The oil drops then fell into an electrici: j senerated by two charged plates in a glass cylinder. The'i:. j was aligned so that the lower plate repelled the oil drop

-:: ; ard, while the upper plate pulled the oil drop upward.

The force upward on the oil drop is thus qV, where q isr: ;harge of the oil drop and V is the voltage of the field.-:: force pulling the oil drop downward is mg, where m isr'r : :nass of the oil drop, and g is the gravitational force

i.stant. If the oi1 drop is suspended, then the two forces are::-a1. and it is possible to determine the charge of the oili:::. The charge of the oil drop must be a whole numberr -itiple of the charge of an electron. The apparatus used in:,. \lillikan oil drop apparatus is shown in Figure 2.

Double filter

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Suspendedcharged oil drop

Figure 2

The hole in the top plate allows for oil droplets to fallinto the chamber. The field between the two plates isassumed to be uniform, so that the force on the oil droplet isequal at any position between the two charged plates. Thefield is adjustable, so that droplets of variable mass can be

suspended. Not all oil droplets are of equal mass, so anaverage value is assumed.

1. An assumption of the Millikan oil drop experiment iswhich of the following?

A . Oil drops are naturally charged.

B. The pull of gravity is equal in all directions at themicroscopic level.

C . The mass of electrons is negligible compared to themass of the oil drop.

D . The magnitude of charge for an electron can vary.

2 . The electron beam bends downward as it passes betweenthe two charged plates in the Thomson experiment.What can we expect for a beam of neutrons andprotons?

A. The proton beam would arc upwards as it travels,while the neutron beam would not arc at all.

B. The neutron beam would arc upwards as it travelswhile the proton beam would not arc at a1l.

C. Both the neutron and the proton beams would arcupwards as they travel between the plates.

D . Neither the neutron nor the proton beam would arcas they travel between the plates.

3. How could a value of 3.2 x 10-19 C in an oil drop inthe Millikan oil drop experiment be explained?

A. The oil drop was doubly ionized.B. The oil drop gained an electron rather than lost an

electron.

C. The oi1 drop lost a proton rather than lost an

electron.

D. The oil drop gained a proton rather than gained an

electron.

.1. The Thomson experiment determined which of the

following?

A . The mass of an electron

B . The mass of a proton

C . The mass-to-charge ratio of an electron

D . The mass-to-charge ratio of a proton

5 . The Millikan oil drop experiment requires knowing all

of the following EXCEPT:

A . the mass of the oil droPlet.

B . the voltage of the electric field.

C. the gravitational force constant.

D . the temperature of the chamber.

6 . In a neutral atom, what can be said about the number ofelectrons relative to the number of protons and

neutrons?

A . The number of electrons must equal both thenumber of protons and the number of neutrons.

B. The number of electrons must equal the number ofprotons only, and the number of neutrons isirrelevant.

C . The number of electrons must equal the number ofneutrons only, and the number of protons isirrelevant.

D . The number of electrons must equal the sum of theprotons and neutrons combined.

7 . The STRONGEST attractive force is felt between:

A. aproton and a neutron.

B. a neutron and an electron.

C. an electron and an electron.

D . an electron and a Proton.

8. If the oil drop were slowly descending once it was in

the electric field between the two charged plates, which

of the following corrections would NOT stop it fromcontinuing to fall?

A. Increase the number ofelectrons on the drop.

B. Increase the mass of the oil droplet.

C. Increase the voltage difference between the twoplates.

D . Decrease the gravitational pull.

Copyright @ by The Berkeley Review@ t2a GO ON TO THE NEXT P

Passage ll (Questions 9 - 14)

To determine the relative abundance of the isotopes

an element, a mass spectrometer is employed. The ele

is first vaporized into its gaseous state, if it is not alreadl

gas form. Because this is carried out in a vacuum, a

temperature may not be required to vaporize the

The cationic compound is generated by high-energy

impact with the neutral atoms. A neutral atom loses

valence electron to ionization upon impact. The cati

species formed is then accelerated in an electric field unnl

reaches a set velocity (v). The particle beam produce'J

projected through a double filter to ensure that no de

particles pass through to the detector.

Once particles are passed through the double filter.

are subjected to a magnetic field perpendicular to the ori

electric field. This curves the pathway of the particles.

force on charged particles moving through a magnetic fieldri

found using Equation 1.

p=qvB

Equation 1

where q is the charge of the particle (assumed to be +1) and

is the magnetic field strength. According to Newton's

law, the force on the particle is F = ma, where m is mass

a is acceleration. The particle is moving along a curved

so it has angular acceleration, which is found using Equari

2.

u=&R

Equation 2

where R is the radius of the arc. Substituting ma for F

Equation 1 and manipulating the variables yields Equation

qB=mvR

Equation 3

The equations support the intuitive perception ofthe path is affected by mass, charge, velocity, and ex

field strength. Figure I shows a basic schematic of a

spectrometer.

Figure 1

Different isotopes have different atomic masses, so

isotope has a unique radius (R). The detector determines

relative quantities of each isotope. This information is

used to determine the average atomic mass' a value that

into account all ofthe isotopes for a particular element.

Detector

fable 1 lists the isotopic abundance for the common isotopes:,f boron, magnesium, and silver.

Element Atomic Mass Percentage

Boron-10 10.013 amu 19.187o

Boron-1 1 I 1.009 amu 8O.22Vo

Masnesium-24 23.985 amu 78.7j%o

Magnesium-25 24.986 amu lO.137o

Masnesium-26 25.983 amu ll.ITVoS ver-107 106.91 amu 5l.80Vo

S ver-109 108.90 amu 48.20Vo

+.

Table L

Why is carbon added to a sample before it is vaporizedin a mass spectrometer?

A . Carbon-12 is used as a standard for mass reference.

B . Carbon-13 is used as a standard for mass reference.

C . Carbon helps to lubricate the atoms at an atomiclevel.

D. Carbon absorbs heat from the atoms and preventsthe sample from overheating.

Which of the following plots represents the outputfrom the mass spectrometer for magnesium?

A. B.

Which of the following calculations accuratelydetermines the average atomic mass for silver?

{.. 0.787 (23.985) +0.1013 (24.986) +0.1117 (25.983)

B. 0.11I1 (23.985) + 0.1013 (24.986) +0.787 (25.983)

C. 0.482 (106.91) + 0.518 (108.90)

D. 0.518 (106.91) +0.482 (108.90)

C)

Mass--> MASS+

qC)

Mass# Mass+

12. When a mass spectrumcompound, the first stepfollowing?

A. Vaporization

B. Condensation

C. IonizationD. Sublimation

is to be obtained for a solidin the process is which of the

13. Identify the TRUE statement(s) from the following:

I. The average atomic mass of an element is alwaysgreater than the mass of the heaviest isotope forthat element.

II. The average atomic mass of an element is alwaysgreater than the mass of the lightest isotope forthat element.

m. When there are two isotopes for an element, theaverage atomic mass is closer to the more abundantisotope than the less abundant isotope.

A. I onlyB. II onlyC. I and III onlyD . II and III only

14. The difference between the two isotopes of boron is:

A. the isotope of boron with mass 11.009 amu has

one proton less than the isotope of boron withmass 10.013 amu.

B. the isotope of boron with mass 11.009 amu has

one proton more than the isotope of boron withmass 10.013 amu.

C. the isotope of boron with mass 11.009 amu has

one neutron less than the isotope of boron withmass 10.013 amu.

D. the isotope of boron with mass 11.009 amu has

one neutron more than the isotope of boron withmass 10.013 amu.

129 GO ON TO THE NEXT PAGE

Passage lll (Questions 15 - 20)

The Bohr model for hydrogen proposes that there are

quantized energy levels in which an orbiting electron may be

found. The theory is derived from the observation that exact

frequencies of light are absorbed by hydrogen gas, no matterwhat the concentration of the sample or intensity of the

incident light source. Given that light frequency directlyrelates to light energy, the idea is that quantized amounts ofenergy are absorbed and emitted as an electron moves between

energy levels. Figure 1 shows the Bohr representation ofhydrogen.

n=4n=3n=2

Orbital levels

Figure 1

Due to Coulombic attraction, the most stable energylevels for the electron correspond to the closest proximity to

the nucleus. This means that the energetics of the orbitingelectron depend on the charge nucleus and the distance fromthe nucleus (or quantum level). The energy level for the

electron ofhydrogen is found using Equation 1.

E=-2.t78* ro-rs14l\n2 /

Equation 1

15. What is the ionization energy for hydrogen?

A. 1.634 x 10-18 J

B. 2.178 x 10-18 J

C. 4.356 x 10-18 J

D. 8.712 x 10-18 J

1 6. The deBroglie equation relates the energy of a photon to

its equivalent mass according to E =mc2. How can the

deBroglie wavelength of an electron be found?

,r- hmc

1_mch

t-2h)mc-)1_mc-

2h

A.

B.

C.

D.

Copyright @ by The Berkeley Review@ 150 GO ON TO THE NEXT P

17. Which of the following transitions relatesGREATEST energy of absorption?

A. From the n = 2level to the n = 4 levelB. From the n = Zlevel to the n = 1 level

C. From the n = I level to the n = 2levelD. From the n = 2 level to the n = 5 level

1 8. Which statements are valid regarding the energy

of hydrogen?

I. As the value of n increases, proximitynucleus decreases, therefore the electron isin a higher energy state.

[I. As the value of n increases, the differenceadjacent energy levels (with n values dione), gets smaller.

m. A larger nuclear charge has no effect onenergies, although it lower all of the energy

A. I onlyB. II onlyC. I andll onlyD . tr and III only

19. For some compound X, which emits orange lighttransition from the n = 4 level to the n = 3 level,

must be TRUE?

A. A transition from the nemits green light.

B. A transition from the nemits red light.

C. A transition from the nemits violet light.

D. A transition from the nemits yellow light.

=5level tothen=4

=4leveltothen=2

=6leveltothen=4

=5leveltothen=3

2 0. Which of the following observation leads toconclusion that energy levels are quantized?

A. A sample always absorbs the same amouillight.

B. A sample always absorbs the same color ofC. A sample always emits different

different temperatures.

D. A sample always emits different frequencierdifferent concentrations.

:assage lV (Questions 21 - 27)

Liquid crystal displays (LCDs) utilize the properties of:.:amagnetic species. When plane-polarized light passes

: rugh a solution containing a paramagnetic species, it- ..-"rles the plane-polarized light. If the paramagnetic species

, .:bjected to a magnetic field, then the particles align in the

' -'.i. and the plane-polarized light is not rotated. The display. .:rmonly seen in calculators and digital watches uses this:::;iple to filter or pass light through templates.

The basic design of the display board involves passing

;rt through a polarizing plate into a cell containing the

:.:a-rragnetic species and then having the light pass through

" i;cond polarizing plate that is rotated 90' relative to the

' :;r plate. If the field is off, the light will rotate and thus be

: -sed through the second polarizer. This results in a bright:':: on the display board. If the field is on, the light will' .: rotate and thus will not pass through the second- ."rizer. This results in a dark spot on the display board.

'' :rre 1 shows a simple display circuit.

Circuit on (Solution is aligned)

Horizontal polarizer(only horizo4fal waves pass)

light passes

. 'gned paramagnetic solution does NOT rotate light plane

Circuit off (Solution is random)

Unaligned paramagnetic solution rotates light plane

Figure 1

-\ paramagnetic species is defined as a compound that'.".' unpaired electrons. Many common examples involve:.-:sition metals with a d-shell count of d1 to d9. Organic. ::.pounds that are paramagnetic are referred to as radicals.'

r.:. sen gas has two unpaired electrons; thus, when exposed

. magnetic field, oxygen molecules will with the field.

Vertical polarizer:ly vertical waves pass)

Vertical polarizer::ly vertical waves pass)

Horizontal polarizer(only horizonJal waves

- -:vright @ by The Berkeley Review@ l5l GO ON TO THE NEXT PAGE

21. Which of the following can be used in an LCD?

A. CdCl2

B. Fz

C. CoClz.6 HzO

D. Hz

22, What are the quantum numbers associated withLAST electron in the nitrogen atom?

_,12

__ I2

=+l2

__12

2 3. What is the shape of Fe(CN)63-?

A. Octahedral

B. Squareplanar

C. Tetrahedral

D. Trigonal bipyramidal

2 4. Which of the following compounds is polar?

A. r'ecl63-

B. Ni(CN)s3-

C. cis-Pt(NH)2C12D. trans-Pt(NH3)2C12

25. Which ion has the MOST unpaired electrons?

A. cr3+

B. Mn2+

C. Cu+

D. Ni2+

the

A. n=2

B. n=2

C. n=2

D. n=2

1=1 ml=+1

l=1 ml =+1

l=1 ml=-1

l=1 ml=-l

ms

rns

IIIg

ms

2 6. Which of the following statements is NOT true?

A. Cr has a higher formal charge in Cr(CN)6 thanCr(NH3)63+.

B. Cu(H2O)O+ has ten d-electrons.

C. Fe(NHr)63* has the same shape as FeCl63-.

D. Scandium is more likely to have +4 formal chargethan titanium.

2 7. Which of the following pairs of molecules does NOThave the central atom (transition metal) with the same

number of d-electrons in both compounds?

A. Fe(NH3)63+ and FeCl63-

B. Co(H2O)63+ and MnCl64-

C. Cr(NH3)63+ and MoC63-

D. Os(NH3)42+ and RhCl:(PRs)t

Copyright @ by The Berkeley Review@ t32

Passage V (Questions 28 - 34)

In vivo, semi-permeable membranes are used bybody to control diffusion and osmosis (entropically fioccurrences). In the kidney, Bowman's capsule functionssuch a manner. Semipermeable membranesmicroscopic pores that allow the passage of someor, in certain cases such as the sodium-potassiumions. They function by allowing only selected ions tofrom the solution on one side of the membrane tosolution on the other side of the membrane.membranes can distinguish by size, charge, or both.

Semipermeable membranes can be applied to benchchemistry Qn vitro) as well. This application can ireaction's selectivity. Semipermeable membranes used

vitro,that separate by size, are often referred to as

4nd atomic sieves. While some molecular sieves diby atomic radius, others differentiate by charge. To study

effectiveness of a semipermeable membrane, twodifferent solute concentrations are separated by theFigure I shows a U-tube with a semipermeableseparating the two solutions.

Side I starts withcannot cross the

solute thatmembrane.

Side 2 starts with solute thatcan cross the membrane.

le membrane

Side 2

After the solute migrates acrossmembrane, there is a concedifference. Flow stops whenpressure equals hydrostatic

Figure 1

Table I the rates of effusion from Solution Isynthetic membrane into Solution II for a generic study-

Cation Rate (M/s)

Li+ 4.7 x l0-2

Na+ 1.1 x 10-2

Ms2+ 2.9 x lO-2

K+ 3.5 x 10-3

ca2+ 8.3 x 10-3

Table IGO ON TO THE NEXT

Note: The synthetic membrane used in the experiment::rresented in Table 1 is believed to be aprotic and:iscriminates by atomic size. It is modeled after a protein::at is found in the lipid bilayer.

It. Based on the data in Table 1, which of the followingconclusions is valid?

A. As cationic size increases, theacross the membrane decreases.

B. As cationic size increases, theacross the membrane increases.

C. The cationic size has no direct effect on the rate ofmigration across the membrane.

D. Because cations are larger than neutral atoms, theeffect of cationic size on the rate of migrationcannot be determined.

The BEST explanation for why potassium (K) has a

lower ionization potential than sodium (Na) is that:

A. potassium is more metallic than sodium.

B. potassium has a larger enthalpy of reaction withwater than does sodium.

C. the 4s-orbital is larger than the 3s-orbital.D. potassium has a larger electron affinity than does

sodium.

The BEST explanation for why fluorine is smaller than

carbon is which of the following?

.{. Fluorine's electrons are at a lower quantum levelthan those of carbon.

B. Filled p-orbitals contract more than half-filled p-orbitals.

C. Fluorine has a larger effective nuclear charge thancarbon.

D . Fluorine is heavier than carbon.

\Vhich of the following is the SMALLEST?

.\. F-B. Ne

C. Na+

D. CI-

rate of migration

rate of migration

- ::'.'right @ by The Berkeley Review@ 133 GO ON TO THE NEXT PAGE

3 2. What is the charge on the MOST stable ion ofmagnesium?

A. +2B. +1

c. -lD. -2

3 3. Which of the following atoms is LARGEST?

A. FB. CIC. Br

D. I

3 4. If the pore of the membrane used in the experimentassociated with Table I were to distinguish by both size

and charge, the SLOWEST rate would occur with whichcation?

A. Li+B. K+

C. Mg2+

D. ca2+

Passage Vl (Questions 35 - 42)

1

1

1

1

1

1

1

1

ooE-tzct)q)ctIJco(5,Nco

IF

/o cl

11

1

Figure 1 shows the first ionization energy for the firsttwenty elements in the periodic table. The first ionizationenergy is defined as the energy required to remove the firstelectron from the valence shell of an element to form amonocation. The general reaction, where E symbolizes anyelement from the periodic table, is shown in Reaction 1.

E(g) -> E+1g; + e-

Reaction 1

To negate solvent effects, ionization reactions are donein the gas phase. The first ionization energy is proportionalto the square of the effective nuclear charge of the element.The effective nuclear charge is the charge exerted upon thevalence electrons by all other charged species in the atom.The effective nuclear charge takes into account the number ofprotons in the nucleus and the number of core electrons(electrons in principal quantum shells lower than the valanceelectrons). Ionization energy depends on the valence shellprincipal quantum number. Ionization energy decreases as

you descend a family in the periodic table. A larger principlequantum number corresponds to a lower ionization energy.

Figure I shows deviations from the behavior predictedfrom effective nuclear charge data. Deviations are seen withelements having valance electronic configurations of ns2np1

and ns2np4, which is attributed to the filled s-level and the

half-filled p-level associated with the cations that are formed.There is additional stability when the p-orbitals have equal

occupancy of electrons. The symmetry associated with the

half-filled state results in cation stability.

35. Which of the following BEST explains the h:dionization energy of helium?

A . Helium has no valance electrons to remove.

B. Helium has an effective nuclear charge greater

two.C. The electron must be removed from the

quantum level, which experiences the g

nuclear charge.

D. Helium requires a great amount of energy toconverted first into the gas phase.

3 6. What can be expected for the second

of lithium, sodium, and potassium?

A. The second ionization energies are only s1i

larger than the first ionization energies, becau:eZ6yhas not changed drastically.

B. The second ionization energies are only slilarger than the first ionization energies, becau-'e

valance shell has changed.

C. The second ionization energies are substanlarger than the first ionization energies, becau:e2"6 has not changed drastically.

D. The second ionization energies are substanlarger than the first ionization energies, becau**valance shell has changed.

Si

89101112Atomic Number

Figure I

.5,9". 'l

s

t

(

T

'\fr

h

ABC

D

Copyright @ by The Berkeley Review@ 154 GO ON TO THE NEXT P

3 7. How can the lower ionization energy for oxygen thannihogen BEST be explained?

A. Nitrogen is more electronegative than oxygen, so itdoes not share electrons as readily.

B. Nitrogen has more protons than oxygen, so itcarries a greater effective nuclear charge.

C. Nitrogen is larger than oxygen, so it is harder toremove a valance electron.

D. Nitrogen has a half-filled p-level, so it does notlose an electron as readily. Oxygen, when it losesone electron, attains half-filled p-level stability.

How can the trend in first ionization energy betweenaluminum, silicon, and phosphorus BEST beexplained?

A. The effective nuclear charge increases as you movefrom Al to P in the periodic table.

B. The effective nuclear charge decreases as you movefrom Al to P in the periodic table.

C. The number of core electrons increases as youmove from Al to P in the periodic table.

D. The number of core electrons decreases as youmove from Al to P in the periodic table.

\Vhy is the first ionization energy of fluorine greater

rhan the first ionization energy of chlorine?

A. Fluorine is less electronegative than chlorine, so itshares electrons more readily than chlorine.

B. Chlorine has more protons than fluorine, so itcarries a greater effective nuclear charge.

C. Chlorine is losing an electron from a shell that isfarther from the nucleus than fluorine, so chlorinecan lose an electron more easily.

D . Chlorine has no electrons in the p-level, so it doesnot lose an electron. Fluorine, when it loses oneelectron, completes its valance shell.

rir it . 1yh'"n of the following elements can be oxidized\IOST easily?

-{. SulfurB. Magnesium

C. Boron

D. Argon

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41. What can be expected for the first ionization energies ofbromine, krypton, and selenium?

A. LE.g1 > I.E.K. > I.E.S"

B. I.E.p > I.E.Br > I.E.Se

C. I.E.5s > I.E.B. > I.E.KtD. I.E.5s > I.E.Kr > I.E.B.

42. The first ionization energy for sodium is 495 kJ/moleand the second ionization energy is 4558 kJ/mole. Thefirst ionization energy of magnesium is 732 kJ/mole,and the second ionization energy is 1451 kJ/mole.How can the change in second ionization energy be

explained?

A . The first and second ionizations of sodium are fromdifferent quantum levels, while the first and secondionizations of magnesium are from the samequantum level.

B. Sodium is less electronegative than magnesium,which implies that the second ionization energymust be greater.

C. Once one electron has been lost from magnesium,the second electron is repelled by the positivecharge of the magnesium cation.

D. Sodium metal gains half-filled stability by losingone electron, while magnesium requires losing twoelectrons to attain half-filled stability

Passage Vll (Questions 43 - 49)

Transition metals do not follow the same periodic trendsassociated with the main-group elements. The reasoningbehind their separate trend is rooted in their valence electrons.The d-electrons of transition metals are not the outermostelectrons, despite the fact that the d-level is being filled as

you move left to right through any row of the transitionmetals. When we look at first-row transition metals, we see

electrons fill the levels according to the Aufbau principle, so

they fill the 4s-level prior to the 3dlevel. However, whenelectrons are lost (when the element undergoes ionization),the electrons are lost from the 4s-level before the 3d-level.Table 1 shows the ionization energy data associated with thefirst row of transition metals. The first three ionizationenergy values are listed for each of the transition metals.

M |./r2+ M3+ lst IE 2ndIF, 3rd IE

Sc 4s23d| tArl 63t 1235 2389

Ti 4s23d2 3d2 3d1 658 1309 2650

V 4s23d3 3d3 3d2 650 t4t3 2828

Cr 4s13d5 3d4 3d3 652 1591 2986

Mn +s23d5 3d5 3& 711 1509 3250

Fe 4s23d6 3d6 3ds 759 1561 2956

Co 4s23d7 3d7 3d6 160 1645 323r

Ni 4s23dB 3d8 3d'7 '736 t75l 3393

Cu 4113610 3d9 3d8 745 1958 35',78

Zn 4123610 3d10 922 1172

Table 1

Periodic trends also include atomic radius, ionic radius,and atomic density at 25"C. Transition metals also show a

deviation from main group elements in terms of trends inatomic size. Table 2 lists radius data for selected metals.

Radius Physical Properties

M }./r2+ M3+ M.P. B.P. Density

K 227 63.7 160 0.86

Ca 197 99 838 I44t t.54

Sc 162 81 1539 2738 3.0

Ti 147 90 11 1668 3259 4.51

V t34 88 14 1903 3447 6.1

Cr 130 85 64 1874 2667 '7.19

Mn 135 80 66 t241 2t52 1.43

Fe 126 77 60 1536 3001 7.86

Co t25 75 64 1495 2908 8.9

Ni r24 69 1453 2'731 8.92

Cu t28 72 1083 2591 8.96

Zrr 138 74 4r9.5 906 '7.14

Table 2

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Trends are predictable when based on the effective nuclercharge and the outermost electron shell (valence shelllHowever, the effective nuclear charge and valence cloud of tbdifferent transition metals do not follow the same pattern ermain block elements. Similar reasons can be employedexplain deviations in trends. Half-filled d-shell stabilityassociated with chromium, and filled d-shell stability iassociated with copper, which accounts for deviations inelectronic configurations from what is expected accordingthe Aufbau principle.

4 3. What is the density of manganese at 1000"C?

A. Between 1.00 and 2.00 g per cm3.

B . Between 2.00 and 7.43 g per cm3.

C . Between 7.43 and9.00 g per cm3.

D. Between 9.00 and 12.00 gper cm3.

4 4 . What is true of the relative secondvalues of Mo, Tc, Pd, and Ag?

A. Mo>Tc>Pd>AgB. Ag>Pd>Tc>MoC. Ag>Tc>Mo>PdD. Ag>Pd>Mo>Tc

4 5. How can the lower melting point and boilingzinc compared to other transition metals be exp

A. Increased interatomic forces with the filledB. Reduced interatomic forces with the fil1ed

C. Increased interatomic forces with the half-shell.

D. Reduced interatomic forces with the half-shell.

4 6. What can be concluded about the ease of removingelectron compared to a 4s-electron?

A. The 4s is more easily removed because it isto the nucleus.

B. The 4s is less easily removed because it isthe nucleus.

C . The 4s is more easily removed because it isfrom the nucleus.

D. The 4s is less easily removed because it isfrom the nucleus.

.17. What is TRUE about changes in density and atomic

radius as one scans across the 3d row of the periodic

table?

A. As atomic radius increases, density increases.

B. As atomic radius increases, density decreases.

C. As atomic radius increases, density increases'except for elements with filled d-shell stability.

D. As atomic radius increases, the density does not

vary in a predictable fashion.

I 8. How can the reduced radii of transition metal cations be

explained?

A. Transition metals lose their outermost electronsfrom the 4s-orbital, when becoming a cation.

B. Transition metals lose their outermost electronsfrom the 3d-orbital, when becoming a cation.

C . Transition metals lose their highest-energyelectrons from the 4s-orbital, when becoming a

cation.D. Transition metals lose their highest-energy

electrons from the 3d-orbital, when becoming a

cation.

+ 9. If the reduction potential of Mn2+ is -1.18 V and Co2+

is -0.28 V, what can be concluded about the reductionpotential of Fe2+?

A . E'reduction of Fe2+ is less than - 1. 18 V.

B . E'reduction of Fe2+ is between -1.18 and -0.73 V.

C . E'reduction of Fe2+ is between -0.73 and -0.28 V.

D . E'reduction of Fe2+ is greater than -0.28 V'

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Passage Vlll (Questions 50 - 56)

The basic principle behind the operation of a laser is the

excitation of electrons from the ground state of a compound

to some meta-stable excited state. The term meta-stablerefers to an excited state where an electron can exist for a

sustained period of time before it falls back to its ground

state. By constantly supplying energy to the compound, the

electrons can be "held" in the excited state. This preference

for a higher energy level is in accordance with Boltzmann's

distribution law of energy for any chemical or physical

system. Boltzmann's distribution law predicts that the energy

of a system will be distributed throughout the different levels

according to a standard probability function that follows a

bell curve. At higher total energy for the system, more

electrons are present in an excited state than at lower total

energy for the system. Figure 1 shows a molecular system

before and after energy has been added. The excited and

ground states are represented as levels capable of holdingmultiple electrons.

Before energy is added After energy is added

hv in

11111111111111M"""rubl""^"ir"dGil

M11 1111 1111 11 1 1

Ground state energy level

Figure IEnergy is most easily introduced into the system by

incident light (photons). A system analogous to the laser is

the electrical capacitor, which can be filled with electrical

energy and then instantly drained, releasing a pulse ofenergy.

Lasers can also store energy (light energy) and then release a

pulse of photons. A laser emits light energy corresponding

to the transition-level changes. Once the crystal or gas in the

laser releases photons, the light is collected and reflected

between mirrors, before it is released through an aperture

emitting the laser beam. Light emitted from a laser is close

to being monochromatic (that is, light of one wavelength),

but not precisely monochromatic. To date, absolutely

monochromatic light has yet to be observed. This isattributed to the fact that the ground state and excited states

are not single energy levels, but are in fact a band ofquantized levels with identical electronic energy' but varying

vibrational and rotational energies.

5 0. Which of the following electronic configurations is the

ground state electronic configuration for aluminum?

A. k22s22pqp3B. rs22s22p63s3

C. 1s22s22p63r2301

D. ts22s21p63r13p2

51. Monochromatic light is NOT observed for which of thefollowing reasons?

A. Lasers are not precise in their emissions.B. Crystals always have some imperfections.C . The laser re-absorbs some of the energy emitted.D. Each electronic energy level has many rotational

sub-levels.

5 2. What is the energy in joules associated with a photonof light with a wavelength of 330 nanometers?

@ = F,where h = 6.6 x lO-34,)"= wavelength in nm)lL

A. 6.0 x 10-28 J

B. 6.0 x 10-19 J

C. 6.0 x 10-17 J

D. 6.0 x l0l5 J

5 3. Which of the following sets of quantum numbers areassociated with the last electron in neutral vanadium(element # 23)?

A. n=3; l=2; m1 =0;

B. n=3; l=2;

C. n=4; l=0;

D. n=4; l=0;

ml=o;

m1 =0;

ml = +1;

ms=+l2

ms=J2

ms={2

m5=J2

5 4. Which of the following elements would have anelectronic configuration of nsl1n-1;d due to half-filledstability?

A. Copper

B. SiliconC. Manganese

D. Molybdenum

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55. According to Figure 1, the photons emitted compare irwhat way to the photons absorbed by the transitionbetween energy levels in the laser system?

A . The light emitted is of higher wavelength.B . The light emitted is of equal wavelengrh.C . The light emitted is of lesser wavelength.D. The light emitted is incomparable.

56. For a laser to emit visible light, the material used mhave an emission in what energy range, knowing

h

Imdimimhilnt

mryrmrlrl

rrwFmturmmdilL

the visible range is approximately 400 nm to 750 nml

1.24 x 10-6 eV(E- - -' - -',wherel,ismeasuredinmeters)L

A, 6.25 eV to 9.90 eVB. 3.10 eV to 6.25 eVC. 1.65 eV to 3.10 eVD. 1.24 eV to 1.65 eV

Passage lX (Questions 57 - 63)

Long before the advent of modern technology, colored:aints existed. The pigments for these paints were extracted

-om natural sources, such as plants and mineral deposits. In:urope and the Middle East, the majority of the pigments

-sed for painting were metal oxides. In the Orient, more

- sanic pigments were used. By tracing the origins of::{ments, art historians have been able to determine the

,,urce not only of paintings, but of stylistic influences as

' :11. Listed in Table I are some of the more common

-':rces of colors used in the early centuries A.D.

P igment Chemical Formula Colori-izarin 1,2-dihydroxyanthraquinone Yellow

-,zurite 2CuCOq.Cu(OH)z Blue

- -nnabar HgS Red

l' f .i:sicot Pbo Yellowl'lrnium PbrO+ Scarlet

niment As253 Yellow

: :algar As2S2 Orange red

Fe2O3.nH2O Brick red

:amarine blue SiS2/Al253 Blue

::digris Cu(CzHtO)2.Cu(OH)2 Greenblue

; rte lead 2PbCO:.Pb(oH)z White

Table 1

-\s a rule, inorganic paint pigments last for a far longer:- :J of time than the organic paint pigments. For this' *, :r. inorganic pigments are a more reliable source used to:-.-ir, the time and region from which many paintings

: rated. The colors in inorganic paint pigments are caused

.:e absorption of light in the visible range of the light" : -,rurrr. The absorption is due to transitions between the

: : ,:a1s, which because of the asymmetry of the ligands are

,:senerate. The splitting of the d-orbitals results in:::nt energy levels within the d-level. Figure 1 shows the

' :'::iic splitting of the d-orbitals in an octahedral complex.

d*z - rzd",

d*, dr"

Figure 1

l-e color we see is the reflected color, the- : :nentary color of the color that is absorbed. If a paint

i;,- :.--. blue, it is because of a paint pigment is present in it,. ; ,bsorbing orange light (the complementary color of., Figure 2 shows an artist's color wheel, which can be

: . determine complementary colors. Complementary-: -ppose one another on the color wheel. Blue opposes

i.-:;. so they are complementary colors of one another'L :.-- j Ereen are also a complementary pair.

dxy

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Figure 2

White light is a combination of all colors, so when one

color is absorbed from white light by a pigment, the other

colors reflect. All the colors are no longer present, so at

least one of the colors has no complementary color present.

The result is that the reflected light takes on a hue of the

color that has no complementary color to cancel it. When

white light reflects off of a pigment, the reflected light can be

analyzed for intensity versus wavelength after a prism splits

the light via refraction. The output is an absorptionspectrum, which appears as a rainbow band with verticalblack lines, due to the absence of the light absorbed.

Absorption spectra are the opposite of emission spectra,

although both require a prism to refract (grate) the light.

5 7. Which of the following pigments absorbs light of the

SHORTEST wavelength?

A. AlizarinB. Azurite

C. Cinnabar

D. Verdigris

5 8. When a sample is heated by flame, and the emitted lightis put through a prism, what result is produced?

A. An emission spectrum of darksome colored bands.

B. An absorption spectrum of dark

background with

background withsome colored bands.

An emission spectrum of colored background withsome black bands.

An absorption spectrum of colored backgroundwith some black bands.

5 9. In the absorption spectrum of realgar, what color band

would contain black lines?

A. Orangeandred

B. Orange and green

C. Blue andred

D. Blue and green

c.

D.

60. Given that the wavelength of visible light has a rangefrom about 740 nm to 390 nm, what is theapproximate wavelength of light absorbed by the copperdication in Cu(C2H3O )2.Cu(OH)2?A.450nmB.550nmC.650nmD. 750 nm

61. White lead absorbs which colors?

A. All colors

B. Green, blue, and violetC. Red, orange, and yellowD. No colors in the visible range

62. All of the following exhibit reflected color EXCEPT:

A. Candle waxB. FabricdyesC. Pen inkD. Gas-filled light tubes

6 3 . Which of the following pigments will decompose first?

A. AlizarinB. AzuriteC. RustD. Verdigris

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Passage X (Questions 64 - 69)

Fluorescence occurs with the absorption of a photona molecule, exciting one of its electrons to a higherstate. Because the material is molecular and notthere are also vibrational and rotational energyassociated with it. The molecule can change its vienergy level (by changing its bond-stretching frequency)give off energy in the form of IR photons or heat. Heatreleased during collisions with less energeticWhen the electron falls back to the ground state, a photonless energy than was initially absorbed is given off.fluorescing compound can thus absorb high-energy li(such as ultraviolet) and give offlesser-energy light (suchvisible light).

Phosphorescence also begins with the absorptionphoton by a molecule, exciting one of its electrons Erhigher energy state. The electron can flip its spin inexcited state, dissipating some of the extra energy. Whenelectron relaxes, because it shares the same spin as

ground state electron, it cannot fall back to its groundIt must relax to an intermediate state, so a lower-photon is emitted than was initially absorbed. Firepresents what goes on during both fluorescencephosphorescence. The Xo potential well representsground electronic state, and X1 represents the excitedThe lines within the well represent the differentmodes (vibrational energy levels) of the molecule.

Bond radius (A)

Figure I

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The wiggly urrrows in Figure 1 represent energy that isdissipated as an excited electron drops in vibrational energy,,rvhich is released in the form of an IR photon. The

',ibrational energy thereby released can be measured as heat

siven off from the system. The amount of heat emittedr aries with different molecules, because the electronic energy

levels are not singular levels (they have both vibrational and

:otational energy levels associated with them). The

:ansition energy is thus the result of a random combinationrfenergy levels. Because the electronic energy levels are notsingular levels and because the transition is random, it is-mpossible for a molecule to absorb or emit light in such a:nanner that all of the photons simultaneously have the same

=equency (or wavelength). For this reason, monochromatic

-rght is not physically possible.

6.1. A spin flip in the excited state is associated with whichof the following processes?

A. Fluorescence onlyB. Phosphorescenceonly

C . Both fluorescence and phosphorescence

D . Neither fluorescence and phosphorescence

6 5. What can be said about the photon absorbed relative to

the photon emitted by a fluorescing diatomic molecule?

A. The energy of the photon absorbed is greater than

the energy of the photon emitted, while thewavelength of the photon absorbed is less than the

wavelength of the photon emitted.

B. The energy of the photon absorbed is less than the

energy of the photon emitted, while the wavelengthof the photon absorbed is greater than thewavelength of the photon emitted.

C . Both the energy and wavelength of the photon

absorbed are greater than the energy and wavelength

of the photon emitted.

D. Both the energy and wavelength of the photon

absorbed are less than the energy and wavelength ofthe photon emitted.

6 6. What is TRUE about monochromatic light?

A. Monochromatic light is not possible, because lightis not quantized.

B. Monochromatic light is possible, because light is

quantized.

C . Monochromatic light is not possible, because there

are vibrational energy levels of lower energydifference than the electronic energy levels withwhich they are associated.

D. Monochromatic light is possible, because there are

vibrational energy levels of greater energydifference than the electronic energy levels withwhich they are associated.

6 7. According to the graph, how do the photons emittedfrom fluorescence and phosphorescence compare?

A. The photons emitted from fluorescence have a

longer wavelength than the photons emitted fromphosphorescence.

B. The photons emitted from fluorescence have a

shorter wavelength than the photons emitted fromphosphorescence.

C. The photons emitted from fluorescence have the

same wavelength as the photons emitted fromphosphorescence.

D. The photons emitted from phosphorescence are ofhigher energy than the photons emitted fromfluorescence.

6 8. Which of the following transitions emits the

SHORTEST wavelength of light?

A. Xt V = I toXg V = 0, whereVenergy level.

B. Xt V =0toX6 V = 0, where Venergy level.

C. XoV = 1 toXl V =0, whereVenergy level.

D. XOV = 0 toXl V = 0, where Venergy level.

is the vibrational

is the vibrational

is the vibrational

is the vibrational

6 9 . Which of the following statements about fluorescence

is FALSE?

A. Fluorescence is possible with molecules.

B. Fluorescence converts visible light into ultravioletlight.

C. Fluorescence can dissipate energy in the form ofheat.

D. There are some atoms for which fluorescence is not

possible.

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Passage Xl (Questions 70 - 76)

A student conducts an experiment where a series ofaqueous solutions of group I cations (alkali metals) or groupII cations (alkaline earth metals) are each heated by a flame.The heat of the flame, if hot enough, can stimulate electronsin the cations to leave their ground state. When they relax tothe ground state, light is emitted. Because the light is anarrow band of photons with predominantly one wavelength,it appears colored. The color is used to identify the cation insolution. Table 1 shows the student's result.

Barium Lime-green ring around flame

Lithium Brilliant crimson flame

Potassium Light pumle ring around flame

Sodium Orange-yellow flame

Strontium Red flame

Table 1

The student then conducts the same experiment usingtransition metals rather than main-group metals. The trendin color is predictable with main-group cations of the samecolumn (family) in the periodic table. The color of thetransition metal cations depends more on the ligands attachedto the metal cation. Table 2 also shows the colors obtained.

Vanadium (tr) Violet flame

Copper (I) Orange flame

Nickel (tr) Red flame

Chromium (tID Red flame

Cobalt (D Green flame

Table 2

With transition metals, the color emitted by the flame isoften the color of the light absorbed when the cation insolution is exposed to white light. The color of the aqueoussolution that is detected by the eye is the complementarycolor of the color absorbed. The color that is associated withtransition metals is due to an electronic transition within thed-orbitals (between the d-levels, which split due to thepresence ofligands). These electronic transitions are referredto as d-d transitions, and they occur only with transitionmetals that are not d0 or d10 in their electronic configuration.

7 0. Which cation is paramagnetic?

A. Na+

B. Sr2+

C. Cu+

D. cr3+

7 1" What is the electronic configuration of chromium?

n. 1ar1+s2:a4B. [Ar]4s13dsc. itul3d6D. tArl3d3

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72. Which of the following transition metals has rir:GREATEST d-d transition energy, given that the coli':observed in the flame test is due to a d-d electron-:transition?

A. Ni2+B. cr3+

c. Co2+

D. V2+

7 3. Identify the TRUE statement(s) from the following.

I. As ionic size increases; the energy of the transitj:increases.

II. The transition energy of K+ is greater than ::rtransition energy of Cr3+.

m. Cu+ and Ni2+ have the same electrol-;configuration.

A. I onlyB. II onlyC. I and II onlyD. tr and III only

7 4. What are the quantum numbers associated of the .:-"

electron of nickel dication?

llrlhtfl

iil'i'llua

linr i

;iii

t,l f,

iir-

A. n=3 l=2 ml=-1

B. n=3 l=2 ml=0

frr=-l2

Ift*=-12

ml=+1 -r=-jml=+1 *r=*f

C. n=3 l=2

D. n=3 l=2

lmm:

iuull

ru:rh{}

,i,r!flr

llll]llilfim:

illtrrrtl!

tNl\ rl

l"@

umr.llt

,iiiillfi.Ilu

x1l'::1r

thin

iil[E,u

l'I!&

,mfr0ri

ilhc

75. Given that lithium cation emits a red flame, sod:-mcation emits an orange-yellow flame, and potass.:mcation emits a violet flame, what color flame wili ruemitted by rubidium cation?

A. YellowB. Green

C. Blue

D. Colorless (and thus the normal orange)

7 6. An emission spectrum of lithium would appear in ',i:ruult

manner?

A. It would be a rainbow with a black line in the :Cregion of the spectrum.

B. It would be a rainbow with a black line in ;:mc'

green region of the spectrum.

C . It would have a black background with a red lin:D . It would have a black background with a green

-Tnc,

Passage Xll (Questions 77 - 81)

Glyphosate, shown in Figure l, is a synthetic compound

.-r:t is currently the world's largest selling commercial:,::bicide. Glyphosate kills vegetation by binding metals,:thin a plant, thus starving the organism of essential.:ta1lic nutrients needed for the transport of other nutrients

..J waste, such as carbon dioxide and oxygen. When

;1,,phosate leeches metals from the plant, its respiration is::.lt down. Glyphosate is shown in Figure 1 in its salt-:rn. as extracted from pH = 7 aqueous solution.

xu* -oy n- r-

*\ c//

t- o- **uHo -i\ -i\

HHHHFigure 1

After the plant has died, Glyphosate decomposes in the

:-:sence of moisture into phosphate, carbon dioxide, and

-,:-nonia. The decomposition products are environmentally

-::. and some are useful nutrients for the soil. In a study of-: decomposition of Glyphosate in vivo, isotopically-::led Glyphosate was applied to vegetation, and the

: rpic abundance was monitored at regular intervals over a

.nn'-day period. Based on kinetic data associated with" :iolysis and decomposition, the conclusion was reached

.-.: ihe decay time of Glyphosate to its inorganic products is

.:::oximately fourteen days. The compound was radio-

-:,:ied in three separate experiments, using a different radio-

-::. in each experiment. The first experiment employed-: ':rr 13H), the second experiment employed 14C, and the- :: experiment employed 32P to monitor the half-life and

";:rv rates. Carbon-l4 was used as a marker, but not for::iic purposes. The half-life of carbon-I4 is far greater

":= the lifetime of a typical lab experiment. The isotopic* ::i3r helps to study the migration of Glyphosate through'- :rYironment.

' - . Which of the following is an isotope of 32P?

.t . 32s

g. 32p-

c. 31p

o. 31si

The molecular geometry about a nitrogen in glyphosate

ls BEST described as which of the following?

-{ . Trigonal planar

B. Trigonal pyramidal

C . Square planar

D. Bent

Tfro

il

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7 9. The half-life of 32P is approximately 14 days. If the

concentration of a32P radiolabeled Glyphosate sample(C3H8NO6P) is found to be 188 parts per millioninitially, how long will it take until the concentrationis 25 ppm?

A. 27 days

B. 33 days

C. 41 days

D. 43 days

8 0. Which of the following molecules have ionic bonding?

I. NHr

II. COz

m. Na2POaH

A. I onlyB. II onlyC. III only

D. I andll only

81. As 32P decays, it changes into a product that is stillphosphorus. This is necessary in a marker so that the

chemical behavior of the labeled species does not

change. What is the MOST likely particle given off inthe decay of phosphorus-32?

A. A neutron

B. An alpha particle

C. AbetaparticleD. A positron

Passage Xlll (Questions 82 - 88)

In order to image vital organs by photon emission,technetium-99 is ingested into the body so that its low-levelgamma radiation (140-keV) may be detected. For brainimaging, 99Tc is introduced via the bloodstream, and itsubsequently diffuses throughout the body, including to thebrain. Technetium-99 has a half-life of approximately sixhours, so scanning and analysis must closely followingestion. For 99Tc to migrate to all of the target organs ina concentration high enough to be detectable requires roughlytwo hours. After two hours, the concentration of activetechnetium-99 is approximately eighty percent (807o) of itsoriginal value. Technetium-99 decays through gammaemission, meaning that its nucleons simply drop in one stepfrom an excited nuclear state to a ground nuclear state whenthey emit photons. The nucleus of technetium does notchange upon the emission of the gamma photon.

For brain imaging, the skull is encompassed by adetection device that collects and records the gamma radiationat arbitrary sites. Radiation escapes more readily from theareas of the brain that have a lower tissue density and fluidthan those that have a greater density, but a complication inthe analysis of brain images is that the radiation dissipates inliquid. This means that slight variations in the texture andcomposition of brain matter can produce different imagingpatterns. The display pattern collected can be converted intoa contour map of the brain. This technique work well withorgans that have a relatively low water content.

8 2. Beta decay of 210Po results in which of the following?

4. 21041

g. 2109i

g. 206p5

p. 20941

83. Alpha decay is observed in all of the followingfr ansformations EXCEPT:

4. 2549, 1o 2509p

g. 2389 6234.1y,g.223pr62219ig,247gn 6243pu

84. Ifthe initial dosage of99Tc gave a reading of 120 pCi,what will the reading be after 12 hours?

A. 60 pCiB. 40 pCiC. 30 pCiD. 20 pCi

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8 5. Electron capture by carbon produces the same elemeilas which of the following processes?

A. Positron capture by boronB. Positron decay of carbon

C . Beta capture by nitrogenD. Alphadecayofoxygen

8 6. Gamma radiation is considered to:

A . weigh 4 amu and have a nuclear charge of +2.

B . be massless with a nuclear charge of -1.

C. weigh 4 amu and have no nuclear charge.

D. be massless and have no nuclear charge.

87. The electronic configurationfollowing?

A. tArl 5s2 3d5

B. tKrl 5s2 4d5

C. tKrl 5sl 4d6

D. tKrl 5s2 4d'l

for 99Tc is which of

8 8. When 99Tc undergoes gamma decay, the finalis which of the following?

A. 99Tc

B. 99Ru

c. 95Nu

D. 99Mo

Passage XIV (Questions Bg - 95)

As of March 1989 physicists and chemists had yet toJarry out a nuclear fusion reaction that did not require more

3nergy input than output. Fusion reactions require high:imperature and strong magnetic fields to be carried out.This is the reason that nuclear power generators are fission:eactors. Reaction 1 is an example of a fusion reaction,:i hile Reaction 2 is an example of a fission reaction.

!r-i * lu --> zlneReaction 1

'3]u * fn -+ 3fn" * 2frlnn * fnReaction 2

Then in March of 1989, B. Stanley Pons and MartinIeischmann announced that they has achieved a nuclear,lsion reaction in a test tube at room temperature. Pons and

:leischmann set up a simple electrochemical cell with a

:alladium cathode in 0.1 M LiOD(DzO) and a platinum

-rode. The design of the cell was to form D2 gas by passing

.n electron flow from the platinum anode to the palladium::thode. The cell generated more heat than expected, so it;i as proposed that a nuclear reaction must have transpired.

The researchers believed that the deuterium migrated into.:re palladium metal of the cathode and gathered in the

:,rckets of the lattice, so they hypothesized that the:euterium fused together in the palladium electrode by eitherR.eaction 3 or Reaction 4.

lu*ln+ln+lHReaction 3

ln *la - fn * ln"Reaction 4

Pons and Fleischmann in their original paper stated that

:.rth tritium and neutrons were observed in the cathode of:eir electrochemical cell, but the concentrations were too

-,rrv to account for the extra heat generated. Since that time,::e hypothesis of a "cold fusion" nuclear reaction taking:race under these experimental conditions has been dismissed

.: improbable by most scientists in the field.

19. What is NOT a product when two deuterium atoms

undergo a fusion reaction?

A. TritiumB. Helium-4

C. Helium-3

D. Hydrogen

9 0. What evidence did Pons and Fleischmann point to that

would indicate that a nuclear reaction had transpired inthe electrochemical cell?

A . Less heat was released than expected.

B. More heat was released than expected.

C . That electrons were emitted by the cell.D. That electrons were absorbed by the cell.

Copyright @ by The Berkeley Review@

91. Cold fusion is BEST described as:

A. an endothermic nuclear reaction capable ofrapidlycooling the environment.

B. an exothermic nuclear reaction capable of rapidlycooling the environment.

C. an endothermic nuclear reaction capable of beingcarried out at room temperature.

D. an exothermic nuclear reaction capable of beingcarried out at room temperature.

9 2. Nuclear power plants employ what reaction to generate

power?

A. Fission, because fusion requires such a great energy

input that it is inefficient, and the net energy

change is unfavorable.B. Fusion, because fusion requires such a great energy

input that it is inefficient, and the net energy

change is unfavorable.C. Fission, because fission requires such a great

energy input that it is inefficient, and the net

energy change is unfavorable.

D . Fusion, because fission requires such a great energy

input that it is inefficient, and the net energy

change is unfavorable.

9 3. Which of the following is a fusion reaction?

^. 'W + [n -r 2lfrn * lu"

n. 23faP" - 23nlu * lu"c. "nlru

* lln -2$lcrn. ferr"- -+ ]errc + rr1

94. What is formed *1t"n 238g emits an alpha particle and

a deuterium nucleus?

;. 2346"g. 2326"g.2369p. 232pu

9 5. Capture of which of the following particles would NOT

result in a change in the atomic number?

A. Alpha

B. Beta

C. Neutron

D. Tritium nucleus

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Questions 96 - 100 are NOT basedon a descriptive passage.

9 6. Which of the following supports the conclusion thatelectrons have quantized energy levels?

A. The existence ofa nucleusB . Scattering of X-rays by a thin sheet of a materialC. The bending of nuclear radiation particles by a

magnetic fieldD. Distinct lines in an electromagnetic radiation

specffa

97. The LONGEST wavelength of light would beassociated with:

A. X-rays.B. violet light.C. green light.D. infrared light.

98. The ionization energy of H(g) is 1312 kJ . A goodmole

approximation for the second ionization potential ofhelium (g) is:

A. 5248 kJmole

B. 2624 kJmole

c. 656 kJmole

D. 328 kJmole

9 9. Which of the following is the electronic configurationfor an excited state of Na+?

A. k21s22p53s2B. ls22s21p63rtC. k22s21p6D, ls22s22p53s1

Copyright O by The Berkeley Review@ STOP, YOU'VE HAD ENOI

100. What is the MOST common shape for a

metal with five ligands attached?

A. SquareplanarB. Trigonal bipyramidalC. Pentagonal planarD. Hexahedral

IL

1,C 2.A6. B 7, D

11. D 12. D16. A 11. C2t. c 22. A26. D 21. B31. C 32. A36. D 31. D41. B 42. A46. C 41. B51. D 52. B56. C 51. A61. D 62. D66. C 61. B71. B 72. D16. C 17. C81. A 82. A86. D 87. B91. D 92. A96. D 91. D

3.A 4.C 5.D8.B 9.A 10.B

13. D 14. D 15. B18. C i9. D 20. B23. A 24. C 25. B28. A 29. C 30. C33. D 34. D 35. C38. A 39. C 40. B43. B 44. D 45. B48. A 49. C 50. C53. A 54. D 55. B58. A 59. D 60. C63. A 64. B 65. A68. A 69. B 70. D13. C 74. B 15. D78. B '.79. C 80. C83. C 84. C 85. B88. A 89. B 90. B93. C 94. B 95. C98. A 99. D 100. B

Atomic Theory Passage Answers

Choice C is correct. If oil drops were naturally charged, then the experiment could not work, because the totalcharge on the droplet would not be generated by the electron beam alone. The charge determined from the

expeiiment would not necessarily reflect the charge of an electron. This eliminates choice A, because it isasiumed in the experiment that the oit droplet is initially uncharged before exposure to the electron beam.

Because the gravititional pull on the oil droplet is calculated as opposing the electric field, it must be assumed

to be in one direction (downward). If the gravitational pull were in all directions, the experiment would not be

possible. This eliminates choice B. The mass used in the calculation is the average mass of an oil droplet, so

ih" *u5 of the electrons is in fact ignored. This can be assumed knowing that protons and neutrons are far more

massive than electrons. The changi in mass by gaining or losing electrons is negligible compared to the rest of

the atomic mass. This makes choiie C the best answeri an answer to be chosen by many, like you. The charge of

an electron is assumed to be constant, if you are solving for an exact value. The experiment could be solved for an

average value, but there is no reasoning behind one electron being of a different charge than another electron.

This is to say that charge is quantized (has an exact value).

Choice A is correct. In the tube in Figure 1, the electron beam is negatively charged, so it bends downwards due

to its attraction to the positively charged plate below and its repulsion from the negatively charged plate

above. This may seem odd that the cathode plate (plate above) is negatively charged and the anode plate(plate below) is positively charged, but the plates form a capacitor, not a battery. For a discharging battery

and charging capacitor, the plate charges are opposite. Because a proton has the charge opposite from an

electron lptotonJ carry a positi.re charge), protons exhibit behavior opposite from an electron. Proton beams

therefore ^bend

upwaids. A neutron is neutral, so its pathway would not arc at all between the two charged

plates. The neutron does not arc, because it is unaffected by the charged plates. The correct answer is choice A.

Choice A is correct. The value of the charge is twice as large as expected. This can be explained in terms of

magnitude of charge rather than sign of charge. A value that is twice as large as expected is attributed to aaouUty charged oil droplet. Choice A, doubly ionized, would explain this. Choices B, C, and D all address sign

of charge and thus are eliminated.

Choice C is correct. The Thomson experiment, as stated in the passage, determined the mass-to-charge ratio of

the electron. It can be used to determine the mass-to-charge ratio for any charged particle, but in this

particular experiment, the particle was an electron. The best answer is choice C. A mass spectrometer is a

mechanical variation of the Thomson experiment.

Choice D is correct. The charge of an electron is determined by equating the mass of the droplet and the

gravitational force constant (mg) with the voltage of the electric field and charge of the oil droplet (qV). To

Jolve for q, the other three variables must be known. This means that choices A, B, and C are all values that

must be known. There is no temperature factor in the equation (mass, charge, gravity, and voltage do not vary

with temperature), so choice D represents the factor that is least important in the calculations.

Choice B is correct. An atom is neutral when it carries no net charge. Electrons caffy a negative charge, while

protons caffy apositive charge. Neutrons are uncharged. In order for an atom to be neutral, the number of

electrons must be equal to the"number of protons. The best answer is choice B. Choice A is not false per se, but

because there exist isotopes whose number of electrons does not equal the number of neutrons, no conclusion can be

drawn about the number of electrons and neutrons within a neutral atom.

Choice D is correct. The strongest attractive force is felt between particles of opposite charge. This is known as

Coulomb's law. An electron aid proton carry opposite charges, so the best answer is choice D. A proton and

neutron exhibit no Coulombic attriction, but because they are held tightly in the core of the nucleus, there must

be some attractive force between them. Taken literally, this question could not have a proton and a proton

Iisted as an answer choice because of the strong force associated with the nucleus that holds protons together.

This is a more advanced topic that you may see in courses in nuclear chemistry (atomic physics) but as it ispresented" in the MCAT, the ambiguity preiented by nuclear attractions (the uncertainty of the course of the

strong force) is not a probable topic for questions based on their past questions'

147 Section II Detailed ExPlanations

8. Choice B is correct' If an electron is descending, a net upward force must be applied to accelerate the particle inthe opposite direction. To increase the net upward force, the force upward Lust be increased or the forcedownward reduced- This would slow the parti"le and eventually stop it from dropping. This means thateither the mass must be decreased, the gravity must be decreased, Ur" charge must be increased, or the voltageof the field must be increased. The elsiesf thing to do is increase the iroltage of the field. In the actualexperiment, this is what is done. Increasing the number of electrons in the drJf increases the charge, whichshould stop (or at least slow) the descent or change the direction of its path io an ascent. Decreasing thegravitational pull, which requires ptacing the apparatus in an anti-gravity environment, would also alleviatethe descending difficulty. The only chbice that would definitely-not stop the descent, but would in factincrease the descent, is to increase the mass of the oil droplet. The best ur,rr"", is choice B.

9.

1"0.

11.

72.

13.

Choice A is correct. All atomic masses in the periodic table are referenced against carbon-12. Carbon-12 is tluestandard traditionally used. The carbon-l2 isotope is defined as having u *ui, of exactly 12.0000 amu. Carbonis-added to a sample to standardize the mass of the peaks given in thJ spectrum. Not all elements have exactwhole number masses, so a reference is necessary. you ut" *ir" to pick A.

Choice B is correct. According to Table L, magnesium has three major isotopes. The isotope with the lowestatomic mass is the most abundant (78.70%). This means that the lowest p"ut Uy mass (the iirst peak from leftto right) ismust be the largest peak. This is true only in the spectrum of choice B.

Choice D is correct. The average atomic mass for any element is the weighted average of all of the isotopes ofthe given element. Choices A and B are eliminated, tecause Table 1 lists"only two isotopes for silver, and thecalculation shows three values being summed. Choice A is the calculation of average atomic massmagnesium. The correct answer is choice D, because that has the correct percentages muttlpUea by the corisotopic masses. You would be a very wisdom-laden soul, if you were to select D as liour answer.

Choice D is correct. In order to send the isotopes of the element through the mass spectrometer, the elenmust-be in the gas phase. This means that anlsample not in the gu" piur" initially must be converted togas.phase to be analyzed u-sing mass spectroscopy. The process of lonverting a solid into a gas is referred tosublimation, thus making choice D correct.

Choice D is correct. Because- the.average atomic mass is an average of the masses of all of the isotopes,average atomic mass must fall within the range of the isotopic musser, meaning that the average atomic nis greater than the lightest isotope but less thah the most mlssive isotope. Whin there are twJ isotopes cthen the average atomic mass is the weighted average of the two. In i titty-titty mixture of two isotopes,average atomic mass lies perfectly between the two masses of the isotopei. If the mixture favors one ofisotopes, then the average atomic mass would be closer to the more abundant isotope than the less abuncisotope. The true statements are II and III, making the best answer choice D.

14. Choice D is correct. The difference between isotopes of the same element is found in the number of neutrons,two atoms have a different number of protons, then they are not the same element. Choices A and Beliminated, because they allude to a different number of piotons. The greater mass associated with the bo11 isotope is attributed to an extra neutron being presenf in the Boron-I1 isotope as compared to the boisotope. The best answer is choice D.

Choice B is correct. The ionization of hydrogen involves an electronic transition from the n = 1 energy lelthe n = - ener8Y level' Using Equation 1 to determine the transition energy yields the following:

2'778x's'B Q\(1-- 1 \r\ni' "eJ'where Z = I,^nt = 1, and nf = x. Having - in the denominator makes the number zero, which results in a vt

2.178 x rc-t9 Gzil. With a value of 2 = 1, tle ionization energy is 2.12t;-10-is-f.' rrr" ionization enehydrogen from its valence level is 2.178 x 10-18 ], which makes tie best answer choice B.

15.

Copyright @ by The Berkeley Review@ t4a Section II Detailed

16. Choice A is correct. This solution requires relating the formula for the energy of a photon to the energyequation given in the question. The relationship is as follows:

E = mC and E = hc ... mC = h. = *. =A =+ l, = -h-l. l,rlrrwhich is choice A.

:[7. Choice C is correct. ChoiceB iseliminated,because fromthen=2energyleveltothen = l energYlevel,thereis a drop in energy, which corresponds to emission of light, not absorption. The best answer is choice C, because

the eneigy levelJ get closer together as the n value incteases, so the transition from the n = 1 energy level to the

n = 2 energy level is always of greater energy than any transition starting at an energy state where n > 1.

However, to be certain, the following mathematical relationship can be derived:

Given that the energy level of an electron is E = -2.178 * tO-t8(*^\1,\''t2 / "

the transition energy is AE = -2.178x 10-18 ,t\(h -r), = 2.178xr0-18 tz1({ #) ,

The relative energies are therefore comparable by the value of "" (# #)*trn

$ i).(i i),$ f)=(i'f'(l #'(l #)/r -r) ,(E--4lt(g-r\ -3 > 27 >3, therefore choice C is the best answer.\ +l \too 100/ \to 76t 4 100 1.6

li. Choice C is correct. Statement I is valid, because as an electron's principle quantum number (n) increases, the

electron is in an energy level (orbit) that is farther from the nucleus. Statement II can be viewed from the

pictorial representation of the energy levels in Figure 1. As the value of n is increasing, the energy levels

beco-e cloier to one another. This makes statement II a valid statement. According to the answer choices, this

makes choice C the best answer. A larger nuclear charge (Z) affects the energy at each level, and thus affects

the transition energies as well. This makes statement III incorrect, and confirms that choice C is the best

answer. Choose C and feel a little giddy.

:q. Choice D is correct. The electronictransitionfromthen = 5 energy level to then = 4 energylevelis of lowerenergy than the electronic transition from the n = 4 energy level to the n = 3 energy level. Therefgle, fhe photon

emitied is of lower energy than a photon corresponding to orange light. Because green light is of higher energy

than orange light, choice A is eliminated. The electronic transition from the n = 4 energy level to the n = 2

energy level is of higher energy than the electronic transition from the n = 4 energy level to the n, = 3 energy

level. Therefore, thJphoton emitted is of higher energy than a photon corresponding to orange light. Because

red light is of lower "t ".gy

than orange light, choice B is eliminated. The electronic transition from the n = 6

"."tg! level to the n = + eiiergy level is of lower energy than the electronic transition from the n = 4 energy level

to the n = 3 energy level. Tirerefore, the photon emitted is of lower energy than a photon corresponding to

orange light. Becilse violet light is of higher energy than orange light, choice C is eliminated. The electronic

transltion from the n = 5 energi level to the n = 3 energy level is of higher energy than the electronic transition

from the n = 4 energy level to itre n = 3 energy level. Therefore, the photon emitted is of higher energy than a

photon correSpondiirg to orange light. Because yellow light is of higher energy than orange light, choice D is

tl're best answer.

ll. Choice B is correct. Choice A is an invalid statement, because according to Beer's law, the absorbance of lightis proportional to sample concentration. Because samples can have varying concentrations, they can in fact

abiorb different utnontrtr of light. The color of light (or frequency) corresponds to transition energy. Because

the sample, independent of concentration, absorbs the same frequency of light each time, the transition energy

must be a fixed value. This leads to the conclusion that energy levels are fixed, and thus quantized. Choice B

is the best answer. The sample emits light of the same frequency, regardless of solution's temperature and

concentration. Temperature can affect the intensity of the light that is emitted, but not the frequency. The

frequency of the light remains constant. This eliminates choice C (temperature effects) and choice D

(concentration effects). Choice B is your answer choice'

-:pvright @ by The Berkeley Review@ 149 Section II Detailed ExPlanations

21. Choice C is correct. A liquid crystal display uses a paramagnetic compound in its polarizing cells to rotate (olnot rotate) light. A paramagnetic species has at least one unpaired electron. Because chlorides are -1 eachcadmium in CdCl2 carries a +2 charge. This is the result of losing two 5s electrons from elemental cadmiur.which results in an electronic configuration of 1s22s22p69"29q64t294104064410 for Cd2+. Each level is full, scthere are no unpaired electrons. Choice A is eliminated. In elemental fluorine, the two fluorine atoms sha-retheir unpaired electrons in the form of a bond. This means that the fluorine molecule has no unpaired electron-iChoice B is eliminated. Because chlorides are -1 each, cobalt in CoCl2.6 H2O carries a +2 charge. This is theresult^of l-osi1g two 4s-electrons from elemental cobalt, which results in an electronic configuration for Co2+ cqb22s22p6g"2gO6g6z. There is anodd number of electrons, so the species mustbe paramJgnetic. The correc:choice is answer C. Just as was the case with F2, molecular hydrogen (H2) has no unpaired electrons.

Choice A is correct. Atomic nitrogen (N) has the electronic configuration 1.s22s22p3. Thu last electron is ttuthird electron to enter the 2p orbital. The principle quantum number is given as 2, and because it is a p-orbita*the angular momentum quantum number (l) is equal to 1. All of the answer choices contain these two values, ;:nothing is eliminated. To obtain the m7 and m, values, the electrons must be filled into their respective r-orbitals. This is drawn below:

i,

1"

Fe(CN) 53-

cis-Pt(NHr)2C12

1"ml= -L ml= 0

Last electron is in the third p-orbital, so 14= +1'Last electron is spin up, so *, = * |

The correct answer is therefore choice A.

23. Choice A is correct. Having six ligands attached results in an octahedral shape. The correct answer is choireA. Drawn below are generic structures for the other answer choices.

N3-C

I

I ....,.*CN

ffi;=TCN

Octahedral

L

I ...*'

T-'L

24. Choice C is correct. Polarity results from the asymmetric distribution of electron density. Cis compoundsalways polar. The correct answer is choice C. The structures for the four choices are drawn below.

Generictetrahedral Generictrigonalbipyramidal

.|r=r"'d\Jh

.1-CI

"ll-z'"s1?trans-Pt(NH3)2C1"

I ..'nrCI

':,/ l"- ''CI

FeC13-

"""tot L

L=7w-r

Generic square planar

.1-CN

I

NC-i"s:ilI

CN

'1"

ml=+7

LI

I

L/\,'L

Copyright @ by The Berkeley Review@ 15() Section II Detailed Explana

25.

cr3*: I I

n\=-2 rr\=-1 n\=0 11\=+1 11\=+2

2 unpaired electrons

Choice B is correct. To determine the number of unpaired electrons (as well as which one has the most), the d-electron count for each transition metal cation first must be worked out. To determine the d-electron count forthe^ four cations, you must consider the electronic configuration. Neutral chromium is [Ar]4s13d5, so when it isCr3+, it has lost the_4s-electron and two 3d-electrons, leaving iJ with an electronic configuration of [Ar]3d3. Werefer to Cr3+ as a d3-cation. Neutral manganese is [Ar]4s23-d5, so when it is Mn2+, it has lost both 4s-electron,leaving it_with an electronic configuration of [Ar]3d5. We refer to Mn2+ as a ds-cation. Neutral copper is[Ar]+st3dru, so when it is Cu+, it has lost its 4s-electron, leaving it with an electronic configuration of [Ar]3dro.We refer to Cu+ as a d10-cation. Finally, neutral nickel is [Ar]4s23d8, so when it is Ni2{, it has lost both 4s-electron, leaving it with an electronic configuration of [Ar]3d8. We refer to Ni2+ as a d8-cation. From here, itis a matter of placing the d electrons into their respective orbitals. The electron filling is as follows:

*., 11, 1L l_L l_L l_Lmr=-2 r\=-1 n!=0 rr\=+1 n!=+2

0 unpaired electrons

lrrfr2*: l_ l_ l_ l_rr\ = +2

*i., 1 I l_L i_L l_ l_

\=2. r\=-1 n\=0 n\=+1 11q=+2

3 unpaired electrons

n{=-1 ntr=0 r\=+15 unpaired electrons

l_mr=2

The most unpaired electrons is found with Mn2+, so choice B is the best answer.

Choice D is correct. The cyano ligand carries a -1 charge, so chromium must have a +6 charge in order forCI(CN)6 to be neutral. The amino ligand is neutral, so chromium must have a +3 charge in order for CI(NH3)6to have an overall positive three charge. The charge of chromium is in fact greater in Cr(CN)6 thanCr(NH3)6, so choice A is valid. The water ligand is neutral, so copper must have a +1 charge in order forCu(H2O)6 to have an overall +1 charge. When copper is neutral its electronic configuration is [Ar]4r1r410, sorvhen copper carries a +L charge, it has electronic configuration [Ar]3dru. This gives copper ten d-electronsn'hich makes choice B valid. In both Fe(NH3)63+ and FeC163-, iron has six ligands attached, so the shape ofboth molecules is the same. Choice C is valid. Scandium has the electronic configuration [Ar]4s23d1, so it canlose only three electrons. It is not possible for scandium to have a +4 charge (at most it is +3). This means thatchoice D is not true. Pick choice D, and feel a little peppier because of it.

Choice B is correct. To determine the number of electrons on the central metal, the formal charge of the metaltirst must be determined. From the charge, the electronic configuration is found, so the d-electron count is found.Fe carries a +3 charge in both Fe(NH3)63+ and FeCl53; so choice A is eliminated, because they both must havethe same d-count. Co carries a +3 charge in Co(H2O)63+, and Mn carries a +2 charge,in MnCl64-. Neutral Co is

[Ar]+s23d7, so Co3+ has a d-electron count of 6 (3d6). Neutral Mn is [Ar]4s23d5, so Mn2+ has a d-electron count of5 (3d5). The d-electron counts are not equal, so cloice B must be the correct answer. Cr carries a +3 charge inCr1NH3;6a+, and. Mo carries a +3 charge in MoCfu3-. Neutral Cr is [Ar]4s13d5, so Cr3+ has a d-electron count of: leae;. Neutral Mo is [Kr]5s14d5, so IiIo3* has a d-electron count of 3 (4d3). The d-electron counts are equal so

choice C is eliminated. Finally, Os carries a +2 charge in Os(NH3)42+ and Rh carries a +3 charge inRhCl3(PR3)3. Neutral Os is [iie]6124114546, s,o Os2+-l'rut u d-electron count of 6 (5d6). Neutral Rh is

IKAS&4y7, so Rh3+ has a d-electron count of 6 (4d6). The d-electron counts are equal so choice D is eliminated'

:S Choice A is correct. The relative sizes of ions from smallest to largest radius is: Li+. Mg2* < Na+ < Ca2+ < K+'The rate of effusion is: Li+ t Mg2* > Na+ > Ca2+ > K+. Li+ (the smallest) is the fastest while, K+ (the largest)is the slowest. The best answer is A. Choice D is eliminated, because cations are smaller than neutral atoms.

:P Choice C is correct. Every trend in the periodic table comes back to the nuclear pull on the electrons. The

ionization potential is the energy required to remove an electron from the outermost shell, which in the case ofsodium and potassium is the 3s and the 4s, respectively. It requires less energy to remove an electron from the

larger 4s-orbital, because the electron is farther away from the nucleus than the 3s-orbital. This makes choice

C the best choice. Choice D is eliminated, because potassium has a lower electron affinity than sodium.

r5l Section II Detailed Explanations

31.

30. Choice C is correct. The size of an ion or element is a result of the nuclear pull on the electrons orbiting thenucleus. This is best explained by choice C. Because C (carbon) has six protons, F (fluorine) has nine protons,and both have the same principle quantum number for their valence shell, the nuclear pull of fluorine is greaterthan that of carbon. Because the electrons are pulled closer to the nucleus in fluorine, and because size of anatom is determined by the electron cloud, fluorine is smaller than carbon.

c:'J,s2zs22p2 F: 1,s2zs2zp5

Choice C is correct. F-, Ne, and Na+ all have 10 electrons total, so Cl- (with 18 electrons) is eliminated. Of thethree choices left, the largest nuclear charge is found on Na+, making it the smallest (the one with theelectrons held most tightly). Choose C to choose correctness. The following chart of the protons and electrorsfor the three choices shows that the greater the proton-to-electron ratio, the smaller the species, assumingthat the outermost electrons are in the same valance shell.

Element Protons Electrons Observation Radius

Na+ 11 10 protons exceed electrons, therefore it contracts 65pm

Ne 10 10 protons equal electrons 70pm

F- 9 10 electrons exceed protons, therefore it expands 136 pm

Choice A is correct. The electronic configuration for magnesium is 1s22s22p63s2. Magnesium must lose lelectrons (the two 3s electrons) to have a filled outer valance shell (the n = 2 shell). This would makemagnesium a +2 cation. Answer choice A is a fine selection if your goal is to be correct.

Choice D is correct. According to periodic trends, the size of an atom increases as you descend a column in tlreperiodic table. I (iodine) is the lowest in the periodic table of the halogen choices, so I is the largest of tlrehalogen choices. Trust periodic trends and choose D.

Choice D is correct. If the pore were to distinguish by charge, then theslower the rate of migration for the cation. For example, if the pore

greater the charge of the cation,were capable of forming attractiwe

interactions with the cations such as polar attraction orhydrogen bonding, then the pore would distinguish hry

charge. The cations with the greatest charges are Mgzi ^nd"Cu2*.

gec"ause Ca2* is larger than Mgf*, Cai-wo.tid migrate more slowly than Mg2+ thiough a pore which distinguishes by both sile and chaige. Th

means that overall , CaZ+ would have the slowest migration rate. This makes choice D the best choice.

35. Choice C is correct. The electronic configuration for helium is 1s2. The two electrons of helium are bothelectrons, so choice A is eliminated. Helium cannot have an effective nuclear charge greater than2, becausecontains only two protons. Choice B is thus eliminated. The ionization energy does not includevaporization energy. Helium is a gas at room temperature, so there is no need to add energy to vaponhelium. This eliminates choice D. The electron must be removed from the first quantum level. The fiquantum level experiences the greatest nuclear attraction, so the Ls-electrons are hardest to remove. It r-s

your best interest to choose C.

36. Choice D is correct. The second ionization energy of the alkali metals (lithium, sodium, and potassiumsubstantially larger than the first ionization energy, because the second electron is being removed from a

octet. After the first electron has been lost and the alkali metals are cations, their electronic configurationtrr2^p6, and they each have a filled valence shell. To remove the second electron would be like removingelectron from a noble gas, only harder, because the nuclear charge is greater for the alkali cation thanneutral noble gas. Given the answer selections, choice D is the best choice.

32.

JJ.

I

i{

0

u

34. (ilfl

fr

(dTfru

Illti

q,*

ilh

Gtuts

Copyright O by The Berkeley Review@ 152 Section Il Detailed Explana

i]i.

Choice D is correct. As stated in the passage, there is half-filled stability for the p-level when each of the p-orbitals contains one electron. Nitrogen as a neutral element has half-filled stability, so when it is ionized, itloses its half-filled stability. Oxygen, on the other hand, has one electron beyond the half-filled state;therefore, when oxygen is ionized, it attains half-filled stability. Losing half-filled stability raises theionization energy of nitrogen (ionization is less favorable), while gaining half-fitled stability lowers theionization energy of oxygen (ionization is more favorable). Nitrogen is less electronegative than oxygen, sochoice A is eliminated. Nitrogen has seven protons, while oxygen has eight, so choice B is eliminated.Nitrogen is in fact larger than oxygen, but a larger radius implies that the electrons are farther from thenucleus on average. Being farther from the nucleus, the electrons are not as tightly held, so the larger radius ofnitrogen would i*ply a lower ionization energy for nitrogen than oxygen. The best answer is choice D.

Choice A is correct. The ionization energies for aluminum, silicon, and phosphorus follow an increasing lineartrend. All three elements are in the same row (period) of the periodic table, so the number of core electrons forall three is the same. This eliminates choices C and D. As you move from left to right across a period of theperiodic table, the atomic number increases, so the number of protons increases, and ultimately the effectivenuclear charge increases. The increasing nuclear charge best explains the trend in first ionization energybetween aluminum, silicon, and phosphorus. Be a hero or heroine by choosing A. Note that there is no half-filled or unfilled p-orbital stability affecting the observed trend.

Choice C is correct. Fluorine and chlorine are in the same column (family) of the periodic table. Fluorine hasits valence electrons in the n = 2 quantum level, while chlorine has its valence electrons in the n = 3 quantumlevel. The larger the quantum number, the easier it is to remove the electrons and thus the lower the ionizationenergy. Fluorine is more electronegative than chlorine, so choice A is wrong and eliminated. The effectivenuclear charge is found from both the number of protons and the number of core electrons. It is true that chlorinehas a greater nuclear charge than fluorine, but chlorine has more core electrons than fluorine. This impliesthat the effective nuclear charge is approximately equal for the two halogens, eliminating choice B. Thelarger radius of chlorine implies that the valence electrons are farther from the nucleus on average than thevalence electrons of fluorine. This makes the ionization energy of fluorine greater than that of chlorine andconsequently makes choice C the best choice. Choice D is nonsensical, so ignore it.

Choice B is correct. The term "oxidation" refers to the loss of an electron. The lower the ionization energy, theeasier it is for an element to lose an electron. The chart lists the first ionization energies, therefore it can beinferred that the element with the lowest first ionization energy is MOST easily oxidized. Of the choices,magnesium has the lowest first ionization energy. Choice B is the choice of wirrners... be a winner.

Choice B is correct. The relative ionization energies of krypton, bromine, and selenium are predictable, becausethey are in the same row (period) of the periodic table. By comparing the three to chlorine, argon, and sulfur,a trend can be determined. The ionization energies of argon, chlorine, and sulfur follow: I.E.a, > I.E.C1 > I.E.S,so the first ionization energies of bromine, krypton, and selenium should be LE.6, > I.E.B, > I.E.S", choice B.

Choice A is correct. \vVhen sodium loses its first electron, it gains a filled octet and thus stability in its valenceshell. If it were to lose the second electron, the octet would be lost and thus it would become an unstable cation.This explains the drastic difference between the first and second ionization energies for sodium. For magnesiumto have a full octet, it must lose two electrons resulting in relatively low first and second ionization energies.This answer is best explained in answer choice A. The following illustrates the point:

1st ionization: Na(g) -----> Na+(g) + e- Na: 1s22s22p63s1 and Na+: 1s22s22p5 Na+ has a filled octet.

2nd ionization: Na+(g) --> Nu2*(g) + e- Na+: 1s22s22p6 and Na2+: 322t22O5. Na2+ has lost filled octet.

Lst ionization: Mg(g) * Mg+1g) + e- Mg: Ls22s22p63s2 and Mg \s22s22p63r1. Mg+ has no filled octet,

2nd ionization: Mg*(g) ------> vtg2+1g; + e- Mg+:1.s22s22p63s1and Mg2+: :f.22s22p6. Mg2+ has a filled octet.

4i Choice B is correct. Table 1 lists room temperature density values of the transition metals. As a metal isheated, it expands. The density of manganese decreases with increasing temperature, because volume increasesas mass remains constant. The density is slightly less than 7.43 grams per cmr, so the answer is choice B.

-:pvright @ by The Berkeley Review@ 155 Section II Detailed Explanations

44. Choice D is correct. The first, second, and third rows of the transition metals follow similar trends as you move

left to right across any particular row. The metals from the answer choices are all in the second-row of the

transitions metals. No information is given for the second-row transition metals, so their relationship must be

extrapolated from the information given for the first-row transition metals. The first-row transition metals

follow the trend Cu > Ni > Cr > Mn for the second ionization energy. According to periodic behavior trends, Moand Cr should exhibit similar properties, Tc and Mn should exhibit similar properties, Pd and Ni should

exhibit similar properties, and Cu and Ag should exhibit similar properties. This means that the correct

relationship between the 4d-transition metals is found by substituting the second-row transition metals inbthe relationship for the first-row transition metals. The relationship is Ag > Pd > Mo > Tc, making choice Dthe best answer.

45. Choice B is correct. The boiling point and melting point of an element increase as the forces holding the at,

together increase. The greater the forces, the greater the energy required to break the forces. Because zinc

u i-o*", boiling and meiling point than other transition metals, it can be concluded that there are weaker fo

holding zinc itoms togethe? than the other first-row transition metals. Because zinc has a filled d-s

([Ar]+s23d10;, there sholld be no covalent interactions between the zinc atoms. This makes choice B the cor

choice.

46. Choice C is correct. All of the transition metals listed in Tabte 1 lose their 4s-electrons prior to losing their

electrons, implying that it is easier to lose the s-electrons. Because the 4s-level fills prior to the 3d-levelcan also be concluded that 4s-electrons are at a lower energy level. hr theory, the 3d-electrons should requi

less energy to remove (since the 3d-level is of higher energy than the s-level). However, the 4s-electrons a

*ot" u*posed (further from the nucleus), so they are lost more easily than 3d-electrons. This makes choice

the best answer.

47. Choice B is correct. Within a row of the periodic table, atomic radius decreases as you move left to right

main-group elements. Table 2 shows that it is nearly true for the first row of transition metals, with zi

showing the only notable deviation. As you move left to right across a row of the periodic table, the mass

the elements increases. The density of an element is measured in terms of mass per volume. As the aton

radius increases, the volume increases, so moving left to right across a row in the periodic table results

greater mass and reduced volume. This means that the density of the element increases from left to right in

ieriodic table. Data in Table 2 confirms this inverse relationship, except in the case of manganese, whele

iadius and density both increase. This eliminates choices A and C. Manganese is an exception, because itIarger radius due to half-fiiled stability, but it follows the same trend with density because its mass is g

thin chromium (the previous transition metal). Zinc follows the trend of greater radius associated with ldensity, so it is not in exception to that rule. Choice D is eliminated, because a general trend is observ

despite the deviation due to manganese. Choice B is the best choice.

48. Choice A is correct. \zVhen a transition metal element becomes a cation, it loses electrons from the 4s-level

electrons are lost from the outer shell (valence shell), the radius must become smaller. As a note, the

nuclear charge is increasing, because the cation experiences less valence electron repulsion with the a

valence electrons. The best answer is choice A.

49. Choice C is correct. A negative reduction potential implies that it is less favorable to reduce the dication

it is to reduce hydrogen ioi (a proton). Th-e unfavorable nature of the reduction correlates to a low ioniz

energy. The easier it ir to ionize (lose two electrons to become the dication), the easier it is to oxirlize

element. The easier it is to oxidize the element, the harder it is to reduce the dication that forms. The sur

the first and second ionization energies of Mn is 2226 kl/mole, and the reduction potential is -1.L8 volts-

sum of the first and second ionization energies of Co is 2405kJ/mole, and the reduction potential is -0.28

The sum of the first and second ionization energies of Fe is 232Okl/mole, which is closer to the value of

(differs by 85 kjlmole) than the value of Mn (differs by 94ld/lnrtole). The reduction potential should be br

-1.18 voltl and -0.28 volts, but closer to -0.28 volts. The best answer is choice C. (The actual value for any

electromotive force trivia buffs is -0.44 volts.) The data are given below:

N&r -+ Mn2+ + 2 e- L lst and 2nd I.E. =2226 kj/mole E'oxidation = L.18 V .'. E"1s4r6tion = -1-

Fe --+ Fe2+ +2e ; 1st and 2nd I.E. =2320kjlmole E'oxidation = ??? Y i' E"reduction= ???

Co + C&+ +ze- L lst and 2nd I.E. =2405 kJlmole E'oxidation =0.28 V ;. E"1s6r6ti.. -

Copyright @ by The Berkeley Review@ t54 Section II Detailed

i0. Choice C is correct. It is grunt work when it comes to figuring out electronic configurations! You can saveyourself some time by eliminating choices with either too many or too few electrons. Aluminum has thirteenelectrons, as do all of the choices. This means that no choices are eliminated. Choice A is out, because an s-

orbital can hold no more than two electrons. Choices B and D are eliminated, because the s-orbital fillscompletely before the p-orbital begins to fill. The correct answer is choice C. If it is needed, the Aufbauprinciple can be applied to determine the filling order.

Choice D is correct. Monochromatic light is light that is composed of photons with exactly one wavelength (orfrequency). This would occur if every photon emitted came from exactly the same electronic transition betweenenergy levels. However, not all electrons are at exactly the same energy level, due to the close proximity ofrotational and vibrational levels. As a result, not every electron undergoes the exact same energy transition,and thus not every photon that is emitted has the same energy (and thus neither the same frequency norwavelength). This is stated near the end of the passage. Choose D to be a stellar student.

Choice B is correct. This question requires determining the photon's energy from its wavelength

(6.6x1"0-3aJ.sec x 3.0 x 1084) _ 19.8 x 10-26J.m r E=+*r1o-1elJ.J330 x 10-9m 3.3 x 10-7m

The more important part of the answer is the power of ten. From that, you must select choice B.

Choice A is correct. The shorthand for the electronic configuration of vanadium is [Ar]4s23d3. For the quantumnumbers of vanadium, we are concerned only with the last electron in vanadium (V). We are thereforeconcerned with the third 3d-electron. Drawn below is the 3d level:

c-hc-l.

1o

3d l_ffil=-z

1" 1'ml= o

Lml= 0

4s

lr

r

n

ffil =-1 ml=+1 fll= +2

Last electron is in the third d orbital, so rn1 = 0; Last electron is spin up, so

lhe last electron is in the n = 3 level, because that is given by gd3. The electron is in the 1 = 2 level, because

that is given by d-orbital in 3d3. The m1 and ms values must be derived by filling electrons into their respective

-er-els. As listed in the box, rrll = 0, ms = +! / 2, which is choice A.

,rJr- Choice D is correct. Half-filled stability would come into play for atoms that can promote one electron from::e lower energy s-orbital up to the higher energy d-orbital to yield a d5 species. The term "half-filleds:ability" is derived from the half-filled d-level (the d-level has a maximum occupancy of ten electrons).:{alf-filled stability is possible for only chromium (Cr), molybdenum (Mo), and tungsten (W), because they are

:-l in the column that should be s2d4. To be a quality chemistry student, pick choice D. Copper can excite one s-

.-ectron to fill the d-level completely (to make the species a d10 atom).

frffiiii ,Choice B is correct. Whatever goes in, must come out for normal behavior of light. There are situations where: high-energy photon is absorbed that results in the excitation of an electron through multiple levels. From:ere, the electron may relax (fall back to the ground state) by a variety of pathways (either gradually::-isipating its energy or releasing the energy all at once). As a general rule, the energy that is absorbed is also

:ritted. Pick choice B. The exception to this rule is phosphorescence. With a phosphorescing species, light:,:.ergy is absorbed, and the compound undergoes a change in its electronic structure. The excited state is:.erefore a different complex than the ground state. \Atrhen the new complex gives off energy to fall to its-ound state, the photon released is not of the same energy as the original photon absorbed. Phosphorescencerd fluorescence were passage topics on a previous MCAT. Quantum numbers were also a passage topic.

*r=+ j

,n- ::fht @ by The Berkeley Review@ 155 Section II Detailed Explanations

55. Choice C is correct. Visible light is found approximately in the range 390 x L0-9 m to 740 x 10-9 m. To be usedthe formula provided, the wavelength must be in terms of 10-6 m. Visible light has a range from 0.39 x 1,0-6 m

58.

0.74 x 10-6 m. The value for this range in terms of eV is 1"24 / 0.74 sy b 1,.24 f g.39 eV. This is a range fromnumber just over 1.5 to some number barely greater than 3.0, a range that fits choice C.

57. Choice A is correct. The reflected color of the four pigments are yellow for alizain, blue for azurite, redcinnabar, and blue-green for verdigris. To determine the light absorbed, you must take the complementaryof the color observed. The absorbencies are thus violet for alizarin, orange fot azutite, green for cinnabar,orange-red for verdigris. The shortest wavelength is associated with the highest-energy light. Violetthe highest energy in the visible spectrum, so it has the shortest wavelength. The correct answer is choice A

Choice A is correct. When a sample is heated, its electrons are thermally excitedWhen they relax back to their ground state, light is emitted at an exact wavelength.specific color, best described by choice A.

to a higher energyThis results in light of

59. Choice D is correct. The absorption spectrum contains black lines where the complementary colors of or(the colors reflected by realgar) should be. The complementary colors are blue-green, so the best ans

selection is choice D.

60. Choice C is correct. The reflected color of verdigris (Cu(C2H3Oz)z.Cu(OH)2) is green-blue, so thecolor is red-orange. Red and orange lie at the low end of the visible spectrum as far as the energy isso the wavelength lies at the high end of the visible spectrum. The spectrum goes redblue-violet, so the wavelength of orange light falls below 700 nm, and the wavelength of red is around 700

This means that the wavelength is less than 700 nm by a small amount. The best answer is 650 nm,makes the best answer choice C. The range of visible light in the EM spectrum is given on page 1L3.

61. Choice D is correct. Because white lead appears white in color, no light has been absorbed in the visibleWhen no light is absorbed, white light (the incident light) is reflected. The d-d transition associatedwhite lead must lie outside of the visible range. The best answer is thus choice D.

52. Choice D is correct. A reflected color can be seen only when white light is reflected off of it. This meansreflected colors cannot be seen in the dark (absence of white light). Candles cannot be seen in the dark, socolored candle is made from a wax that contains a dye that exhibits reflected color. Clothes cannot be seen ithe dark, so a fabric dye exhibits reflected color. If you wish to argue that there are certain glow-in-fabric dyes, you're absolutely right. Glow-in-the-dark dyes exhibit emitted color. You are wise in the wavstrivia. Unfortunately, you get zero credit on this question, because it is not the best answer. Knowing a

special cases like that is a great way to impress your peers, but it hurts you on a standardized exam. Inkpens cannot be seen in the dark, so ink exhibits reflected color. Gas-filled light tubes (i.e., neon lights) can

seen in the dark, so a glowing gas-filled light bulb exhibits emitted color. The best answer is choice D.

Choice A is correct. The fastest decomposition is observed in the organic pigment. This is stated inpassage. The inorganic pigments are already oxidized, so they should remain air-stable for some time.organic pigments can oxidize in air, so they do not last as long. The only organic compound among the choi

(and in Table 1) is 1,2-dihydroxyanthraquinone, which is the pigment of alizarin. This makes choicecorrect. Note that organic pigments contain conjugated n-networks and are found in oil-based paints.

U. Choice B is correct. As stated in the passage, a spin flip is associated with the process of phosphorescence.the ground state, two electrons in the same orbital must be spin-paired (have opposite spins). Once an e

has been excited to a higher electronic level where it occupies the orbital alone, it is free to flip itsEqually, the electron that remains in the ground state may also flip its spin. A spin flip changes theenergy of the system. The excited electron may not be able to relax back to its original level, since it sharessame spin as the lower level electron now. \A/hen the electron falls back to a lower level, it falls to a difenergy level, which emits a different frequency of light than it absorbed. Whether or not you knowphosphorescence is, the answer is given in the passage. The best answer is choice B.

53.

IItt

(f€

t1

-F

e

(ue!'!

B

Copvright @ by The Berkeley Review@ 156 Section II Detailed

55. Choice A is correct. A diatomic molecule can gain and lose energy in several ways, including changingelectronic levels, vibrating at different frequencies, and rotating at different frequencies. In a diatomicmolecule; when an electron is excited, the valence shell increases, which affects the bond length and bondstrength. The diatomic molecule is more susceptible to changing its vibrational frequency. This is the physicalcause of fluorescence. \Atrhen an electron in the diatomic molecule is excited, a photon is absorbed. Becauseenergy is dissipated from the excited state in the form of rotational and vibrational energy, the molecule is in alower energy state. When it finally relaxes back to its ground state, the energy of the photon emitted is lessthan the energy of the photon absorbed. The energy of the photon absorbed is equal to the energy of the photonemitted plus the dissipated energy (vibrational and rotational transitions). This eliminates choices B and D.As the wavelength of a photon increases, it has less energy. This implies that the photon emitted @eing ofless energy) has a longer wavelength than the photon absorbed. This makes choice A correct.

Choice C is correct. "Because the electronic energy levels are not singular levels and because the transition israndom, it is impossible for a molecule to absorb or emit light in such a manner that all of the photonssimultaneously have the same frequency (or wavelength). For this reason, monochromatic light is notphysically possible." These last two sentence from the passage state that monochromatic light (light of onewavelength) is not possible, and give the reasoning for that. Because monochromatic light is not possible,choices B and D are eliminated. The reasoning has to do with energy levels, not the quantization of light, sochoice C is the best answer. This question about a difficult concept is actually an easy question to answer, if youdon't get intimidated. Leaming to manage the intimidation factor associated with seeing new information is apart of your MCAT preparation.

For the sake of learning the concept, we shall look at what choice C is stating. The diagram below shows twoscenarios, one where electronic transitions are not coupled with vibrational transitions (on the left) and theother one where electronic transitions are coupled with vibrational transitions (on the right).

Electronic Transition (no vibrational transitions) Electronic Transition (with vibrational transitions

EExcitedEExcitedV2EExcitedVtEExcitedVo

EGroutrd

EGrou,.dV2EGrorrr,dVlEG.orr.rdVo

l'ti.

Electronic transitions occur without vibrationaltransitions, so a single energy transition is possible.Only one photon is emitted, which would result inmonochromatic light.

Electronic transitions occur withvibrationaltransitions, so a multiple energy transitionsare possible. Multiple photons are emitted, sothe emission is polychromatic light.

From the diagram, it can be seen that when vibrational energy levels are closer together than electronic energylevels, the transitions can couple. Single transitions between electronic levels are not possible, although singletransitions between vibrational levels appear to be possible. However, vibrational transitions couple withrotational levels, so inJrared emissions are not of single wavelength. The best answer is choice C.

r.f Choice B is correct. The arrow in Figure 1 that represents emission due to fluorescence is longer than the arrowthat represents emission due to phosphorescence, so the energy associated with fluorescence is greater than theenergy associated with phosphorescence. This means that the light from fluorescence is of shorter wavelengththan the light from phosphorescence. Choice B is the best answer. Choices A and D are the same answerworded differently (if fluorescence emission is of longer wavelength than phosphorescence, thenphosphorescence must have emission of higher energy than fluorescence), so both choices should have beeneliminated (assuming as we do that there is only one best answer per question).

rlrrlt' Choice A is correct. The shortest wavelength of light belongs to light of the greatest energy. This occurs withthe transition from the highest excited state relaxing to the lowest ground state. Choices C and D areeliminated immediately, because the transitions they represent are increases in energy, which absorb light,not emit light. The best answer is choice A, because the excited state is the highest of the choices left (A andB), and both of the choices left drop energy to the same ground state.

157 Section II Detailed Explanations

59- Choice B is correct. Fluorescence is possible with molecules, because molecules can exhibit vibrational energltransitions. This allows for the dissipation of energy via of heat. Choices A and C true, so they areeliminated. Because atoms do not have vibrational energy transitions (they have no bonds, so they have ncbending and stretching modes of their bonds), atoms may not necessarily exhibit fluorescence. It is possible tcconvert ultraviolet light to the lesser-energy visible light by fluorescence, but it is not possible to convervisible light to the higher-energy ultraviolet light by fluorescence. This makes choice B the false statement.

Choice D is correct. The definition of paramagnetic is having at least one unpaired electron. The electron:rconfiguration for Na+ is 7s22s22p6. All electrons_ are, pqire{, b-ecause the octet is complete, eliminating choi:eA. The electronic configurationior Sr2+ is7s22s22p6it29064t236704r6. Ail electrons are paired, because tluoctet is complete, elimiiating choice B. The electronic configuration for Cu+ is 1,s22s22p6is23063410 (copF*shas filled d-shell stability, and it loses its 4s-electron before its 3d-electrons). All electrons are^paireabecause each level is filed, ehminating choice C. The electronic configuration for Cr3+ is k22s22p6gs2gp6aO3(first-row transition metals lose their s-electrons before losing their 3d-electrons). Not all electrons can repaired, because there is an odd number of electrons. Because Cr3+ is paramagnetic, choice D is the best answer.

Choice B is correct. Chromium has half-filled d-shell stability, giving it an electronic configuration cn

7s22s22p69t291647965. The correct choice is answer choice B. Without the half-filled stability, the ansrrmwould have been choice A. Molybdenum (Mo) and tungsten (W) also exhibit half-filled stability.

Choice D is correct. The largest transition is associated with the greatest energy. Vanadium dication (\-:-rproduces violet light, therefore the highest energy is associated with vanadium dication. Pick choice D.

Choice C is correct. It can be observed from the data in Table 1 that as the first or second columns in rPperiodic table are descended, the light emitted frorn the transition is of progressively higher energy. Ti,:m

makes statement I a true statement. Because statement I is not included in choices B and D, choices B and D aeeliminated. It can be deduced from the answer choices that rernain that statement III must be false. To verly'this, copper has the electronic configuration[Ar14s13d10, so Cu+ has the electronic configuration [Ar]3d:r'

f'F.il-

70.

71.

72.

/5.

Nickel his the electronic configuratioir [Ar]4s23d8, so-Ni2+ has the electronic configuration [Ar]3d8. StateneIII is in fact a false statement, because Cu+ and Ni2+ do not have the same electronic configuration. Itransition for potassium cation (K+) emits purple light, while the transition for chromium trication (Cr-

yields red ligtrt. The transition for K+ is of high"t "r"rgy

than the transition for Cr3+, making statemen:true and choice C correct.

74. Choice B is correct. Nickel dication (Niz*, has the electronic configuration"J,s22s22p63s23p636s. *"electron is the eighth electron in the 3d-orbital. The principle quantum number is given as 3 and being in a

orbital makes I equal to 2. From the answer choices, this is already known. To obtain the m1 and mr values,electrons must be filled into their respective d-orbitals. This is drawn below:

N,2*, 'lft\=A'

"1rn\=-1

'lfrr\=0

'l_n\=+1 n!=+2

"1

Last electron is in the third d orbital, so n! - 0; Last electron is spin down, so ms =

The correct answer is therefore choice B.

75. Choice D is correct. Because violet light is higher in energy than orange-yellow light, which is in tumenergy than red light, the electronic transitions for the group I cations increase as the column is descended.

transition for rubidium should therefore be of greater energy than violet light, which makes the tran-':emit ultraviolet light. Ultraviolet light is not detected by the human eye, so the flame from heatingshould appear coloriess. The best answer is choice D.

I2

I

jJ

il

'n

Copyright @ by The Berkeley Review@ l5a Section II Detailed Ex

76, Choice C is correct. An emission spectrum shows just the color emitted by the compound after it has beenexcited. The color observed in the flame test is an emitted color, so it is present in the emission spectrum.Crimson was emitted by lithium cation, so the emission spectrum is simply a bright red line. The best answer ischoice C. Other minor emissions may be seen, but they won't be as intense as the red emission.

:3.

Choice C is correct. Two isotopes (of neutral elements) have the same number of protons and electrons, but adifferent number of neutrons. Choice A has one more proton than phosphorus-32, choice B has one more electronthan phosphorus-32, and choice D has one electron less than phosphorus-32. This eliminates choices A, B, andD. Choice C has a mass that is one less with the same number of protons as phosphorus-3Z, so it must have oneneutron less. This makes choice C the correct answer.

Choice B is correct. Nitrogen in neutral molecules makes three bonds total (in this case, the three bonds are allsigma bonds), and has one lone pair of electrons. The lone pair of electrons repels the electrons in the threesigma bonds to form a trigonal pyramrj{al orientation about the nitrogen. This can be confirmed when lookingat the hybridization of nitrogen (sp3). The best choice is therefore answer B. Drawn below is a threeldimensional picture of Glyphosate with the nitrogen isolated:

ru.

H/dXt*Co'-Na+-crqoeorFf Nu*

Choice C is correct. The concentration of 32P label at consecutive half-life points along the first-order decay is:188 -+ 94 -+ 4 -+ 23.5. Each arrow represents one half-life, so after three half-lives the concentration is lessthan 25 ppm (it has decayed to 23.5 ppm). To reach a concentration of 25 pprn, it takes a little less than threehalf{ives. The best answer is a little less than 42 (3 x 14) minutes. Choice C, 41 minutes, is the best answer.

Choice C is correct. An ionic compound is made up of ions. The quickestway, without just knowing the answer,is to look for metals such as sodium. Ammonia is held together by three covalent bonds (sigma bonds). Carbondioxide is held together by two covalent bonds (two doubles bonds made up of one sigma bond and one n-bondeach). It is only in compound III, Na2PO3H, that we find ionic bonds. Choice C, III only, is the best answer.

Choice A is correct. Because 32P remains chemically equivalent after nuclear d.ecay, it is phosphorus. Thismeans that a proton was neither gained nor lost in the process. \Atrhen a neutron is lost, a phosphorus-32 isotopebecomes a phosphorus-31 isotope. This makes choice A the best choice listed. An alpha particle contains twoprotons and two neutrons, so the loss of an alpha particle would form aluminum, eliminating choice B. The lossof a beta particle converts a neutron into a protory which would form sulfur, eliminating choice C. The loss of apositron converts a proton into a neutron which would form silicon, eliminating choice D. The answer choicesdid not list a gamma ray. A gamma ray is a high-energy photon that, when given off, does not change thechemical behavior, either. This was not listed as a choice, but it is food for thought.

m- Choice A is correct. Beta decay is the loss of an electron from the nucleus. No mass is lost, therefore the massshould not change. Choices C and D are eliminated. Charge must be conserved, so losing a negative chargeshould increase the atomic number by 1. This makes choice A the best answer. The reaction is shown below:

2$!ro+-!e+2$$et

ffi, Choice C is correct. An alpha particle is a helium nucleus (mass = 4 amu and z = 2), so the loss of an alphaparticle decreases the mass by 4 and the atomic number by Z. 250Sp is 4 mass units less than 254Es, so choice Ais eliminate4. 23a11t is 4 mass units less than 238U, so choice B is elimina L4. 22rgi is only 2 mass units lessttran 223Fr, so choice C is the best answer . 243pu is 4 mass units less 11'rvyr247grr, so choice D is eliminated.What else but berkeliuru could be right?

tr{il.

:qi.nght @ by The Berkeley Review@ 159 Section II Detailed Dxplanations

84.

85.

Choice C is correct. From the passage, the half-life is given as six hours. The twelve-hour duration is

therefore a total of two half-lives. The initial concentration should therefore be cut in half two successivetimes to determine the final concentration. The math is as follows: 120 pCi -> 60 pci

-> 30 pCi. The finalreading is 30 pCi, so the correct choice is choice C.

Choice B is correct. Electron capture by a nucleus decreases the positive charge (converting a proton into a

neutron), which reduces zby 1,. This converts element # 6 (carbon) into element # 5 (boron).

tF * -or" -+ 1s2B

The question is asking for a nuclear process that forms boron. Positron capture increases the nuclear charge brone, so positron capture by boron cannot yield boron. It yields carbon, one atomic number higher, whicheliminates choice A. Positron decay decreases the nuclear charge by one, so positron decay by carbon yield.boron, one atomic number lower. This makes choice B the best answer. Beta capture decreases the nucleecharge by one, so beta capture by nitrogen yields carbon, one atomic number lower. This eliminates choice C

Alpha decay decreases the nuclear charge by two, so alpha decay by oxygen yields carbon, two atomic numbers

lower. This eliminates choice D. The best answer is choice B, the only choice that didn't form carbon. The

processes in choices A,B, C, and D are shown below.

ChoiceA:1!B+!s+1[C;ChoiceB,lp.-+13B+t8;Choicec:1fN+-fe+1fC;Choiceo:1$o+1frC+3n

Choice D is correct. A gamma ray is high-energy electromagnetic radiation, not a particle. The energlassociated with a gamma ray is greater than the energy associated with an x-ray. Because it is a photc:,(energy) and not a particle, a gamma ray is massless and without charge. \A/hen the nucleus of what emit-' a

gamma ray, it drops from a nuclear excited state to a lower (and possibly ground) state, as mentioned in tl,*pu"rug". No -uri is lost by gamma emission. Choice D is the best answer. Choice A describes an alp:;particie, and choice B sort of describes a beta particle (the charge is negative one, but it is not necessarily tF*

nuclear charge). Choice C is not a common nuclear particle and is probably a conglomeration of subatomrparticles.

Choice B is correct. The electronic configuration for 99Tc is no different than for any other isotope :rtechnetium. Although isotopes have a different number of neutrons, they have an identical number of protc:,r

and an identical number of electrons in their neutral state. Technetium is element number 43, so it has 43 protcrsand 43 electrons as a neutral element. The filling of electrons follows standard Aufbau principle rules, wh-:m,

makes answer B the right choice. It is sometimes a shortcut to look at the periodic table and see where the I'amtt

electron falls. In the case of technetium, it is directly below *ungunur", so its last electron should be a ;i*electron (like manganese). The last electron is in a 4d-orbital, so choice B is best.

86.

87.

88. Choice A is correct. Gamma decay just involves the loss of nuclear energy, and not a particle. Technetium 5;from an excited nuclear state to a iow"r nuclear state after it undergoes gamma decay. No nuclear particles ;

90. Choice B is correct. We see in the passage, that the major piece of evidence for believing a nuclear reactiorr

transpired was the release of mole heat than can be explained by the electrochemical cell reaction.

expecied nuclear products were observed only in low concentrations however. The best answer is choice B'

abiorption or emission of an electron (beta particle) had no effect on the reaction.

Choice D is correct. As described in the passage, cold fusion is a nuclear fusion reaction that can be carrie;

at room temperature. The example in the passage takes place in an electrode within a test fube. Because ienergy is released than expected, it can be assumed that the fusion reaction is exothermic (because of the

heat tirat was released). The correct answer is choice D. Pick D, and you'll smile brightly.

lost or gained by technetium, so its mass and atomic number remain the same. This means that the nuc-;

.o*poftion of iire element remains the same, so the element remains the same. The correct choice is thus A.

89. Choice B is correct. Reaction 3 and Reaction 4 show fusion of two deuterium atoms. The fusion of two deuter:

atoms can generate either a tritium (3H) and hydrogen 1tH) (as shown in Reaction 3), or one neutron and hel:

isotope 13tie; 1ut shown in Reaction 4). The only particle in the answer choices that is not shown as a proiis helium-4, so the correct answer is choice B.

{

.

.

(

"rl

:]:l

91.

Copyright @ by The Berkeley Review@ 16() Section II Detailed

92. Choice A is correct. In the passage, we read that nuclear power plants employ the fission reaction, because todate, fusion reactions require both high temperature and strong magnetic fields. Fusion reactions require moreinput energy than the fusion reaction releases, making them endothermic and unfavorable as a source of energy.The best answer is choice A.

93. Choice C is correct. A fusion reaction results in the combination of the nuclei of the two particles undergoingfusion. Answer choice C involves the combining of two particles to form one product (with an atomic numbergreater than the two incident particles). This defines fusion. Choices A and B are both fission, and choice Dinvolves the emission of a gamma photon via the drop from a nuclear excited state to the nuclear ground state.

94. Choice B is correct. An alpha particle has a mass of four and two protons, and a deuterium nucleus has a mass oftwo and one proton. After losing an alpha particle and deuterium nucleus, the mass of an element drops by sixand the number of protons drops by three. Choices A and C are not six mass units less, so they are eliminated.Choice D is not three atomic numbers less, so it is eliminated. The process is shown below.

,t&u -+ la + lu* 2T#"

Choice C is correct. A change in the atomic number results from a change in the number of protons in the nucleus.The alpha particle has two protons, so the capture of an alpha particle increases the atomic number by two.This eliminates choice A. The capture of a beta particle converts a proton into a neutron, so the atomic numberdecreases by one. This eliminates choice B. The tritium nucleus carries one proton, so capture of a tritiumnucleus increases the atomic number by one. This eliminates choice D. A neutron capture increases the mass byone, but does not affect the atomic number. Choice C is the best answer.

Choice D is correct. "Quantized energy levels" refer to states of finite energy where electrons may exist. Forthis exam, you should know conceptually what behavior is expected. The existence of a neutron or proton at thenucleus may exert Coulombic forces on an orbiting electron, but it does not have any bearing on the quantizationof energy levels. Choice A is therefore eliminated. The scattering of x-rays by thin sheets of material (metalfoil in the Rutherford experiment) shows that matter is mostly empty space, with dense uniformly spacednuclei. Choice B is eliminated. The bending of any particle when moving through a magnetic field simplyindicates that the particle in motion has a net charge of some kind, and that the direction of motion is not inline with the field. Choice C is eliminated. Distinct lines (which can be reproduced in separate trials) showthat the same amount of energy is absorbed when an electron is excited. If the transition between levels is aquantized value (an exact quantity), then it seems logical that the energy levels are also quantized. Thismakes choice D the correct answer.

Choice D is correct. As the wavelength of a photon increases, the energy of the photon decreases. This questionis testing your recall of relative energetics of electromagnetic radiation. The lowest energy of the choices givenis associated with infrared light. This makes D the best choice.

Choice A is correct. The value oI Z"Sy for H is +1. The value of Zsff for He+, the starting point for the secondionization of helium, is +2. Because the ionization energy is proportional to 22, the second ionization energy ofhelium should be four times as great as the ionization energy of hydrogen. Both electrons in question are beingionized from the 1s-orbital, so n = 1 for both ionization energies. You need consider only the effective nuclearcharge. To feel the sensation of correctness, pick choice A. Drawn below are the respective ionizationreactions:

+ Le

\-

%,A.n=1/ra\g/

H"* --> ""t*

-1

\"

@H -----> tt*

ight @ by The Berkeley Review@

+ 1e

161 Section II Detailed Explanations

99. Choice D is correct. Sodium cation has ten electrons, which eliminates choices A and B, both with elevenelectrons. Choice C is the ground state (all electrons fill sequentially) for sodium cation, given that it has ttreten electrons filled in order. In choice D, there are ten electrons and an electron has been excited from the 2p.level to the 3s-level. This leaves choice D to be the best answer.

100. Choice B is correct. The most common shape for a transition metal with five ligands is trigonal bipyramideilThe best answer is choice B. Square planar has only four ligands attached, so choice A is eliminated-Hexahedral does not exist; and if it did, hex is Greek for six, and only five ligands are attached. This wouMalso eliminate choice D. Drawn below is a chart for determining molecular shapes:

Coordination Number = 2

no lone pairs (sp hybr.)

Linear

Bond Angle - 180"

Coordination Number = 2ao

/A:--L-Lone lone pair (sp 2

t-rybt.1

Bent

Bond Angle <120'

CoordinationNumber = 2.A.

,/"\LLtwo lone pairs (sp3 hybr.)

Bent

Bond Angle < 109.5'

Coordination Number = 3

L

I

,-zA:-,no lone pairs (sp 2 hybt.lTrigonal Planar

Bond Angles = 120"

Coordination Number = 3

ro.7^\,L

one lone pair (sp 3 hyur.;

Trigonal Pyramidal

Bond Angles < 109.5'

CoordinationNumber=3 L

lP';-' or '-{,'two lone pairs (dsp3 hybr.)

L

Trigonal Planar or T-shaped

Bond Angles = 90', 120", 180'

Coordination Number = 4

LI

,*/\,.

no lone pairs (sp 3 t ytrr.;

Tetrahedral

Bond Angles = 109.5"

Coordination Number = 5

LI| -"'rrtl L'-ia'L

m lone pairs (dsp3 hybr.)

Trigonal Bipyramidal

Bond Angles = 90'& 120"

Coordination Number = 6

LI

| '"'s Lr,--rl- u,/l"l

L

m lone pairs (d2sp3 hybr.;

Octahedral

Bond Angles = 90'

Copyright @ by The Berkeley Review@ t62 Section II Detailed Expl

Section ItrEquilibrium

by Todd tsennett

A(S) + 6191 ==-

C(S) + D(g)

r{ products (Pc)(Po)' 'e9 neactants (Pn)(Pn)

Uncatalyzed Catalyzed

Fundamentals of Dguilibriuma) Definitions and Terminologyb) Dquilibrium Constant (Keq)

c) Keaction Quotient (Qrx)

d) Case Specific K-Valuese) Keq Calculationsf) Using Keq to Calculate Shiftsg) Complex trquilibriumh) Experimental Determination of K

Le Chdtelier's Principlea) Effect of Stressb) Perturbations and Shifts

i. Direction of Shift

Solubilitya) Definitionsb) Solubility Rulesc) Ionic Structures

i. Nomenclature of Saltsii. Polyatomic lonsSolubility Product and lvlolar SolubilityRelative SolubilitySolubility trxperimentsCommon lon EffectSeparation by PrecipitationIon Exchange Columns

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Equilibrium Section GoalsKnow how to determine an equilibrium constant from exDerimental data.ov There are experiments that are designed to determine a change jn total pressure over time for a gasequilibrium system. The change in pressure (from the time the reactants are mixed until the reactiorrreaches a steady equilibrium pressure) can be used to calculate an equilibrium constant. The differencein pressure is the shift requiied to reach equilibrium

Understand Le Chdtelier's principle and its effects on eguilibrium svstems.Le ChAtelier's principle predicts certain behavior in an equilibrium system once a stress is appliedto the system. Because changing one variable can affect another variable (for instance, changihg thep_ressure may also change the volume), you must evaluate how the chemical reaction will a-djuit toalleviate the applied stress in order to reestablish equilibrium.

Understand the mathematical arrangement of the equilibrium expression.The equilibrium expression, simply put, is the ratio of products over reactants. The mathematicalrules require that you do not put pure liquids (solvents) or solids into the expression. If more thanone mole of product or reactant are involved in the reaction, then the stoich:iometric coefficient inthe balanced equation becomes an exponent in the equilibrium expression. The value of theequilibrium constant changes only with temperature.

Know the effects of the system variables P, V, n, and T on reaction equilibrium.It is not possible to change just one variable in an equality. Changing the conditions of the systemshifts an equilibrium, but it does not necessarily change the equilibrium constant. The nurnericalvalue of the equilibrium constant changes only with temperature.

Know the relationship between molar solubility and solubility product.The molar solubility of a salt is the concentration of the salt (as measured in molarity) needed tosolvate an aqueous solution completely. The solubility product is an equilibrium consiant for thedissociation reaction. What makes solubility products unique is that the reactant is alwaya a solid,so the equilibrium expression has no denominator. Molaf solubilitv is a more useful quantity toknow thin solubility product, because it measures the amount of sali in solution.

Understand separation by precipitation and the chelation effect.Chelating is the formation of a Lewis acid-base bond between a lone pair-donor (ligand) and a lone-pair acceptor (central atom). Chelation changes the solubility of a salf by changing the concentrationof free ions in solution. When a ligand binds a central metal, there is a foimation constant thatmeasures the strength of the chelation. This allows for specific ions to be removed from solutionby binding them to form a more soluble complex ion. Ions also can be removed from solution bvadding counter-ions that form an insoluble salt.

Understand the common ion effect.The common ion effect is a twist on Le ChAtelier's principle as it applies to solubility. The additionof products to an equilibrium mixture shifts the reaction in the reverse direction, which in the caseof solubility results in precipitation. This means that addition of an ion to solution or the presenceof an ion iri solution rehucei the solubilitv of a salt.

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General Chemistry Equilibrium Introduction

Equilibrium, as you have leamed it, involves the balancing of a chemical reaction

between reactants and products. Equilibrium is a state that is achieved when the

reactants go on to form products at the same rate that the products go back toform reactants. Equilibrium reactions take place in both the gas phase and

solution state, where reactant and product molecules are free to migrate and

collide. The kinetic theory model states that molecules must collide to react.

Equilibrium occurs only in a closed system, although steps in a pathway (an

open system) may be an equilibrium reaction.

Equilibrium is an odd yet obvious thing. It is the essence of nature and the foe ofpermanence. We can't beat it... EVER! The concept of equilibrium pervades notonly chemistry, but politics, economics, sociology, health..-even ourrelationships. A bank represents a good illustration of this concept. There are

deposits and withdrawals continually going on, but on the whole, the amount ofmoney in the bank remains essentially constant (except following a bankrobbery, after which a new equilibrium will be established). Everything lies in a

balance, and all one has to do is read the scale. Fortunately, in chemistry it is easy

to read the scale. To understand equilibrium better, consider the following saga:

In a certain house with a large backyard lived an elderly man with a plum tree.

Next-door to him lived a young boy who also had a plum tree. Given thatneither of these two were farmers by any means, the fruit generally fell to theground and rotted. One day, while out in his back yard cleaning up the plums,Ere elderly man was struck by a flash of insight; instead of using all his energy to;onsolidate the plums in a trash container, he would instead simply fling themover the fence. The boy next-door saw this and decided that two could play atthis game. The next time he was out in the back yard, he flung a few plums overthe fence. This soon evolved into a daily contest with the advantage going to the

-lttle boy, for he was younger and quicker and could move around the yard:aster. He was therefore able to fling the unlikely projectiles over the fence at-Jrree times the rate of the old man. At first, there were roughly equal amounts of:lums in each yard, but due to ihe boy's greater flinging Prowess, the old man's.;ard gradually accumulated more. Finally, when the point was reached where:he old man's yard had three times as much as the boy's, the overall amount ineach yard became stable and didn't change. Because the boy had so few plums inris yard, he spent the majority of his time running around collecting them. The

nan, however, could simply stand in one place, scooping and flinging. From the

reighbors' perspective, for every one that would sail through the air to the left,jreie was one that would sail through the air to the right. The plum-flinging had:eached its equilibrium. The ratio of the plums in each yard was equal to the

:atio of the rates at which the two could fling plums (known as the "plum-:Jnging rate"). This in essence is as deep as equilibrium gets.

This example may not be the most eloquent, but it serves the purpose of:enerating a memorable analogy to chemical reactions. It is the external:onditions that affect the equilibrium between products and reactants, and a

:killed chemist knows how to manipulate this relationship. We will address how.quilibrium plays out in a solvent environment and in the gas phase' We shall,'iew the effect of factors governing the system, such as volume, pressure,

=mperature, and concentration. We will finish by looking at the equilibrium of a

'alt dissociating into aqueous solution, and the factors that are involved.lhroughout all of the discussion, Le Chatelier's principle will play a ro11.

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General Chemistry Equilibrium Fundamentals of Equilibrium

uriufi|ffiffiffi*ibfiumDefinitions and TerminologyEquilibrium occurs in a closed system when the rate of the forward reaction isequal to the rate of the reverse reaction (rate forward = rate reverse). The resultis that from a macroscopic perspective, the system appears not to change.However, a net change of zero does not automatically mean the system is rnequilibrium. Reactions may stop when there is no reactant present, which is a

case of no reaction, not equilibrium. As an introduction to equilibrium, considerReaction 3.1, where R = reactant, P = product, kf - forward rate constant, and k1

= reverse rate constant.

kt$r\

-

rkr

Reaction 3.1

The forward rate for this reaction is based on the amount of reactants and theforward rate constant (rate forward = kf[Reactants]). The reverse rate for thrsreaction is based on the amount of products and the reverse rate constant (ratereverse = kr[Products]). Equilibrium is achieved when:

k1[Reactants] = krlProducts] (3.11

Example 3.LWhich of the following graphs represents what is observed over time for a

reaction starting with all reactants?

SolutionChoices A and B are eliminated, because the rates must be equal at equilibriurr,and neither graph reflects this. Choice D is eliminated, because straight lines arrnot very common for graphs in chemistry. A straight line would imply that tLoreaction abruptly stopped, once equilibrium was reached. The reality is that tLrreaction gradually slows until it reaches equilibrium. The best answer is choic"C, which shows equal rates after time and constantly changing rates unti-equilibrium is reached.

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General Chemistry Equilibrium Fundamentals of Equilibrium

Equilibrium Constant (Kgq)The equilibrium constant is a mathematical quantity that isreaction at equilibrium. By definition, the equilibriumconcentration of products at equilibrium divided by thereactants at equilibrium. Equation 3.2 shows this relationship.

calculated for a

constant is theconcentration of

K"n = [Prodr.tt] /[Ruu.tur,tr] (3.2)

By applying Equation 3.2 to Reaction 3.1, we find that the equilibrium constant isequal to the forward rate constant divided by the reverse rate constant.

Example 3.2For Reaction 3.1, if the forward rate constant is four times the reverse rateconstant, what is Ksq after a catalyst has been added that doubles the rate of theforward reaction?

A. Ksq=2B. 3*q=4C. B*q=BD, Xbq = 16

Solution-ddding a catalyst lowers the activation energy, so the reaction speeds up. In this.rample, the forward rate is doubled, because the forward rate constant isioubled. However, the activation energy is lowered for the reverse reaction asi.;ell. The reverse reaction rate is also doubled. The ratio of the forward rate tole reverse rate remains the same. This means that equilibrium is the same, so-:.e equilibrium constant (K"q) is the same. Equilibrium is achieved sooner, but

=.e same equilibrium conditions are reached. The forward rate is four times the:er,-erse rate, so the equilibrium constant is equal to 4, choice B.

l:ble 3.1 lists some rules about the equilibrium constant you must know.--:rong them is the rule that the numerical value of K"O changes only with::-=Lperature. Knowing this rule would have made Example 3.2 easier to solve.

Table 3.1

'"hen multiple reactants or multiple products are present in the reaction, which. usually the case, rule # 1 applies. Because the concentration of a solid or pure

-:-urd (solvent) does not change, their values are constant' For this reason, they::e ignored in the K"n determination. As a rule of thumb, only molecules that::e free to move and aie rarely in contact with other reacting molecules affect the:,;uilibrium expression. Lastly, because the equilibrium constant is a measure of::,ergy distribution, only a change in temperature (a measure of the system's

=:,ergy) changes the value of K"q.

Stoichiometric values from the balanced equation become exponentsin the Kgq expression.

Do not include solids or pure liquids in the Ksq expression, onlysolutes (for K6) and gases (for Kp).

The numerical value of Ksq varies only with changing temperature,not with catalysts, pressure, volume, or moles.

-:pvright @ by The Berkeley Review 167 Exclusive MCAT Preparation

General Chemistry Equilibrium Fundamentals of Equitibrium jConsider Reaction 3.2,

k.2 A(s)

== B(g) + C(s)

Kr

Reaction 3.2

Applying the three rules to Reaction 3.2 generates the correct expression for Kgq.

We start by plugging in products over reactants: K"o = PllgltAl

Invalid, because stoichiometric coefficients must be exponents .'. K"o - tB]t^q]

' IAIZ

Invalid, because Kgq does not include solids or liquids.'. K"q =l9l-. _.1 [A]2

K"o can be written as either K. = JEI tr4-i orKo = Pu =

atm.-l- tAl2 ' (po)t

Reaction Quotient (Qrx)\.Alhen a reaction is not at equilibrium, the expression of products over reactantsis said to be the reaction quotient (Q.*). The relationship between K"O and Q.*dictates the direction in which a reaction proceeds to reach equilibrium. \Atrhen K> Q, the denominator of Q (reactants) is too large and the numerator of Q(products) is too small. To establish equilibrium, the reaction must shift to theright. The reaction shifts to the right to increase the products (numerator) anddecrease the reactants (denominator). When K < Q, the denominator of Q(reactants) is too small and the numerator of Q (products) is too large. Toestablish equilibrium, the reaction must shift to the left. The reaction shifts to ttaeleft to decrease the products (numerator) and increase the reactants(denominator). A shortcut to determine the direction the reaction proceeds breach equilibrium involves drawing the relationship of K and Q alphabeticall'r-.and then converting the < or > sign into an arrow. For example: K > Q becomesK -> Q, so the reaction moves right to reach equilibrium, because the arrorr lspointing to the right.

Example 3.3When the reaction quotient is greater than the equilibrium constant, which of thefollowing is NOT true?

A. The system has too many products and too few reactants.B. The reaction is displaced from equilibrium.C. The reaction must shift in the forward direction to reach equilibrium.D. The reverse reaction rate is greater than the forward reaction rate.

SolutionWhen the reaction quotient is greater than the equilibrium constant, the syshas an excess of products and shortage of reactants, relative to equilibriChoice A is a valid statement. The system is not at equilibrium, so choice B isvalid statement. To reach equilibrium, the reaction must have a net shift inreverse direction to reduce the amount of products and increase the amountreactants. This means that the reverse reaction rate is greater than the fonrreaction rate, making choice D valid. The system cannot have a net shift inforward direction, so choice C is an invalid statement.

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General Chemistry Equilibrium Fundamentals of Equilibrium

Case Specific K-valuesAll equilibrium constants obey the same rules, but depending on the reaction,there may be special features that recur. Different reactions have special K-values. Table 3.2 lists the six types of K"O values we shall address'

K type Type of reaction to which the K applies

Kp K"o for the reaction of gases. Values are in pressure units.

K" Kgq for the reaction of solutes. Values are in concentration units.

Ksp K"o for salts dissociating into ions. Measures the solubility.

Ka Kuo for acids dissociating in water. Measures the acidity.

K6 K"o for bases hydrolyzngin water. Measures the basicity.

Kw Ksq for autoionization of water into hydronium and hydroxide.

Table 3.2

Understanding that the rules are the same for all types of K-values will enhance-,-our journey through the wonderful world of equilibrium. Knowing that the:ommon ion effect is nothing more than Le ChAtelier's principle applied to

'olubility systems is a perfect example of how the rules apply to all equilibriumsr-stems. There are just different names to describe the reaction conditions.

K"O Calculations\ow comes the math part, which we will handle through practice! Getting:eacquainted with equilibrium math is a matter of repetition and practice. Once-,'ou feel sufficiently familiarized, then move on to the next topic.

Lxample 3.4it 650K, the partial pressures of the component gases were determined for::liowing reaction:

Hz(g) + Iz(g) + 2HI(g)

PHz 0.20 atm., PIz = 1.50 atm., and PHI = 3'00 atm' Ve'

',hat is the KO for this reaction?

t 10.0 atm.ts. 15.0 atm.: 30.0 atm.l. 45.0 atm.

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S,olution::'ssible errors with this problem stem from forgetting to square the numerator:r :gnoring 12, because you have seen it as a solid before (note that this reaction is

l: r50K). In this case, forgetting to square the numerator would yield an answer::at is too small by a factor of 3. As long as you don't forget these things, the

:::,' lem merely involves doing your math quickly. The math setup is as follows:

r^= (Pur)2 _ (3)2 -9 _D=30" (PHzXPrz) (0.2X1.5) 0.3 3

l-,rice C is the correct answer. To make this a more conceptual type of question,,J'€"'' may give the answer choices as ranges rather than exact numbers.

- -:r-right @ by The Berkeley Review 169 Exclusive MCAT PreParation

General Chemistry Equilibrium Fundamentals of Equitibrium

Example 3.5At 323'C, there are 0.10 moles HZG),0.20 moles HCIOa(g), 0.10 moles HZO(g),and 0.36 moles HCI(g) at equilibrium in a 400-mL flask. what is Kuo for thisequilibrium mixture?

r HCIOa(s) + 4H2@)

t *rr )'

' iti('"

SolutionIn this problem, the system is already at equilibrium, so no determination otchanges (x-values) is required. This problem is of the plug-'n'-chug (simpiesubstitution) type. Moles can be used directly in the equilibrium expression.because although it is technically correct to use concentrations, in this casevolume cancels out of both the denominator and the numerator, leaving just themoles. This is true any time that the number of reactants is equal to the numberof products in either the gas phase or as solutes. The question is really just amath problem, solved in the following way:

(0.36X0.10) 4-0.36-0.18-ro

0.20 0.10(0.20x0.10) 4

This makes choice C the correct answer.

Example 3.6At srP, the partial pressure of No is 152 torr and the partial pressure of 02 is 3-$torr. If the mixture is at equilibrium, what is the K"o at STP for the follora'ingreaction?

No(g)+Oz(g) + NO3(s)

A. 8.333 atm.-1B. 4.167 atm.-IC. 1.000 atm.-1D. 0.240atm.-1

SolutionThe key piece of information in this question is STP (standard temperaturepressure). This implies that the total pressure of the systems is 760 torr. Theof the partial pressures is the total pressure, so 760 = PNOg + PruO + PO2.

substitution , 760 = PNOg + \52 + 228, so PI.JO, = 380 torr. Because the ansrrelisted in atmospheres, the values in torr must be converted to atmospheres betthey are useful. The conversion is 760 torr per atmosphere. The equation d

calculating K"O is:

K-^= PNo: = 3Solzeo = tl, =tl,-So.'e(pNoXpor) (rurlruo\rrtlruo) (%X%o) 3lso 6

Of the answer thoices given, only choice A is greater than 8. If you pick Awill definitely be a star!

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A. 0.0362B. 0.0724c. 1.8000D. 44.100

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General Chemistry Equilibrium Fundamentals of F,quilibrium

Example 3.7For the following reaction, 0.20 moles Bi2S3(s) are mixed with 0.50 moles Hz(g).Once equilibrium is established,0.225 moles of H2(g) remain. What is the value _ 1,

of K"O for this reaction?

1Bi2S3(s) + 3H2(g) ==:==- 2Bi(s) + 3H2S(g) o 1r-+.-)

A. 0.57B. 0.82c. 7.22D. 1.81

SolutionThis question is best solved by determining the equilibrium concentrations forboth of the gases. Solids do not affect the equilibrium, so be sure not to includeany solids in the equilibrium expression. The equilibrium expression involvespartial pressures, but the answer is the same whether you use the partialpressures of the gases or the concentrations of the gases. This is because there isthe same number of gas molecules on each side of the reaction. The solids areignored altogether as long as they are not the limiting reagent. Because you needthree times as much H2(g) as Bi2S3(s) and you have only 2.5 times as much,hydrogen is the limiting reagent in this reaction, if it were to go to completion.This means that to determine the value of Kgq, you must determine the ratio ofthe two gases. The values are found in the following way:

2.

i c. " )

Reaction:

lrLitially:

Shift:

Equilibrium:

0.5

-3x0.5 - 3x

---+

Bi2S3(s) 3 Hz(g) :t' 2 Bi(s) 3 H2O(g)

00+2x + 3x

2x 3x

0.2

-X0.2-x

-n this case, we can solve for the value of 3x from the information given. Atequilibrium, there arc 0.225 moles of hydrogen gas remaining, so 0.5 - 3x = 0'225.This means that 3x = 0.275. There is no need to solve for x, because 3x is presentn the gas terms, and solids are not going to be considered. Plugging1.271 tnfor-r'r nto the equilibrium line of the reaction chart yields:

Reaction: Bi2S3(s) 3 Hz(g) -$ 2 Bi(s) 3 H2O(g)

don't care 0.275:quilibrium: don't care 0.225

These numbers work nicely in determining the equilibrium constant. The math*. shownbelow:

v -(Puzs)3 -(molessrs)3 -(0.2713 =p.zzs\3 =lU'1, =7.223 =7.22+"'"q=F;f = ,-'ru*F= $nqt=\o.nsl

-\ g | -'--

Jrr1y choice D is a value that is greater than L.22, so that is the best answer. If-"-ou forget to cube the value, it is easy to choose answer choice C by mistake.lhoices A and B are eliminated, because there are more products than reactants,;r the value of K"n must be greater than 1,0. Be sure to use common sense to

=jminate incorrect answer choices. Developing intuition and learning to trust'.-our common sense is more important in MCAT preparation than honing yourJgebra and multiplication skills.

-opyright @ by The Berkeley Review t7t Exclusive MCAT Preparation

General Chemistry Equilibrium Fundamentals of Equilibriun

Using K"O to Calculate ShiftsThe equilibrium constant is used to calculate the quantity of the products and thereactants present at equilibrium. This is achieved by following the moles of eachspecies from initial conditions to equilibrium conditions, specificalty by settingup a table to keep track of the components systematically over the course of tlnreaction. There are three stages to consider: the initial stage, the shift, andequilibrium. Consider Reaction 3.3:

Co(g) + HzO(g) COz(g) + Hz(g)

Reaction 3.3

At375'c, the equilibrium constant for Reaction 3.3 is 2.51 x 101. The equilibriuurconstant is unitless when the number of products equals the number of reactantsConsider the reaction to start with 1.0 atm of Co and 1.0 atm. of H2o. Figure 91shows the setup for determining the shift and final pressures in Reaction 3.3.

Reaction:

initially:shift:Equilibrium:

Co(g) Hzo(s)

Figure 3-1

As mentioned, there are three considerations in the setup. The first line is what isinitially given. The second line shows the direction of the shift and thestoichiometric consequences. You must be able to determine the direction of theshift by comparing the initial concentrations to the equilibrium distribution. Inthis case it was easy, because there are no products, so the reaction must shift inthe forward direction. The third line accounts for what is present onceequilibrium is established. Values from line 3 are plugged into the equilibriur::rexpression, to solve for x, the shift in the reaction. The solution is as follows:

Coz(g)

0

HzG)

01.00 1.00

-X -X1.00-x 1.00-x

--+ +X +XXX

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r"n =.(P.orXp"r) - (x)(x) = 2s.r = 2s.L = xz

' (PcoIPHzo) (1 - x)(1 - x) (1 _ *)2

25.7= x2 - =95= x +5-5x=x.+5=6x...x=5=0.83(1-*)2 1-x 6

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The math was simpler that it first appeared in this case. The MCAT does notseek to test your algebra skills, as much as it tests your reasoning ability. Beingable to estimate the magnitude of the x (the shift) relative to the initial values isimportant.

Example 3.8At 773K, the KO for the following reaction is 3.0 x 10-5. If the partial pressure ofN2(g) is initially 3.75 atm., and the partial pressure of H2(g) is initially 2.0 atm,;what is the partial pressure of NH3(g) once equilibrium is established, assumingthere is no ammonia in the system initially?

Nz(g) + 3Hz(g) T+ 2NHg(g)

A. 0.0900 atm.B. 0.0300 a!m.C. 0.0100 atm.D. 0.0010 atm.

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General Chemistry Equilibrium Fundamentals of Equitibrium

SolutionThis is a straightforward example. Because the reaction starts with all reactants,it shifts forward to reach equilibrium. The reaction has a very small KuO andstarts with all reactants, so the shift is small. The x-term can be ignored when itrs subtracted from or added to numerical values. The setup is as follows:

Reaction: NZ(g) 3 HZ(g)

-Initially: 3.75 2.00

Shift: -x -3x

->Equilibrium: 3.75 - x 2.00-3x

NHs(e)

0

+2x2x

Ignoring the x and 3x portions of the reactant quantities yields the following:

a-^ - (PNSJ2 - Q*)2 = 3x 10-5 = 3x \0-s = --4*- =4x2

(prurXpnr)3 (s.7s) e)3 3.75 x 8 30

x2 =30 x3x 10-5 =22.5x10-5+ x2=2.25x10-4...x =r[LzS x1,0-2=1.5x 10-24

The concentration of NH3 at equilibrium is 2x, which is 3.0 x 10-2 M. The best

answer is choice B. This is considerably more math than the MCAT requires youto use, so think of this example as merely a step towards getting re-acquainted.

-\ concern you may recall from your general chemistry class involves whether itis safe to ignore x when it is either added to or subtracted from a numerical term.The x-value is ignored when the initial conditions are like the equilibriumconditions, because the shift is minimal and the value of x is trivial. In Example

3.8, Keq is less than 1.0, so there are fewer products than reactants at equilibrium.This m-eans that hardly anything shifts over to the product side, so x can be

ignored. Table 3.3 shows cases when x can and cannot be ignored.

Table 3.3

Understanding the math associated with equilibrium is important. KeQ can be

used to find concentration values at equilibrium, and equilibrium concentrationr.alues can be used to find KuO. This fits nicely into an MCAT experiment,because equilibrium constants r6sult from experiments conducted to determineequilibrium concentrations, in several trials with different starting conditions.

InitialConditions

EquilibriumConstant

shift x-term?

AllReactants

Keq < 1o-3 Q and K are SIMILAR .'. SMALL xSMALL shift in the forward direction

Ignore

AllProducts Keq < 1o-3

Q and K are DIFFERENT .'. LARGE x

LARGE shift in the reverse directionConsider

AllReactants Keq > 103

Q and K are DIFFERENT .'. LARGE xLARGE shift in the forward direction

Consider

AllProducts Keq > 103

Q and K are SIMILAR ".

SMALL xSMALL shift in the reverse direction

Ignore

Reactants &Products K"q=1 Q and K are SIMILAR .'. SMALL x

SMALL shift in either directionIgnore

Reactants &Products Keq )) or << L

Q and K are DIFFERENT .'. LARGE xLARGE shift in either direction

Consider

Copyright O by The Berkeley Review t73 Exclusive MCAT PreParation

General Chemistry Equilibrium Fundamentals of Equilibrium

Example 3.9At92.2'C, the KO for the following reaction is 0.2000 atm.-1. If you were to placeexactly 0.200 atm. of N2Oa(g) into a 1.00 liter vessel, what would the partialpressure of NO2(g) be once equilibrium was established?

2 NO2(s)

A. 0.025 atm. NO2(g)B. 0.200 atm. NOz(g)C. 0.350 atm. NOz(g)D. 0.400 atm. NO2(g)

Reaction:

Initially:Shift:

Equilibrium:

SolutionTo solve this precisely, it is necessary to use the quadratic equation. on theMCAT, it is unlikely that you will ever use the quadratic equation. Th_.question emphasizes the technique of eliminating answer choices throug:rapproximation. First, you must use the equilibrium expression to estimate tl:emagnitude of x. In this case, Ksq is less than 1.0, and the reaction starts with a-products. The value of x is goingto be significant (more than half shifts over). Lhalf of the 0.200 atm. of Nzo+ shifts over, then the partial pressure of nitroge:dioxide is 0.200 atm. Considering that more than half of the N2o4(g) is going icshift, the value of NOZ(S) is greater than 0.200 atm. However, not all of theNZO+(g) can shift over (which would result in 0.400 atm. of NO2), so the ans\\-e1must be less than 0.400 atm. Only choice C falls within the range of 0.200 tc0.400. In a multiple-choice format, this question is rather easy to answer.

In the interest of developing an alternative method to answer the questionsubstitute one of the four answer choices (the one closest to the x youapproximate) into the equilibrium expression and then compare the answer youget to the given value of K"n. Your answer either will equal X.eq or it won't equaiKeq. If it equals Keq, you picked well (go to Vegas with those'skills). If it doesnot equal K, the error can be used to zero in on the correct answer/ depending onwhether your value is too high or too low. For this reason, always start br-substituting a middle value.

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2No2(g) :0

+2x *+2x

Nzo+(g)

0.200

-X0.20 - x

:-

u,

C,

D

These numbers should be plugged into the equilibrium expression.

v - PNzo+ -0.2-x -o.z-xt (p*or)' (2*)2 4x2

Be sure not to substitute the answer choice values directly for x, because thevalues in the answer choices are for Ppg, which is 2x. Choices B and C fall in themiddle of the range, so choose either one to plug in. For numerical ease, let's usechoice B. If PNg, is 0.200 atm., then x is 0.100.

v -0.2-x -0.2- 0.1 - 0.10 - 10 - o r^P - (2# - (orf -o-04 - 4 - '''This number (2.5) is too large (greater than 0.2000), which means that the x u'echose was too small, so we need to choose a larger value for x. Choice C is best.

Copyright @ by The Berkeley Review 174 The Berkeley Revieu'

General Chemistry F.quilibrium Fundamentals of Equilibrium

Reaction iI: CaO(s) + SOZ(g)

Cornplex EquilibriumA complex equilibrium is a balance between two separate reactions that share acommon reagent. In essence, it is two reactions whose equilibrium states dependon one another. In a complex equilibrium, a product in one reaction is a reactantin the other reaction. As a result, if one reaction is displaced from equilibrium,then the other reaction is also affected. When adding the component reactions,the common reagent is absent in the overall reaction and is considered to be anintermediate. The equilibrium constant for the overall reaction is found bymultiplying the individual equilibrium constants of the component reactions.This is because the reagent that disappears when you add the reactions is in thenumerator of one reaction and in the denominator of the other reaction. Toeliminate it, K"Os are multiplied. Figure 3-2is a sample of complex equilibrium:

Reaction I: 2 So3(g)

+ CaSO3(s) KeqZ

Overall: 2CaO(s) + 2SO3(g): 2CaSO3(s) + Oz(g) Keqlx(K*qD2

Figure 3-2

Reaction II is multiplied by 2 to balance SO2 when you add the reactions.Because Reaction II is multiplied by 2, its KuO value is squared. The addition ofcalcium oxide to an equilibrium mixture of C,2, SO2, and SO3 reduces the partialpressure of SO3. This occurs despite the fact that sulfur trioxide and calciumoxide do not react directly. Complex equilibrium questions are essentially justmanipulations of the overall equilibrium and of Le Chdtelier's principle.

Example 3.10Consider the following complex equilibrium:

2NO(g) + 1Oz(g) :1-2 NO2(g)

2 NO2(g)

l NzO+(g)

K"q1

Keq2

2 NO(g) + 1 Oz(S) ,- 1 N2O4(g) Keql x B.eq2

i\hen NzO+ gas is removed, how are the partial pressures of NO gas and NO2:as affected?

-{. PXO and Pyg, both decrease.

B. Pyg and P111g, both increase.

C. Ppg decreases, while P11g, remains the same.

D. PruO increases, while P116, remains the same.

Solutionlhe second reaction shifts to the right to compensate for the loss of N2O4, so the:artial pressure of NO2 decreases. A decrease in NO2 causes the first reaction to.hiJt to the right as well, resulting in a decrease in the partial pressure of NO gas:nd an increase in the partial pressure of NO2. However, the increase in NO2:om the forward shift of the first reaction is less significant than the decrease in\Q caused by the forward shift of the second reaction. This is because the shiftn the first reaction cannot completely replenish the lost NO2 without losing so:'Luch NO that the reaction is beyond equilibrium. A shift never regenerates as

:uch as was lost. Both NO and NO2 decrease, making choice A the best answer.

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General Chemistry Dquilibrium Fundamentals of Equilibrium

Experimental Determination of KIn this experiment, a 500-mL closed glass flask is employed, so the system isisochoric (has V"6rrr1211). The pressure of the system is monitored continuallyover the course of the experiment. The researcher tests Reaction 3.4by observingthe total pressure of the system during a fixed interval.

H2(g) + I2(s) + 2Hi(g)

Reaction 3.4

The researcher mixes together 1.00 atm. of hydrogen gas (H2) with excess iodinesolid (I2). The total pressure of the system is monitored by a small detector in thewall of the flask throughout the reaction, which is said to be in equilibrium oncethe total pressure of the system stabilizes. Ample time is given to allow thesystem to reach equilibrium. The setup for the reaction is shown in Figure 3-3.

Reaction:

h"ritially:

shift:Equilibrium:

Iz(s)

excess

2 HI(g)

0

Hz(g)

1.00

_x _x ------.> +2x1.00 - x who cares? 2x

Figure 3-3

The reaction proceeds forward to reach equilibrium. It is unknown how mudrthe reaction proceeds, so the shift value is assigned the term x. Hydrogen gasdecreases by x, while hydroiodic acid gas increases by 2x, because of thestoichiometry. Because the number of gas molecules increases, so does the totalpressure of the system. Figure 3-4 shows the total pressure of the system as efunction of time. The pressures for H2 and HI are inferred from the totalpressure graph.

Pto,ulincreases from 1.00 to 1.00 + x

Psrincreases from 0 to 2x

P",decreases from 1.00 to 1.00 - x

Figure 3-4

K"O is calculated from P1o1u1, because the shift in the reaction (x) is equal toFor instance, if P16131 is 1.60 atm. and Pir,itiul is 1.00 atm., then AP is 0.60 atrn-AP is 0.60 atm., then at equilibrium PHt is 1.20 atm. and Pg, is 0.40 atm.

equilibrium expression for the reaction does not include iodine, because it issolid. Plugging into the equilibrium expression yields a numerical value for

Copyright @by The Berkeley Review 176 The Berkeley

General Chemistry Equilibrium Fundamentals of Equilibrium

Determination of K

Ko. = ( PHr)2 - (2*)2 - (r.z)2

-o.2)(r.2) - t.2 6.2) = (3)(1.2) = 3.6-r PH, 1.0 -x 0.4 0.4 0.4

This experiment determines the value of Kgq under the ambient reactionconditions. Changes to the system are made to study equilibrium further.

Effect of the Addition of Reactant

Once the pressure stabilizes at 1.60 atm., the researcher alters the system byadding 0.10 atm. of isotopically rich hydrogen gas (predominantly moleculardeuterium, 2H-2H). The total pressure of the system increases as hydrogen gas

is added to the system. After the addition is complete, the total pressurecontinues to increase, gradually slowing until it reaches a stable value of 1-.76

atm. P1s1n1 for the system increases by 0.16 atm., more than the 0.10 atm. added.Analysis using infrared spectroscopy confirms that both 2H-I and 2H-H areformed. From these observations, it is concluded that addition of a reactant gas

shifted the reaction in the forward direction to a new distribution of compounds.Substituting the values for Pg, and Pg1 in the equilibrium expression reveals

that K"o remains the same, although the partial pressure of each gas haschanged. This means that addition of the reactant gas displaces the reaction fromequilibrium, following which the reaction shifts in such a way as to reestablishequilibrium. In addition, the equilibrium is dynamic, given that the mixedhydrogen eH-H) forms by the reverse reaction.

Following the addition of the labeled hydrogen, the researcher makes a second

addition to the system, after the pressure stabilizes at 1.76 atm., by adding 10.0 g

of isotopically rich iodine (predominantly 1271-tzzt;. the system is continuallymonitored, but no change in pressure is detected. Again, the gases in the systemare analyzed by infrared spectroscopy The labels appear to be completelyscrambled, with H-1271,29-7271,H-[ and 2H-I all being observed. Fromthis, it is concluded that the addition of a solid reactant does not shift the reactionrn either direction. It reconfirms that the equilibrium is dynamic, given that themixture of isotopically labeled compounds.

-\ major point of this experiment is to support the concept of dynamic equilibrium,a state where the system is continually reacting in both the forward and reversedirections. On the macroscopic level, there is no net change. This implies thatthe forward and reverse reaction rates are equal and that the concentration of theleactants and products remains constant. The scrambling of the isotopic labelssupports this idea. If this were a case of static equilibrium, tlne isotope would notbe incorporated into the product or back-react to form the mixed isotopereactant. Any set of questions on the MCAT that accompanies the description ofan experiment like this, would include a question on dynamic equilibrium.

Such an experiment is typical for an MCAT physical sciences passage. It is to.,-our advantage to ponder what types of questions might be asked. Questions:ould involve total pressure and its relationship to equilibrium through partial3ressure. The shift can be rationalized using Le Chdtelier's principle, so

questions on Le ChAtelier's principle are probable. To round out the questions,*rere is the possibility of pH questions, solubility questions, and the relationship':etween

the reaction quotient (Q) and the equilibrium constant (K). It is morehan worth your time to make up some multiple-choice questions to accompany--his passage. If you know how to write a multiple-choice test, you will'-nderstand better how to take one.

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General Chemistry Equilibrium Le Chdtelier's Principle

ffi1G :$liffiH6i lLe Chatelier's principle is justified both mathematically and theoretically. Thebasic rule is that whatever change you make to a system that is in equilibrium,the reaction mixture will react in a way to undo the change and ieestablishequilibrium. The formal definition of the principle is as followi:If an external stress is applied to a system at equilibrium, the system wiII shift itself insuch a way that the stress is partially relieaed and equilibrium is reestabtished.

Effects of StressAn external stress on the system includes changing moles (concentration of acomponent), pressure, volume, and temperature. In all cases, except changingtemperature, the numerical value of K remains constant, and the system willcontinue to react until the K value is obtained once again. When a reactionsystem is not in a state of equilibrium, the same calculation for the equilibriumexpression (K) is used, but it is referred to as the reaction quotient (e). If K is notequal to Q, then the system is not in equilibrium. It is important to note thatchanging an equilibrium system by adding a pure solid or u p,rr" tiquid does notdisrupt the equilibrium nor change the equilibrium constant.

This is why in the sample experiment on the preceding page, when iodinecrystals were added (a solid), the system was not moved from equilibrium.However, when molecular hydrogen was added (a gas), the system was movedfrom equilibrium, and it shifted accordingly, to reestablish equilibrium. It isimportant that this conceptual view of equilibrium makes sense to you, becausethe MCAT is a conceptual exam that evaluates your understanding at this levelWe shall look at Le ehatelier's principle in terms of mathematics and intuitiveproblem-solving. We shall address the mathematical relationships only to theextent necessary to support our observations. In generic Reaction 3.5, themathematical aspects of Le ChAtelier's Principle are considered.

A(g) + B(g) --

C(s)

Reaction 3.5

Let's assume that K"o for the reaction is 1.00 and that initially, the partialpressure of all three components is 1.00 atm. BY substitution into the equilibriumexpression, we see that the value is 1.0, meaning that the system is inequilibrium. upon doubling the external pressure (i.e., by applying a stress tcrthe equilibrium), the system is displaced from equilibrium. According to leChatelier's principle, it will shift to reduce the pressure of the system, which rr epredict will be in the forward direction. By shifting in the forward direction, thereaction goes from the side with two molecules to the side with one molecule.thereby reducing the number of molecules and thus reducing the pressure. Tfumath to support this prediction is as follows:

c(g) t / a.u State

t

tjl:

:ftW@

Reaction: A(g) +

Initially: 1.0

After Stress 2.0

Shift: - x

Final: 2 - x

B(g) +

1.0

2.0

_x __+ +x2-x 2+x

1.00 equilibrium

0.5 not equilibri''rrn

reacting

1.00 new equilibrium

The external pressure increased, causing internal pressures to increase. Thitgenerated a system where K > Q, so to reestablish equilibrium, the reaction shi-fisforward until once again Q = 11.

1.0

2.0

Copyright @ by The Berkeley Review t7B The Berkeley

General Chemistry Equilibrium Le Chitelier's Principle

The numerical value of K does not change with the change in pressure, although*re equilibrium distribution does change.

K^^= PC - 2+x - x+2 =L=x2-4x+4=x+2)*2-5x+2=0(pnXps) e-*)2 x2-4x+4

L-sing the quadratic equation yields a value of roughly 0.44, which means that

'J1e new partial pressules are P4 = 7.56 atm., Pg = 1.56 atm., and PC = 2'44 atm.Table 3.4 shows different equilibrium distributions for Reaction 3.5. You shouldrotice that the ratio of the compounds varies, but the value of K"O is constant.

P4 Pg Pq Pc/pn Pc/ra.r,g

0.80 0.80 0.64 0.80 1.000

0.90 0.90 0.81 0.90 1.000

1.00 1.00 1.00 1.00 1.000

1.10 1.10 1..21, 110 1.000

1..25 7.25 1.56 1.25 1.000

1.50 1.50 2.25 1.50 1,000

1.56 l.5h 2.44 1..56 1.003

0.80 7.25 1.00 t.25 1.000

t.20 1.50 1.80 1.50 1.000

1.33 0.50 0.67 0.50 1.000

Table 3.4

Example 3.11\\4rat is the observed result of increasing the total pressure under isothermalconditions in the following system?

nCl3(s) + C12(g) -- nC15(s)

A. No change in the Pp61, to Ppg1, ratio

B. An increase in the Pg1, to Ppglu ratio

C. A decrease in the ratio of Ppg1, to (Ppg1r) x (PCI2)

D. An increase in the Ppg1, to Ppq1, ratio

Solutionlncreasing the external pressure shifts the reaction to the right, so it can reduce

the pressure to counteract the stress associated with increasing the pressure ofthe system. An alternative way to look at this is to say that when the totalpressure increases, the number of collisions between molecules increases, forcingthe reaction to proceed in the forward direction (only the forward reactiondepends on collisions). If the system shifts to the right, the partial pressure ofPC13 decreases, the partial pressure of Cl2 decreases, and the partial pressure ofPCl5 increases. This eliminates choice A, because choice A could be true only ifthe reaction did not shift. Choice B is eliminated, because an increase in the C12-

to-PCl5 ratio results from a reaction shifting in the reverse direction. Choice C is

eliminated, because the equilibrium constant does not change unless the

temperature changes. The change is said to be isothermal, so temperature didnot change during the shift. Choice D results from the reaction shifting in the

forward direction, so it is the best answer.

Copyright @ by The Berkeley Review r79 Exclusive MCAT PreParation

General Chemistry Dquilibrium Le Chdtelier's Principle

Example 3.12Which of the following reactions at equilibrium would NOT shift as the result oian increase in pressure?

A. CaO(s) + CO2(g) -L

CaCO3(s)B. PC13(g) + Cl2(s) + PCl5(g)C. Nz(g)+3Hz(g)

-

2NH3(g)D.Hz(g) +Iz(g):2HI(g)

SolutionThe equilibrium reaction does not shift after an increase in pressure when there L.an equal number of gas molecules on each side of the reaction. This is observedin choice D. As a point of interest, CaCo3 cannot be stored in an open space forthis reason, because it will fully dissociate into Cao and Co2 over time, if thecarbon dioxide partial pressure is not high enough. Also take note that 12 is a ga-"in choice D, which means that the reaction must be at a temperature abor-eambient conditions.

Perturbations and ShiftsBefore conducting an experiment, it is important to consider the properties of thecontainer in which a reaction transpires. Two common containers are the closeciflask and the closed piston. Flasks are made of a rigid material, so their volumeis fixed (Vconstant). The pressure varies when using a flask (Pvariable). Piston-oare flexible (a wall of the piston is free to move), so the volume can varr-(Vvariable). In piston reactions, the initial pressure equals the final p."rr.riu(Pmitlut = Pfinal), if the volume of the piston exhibits no instantaneous change.This is to say that for a piston that starts with a stationary lid, once the lid isstationary again, the internal pressure equals the external pressure. As a resu,ltreactions are susceptible to environmental changes (perturbations) depending or:the container. Temperature may change in each container, but its effect onpressure and volume depend on the container itself. Le ChAtelier's principleshould be applied taking into account the features of the reaction container.

Le ChAtelier's principle deals with changes to a system that starts in equilibriunr-.When you have a system in balance, a change in environment results in thedisturbance of equilibrium. To compensate, the system will shift either left o:right (increasing and decreasing concentrations in doing so) to reestablish"equilibrium. The general rule is that the system will do whatever it takes to undcor compensate for what you have done to disturb it. Table 3.5 lists the genencshifts a reaction undergoes to alleviate an applied stress.

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Applied Stress System's Adjustment Direction of ShiftAdd reactant Remove reactant To the RightAdd product Remove product To the LeftDecrease volume Decrease gas volume To the side with fewer moleculesIncrease volume Increase gas volume To the side with more molecules

Decrease temperature React to generate heat ln the exothermic directionIncrease temperature React to absorb heat In the endothermic direction

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Table 3.5

The Berkeley Revier

General Chemistry Equilibrium Le Chitetier's Principle

Direction of ShiftTo observe how these shifts work, each is applied to Reaction 3.6.

N2Oa(s) L}{= +146 k]/mol

Reaction 3.6

Situation #1 Add NzO+(g): This perturbation results in a system with too manyreactants. The value of K is greater than Q. To reestablish equilibrium, thesystem reacts in the forward direction to absorb some of the excess reactants andform more products. The system shifts to the right. The same result would haveoccurred had NO2(g) been removed.

Situation #2 Add NOZ(g): This perturbation results in a system with too manyproducts. The value of K is less than Q. To reestablish equilibrium, the systemreacts in the reverse direction to form more reactants and absorb some of theexcess products. The system shifts to the left. The same result would haveoccurred had N2Oa(g) been removed.

Situation #3 Increase the external pressure: This perturbation results in a systemrvhere the partial pressures are too high. This means that the space in which thegases coexist has been reduced, so their molecules are more crowded and collidemore often. To reduce this crowding, the system goes from products (twomolecules) to reactants (one molecule) and reduce the total moles of gas in thecontainer. Also, if they collide more often, this forces the NO2 molecules to formbonds and thus dimerize to N2O4. The system shifts to the left.

Situation #4 Increase the volume: This perturbation results in a system wherethe concentrations are too low. This means that the space in which the gases

coexist has increased, so they are less crowded and collide less often. Using theinverse of the reasoning from situation #3, the system reacts to make twomolecules of NO2(g) from N2O4(g). The system shifts to the right.

Situation #5 Heat the system: This perturbation results in a system where thereis too much free energy. By heating the system, we have added energy to thereaction. To consume most of this additional energy, the system reacts in theendothermic direction, which for Reaction 3.6 is the forward direction. Thesystem shifts to the right.

Situation #6 Cool the system: This perturbation results in a system where thereis too little free energy. By cooling the system we have taken energy away fromthe reaction. To regenerate energy to balance this loss of energy to some degree,

the reaction moves in the exothermic direction, which for Reaction 3.6 is thereverse direction. The system shifts to the left.

These scenarios all address a gas-phase equilibrium. The shifts are similar for a

solution-phase equilibrium, except that concentrations are considered, ratherthan volume changes of the container. Changes in the concentration can resultfrom changes in the volume (quantity) of solvent. The only effect is thatsituations #3 and #4 arc now dilution and evaporation of solvent. The systemstill reacts by asserting the inverse of the stress done upon it. When diluted, itreacts to increase its concentration. When solvent is removed, increasing theconcentration, it reacts to reduce the concentration. The rules of Le ChAtelier'sprinciple work very well, if you apply them correctly.

Copyright @ by The Berkeley Review lal Dxclusive MCAT Preparation

General Chemistry Equilibrium Le Chitelier's Principle

Example 3.13\zVhich of the following is the result of cooling an endothermic reaction that startsat equilibrium?

A. The amount of products decreases.B. The amount of reactants decreases.C. The equilibrium constant increases.D. A11 partial pressures increase.

SolutionCooling the system causes it to move in the direction that generates heat. Heat isreleased when a reaction that is endothermic as written goes from products toreactants. Endothermic reactions are viewed as follows:

Reactants + heat --:: Products

Heat can be treated as a reactant. As the reaction shifts to the left, the amount ofproducts decrease, the amount of reactants increase, and the equilibriumconstant decreases. This eliminates choices B and C, and makes choice A thecorrect answer. Cooling a system decreases the pressure, so in all likelihood, al1partial pressures would decrease, not increase. Even if the reaction shifts enoughto offset the pressure decrease due to reduced temperature, that is true only forreactants. Partial pressure of products must decrease. This eliminates choice D.You should recall that AH is positive for an endothermic reaction.

Example 3.14If the following reaction represents a system at equilibrium, indicate whichstatement is NOT true.

lC13(s)+CI2(g): PCl5(s)

A. Increasing the pressure would cause a decrease in pCl3.B. Adding PCl5 would cause an increase in PCl3.C. Increasing the volume would cause an increase in pCl3.D. Removing Cl2 would cause a decrease in PCl3.

SolutionThe term "Nor true" means false. Do not forget halfway through the problem.that you're looking for the false answer choice. A common mistake on "NoIproblems is to forget that the correct answer is a false statement. To avoid thr-.write either "T" or "F" next to each answer choice as you run through them. Therchoose the answer choice with the unique letter.

To compensate for increasing external pressure, the reaction shifts to the right, su.

it can reduce the pressure to counteract the increase in pressure felt from th.change to the system. If the system shifts to the right, PCl3 decreases, so choice -iis true. Adding a product to an equilibrium mixture shifts the reaction to th;reactant side in order to counteract the increase in PC15 (the product). If thrsystem shifts to the left, PCl3 increases, so choice B is true. Increasing the volurntdecreases the crowding in of the system. To compensate for increasing volumEthe reaction shifts to the left, so it fills in the empty space created by the volumeincrease. If the system shifts to the left, PC13 increases, so choice C is trueRemoving a reactant shifts the reaction to the reactant side in order to make u:for the lost reactant (Clz). If the system shifts to the left, PCl3 increases, so choi;:D is a false statement, and is thus our answer choice.

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Copyright @ by The Berkeley Review t82 The Berkeley Revier

General Chemistry Equilibrium le Chdtelier's Principle

Example 3.15For the following equilibrium reaction, what is the effect of increasing theexternal pressure?

ZSO2(g) + rO2(s) 4 ZSO3(g)

A. A decrease in the moles of sulfur dioxideB. An increase in the moles of oxygenC. A decrease in the moles of sulfur trioxideD. An increase the equilibrium constant

SolutionIncreasing the pressure crowds the molecules in a reaction and forces the reactionto shift to the side of the equilibrium with fewer gas molecules (in this case, theproduct side). This means that the moles of products (sulfur trioxide) increaseand the moles of reactants (sulfur dioxide and oxygen) decrease. This eliminateschoices B and C, and makes choice A the best answer. The equilibrium constantdoes not change with pressure changes, so choice D is eliminated.

Example 3.16NO2 is a brown gas, while NZO+ (the product of dimerization) is a clear gas. Thetwo are in equilibrium in a 1.00 liter flask. Upon heating, the contents of the flaskbecame darker brown. \rVhat can be said about the following reaction as written?

2 No2(e)

-

NzOa(g)

A. The reaction is endothermic as written,B. The reaction is exothermic as written.C. The reaction is isothermal as written.D. The reaction is adiabatic as written.

SolutionBecause the reaction produced a darker mixture with the addition of heat, thereaction must have shifted to the left, since it formed more of the brown gasf'lo21g1;. This means that heat is acting as a product in this reaction. If heat is aproduct, then the reaction is exothermic as written. This makes choice B correct.The term isothermql means that there was no change in temperature during thereaction, and the term adiabatlc means that there was no change in heat during:he reaction. Both of these statements (choices C and D) are incorrect.

Example 3.17i\hat is the effect of adding CaO(s) to the following equilibrium mixture?

CaCO3(s) + COz(g) + CaO(s)

A. The products will decrease.B. The reactants will increase.C. The equilibrium constant wiII increase.D. There is no change in the equilibrium.

Solution{dding a solid to a reaction mixture that is already at equilibrium has no effectrn the equilibrium. This means that the correct answer is choice D. A solid:tfects a reaction mixture only if the reaction is not yet at equilibrium. This is to

'av that the solid is involved only when it is the limiting reagent for the reaction.

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General Chemistry Buffers and fitration Indicators

SolubilitySolubility is a popular topic on the MCAT. However, for many people, it was atopic covered only briefly during their general chemistry course. For this reason,we shall look thoroughly at solubility. Conceptually, solubiiity is the breakingapart of a lattice to allow particles to move freely in solvent. In this section, wefocus on the dissociation of a salt into water. When a salt dissociates into water,the ionic forces holding the charged species together break, and interactions withthe dipoles of the water form. The fundamental rationale behind solubilityinvolves lattice energy (the interionic forces in the crystal form), the solvationenergy (the strength of the attraction between solvent and dissociated ions), andentropy (the solute form is more disordered than the crystalline form). Whethera salt remains in its crystalline form or dissociates into solution dependscompletely on the relative energetics of the three key features. Figure 3-5 showsthe process of dissociation into solvent (water) for sodium chloride.

Figure 3-5

The base of the container holds ihe undissociated salt, while above the salt areions in solution. Salts dissociate from the surface, where they have the least ionicinteractions with neighboring ions in the lattice, and thus are held least tighth'.Surface ions have the greatest contact with solvent, so they are more susceptibleto solvation. Corners dissolve away fastest, followed by edges and then faces.

DefinitionsFor every solute, there is a maximum amount that dissolves into a given volumeof solvent at a set temperature. The solubility of most salts has been measured at25"C and are compiled in iables (like Table 3.8). You should know how to extractinformation from a table. Information listed in solubility tables measures thedegree to which a salt dissociates into water. As a rule, some salts have a vervlor.i solubility, leading to a low numerical value for the equilibrium constan:associated with dissociation (KrO). The equilibrium constant associated with.dissociation is knor,r,'n as the solubility product. However, the relative KrO valuesfor a group of salts is not always a good indicator of their relative solubility. Thisis because the units of solubility product (KrO) vary with the number of ions.

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Copyright @ by The Berkeley Review ta4 The Berkeley Revie*

General Chemistry Equilibrium Solubility

The subject of solubility requires being familiar with several terms. Definedbelow are eleven that are regularly encountered in this subject.

Dissoloing: The breakdown of intermolecular forces between molecules as a solidbecomes a solute within a solvent. The molecule remains intact when dissolvinginto solution. An example is the dissolving of sucrose into water, where atoms inthe sucrose molecule remain covalently bonded, but the forces between sucrosemolecules are eliminated.

Dissociqtion: The breakdown of ionic bonds between atoms within a latticestrucfure as a salt turns into a solute within the solvent. The crystal lattice of thesalt breaks apart when it dissociates into solution. An example is sodiumchloride dissociating into water. The ionic bonds between sodium cations andchloride anions break, and the ions are stabilized by the partial charges of water.

Soluent: The species in greatest concentration into which the solute dissolves, orsalt dissociates. A solvent must be a fluid (have the ability to flow).Solute: The species not in highest concentration that dissolves into the solvent, orin the case of a salt, dissociates.

Solubility: A measurement of the degree of dissolving that a solute undergoesrvithin a particular solvent. The driving force for solubility is a preference forsolvation of molecules (or ions) over the lattice strength of the solid. In addition,entropy favors the dissolving process. As the solubility of a compound increases,it is deduced that either the lattice energy of the solid is decreasing, the solvationenergy of the solute form is increasing, or both effects are taking place.

Saturated: Describes the state of a solution at the point where no more solidi,solute) can dissolve into solution. When an aqueous salt solution is saturated,the rate of dissociation of the salt equals the rate of precipitation.Supersaturated: Describes the state of a solution where the amount of solidisolute) that is dissolved into solution is beyond the maximum amount at a giventemperafure. The solution is actually a suspension that when disturbed can forma precipitate rapidly. This state can be achieved by first heating a solvent, thenadding solute to the solution until the solution is saturated at that temperature.Slowly cooling this solution causes the amount of solute in it to exceed whatshould dissolve at the reduced temperature.

Solubility product (Ks/: The equilibrium constant for a dissociation reaction,Cetermined from the molar solubility according to standard rules for calculating:quilibrium constan ts.

ilolar solubility: The quantitative measurement of the maximum number ofmoles of solid (solute) that can dissolve into enough solvent to make one liter ofsolution under standard conditions. For all practical purposes, the solvent isahvays water in inorganic chemistry and thus the calculations are similar inrearly every example. Molar solubility can be thought of as the x-value in the:alculation of the solubility product (Krp).

3ram solubility: The quantitative measurement of the maximum number ofirams of solid (solute) that can dissolve into enough solvent to make onehtindred milliliters of solution under standard conditions.Jommon ion effect: This results in a reduction in the amount of solid (solute) that;an dissolve into solution due to the presence in the solution of an ion that is alsouesent in the solid. This concept is similar to Le ChAtelier's principle, except that

"-,-ith Le ChAtelier's principle, the addition of one of the products (ions) causes

:recipitation (reduced solubility). With the common ion effect, the ion causingIe reduced solubility is present in solution at the beginning of the reaction,iather than being added once the solution has reached a solubility equilibrium.

Copyright @ by The Berkeley Review ra5 Exclusive MCAT Preparation

General Chemistry Bquilibrium Solubility

Solubility RulesYou may recall from general chemistry a list of general rules for predictingsolubility. We will emphasize using data from tables to determine solubility, buthere is a paraphrased version of the solubility rules. The rules as they aretypically written in general chemistry textbooks are listed on page 26 of thisbook. These rules may be used to give a good estimate of solubility and shouldbe used when no solubility data are available.

1. Most salts composed of a +1 cation (excluding transition metals)and -1 anion are soluble in water at room temperature.

2. Nitrate (NO:-) is a large anion that forms weak latticeinteractions and forms strong hydrogen bonds with water, somost nitrate salts are water-soluble.

3. Most salts containing sulfate anions (SOnz-; with +1 cations(excluding transition metals) are water-soluble.

4. Most salts with -2 or -3 anions are insoluble in water, excludingthe sulfate salts.

5. Most oxide 1O2-; and hydroxide anion (OH-) salts are onlyslightty water-soluble. KOH and NaOH are notable exceptionsthat are substantially soluble.

Ionic StructuresComposed of ions (charged species) held together by ionic bonds (electrostaticforces). In ionic structures, electrons are not shared; they are transferred betweenatoms, so that atoms with a deficiency of electrons (cation) and atoms with an

excess of electrons (anions) are formed. A typical example of an ionic compoundis sodium chloride (NaCl).

Cation: An atom in which the number of protons exceeds the number ofelectrons, thereby resulting in an excess of positive charge. A cation is apositively charged atom. Potassium cation (K+) carries a +1 charge, implyingthat there is one more proton than the number of electrons in the atom"Potassium has nineteen protons, so potassium cation has eighteen electrons'

Anion: An atom in which the number of electrons exceeds the number otprotons, thereby resulting in an excess of negative charge. An anion is anegatively charged atom. Fluoride anion (F-) carries a -1 charge, implying thatthere is one more electron than the number of protons in the atom. Fluorine has

nine protons, so fluoride anion has ten electrons. Note that the name changes

from "fluorine" (used when the atom is neutral) to "fluoride" (used when the

atom carries a negative charge). Negatively charged species are given the "-ide'suffix.

Nomenclature of SaltsScientific convention says that when you name a salt, the name of the cationprecedes the name of the anion. That is why we refer to NaCl as sodiuqtchloride, rather than chloride sodium. The rules for naming a simple binary saltare as follows:

O Name the cation before the anion in the salt.

@ The cation name is derived from the element forming the cation (oftmending in "-ium")

O The anion name is derived from the element forming the anion with an'-ide" suffix added.

{

r

n

Ifl

ild

fi

Copyright O by The Berkeley Review la6 The Berkeley Revier

General Qhemistry Equilibrium Solubitity

Example 3.18\Alhat is the proper chemical name for the salt BaF2?

A. Barium difluorineB. Barium fluorideC. DifluorobaraneD. Fluorinium baride

SolutionThe cation of this salt is derived from barium, and the anion is derived fromfluorine. Thus, the name begins with barium and ends with the "-ide" form oftluorine (fluoride): barium fluoride, which makes choice B the correct answer.The subscript two in the formula (showing that there are two fluoride anions)need not be named, given that fluoride can be only -1 and barium is +2. The ratiois implied. It was fun to write wrong answers for this question. Granted, I am achemistry-loving geek, so my credibility when it comes to determining what is;ool and what isn't is dubious, but nevertheless, choice C is a cool wrong answer.

Polyatomic Ionsor,'er the course of your academic science career/ you may have seen the same:ons recurring in a variety of contexts in different subjects. In biology, thenteractions of bicarbonate with carbonate and dihydrogen phosphate withivdrogen phosphate are essential in blood buffering and in the action of the<idney. Kidney stones result from the precipitation of calcium with various:olyatomic ions. Because of the frequency with which molecular ions appear as: factor in many problems, the MCAT test writers expect you to be familiar withsome common ones. Table 3.6 lists a few that you should know.

-1 Anions -2 AnionsAcetate C2H1O2- Carbonate Co32-

Bicarbonate HC03- Chromate CrOn2-

Bisulfite (hydrogen sulfite) HS03- Dichromate Cr2072'Bisulfate (hydrogen sulfate) HSOa- Hydrogen phosphate HPc)42-

Dihydrogen phosphate H2POa Oxalate C2O42-

Hypochlorite CIO- Oxide 92-

Chlorite ClO2- Peroxide oz2'Chlorate ClO3- Sulfite Sos2

Perchlorate ClOa- Sulfate Sonz-

Cyanide CN.Hydroxide oH- -3 AnionsSuperoxide O2- Phosphate PO43-

Nitrite Noz-Nitrate Nos- +L CationsPermanganate MnO4- Ammonium NH+*

Table 3.6

r -.; should know these twenty-six ions by charge, structure, and nomenclature.t,:.oving a little about their common applications may help, too.

187 Exclusive MCAT Preparation

General Chemistry Equilibrium Solubility

Solubility Product and Molar Solubility (Mathematical Applications)With solubility, calculations typically involve converting between molarsolubility and solubility product (KrO). Molar solubility and solubility productshow up together in calculations, bul they are very different terms. If one salt

has a lower KrO value than another salt, it does not mean that it is less soluble(has a lower molar solubility). Solubility is determined by the molar solubility ofa salt, not the solubility product. KrO values are calculated as products overreactants, but because solids are ign6red, there is no denominator. Table 3.7

shows different ion combinations for salts, where x represents molar solubility.

Solubility Reaction Kro Expression Kro Calculation

MXls; --: M"(uq) + X-(aq) Ksp = M*llxl KsO =(x[x) = x2

MX2(s)

- M2*(uq) + 2X-(aq) Ksp = M2*l[xl2 Kro =(x{2xY =q*3

MX3(s)

-

M3*(uq) + 3X-(aq) Ksp = M3*l[r]3 Kro =(x{3x)3 =27*a

M2X(s) .* 2M+(aq) + X2-1aq) t<ro =1tvl+121x2-1 Ksp =(2xf(x) =4x3

M1X(s) + 3M+(aq) + X3-1aq1 rro =1M+131x3-1 rcro =(ex)31x)=27x4

Table 3.7

Example 3.19What is the molar solubility for the hypothetical compound MX, if it is known to

have KrO = 4.0 x 16-10 142 in water?

A. 1.0 x 10-5 MB. 2.0 x 10-5 MC. 4.0 x 10-5 MD. 2.0 x 10-10 M

SolutionFor an MX salt, K5p = [M+][X1 = x2, where x is the molar solubility. This means

that the molar solubility is the square root of the KrO value.

Krp =*2 .'.*=y'Krp ={cotlo:ro =,{4n */torTO* =2.0x 10-5

This makes choice B the best answer.

Example 3.20\zVhat is the molar solubility for the hypothetical compound MX2, if it is knor'r'n No

have Ksp = 1.08 " 10-7 Nts in water? ,,

A. 1.65 x 10-4 MB. 2.10 x 10-3 MC. 3.00 x 10-3 MD. 4.80 x 10-3 M

This makes choice C the best answer'

?^ '' '\t, 1 - \C6z: "

Y-' "''-1 - ''-"."\

t' '.' ta: '

t r)

SolutionFor an MXI salt, Kso = [M2+][X]2 = 4x3, where x is the molar solubility'means that the molai solubility is the cube root of one-fourth of the KrO value

'/ro8;lnq=ntl n

.t- a-=4127 ,"^'{rc'e = 3.0x 10-3

4

Copyright O by The Berkeley Review l8a The Berkeley

General Chemistry Equilibrium Sotubility

It is good to understand the hypothetical examples, because actual salts are donethe same way, except the numbers are not as clean. The key fact to recognize isfound in the units. If Kro is M2, then square root must be found. squaie rootsare easier to solve if the ljower of ten is divisible by two. if Ksp is M3, then cuberoot is necessary. Cube roots are easier to solve if the power often is divisible bythree. From that point it requires approximating square roots and cube roots byranging the values based on squares and cubes you know.

Example 3.21\zVhat is the molar solubility of lead(Il) iodide, PbI2,7.42 x 70-8IrzF in water?

.:Ii.>"

,.1

if it is known to have KrO =

jii

..,tLl I

A. 1.19 x 10-4 MB. 8.83 x 10-4 MC. 1.52 x 10-3 MD. 2.42x 10-3M

SolutionFor PbI2, KrO = [Pb2+][I-]2 = 4x3, where x is the molar solubility. This means thatthe molar solubility is the cube root of one-fourth of the KrO value.

Krp =4"3 "'t -

"- 1-

=ilUZ *",llO-e =Vl.SS x 10-3 = 1.5 x 10-3+Vqrhe cube root of 3.55 is not a common piece of knowledge. Flowever, the cube:oot of 1 is 1 and the cube root of 8 is 2. This means that the cube root of 3.55 falls:etween 1 and 2. This makes the best answer choice C.

-:- the previous three questions, you were asked to derive the molar solubility,':om the solubility product. An alternative way to pose this question is to askru to derive the solubility product from the molar solubility. Before you attack

----ese problems, it is important that you make a conscious note of potential.-,:stakes. Remember to multiply the ions by the coefficient from the balanced

e quation! This is the most common error on these problems. The second trick is:, remember cube roots by thinking of common numbers cubed, such as 23 = 8,',1 = 27, and 43 = 64.

:rample 3.22: at is the solubility product of an M2X salt with molar solubility = 5.0 x 10-2 M?

r". 1.25 x 10-4 M3: 2.50x10-4M3 (l'- (i

''

-- 1.00 x 10-4 M3- 1.25 x 10-3 M3

: - lution: : ar1M2X salt, KrO = [M+]2[X2-] = 4f, where x is the molar solubility. This-' :;r1s that the KrO value is four times the molar solubility cubed.

1 I rr?Krp =4xj = +x(S.0 rt0-2)t = 4x725x 10-6 = 500x 10-6 = 5.0x 10-4

--,; makes choice C the best answer.

5!P4

:-. right @ by The Berkeley Review la9 Dxclusive MCAT Preparation

General Chemistry Equilibrium Solubility

Relative SolubilityIt is difficult to prldict relative solubility. The only way to comPare is through

experiments or data analysis. We'll start by comparing the solubilities of calcium

carbonate (CaCO3) and calcium fluoride (CaF). When a question asks for the

highest solubility, it refers to the compound that produces the greatest amount of

dissociated salt, which refers to greatest molar solubility, not solubility product!

The solubility product of CaCO3 is 8.7 x 70-9 \,/P, and the solubility product for

CaFr is 4.0 ri0-11M3. whlch of the two salts is more soluble?

Witli questions like these, keep in mind that you never compare numbers with

unlike units. Solubility products are used to determine the molar solubility'

CaCo3 is an MX salt, sJ iro = "t, while CaF2 is an MX2 salt' so KsP = 4x3' The

determinations of molar solibility for both calcium salts is shown below.

CaCo3: * = lR* =t[ei;i'ug ={87 xfio-1o = 9'? x10-s M

CaFr: x =: E9Y- V4af-

=Vro "{tr'FT =2.?x1o-4M

The molar solubility for CaCO3 is 9.3 x 10-5 M_ and for CaF2 it is 2.2 x 10-4, but

exact values are not necessary, b".urrr" 9.? x 10-5 M is less than2.? x 10-4 M. CaF:

is more soluble than CaCO3, even though it has a smaller solubility product'

Relative solubility questions are asked in many ways such as: "\Alhich salt

exhibits greater solvation?" and "Which salt precipitates first?-'' Use molar

solubilitylo answer questions that address relative solubility' Solubility product

is employed only for calculation-based questions'

Example 3.23

Vr4rich of the following salts yields the LEAST amount of silver ion in water?

A. AeBr(s) K". = 7.9 * 1g-13142

B. a[cri'i r,i = i.o * 1s-10 142

C. a[zcto+t'i Krp = 9.0 * 19-12 tr43

o. a!3ro41r1 frf'= 1.6 * 19-18144

SolutionYou must recognize that the choices ale not all the same type of salt. Choices A

and B are MX ialts, so their solubility products may be compared^ AgBr has a

lower solubility product than AgCl, soAgBr also has a lowerrnolar solubiLit''-

This eliminates choice B. Choice C is an M2X salt, and choice D is an M3X sai:

To solve the question, you need molar solubility values- The calculation of th'e

molar solubiliiy for the three remaining salts is shown below'

Choice A: *=y'K; ='[zs *taB ='[7g x^tlT{ = 8'9x 10'7M

l

x

(

I

Choice C: x =

Choice D: x =

I Rr" z ffi" 1g-TT

V4 V 4

4 R; o,fs t lcrlF4l ' =4t1'=-Yzz Y zz

= 1.3 x 10-4 M

= 1..6 x10-5 M

=1'lri ,

=\ (,.7 x)tr

d

s

dll

AgBr, choice A, has the lowest molar solubility' However, the question asked fllloivest [Ag*], not lowest molar solubility. For choice C, Ag2CrO+, there are tf i:sltver cations, and choice D, Ag3PO4, has three silver cations, so the [Ag*] :-'

actually double and triple their iespective molar solubilities, making them mu::too high to be the actual answer.

t0-12

10-20

Copyright O by The BerkeleY Review 190 The BerkeleY Revier rf,

General Chemistry Dquilibrium Solubility

Solubility ExperimentsThe molar solubility of a salt is an empirical value, determined from experimentsinvolving saturated solutions. While forming a saturated solution is easy to do,determining the ion concentration is not necessarily an easy task. To form asaturated solution, enough salt is added to water so that precipitate remains onthe bottom of the container. Determining the amount of dissolved salt may beaccomplished in many ways, of which we shall discuss three. The first methodinvolves spectroscop!, where the amount of absorbed light at a fixed wavelengthdepends on the concentration of ions, according to Beer's law (Equation 1.3).This method works only if one of the ions absorbs electromagnetic radiation. Thesecond method involves measuring the amount of salt added and then collecting,drying, weighing, and then subtracting the mass of precipitate from this value.This method works only if the salt is highly soluble. The third method involvesusing an ion exchange column to exchange the cation in solution for hydronium.The concentration of hydronium is determined via pH. This method works onlyif the salt does not exhibit acid-base properties and has a cation concentration ofat least 10-5 M. You may encounter passages about all three methods.

Example 3.24What is the molar solubility of calcium carbonate in water given that the additionof 4.00 mg of CaCO3(s) to enough water to form exactly 500.0 mL of an aqueousCaCO3 solution yields 0.55 mg of anhydrous precipitate?

A. 6.90 x 10-3 M CaCO3B. 3.45 x 10-3 M CaCO3C. 6.90 x 10-5 M CaCO3D. 3,45 x 10-5 M CaCO3

SolutionYou must keep in mind the units of the question and then solve for each unit. Inother words, the answer is in terms of moles/liter, which implies that you mustfind both moles and iiters to solve the question. The liters of solution are simple,in that 500.0 mL is equal to 0.500 liters. The amount of calcium carbonate thatCissociates into solution is 3.45 mg. The amount of calcium carbonate inmilligrams is converted to grams by multiplying by 10-3, and then to moles byCividing by the molecular mass of calcium carbonate (100 grams per mole). Themathematical set-up is shown below

Moles: 3'45 x 10-3 gramsCacoa = 3.45x 10-5 molesCaCo3

1oo gramTmole

Volume: 500.0 mL . *#h = 0.500L CaCo3(aq)

Concentratio',. 3.45 x L0-5 molesCaCO3 = 6.90x

0.500 L solutionThe correct answer is choice C.

1o-5 MCaCo3laqy

>olubility looks daunting, but is actually quite simple conceptually. Don't be-ntimidated by difficult-looking topics. They're usually pretty easy. Someapplications of solubility include ion exchange (ion exchange columns), selectivesolubility (precipitation of a selected cation or anion), and the common ion effect.iJf ion exchange, selective solubility, and the common ion effect, only common-on effect involves heavv mathematics.

-opyright O by The Berkeley Review t9t Exclusive MCAT Preparation

General Chemistry Equilibrium Solubility

Common Ion EffectThe common ion effect is the reduction in solubility of an ionic solid (salt) in

solution, because one of the ions in the salt is already present in solution. In a

way, the concept is similar to Le Chatelier's principle. According to Le

ChAieter's principle, when you add one of the ions to a saturated salt solution,

the reaction shifti to the lefi (reactant side). This results in the formation of a

frecipitate (more solid), so the salt is less soluble. Addition of an ion to solution

,"d.r""d the solubility. According to the common ion effect, if one of the ions is

already present in iolution, then less salt is capable of dissociating, so the

sotufiiity is reduced. The results are the same when an ion is added to solution;

the difference between Le Chdtelier's principle and the common ion effect is the

timing of when they are added. Consider Reaction 3'7'

AgCl(s) + HrO(l) =- Ag*(^q) + Cf(aq)

Reaction 3.7

with the addition of either Ag+ or Cl- to solution in Reaction 3.7, there is a

reduction in the amount of Agcl that can dissociate into it, due to the presence in

the solution of one of the ions-constituting the salt (Ag+ or Cl-). This is similar to

Le Chatelier's principle, except that either Ag+ or Cl- is present in solution in the

beginning rather than being added once the solution has reached a solubilitl

uqii6U.i i*. euestions ot ih" MCAT may be asked from either a common ion

perspective or a Le ChAtelier's principle's perspective'

Le Chatulier's style question: which way will the reaction shift when Ag+(aq) is

added to a saturated aqueous silver chloride solution? According to Le

Chatelier's principle, adding a product shifts the reaction to the reactant side' so a

precipitate ?ortnt. This means that the salt becomes less soluble.

Common ion effect style question: Is silver chloride more soluble in pure water or a

0.10 M Naciiaq) sbluiionl According to the common ion effect, because the

sodium chloride solution has chloride i-ons (also found in silver chloride) already

present in solution, the solubility of the salt is reduced. we shall look at this'question from a mathematical perspective, later in the discussion of this topic'

The common ion effect, simply put, says that the presence of a mutual ion

already dissolved into solution tldn."t the solubility of a second salt sharing that

mutual ion. For instance, stannous fluoride is less soluble in a solution of sodium

fluoride than distilled water because of the presence of the fluoride anion'

Common ion effect questions can incorporate pH and hydroxide concentration.

Example 3.25

rurufirir- hydroxide is MoST soluble in which of the following solutions?

A. Vinegar (PH = 2.5)

B. Citric acid (PH = 4.5)

C. Los Angeles taP water (PH = 5'5)

D. Distilled water (PH = 7.0)

Solutionihis question involves the common ion effect, where the common ion is

hydroxide anion. Magnesium hydroxide is most soluble in the solution with the

least hydroxide anion)which corresponds to the most acidic solution. The most

acidic solution has the lowest pH, wirich makes choice A the best answer' This is

the reason that acid rain is tr.h u serious problem: it dissolves basic salts'

Copyright @ by The BerkeleY Review 192 The BerkeleY Review

General Chemistry Equilibrium Solubility

Example 3.26At what pH is Ca(OH)2(s) most soluble in water?

A.38.5c.9D. 11

SolutionThe pH at which Ca(OH)Z(s) is most soluble is the pH at which there is the least

common ion (OH-) present in solution. There is the least OH- present at lowerpH values, where the solution is acidic. The best answer is choice A. Note thatbasic salts are most soluble in acidic solutions.

Example3.27(NHa)3(CoCl6)(s) is MOST soluble at which pH?

A.38.5c.9D. 11

SolutionThe pH at which (NHa)3(CoCl6)(s) is most soluble is the pH at which there is the

most OH- in solution. This is because of a complex equilibrium. The OH- insolution can deprotonate the ammonium cation (NHa+) to form ammonia, whichreduces the amount of ammonium present in solution. As the amount of NH4+in solution decreases, the (NHa)3(CoCl5)(s) equilibrium shifts to products togenerate more NH4+. The most OH- is present at higher pH values, where thesolution is basic. The best answer is choice D. Note here that acidic salts aremost soluble in basic solutions. The two equilibrium equations are as follows:

(NHa)3(CoCl5)(s) -_- 3 NH4+(at) + CoCl63-(at)

NH4+(aq) + OH-(aq) =- NH3(at) + H2O(aq)

-{s the second reaction shifts to the right, the first reaction is displaced fromequilibrium, so it too shifts to the right. This is a complex equilibrium. The two:eactions are flependent on one another.

Be able to make qualitative predictions about solubility based on both the;onunon ion effect and complex equilibrium. Example 3.25 and Example 3.26 are:ramples of the common ion effect. Example 3.27 is an example of increased

'olubility due to the presence of a complexing ion in solution. Both of these:oncepts play a role in determining solubility. \Alhen doing calculations with the:rmmon ion effect, be sure not to plug variables into the solubility product::rmula blindly. In the case of an MX salt, for instance, KrO still is equal to\l+l[X], but the values of [M+] and [X-] are not just x with the common ion

-ifect. The concentration of the ion that is already present in the solution is:rund by summing the initial concentration and the additional ion formed from-:.e dissociation of the salt (x). This is [X-]61g61 + x. In most cases, the x will be

:significant relative to the [X-]nitiul, so it can be ignored. Most common ion:tect questions should be simple and fun after you have done enough of them.

*rpvright @ by The Berkeley Review 193 Exclusive MCAT Preparation

General Chemistry Equilibrium Solubitity

A.B.c.D.

Example 3.28What is the molar solubility of CaCl2(s) in 0.01 M NaCl(aq) solution?

CaCl2(s) 4 Ca2+1aq; + 2 Cl-(aq) Ksp = 2.5x 10-10 M3

2.5 x 10-4 M2.5x10-6M . ." '' :'l t"i";

2.5 x 10-8 M2.5 x 1o-10 M

SolutionThis question is purely mathematical. The key fact is recognizing that there0.01 M Cl- in solution initially. The setup and solution are as follows:

Reaction: CaCl2(s) : Ca2+1aq; + zCI-

Initially: excess

Shifl - x ---+Equilibrium: whocares?

Kro = pa2+ltcrl2 ...Kro =1x{0.01. +zxf

If weignore2x: Ksp =2.5x1g-10 =1x{0.01f = 0.0001(x) .'.x = 2.5x 10-6M

The correct answer is choice B.

Example 3.29\Alhat is the molar solubility of Ca(OH)2 in an aqueous solution at pH = 13, giveur

that the KrO for Ca(OH)2 is 6.5 x 10-o Mr?

A. 6.5 x 10-5MB. 6.5 x 10-4MC. 'J..2x10-2M

D. 6.5 x 1,020 M

SolutionThis question is purely mathematical as well. You must recognize that there is

0.10 M OH- present in a pH = 1?_solution, because pH + pOH = 1.4 fot anaqueous solution and [OH-] = 10-PoH. The equation, set-up, and solution are as

follows:

Reaction: Ca(OH)z(s)

Initially: excess

Shift: - x

Equilibrium: whocares?

0 0.01

+x +2xx 0.01 + 2x

-

Ca2+1aq; + 2OH-

0 0.10

--+ + x +2xx 0.10 + 2x

Kro = [Ca2+]toH-12 .'.Kro =(x{0.10 +zxf

Upon ignoring 2x: Kro = 6.5x 19-6 =1x{O.t0P - 0.0{x) .'.x = 6.5x 10-4 M

The correct answer is choice B.

Calculation questions involving the common ion effect are actually rather simpL+but in most cases the average student is not familiar with the process. Si-pSplug in the value for the preexisting concentration, and solve for x from the K*expression.

Copyright O by The Berkeley Review 194 The Berkeley Revicn

General Chemistry Equilibrium Solubility

Separation by PrecipitationRemoval of certain ions (cations or anions) from solution can be done in a

selective fashion, by precipitation of cations with various anions, according totheir relative solubility values. The least soluble salt precipitates from solutionfirst. Questions about selective precipitation involve comparing molar solubilityvalues. Be careful not to fall into the trap of thinking that the smallest KrO is theleast soluble. You must consider what type of ionic system it is. The compoundwith the smallest value for x (molar solubility) precipitates from solution first.

Table 3.8 lists the solubility products for some common salts at standardtemperature. The values are listed without units, as is conventional in generalchemistry textbooks, but you should always consider the units when they areprovided.

Salt Ksp Salt Ksp Salt Ksp

BaF2 2.4x10-5 CaF2 4.0 x 10-11 MsFr 6.4 x 10-9

PbF2 4.1x 10-8 SrF2 7.9 x10-10 ZnF2 2.5 x 10-8

BaS04 1.5 x 10-9 CaSOa 6.1x 10-5 MgSOa 7.2x\0-5PbSOa 4.1x 10-u SrS04 7.9 x10-10 ZnSO4 8.3 x 10-9

BaC03 L.6 x 10-9 CaCO3 6.1x t0-9 MgCOs 1.1 x L0-15

PbCO3 1.4 x 10-15 SrCO3 8.4 x 10-10 ZnCOs 2.2x 10-12

Ba(OH)2 5.3 x 10-3 Ca(oH)2 7.4x 10-6 Mg(oH)2 L.2 x L0-11

Pb(oH)2 1.3 x 10-15 Sr(OH)2 4.7 x 1,0-4 zn(oH)2 2.1 x 1O-16

Table 3.8

Figure 3-6 shows a precipitation flow chart from a qualitative analysisexperiment designed to identify cations in solution. The cations are selectedfrom Table 3.8. All of the cations are +2 cations, so they all form the same type ofsalts. As such, solubility products can be directly compared. Each step showsthe salt that precipitates.

SrSOa(s) Ba2*, ca2+a small amount of BaSOa(s)

also forms a precipitate

CaF2(s)

Figure 3-6

Ba2*,ca2*,Pbz*,srz*

Ba2*,ca2*,sr2r

Copyright @ by The Berkeley Review 195

Ba''

Exclusive MCAT Preparation

General Chemistry Equilibrium Solubility

.1

,,, x

r \ t.

t^

| ..) -, ,

t,)

1-J

- (D^ltL

Example 3.30According to data in Table 3.8, to remove strontium by precipitation from asolution with Ca2l Mg2*, and Sr2+, it is BEST to add:

A. F-B. SO42-c. Cotz'D. OH-

SolutionThe best anion to add to solution to remove a cation is the one that forms a saliless soluble than the salts of the other cations in solution. Of the three cations(Cu2*,}y'rg2*, and Sr2+), the least soluble fluoride salt is CaF2, eliminating choiceA. Of the three cations, the least soluble sulfate salt is SrSO4, so choice B is thebest answer. Of the three cations, the least soluble carbonate salt is MgCO3, andthe least soluble hydroxide salt is Mg(OH)2, eliminating choices C and D.

Example 3.31Mercury II cations are MOST soluble in which of the following solutions?

A. 0.10 M NaClB. 0.10 M Na2SC. 0.10 M NaID. 0.10 M NaNO3

SolutionThis is a question of relative solubilities. This question is asking what the relativesolubilities of HgCl2, HgS, HgI2, and Hg(NO3)2 are. The key fact to keep inmind is that nitrates are infinitely soluble in water (one of the solubility rulesi.Of the anions, only nitrate can form hydrogen bonds with water, so it has thegreatest solvation energy of the four anions. It is also the largest of the anions, scit has the weakest electrostatic forces in its lattice structure. The conclusion isthat nitrate salts are the most soluble, so the best answer is choice D. This ispurely a qualitative argument for the solubility. There is mathematical evidenceto back this up as well, but the numbers were not given in the problem.

Example 3.32Given a solution that combines Ag+(aq), Pb2+1aq;, Sr2+1aq;, and. Zn2+1aq), what i-.

the sequence of precipitation when NaCI(aq) is added to solution?

Ksp AgCl =2x70-10,Ksp PbClz =2x10-5,Ksp SrCl2 =9 x10-3,Ksp ZnCl2 = 1 x 104

.K.. 1.st: AgCl, 2nd: SrCl2, 3rd: ZnCl2,4th: PbCl2

E. 1sl AgCl, 2nd:PbCl2,3rd: ZnCl2,4th: SrCl2

,C. 1st: SrCl2, ?nd ZnCl2, 3rd: PbCl2, 4th: AgCl

"8. 1-st: PbCl2,2nd: ZnCI2,3rd: SrCl2, 4th: AgCl

SolutionThe salt with the lowest molar solubility forms a precipitate first. AgCl has tlelowest solubility product, and the molar solubility is the square root of KrO. Fcnthe other three salts, use the cube root of KrO to obtain molar solubility. Thlsmakes AgCl the least soluble, and the first to-precipitate, eliminating choices Cand D. The remaining salts have +2 cations, so the order of precipitation is frcmthe lowest KrO value to the highest KrO value. This makes choice B correct.

Copyright @ by The Berkeley Review 196 The Berkeley Reviw

General Chemistry Equilibrium Solubility

Ion Exchange ColumnAn ion exchange column exchanges one ion in solution for another (which isinitially bound to the column), by precipitating out the ion that forms the lesssoluble salt and releasing a more soluble ion into solution to_replace it. A watersoftener is an ion exchange column. Hard water (rich in Ca2+1aq;) travels downthe column where the anion in the ion exchange resin binds calcium cations toform an insoluble salt while releasing sodium cations (Na+(aq)) into solution.Aqueous sodium cation is referred to as soft water. As calcium precipitatesthrough the ion exchange column, it is filtered from the water, preventing it fromforming precipitates in the plumbing lines of appliances (such as a washer orwater heater). To evaluate the viability of an ion exchange column, the molarsolubility values are compared. Figure 3-7 shows an ion exchange column.

Ca2*1aq; enters at the top ofthe ion exchange column.

The column is filled with an ionexchange resin. The cation to beprecipitated must have a lowermolar solubility with the anion ofthe resin than the salt in the resin.

Na"(aq) exits from the base

of the ion exchange column.

Figure 3-7

ln the water softening example, the removal of calcium ions from tap water isaccomplished by passing the water across rock salt (large chunks of NaCl) andthen filtering out the calcium chloride. Reaction 3.8 is the exchange reaction in aivater softener.

2 Na+(aq) + CaCl2(s) =+ Ca2*1aq; + 2 NaCl(s)

Reaction 3.8

Rock salt is employed to minimize surface area, and to prevent sodium chloridetrom dissociating into solution too rapidly. Sodium chloride is used because itsions are non-toxic, and it does not change the pH of the water.

Selectiae Precipitntion and lon Exchtange: The addition of an ion to solution with thei:rtention of precipitating an existing ion out from the solution. This is achievedivhen the salt being precipitated is less soluble (has a lower molar solubility) thanany other salt combination in solution.

2AgCl(s) + Hg2*{ug)

-

2Ag+(aq; + HgCl2(s)

Reaction 3.9

Because mercury chloride is less soluble than silver chloride, the addition ofsilver chloride to the mercury solution precipitates mercury out of water. This is-he same principle applied in ion exchange columns and can also be applied inqualitative analysis when we look for ions in solution.

I

Copyright Oby The Berkeley Review 197 Exclusive MCAT Preparation

General Chemistry Equilibrium Solubility

Complex lon Formation: Complex equilibrium occurs when two reactions aresummed to form one overall reaction. This is possible when the product of onereaction is a reactant in a second reaction. A great example involves the additionof ammonium to salt buildup in sinks in order to remove the deposit. To checkthis, take a small whiff of a bathroom cleaner, and you will detect ammonia.

CaCo3(s) + H2o(t) =:> Ca2*1aq; + Co.?2-(aq)

Reaction 3.10

Ca2*1aq; + 4NH3(aq) + Ca(NHr)n2+1aq;

Reaction 3.11

Because ammonia forms hydrogen bonds with water, the calcium-ammoniacomplex is more soluble than the calcium cation. By the addition of ammonia tosolution, free calcium cation forms a complex ion (the result of Reaction 3.11shifting in the forward direction), which reduces the amount of free calcium andforces the Reaction 3.10 to proceed in the forward direction. As mentioned prior,when reactions are added together, the equilibrium constant for the overallreaction is the product of the equilibrium constants of the component reactions.

Complex lons (Coordination complexes): Complex ions form when a ligand (lonepair donor or Lewis base) donates a pair of electrons to a central atom (Lewisacid), which is typically a metal, to form a coordinate covalent bond. Thisprocess is referred to as chelation, where the Lewis base is a chelating agent. Aprime example is hemoglobin whose porphyrin ring serves as a polydentateligand by chelating the central iron.

Example 3.33Silver chloride is MOST soluble in which of the following solutions at 25"C?

A. 0.10 M HgNO3(aq)B. Pure waterC. AgNO3(aq)D. NaCl(aq)

SolutionSilver chloride (AgCl(s)) is most soluble in a solution where Hg+ is present.is because of a complex equilibrium. The Hg* i.t solution can bind the chanion (Cl-) to form a precipitate. As the amount of Cl- in solution decreaserlmore AgCl dissolves into solution to replace it. The best answer is choiceBoth choices C and D are eliminated, because of the common ion effect.

Copyright @ by The Berkeley Review r9a The Berkeley

I. Carbon Dioxide and Carbon Monoxide Equilibrium

II. tlydrogen and Bromine Equilibrium

III. Equilibrium Constant Magnitude

IV. Qas-Phase Equilibrium Constant Experiment

V. Equilibrium Reactions

VI. Le Chatelier's Principle

VII. Solubility and Qualitative Analysis

VIII. Solubility Experiment

IX. Qualitative Analysis Experiment

X. Solubility Chart

XI. Calcium Salts Solubility

XII. Complex Equilibrium

XIII. Hemoglobin and Acclimation

XIV. Equilibrium Keaction of NO2 and NzOA

Questions Not Based on a Descriptive Passage

Dquilibrium Scoring Scale

Kaw Score MCAT Score

84 - IOO 15-1566 B5 10 l247 65 7 -934 46 4-6l-33 t-3

(r -7)

(8 - 15)

(r7 - 22)

(23 - 29)

(5O - 56)

(37 - 42)

(45 - 4e)

(5O - 56)

(s7 - 62)

(65 - 6e)

(7O - 75)

(76 - 82)

(85 - Be)

(eo - e6)

(e7 - 1OO)

Passage I (Questions 1 - 7)

A researcher studies the equilibrium of carbon monoxide,'. ith carbon dioxide in the presence of oxygen gas at various

::nperatures. The relationship is shown in Reaction 1.

2 CO(g) + Oz(g)=+ 2 COzG)

Reaction L

This is accomplished by monitoring the total pressure of.- 3 system as it depends on temperature. There are different:-'ichiometric amounts of gases on each side of Reaction 1'

: ':he reaction can shift with changes in pressure or volume.

. : quantify the contents in the flask at equilibrium, the

::3ssure for the system is analyzed. It is assumed that

::3ssure changes are due to shifts in equilibrium and changes- :he system that obey the ideal gas equation (PV = nRT).

- -:anges in the total pressure are evaluated for the system and

:: equilibrium constant is derived from the values for::essure. The reaction is monitored for three trials at each

::rlperature over a wide range of temperatures. The results

.:: summarized in Table l.

Temperature (K) Ko (atm-r)

375 09.0425 20.8500 32.2600 60.5

Table 1

Data in Table I demonstrate that the value of the

: luilibrium increases with increasing temperature. Although::e equilibrium system shifts upon changes in the moles of-3actant, moles of product, external pressure, and volume, the

:quilibrium constant remains constant.

1. At 102'C, PCo = 2.0 atm. and Pg, = 1.0 atm. What is

the partial pressure of CO2 at this temperature?

A. 2.2 atm.B. 6.0 atm.C. 9.0 atm.D. 18.0 atm.

Z . Addition of COz(g) to the equilibrium mixture results in

which of the following?

A . A decrease in the moles of O2(g).

B . A decrease in the moles of CO(g).C . An increase in the value of Ko.D. No change in the value of Ko.

Copyright @ by The Berkeley Review@ 20r GO ON TO THE NEXT PAGE

3. According to Table 1, the reaction is which of the

following?

A. Exothermic as writtenB. Endothermic as writtenC . Adiabatic as carried outD. fsothermal upon reaction

4. The equilibrium constant (K) varies only with which ofthe following?

A. A change in the pressure ofthe system

B. A change in the volume of the system

C. A change in the concentration ofthe system

D . A change in the temperature of the system

5 . If a flask were filled with pure COZG) to a total pressure

of 1.00 atm., then once equilibrium is reached' the total

pressure of the system is which of the following?

A. Less than 1.00 atm.

B. Exactly 1.00 atm.

C. Between 1.00 atm. and 1.50 atm.

D. Exactlv 1.50 atm.

6. According to Reaction I, compressing the reaction

vessel leads to which of the following?

A . An increase in the partial pressure of COZ(g)

B. An increase in the moles of OZ(g)

C . An increase in the mole fraction of CO(e)

D. An increase in the equilibrium constant

7 . Based on the following reaction, the addition ofCa(OH)2(aq) to Reaction I at equilibrium results in

which of the following?

ca(oH)2(aq) + CO2(g):- caco:(s) + HzO(l)

A. No change in the moles of COZ(g).

B . A decrease in the concentration of O2@).

C . An increase in partial pressure of CO(g).

D. An increase in the total pressure'

Passage ll (Questions 8 - 15)

Hydrogen bromide gas is one of the strongest acidsavailable in gas form. In aqueous solution, HBr has a pKuvalue of approximately -12. A pKa value of -12 is indicativeof a very strong acid. By comparison, HCI has a pK6 ofapproximately -9. Hydrobromic acid (HBr) can be producedby Reaction 1.

H2(g) + Br2(l)

Reaction 1

A student decides to study the equilibrium distributionfor Reaction 1 Under ambient conditions, Reaction 1

reaches equilibrium at some temperature T and pressure p.Once at equilibrium, measurements are taken for the partialpressure of each gas and the mass of bromine liquid. Thevalues are listed in Table 1.

B12(l) 10.0 grams

Hz(e) 1.36 atm.

FlBr(e) 2.'72 atm.

Table 1

The student takes notice that Reaction I shifts withchanges in the conditions of the system, such as the additionof bromine liquid, the removal of hydrobromic acid, andincreases in volume. The student irreversibly increases thevolume of the system by opening a valve on the reactionflask that is connected to an evacuated column. Not allchanges shift the reaction. Changes in the equilibriumconcentrations are not accurately recorded.

8 . What is the value of K at ambient temperature, T?

A. 0.20 atm.B. 0.544 atm.C. 2.00 atm.D. 5.44 atm,

9. Addition of HBr(g) to an equilibrium mixrure ofReaction 1 results in which of the fbllowing?

A . An increase in the partial pressure of H2@.1

B. A decrease in the partial pressure ofH2(g)C . An increase in the concentration of Br2oD. A decrease in the concentration ofBr2(1)

10. Which of the following starting conditions results inthe GREATEST amounr of H2(g) at equilibrium?

A. 0.80 atm. Hz(g) and2o g Br2(l)B. 1.00 atm. H2(g) and 20 gBr2e)C. 0.80 atm. H2(s) and 30 g Br2(l)D. 1.00 atm. HBr(g) and 30 g Br2(l)

Copyright O by The Berkeley Review@

I 1. Subsequent experiments show that as the temperatureincreases, the mole percent of the hydrogen bromide ga.increases. Which of the following is TRUE about theequilibrium reaction as written?

A . lt is endothermic.B. tt is exothermic.C. It is isobaric.D. It is not an equilibrium mixture.

I 2. How can it be explained that a reaction mixture of f .iratm. HBr(g) and 1.0 atm. H2(g) results in no change iuhydrogen gas partial pressure?

A . The reaction is in equilibrium.B. The reaction cannot move in the forward direc

due to a limiting reagent in zero concentration.C . The reaction cannot move in the reverse di

due to a limiting reagent in zero concentration.D . The reaction is moving forward and backward ar

same rate.

13. Mixing 3.0 atm. HBr(g) with 1.5 atm. H2(g) orwBr2(l) shows which of the following in time?

A . A large increase in the partial pressure of HBr@:B . A small increase in the partial pressure of HBrtgC. A large increase in the partial pressure ofH2(glD . A small increase in the partial pressure of H2(g

I4. A flask is inirially filled with pure HBr(g). Whichthe following graphs represents the partial pressureH2(g) over time?

A. B.

$fi

m

tr

m

l{u

r!um

time

C. D.9)

()!AN

+r

gLFbtrme

elnl-time

15. Addition of which of the following to the equilimixture will NOT affect the partial pressure of H2(gri

A. NaOH(aq)B. Br2(l)c. Hz(e)D. HBr(g)

202 GO ON TO THE NEXT PAGS

Passage lll (Questions 16 - 22)

Chlorine gas has a high reduction potential (largepositive voltage for its reduction half-reaction). Chemicallyspeaking, chlorine gas is very easily reduced. Chlorine gas is

a strong oxidizing agent that can dissolve away most metals

in short periods of time. The dissolving is the result of an

oxidation-reduction reaction. The relative ease with whichmetals are oxidized accounts for this strong tendency to lose

electrons to chlorine. Chlorine gas is not as successful at

oxidizing non-metals as it is at oxidizing metals. This isattributed to the decreased desire for non-metals to be oxidizedrelative to metals. Non-metals have a higher ionization3nergy and electronegativity than metals. With non-metals,

;hlorine may only partially oxidize the correspondingreducing agent. Reactions 1 and 2 are prime examples of the

rxidation capability of chlorine on non-metal compounds.

Pcbtel + Cl2G) =-Reaction 1

2 NO(e) + Cl2(g) =+

PClste)

2 NOCI(g)

Reaction 2

The equilibrium constant for Reaction 1 is 1.95 atm.-1rr 100"C, while the equilibrium constant for Reaction 2 is1.1 x 104 atm.-l at i00"C. Both reactions favor products as'.',ritten, but not to the same degree.

In these examples, the non-metals (phosphorus inReaction 1 and nitrogen in Reaction 2) are already in a

:ositive oxidation state in the reactants. The greater the. alue of the equilibrium constant, the more favorable the

.:action, and thus the more product that is formed.

1 6. According to the equilibrium constants associated withReaction 1 and Reaction 2, which compound is mostreadily oxidized by chlorine?

A. PCI:B. PCl5

C. NOD. NOCI

[ 7. If a flask were filled with PCl3 and C12 to a totalpressure of 1.50 atm. such that the mole fraction ofPCl3 is twice that of Cl2, then what is the totalpressure of the system once at equilibrium?

A. Less than 1.00 atm.B. Between 1.00 atm. and 1.25 atm.C. Between 1.25 atm. and 1.50 atm.D. Greater than 1.50 atm.

-opyright @ by The Berkeley Review@ 203 GO ON TO THE NEXT PAGE

18. What is the equilibrium constant for the followingreaction?

2 NOCI(g) =- 2 NO(e) + Cl2(g)

A,. 4.9 x 10-17 atm.B. 4.9 x 10-5 atm.C. 1.4 x 102 atm.D . 2.1x 104 atm.

19. If there is initially 1.00 atm. of NOCI in a flask, what

is the partial of NO gas, once at equilibrium?

A. 0.0230 atm.B. 0.0460 atm.C. 0.0025 atm.D. 0.0050 atm.

20. The GREATEST concentration of CIZ(g) is in which ofthe following systems, once equilibrium is reached?

A. A system initially with 1.0 atm. PCl3

B. A system initially with 1.0 atm. PCl5

C. A system initially with 1.0 atm. NOD. A system initially with 1.0 atm. NOCI

21. Which of the following statements is true about the

effect of adding chlorine gas to each equilibrium?

A . It will cause the greatest change in Reaction l.B. It will cause the greatest change in Reaction 2.

C. It will cause equal changes in both reactions.D . It will change the equilibrium constant.

22. A flask at 100"C is charged with 0.40 atm. each ofPCl3, PCl5, and Cl2. In which direction will it react

to reach equilibrium, according to Reaction I ?

A . It is already at equilibrium.B. To the left (reactant side)

C. To the right (product side)D . It varies with a catalyst.

Passage lV (Questions 23 - 29)

The equilibrium constant for a reaction involving gases

can be determined by experimentation, using a glass column

as a closed system. The apparatus used in equilibrium

experiments, shown in Figure 1, is connected to a vacuum

pump at one end and is sealed at the other' On the column

u." ih."" stopcocks with air tight fittings, to which flasks

containing gut"t -uy be attached. At the sealed end of the

tube is a pressure gauge. The volume of the glass column is

exactly 1.00 liters, when all stopcocks are closed' Each flask

and aitached stopcock contain volumes of exactly 250 mL'

This means that when all three stopcocks are open in a

manner where gas is free to flow between the glass column

and any of the ;ttached flasks, the total volume for the closed

system is exactly 1.75 liters. Figure I shows the apparatus'

Pressure gauge

Flask I Flask 2

Connectedto vacuum

Flask 3

Figure 1

During an experiment, the closed glass cylinder is filled

with 1.00"ut-. of hydrogen (H2) according to the gauge in

the cylinder. Flask I is filled with 1.0 atm' carbon disulfide

(CS;). The reaction starts when the stopcock is opened'

allowing the two gases to mix. The temperature of the glass

column is maintained at 25"C using an external heat sink'

The internal pressure is monitored until it stays constant'

Figure 2 shows the internal pressure over time, where t = 0

represents the time at which the two gasses were mixed'

Figure 2

The pressure of each gas in the reaction mixture can be

calculated from the change in internal pressure' The initial

partial pressure of hydrogen gas is 0'8 atmospheres in the

l.ZS tit"t closed system. The decrease in partial pressure of

hydrogen gas is double the decrease in the internal pressure'

based on the stoichiometry of the Reaction 1, which shows

the reactivity of the comPounds.

Copyright @ by The Berkeley Review@

Time(minutes) €

1 CSz(g) + 4H2G) =: 1 CH+(e) + 2 HzS(er

Reaction L

The final internal pressure is the sum of the parrr:zu

pressures. The total pressure can be found using Equation 1

Ptotal = PCSz + Pg, + Pggo + PHzS

Equation 1

2 3. Which of the following is the BEST explanation a-< uwhy the reaction temperature must be held at 25'C?

A . A temperature change can cause the carbon disul6ja

to decomPose to carbon dioxide'

B. The reaction will not proceed unless ma

temperature is exactlY 25"C.

C. The glass will begin to melt at temperatures ru

excess of 25"C.

D . The equilibrium constant can change rr----m

temperature'

24. The temperature of the closed reaction vessel m'rs

increased. This resulted in the pressuls ef ths qlon:di

system decreasing, rather than increasing as expeci*ix

Which of the following explanations BEST expla:mn

this observation?

A. The pressure of a closed system decreases a-i futemperature increases according to Charles's laq

B. The reaction is endothermic as written'

C . The reaction is exothermic as written'

D. Heavier molecules are being formed causing

number of collisions to increase'

25. How can the partial pressures of CH4 and H'S

equilibrium be determined from the experiment?

A. At equilibrium, the partial pressure of CH4

the value for the change in the pressure ofsystem (APtotut), while the partial pressure of X{

equals twice the value for the change tn

pressure of the system (2AP16161)'

B. At equilibrium, the partial pressure of H2S eq

C.

the value for the change in the pressure ot

system (APtotut), while the partial pressure of (

equals twice the value for the change tn

pressure of the system (2APtotal).

At equilibrium, the partial pressure of CH4 eq

the value for the change in the pressure ot

system (APtotuil, while the partial pressure oI

equals half the value for the change in the pre:

of the system (lAPtotat).

At equilibrium, the partial pressure of H2S

the value for the change in the pressure o;

system (APtotat), while the partial pressure ot {

equals half the value for the change in the pt

of the system (lAPtotat).

204

D.

GO ON TO THE NEXT P.

26. Which of the following relationships between partialpressure for the gases in the mixture at equilibrium willALWAYS be true for the reaction as it is set up?

4. PCSz > PCH+

3. PCH+ > PgzS

C. PH2 > PCS2

D. Pgr5 > Pg,

27. Which of the following equations represents theequilibrium expression for the reaction?

A. Ko=(PcH/(PHrs)2

(P65r)GH2)a

B. Kn- (Pcsr)(PHr)a-v

lP6goXPgr5 )'

C. Kn - (Pcno)tPu's)

' (P6grXPs2)

D. Kn - (Pcs,xPH,)' (pggoXpn2s)

I8. The final total pressure of the system must be:

A. greater than 1.6 atmospheres.B. greater than 1.0, but less than 1.6 atmospheres.C. greater than 0.6, but less than 1.0 atmospheres.D. less than 0.6 atmospheres.

I9. Which of the following stresses on the system willchange the value of the equilibrium constant?

A . Opening the stopcock to Flask 2, which was filledwith hydrogen gas.

B. Turning on the vacuum pump to remove some ofthe gas in the column.

C . Opening the stopcock to Flask 3, which was filledwith hydrogen sulfide (H2S) gas.

D. Heating the gas in the glass column by heating theglass with a hot air blower.

-opyright @ by The Berkeley Review@ 205 GO ON TO THE NEXT PAGE

Time

Passage V (Questions 30 - 36)

Chemical equilibrium obeys the rules of Le Chdtelier'sprinciple. Le Chatelier's principle states that when a systemat equilibrium is stressed, the reaction will shift in a mannerto partially relieve the stress and reestablish equilibrium. Ina gas phase equilibrium, stresses that can be applied includechanges in pressure, volume, temperature, and moles ofeither reactant or product. It is not possible to change justone of the variables, so a change in one variable will lead toin a change in another variable. For instance, if the volumeis increased, the pressure must decrease. Written below are

three reactions and their respective equilibrium constants at

500'C:

Reaction l:2 SO2(g)+ I O2(g) -$ 2 SO:(e)

Kp = 5.82 x 102 atm.-l at 500"C

Reaction 2:

1 H2O(g) + I Cl2O(g) -$ 2 HOCI(g)

Kp = 8'61 x 10-3 at 500'C

Reaction 3:

1 HCI(g) + I CO(e) + I HCOCI(g)Kp = 3.26 x 10-6 atm'-l at 500'C

The equilibrium constant is derived by dividing thepartial pressures of the product gases by the partial pressures

of the reactant gases. In Reaction I and Reaction 3, theequilibrium constants were determined using a piston systemwhere the total pressure and volume were monitored, and anychange in pressure or volume can be attributed to a shift inthe reaction equilibrium. Because the number of productsequals the number of reactants, the total pressure and volumecannot change in reaction 2, so equilibrium concentrationswere determined using infrared spectroscopy.

30. Given the initial partial pressure for Pgrg is 0.5 atm

and for P61rg is 0.5 atm, which graph accurately shows

the rate as a function of time for the reaction?

A. B.

(.)

d!

ooo&

D.

0)

!

oo6)&

.9!

oodo

C.

C)2H

oo&

Time

Time Time

33.

31. Which reactions will shift when the pressure changes?

A. Reaction 1 onlyB. Reactions I and 2 onlyC. Reactions 1 and 3 onlyD. Reactions 1, 2, and 3

3 2. Upon increasing the volume at constant temperature ofa vessel holding Reaction 3, how are the partialpressures affected?

4. Pucoct increases; Pgcoct

increasesPSCT.PCO PHC1

g. PHcoct decreases;

PHcoct decreases

PHcr'Pco PHcr

g. Pucoct remains constant; PHcoct

increasePUCT.PCO PHC1

p. Pgcoct remains constant;

Pgcoct decreases

Pucr'Pco PHcr

Which graph depicts partial pressures as a function oftime after 0.75 atm SO2 is mixed with 0.50 atm 02?

-

SO2 ------ 02 -So:B.

Time

3 4. Which IGq value does NOT match the reaction?

A. 2 HOCI(g) =- I HzO(e) + 1 ClzO(e)Kp = 8.61 x 10-3 at 500'C

n. Iszo(ey+|ct2o1g) --

1HoCl(g)2

Kp = 9'28 x l0-2 at 500'C

c.;Hcocl(e) * |Hcrlg;+f co1g;

Kp = 5.54 x 102 at 500"C

D. 2 HCI(g) +2CO(s)

-

2HCoCl(e)Kp = 1.06 x 10-11 at 500'C

Copyright @ by The Berkeley Review@

A.

o!aq

tJr

D.

C)La6)!o.

C.

()k

a0-)

H

I60)k

Or

3 5. To convert a Kp_ value in terms of torr-l to a Kn valrcin terms of atm-1, you should:

A. multiply the Kp value by 760.B. divide the Kp value by 760.C. multiply the Kn value by 0.76.D. divide the Kp value by 0.76.

3 6. Cooling an endothermic equilibrium reaction resultswhich of the following?

A . The mole fraction of products will decrease.B . The mole fraction of reactants will decrease.

C . The mole fraction of products will increase.

D. The ratio ofproducts to reactants will increase.

206 GO ON TO THE NEXT

Passage Vl (Questions 37 - 42)

Le ChAtelier's principle is invoked to explain reaction

rrifts following the disruption of equilibrium. A system in:r namic equilibrium is a closed system where the forward::action rate and reverse reaction rate are equal. Components

,i the mixture maintain fixed concentration. Equilibrium is

:ifferent from steady state. In steady state systems' an

:termediate has a static concentration, because the formation

::action and the consumption reaction have equal rates.

A system can be disturbed from equilibrium by applying

. stress, such as changing the temperature, volume, pressure,

: moles of a reagent. According to Le Chdtelier's principle,

:re system reacts in a manner to partially alleviate the stress

.rd reestablish equilibrium. Reaction 1 starts with three

::iferent starting concentrations on its way to equilibrium.

CO(e) + HzO(e)

- COz(e) + Hz(e)

Reaction ITable 1 lists the starting and equilibrium concentrations

ri each of the three trials. The reactions are observed at

-r00'C and the equilibrium constant is found to be 0'569'

Table 1

Once at equilibrium, a separate stresses were applied to

,.;h system. To reaction System I, an additional 0.10 moles': CO was added to the 1.00 liter vessel in which the reaction,.:s housed. To reaction System II, all of the H2O was':noved from the 1.00 liter vessel in which the reaction was'rused. The 1.00 liter vessel containing reaction System III

:s cooled to 500'C. Once equilibrium was reestablished in.e three reaction vessels, the partial pressure of CO2 had

-:reased in reactions systems I and Itr and decreased in II.

r I . All of the following affect the reactant/product ratios ofthe equilibrium mixture EXCEPT:

A . increasing the volume of the reaction vessel.

B. heating the reaction vessel.

C. cooling the reaction vessel.

D . adding reactant to the reaction vessel'

I 8, If 1.00 moles of both CO2 and H2 were mixed in a

1.00 liter container at 1000'C, what will theequilibrium concentrations be for both CO2 and H2?

A. [COz] = 0.57 M; [Hz] = 0.57 MB. [COz] = 0.57 M; [Hz] = 0.43 MC. [COz] = 0.43 M; [Hz] = 0.57 MD. [COz] = 0.43 M; [Hz] = 0.43 M

\folarit

Trial IIIInitialInitial

-opyright @ by The Berkeley Review@ 207 GO ON TO THE NEXT PAGE

Time

3 9. As heat is added to Reaction l, what occurs?

A . Both the equilibrium constant (K"O) and the partial

pressure of CO2 decrease.

B. Both the equilibrium constant (K"O) and the partialpressure of CO2 increase.

C . The equilibrium constant (K"O) remains constant

while the partial pressure of CO2 increases'

D. The equilibrium constant (K"q) remains constant

while the partial pressure of CO2 decreases.

4 0. What effect would an increase in pressure have on the

equilibrium system of Reaction 1?

A. The reaction would shift to the right and the

equilibrium constant would increase.

B. The reaction would shift to the left and the

equilibrium constant would decrease.

C. The reaction would shift to the right and the

equilibrium constant would remain constant.

D . The reaction would not shift in either direction and

the equilibrium constant would remain constant'

4 1. Which of the following graphs represents what wouldbe observed over time if equal parts of CO(g)' COZ(g)'

and H2(g) were mixed in a closed system?

Time

-

= Total Pressure

= HzO Pressure

= CO Pressure

42. For an exothermic reaction at 100'C, the equilibrium

concentration of the products equals the equilibriumconcentration of the reactants. What is true at 25"C?

A . Keq > l; fProducts] > [Reactants]B . Keq > l; lProducts] < [Reactants]C . Keq < 1; lProducts] > [Reactants]D . Keq < 1; lProducts] < [Reactants]

A. R'-

,l3lnv'

()

ok

c-)ka6)

Ai

6)!r

Iq

Time Time

Passage Vll (Questions 43 - 49)

The solubility of a salt in water is measured by its molarsolubility. The molar solubility is defined as the moles ofsalt that dissociate into one liter of aqueous solution. Thesolubility product (KrO) is the equilibrium constant for thedissociation reaction. Table 1 lists solubility data for varioussalts of siiver, lead, strontium, and zinc.

Compound Krp Molar solubilityAgCl 17*19-10y2 1.3 x 10-5 MPbClr 1.6 x o-5 u3 1.2 x 10-2 MSrCl2 8.8 x o-3 u3 1.3 x 10-1 MZnCl2 1.1 x o-4 lt3 3.1 x 10-2 MLg2CO3 8.1 x g-12 y3 1.2 x 10-4 MPbCO3 8.3 x 0- l1 vp 9.1 x 10-6 MSrC03 7.1 x 6-10 y2 2.6 x 10-5 MZnCO3 2.0 x g-10 y2 1.4 x l0-5 MAgOH 2.0 x o-8 M2 1.4 x 10-4 MPb(oH)z 1.2 * 19-15 143 6.5 x 10-6 MSr(OH)2 3.2 x lT-a M3 4.3 x 10-2 MZn(OH)2 2.1 * 19-16 y3 3.8 x 10-6 MAgzS 1.6 * 19-49 143 3.4 x 1o-17 tvt

PbS 2.6 * 19-27 YP 5.1 x 10-14 MSrS 4.0 x 10-6 M2 2.0 x l0-3 MZnS 2.5 * 1g-22 YP 1.6 x t0-1 I MAg2SOa 1.4 x 10-5 M3 1.3 x 10-2 MPbSOa 1.7 x 10-8 M2 1.3 x 10-4 MSrS04 3.2 x l0-1 M2 5.6 x 10-4 MZnSO4 7.6 x l0-8 M3 2.1 x I0-4 M

Table 1

The relative molar solubility values can be used todetermine which salts precipitate out from solution whenvarious cations or anions are added. For instance, PbSOaprecipitates first when sodium sulfate is added to a solutioncontaining lead, strontium, and zinc cations, because leadsulfate is the least soluble (has the lowest molar solubility)of the sulfate salts.

4 3. Which of the following statements are valid?

L Sulfates are more soluble than carbonates.

II. Insoluble cations can be dissolved into solution byadding a complexing agent such as EDTA.

m. The solubility of chloride salts show pHdependence, while the solubility of sulfide saltsshow no pH dependence.

A. I and II onlyB. I and III onlyC. II and III onlyD . I, II, and III

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44. The Kro value for which of the following salts is theMOST accurate in terms of smallest percent error?

A. Ag2SOaB. Pb(oH)2C. SrCO3

D. ZnS

4 5. Which anion should be added to selectively remoleZn2+ from an aqueous solution that contains Ag-Sr2+, and Zn2+?

A. Cl-B. COr2'C. OH-D. so42-

4 6 , If a precipitate forms when NaCl is added to an aqueor$salt solution, what conclusion can be drawn?

A. There is no Zn2+1aq) present in solution.B. There is Ag+(aq) present in solution.C. There is both Sr2+1aq; or Pb2+1aq; presenr r!

solution.D. There is no Sr2+1aq) present in solution.

:!t!,

lm:.

hrl

lml

fr$uuilI

WgmU,l

wllr

jT

m

ilh

@

hr

$ii

T1.

W;ilo

,{h

mflh&Wd

flp!'!

47. If an aqueous solution containing Ag+, Pb2+, Sr2*. iZn2+ is treated first with chloride anion, secondh'n

48.

sulfide dianion, and lastly with hydroxide anion.what cation is most likely still present in solutionl

A. Ag*B. Pb2+C. sr2+D. zn2+

Adding acid to an aqueous solution containing a

salt has what effect?

A. It will increase the solubility.B. It will decrease the solubility.C . It will have no effect on the solubility.D . It will lower the temperature of the solution.

49. Silver hydroxide is MOST water-soluble at $'llthe following pH values?

A. 3

B. 5

c. 9D. 11

Passage Vlll (Questions 50 - 56)

The solubility of salts is often measured in terms ofmoles solute per liter solution in a saturated solution. I tmay also be measured in terms of grams solute per 100 mLsolution. The term saturated refers to a solution in which the

maximum amount of salt that can dissociated into solution is

dissociated. The more salt that dissociates into solution, the

more scluble the salt. Solubility measurements cannot be

used interchangeably when comparing the solubility of two

salts without first considering the molecular mass of each

salt. This can be observed in the following experiment.

Experiment I

Water at 25"C is added to a calibrated burette, to the 25-

mL mark. A salt is added in 1.0 grams increments, untilthe salt no longer dissociates into solution. At thispoint, a precipitate should settle to the bottom of the

burette. The temperature of the solution in the burette is

slowly increased until the salt completely dissociates.

The solution is slowly cooled until the first signs ofprecipitate are observed. At this point, the temperature

and volume of the solution are recorded.

Experiment 2

The solution continues cooling until it reaches 25"C. Atthis point, pure water at25"C is added quantitatively untilthe last detectable sign of the precipitate dissociate intosolution. The volume of the solution is recorded as

precisely as the burette reads.

Table 1 lists data for this experiment using hypothetical.alts MX, MY, and MZ. Column 3 lists data fromErperiment 1 and column 4 lists data from Experiment 2.

Salt Mass Volume when heated Volume w/ H2O added

\x 7.0 e 26.2mL (@ 31.6'C) 28.9 mL (@ 25.0'C)

\IY 9.0 e 25.3 mL (@ 27.4'C) 26.1mL (@ 25.0'C)

\12 4.0 c 25.7 mL (@ 30.3'C) 28.7 mL (@ 25.0'C)

Table 1

The molecular mass for MX is 100 grams per mole, for\[Y is 120 grams per mole, and forMZ is 150 grams per

:ole. The concentration of each solution is measurable in,:11' standard units, including molarity (moles solute per literr:lution), grams per mL, mass percent, and density.* ucentration units can be inter-converted. For instance, the

:,3rcent solution by mass can be multiplied by the density to

::termine the mass solute per volume of solution, which can

:e converted to molarity using molecular mass.

Adding solvent to a solution dilutes the solution and

::rs reduces the concentration of the solution. All of the

:.iasurements of concentration decrease upon the addition of;,:1r.ent, with the exception of the density. The density:.1f,nge depends on the relative density of the solvent and

s, lution. In these experiments, water is the solvent for each

;.:lution, and the density of water at 25"C is found to be

, 9971 grams per mL.

*rpyright @ by The Berkeley Review@

5 0. How do the solubility products for the three salts in the

passage compare to one another?

A. KspMx > KspMy > KspMZB. KspMy > KspMX > KspMZC. KspMZ > KspMX > KspMyD. KspMZ > KspMy > KspMX

51. Which of the following solutions has the GREATESTmolarity?

A. Saturated MX at 31.6 "C

B. Saturated MX at 25.0'CC . Saturated MY at 27.4 "C

D . Saturated MY at 25.0 "C

5 2. Roughly how many grams of MX will dissociate into25 mL of water at25.0'C?

A. Less than 4.0 grams.

B. Greater than 4.0 grams, but less than 7.0 grams.

C . Greater than 7.0 grams, but less than 9.0 grams.

D . Greater than 9.0 grams.

5 3. What is the molarity of a saturated solution of MY at

25"C?

9 grams

12g Srams x 0.0261 litersmole

U. 9 grams

120 grams x 0.0253 litersmole

9 srams x 0.026 I liters

120 grams

molen 9 grams x 0.0253 liters

120 grams

mole

5 4. As temperature increases, what happens to the various

concentration measurements of a solution, assuming no

evaporation of solvent and no addition of solute?

A. The density and molarity both increase.

B. The density increases, while the molarity decreases.

C . The density decreases, while the molarity increases.

D . The density and molarity both decrease.

A.

209 GO ON TO THE NEXT PAGE

5 5. Which of the following relationships at 25"C accuratelyshows the relative density values of saturated solutionsof MX(aq), MY(aq), andMZ(aq)?

A. 1.00 > MX(aq) > MY(aq) >MZ(aq)B. 1.00 > MY(aq) > MX(aq) >MZ(aq)C. MX(aq) > MY(aq) >MZ(aq) > 1.00D. MY(aq) > MX(aq) >MZ(aq) > 1.00

5 6. Which solution has the HIGHEST boiling point?

A . 1.0 grams MX with 10 mL water at25"CB. 1.0 grams MY with 10 mL warer at25"CC. 1.0 grams MX with l0 mL water at 50'CD. 1.0 grams MY with 10 mL water at 50'C

Copyright @ by The Berkeley Review@ 2to

Passage lX (Questions 57 - 62)

The qualitative analysis of an aqueous salt solinvolves the systematic addition of reagents designedidentify certain component ions. For instance, mercurybe distinguished from alkaline earth metals by adding su(S2-) to solution by the formation of an insoluble precipiwith mercury dication. Silver can be distinguishedalkali metals by the addition of either chloride, bromide,iodide. The precipitate that forms between chloride andcan be re-dissolved by adding ammonia to solution.

Once a precipitate is formed, it can be removedsolution by centrifuging the mixture and decanting awaysupernatant or by filtering away the filtrate. The solutican then be further analyzed for other ions. Table 1

the results of a matrix involving the mixture of four cainitrate solutions mixed with five potassium anion solutimixed one at a time. All solutions are 0.05 M, withexception of KOH(aq), which has a concentration of 0.10Any compound that forms a precipitate is assumed tomolar solubility less than 0.001.

Tahle 1

Table 2 shows the results of a similar experimentanions were exposed to a sequence of multiple reactions-anions selected for testing are chloride, iodide,sulfate. The test solutions are silver nitrate, bariumammonia, and nitric acid. The matrix of Table 2 lisrrobservations of sixteen different test tubes. Nitric acidwith carbonate anion to form carbon dioxide gas, whichbubbles out of solution.

Table

GO ON

)

TO THB NEXT

5 7. What anion should be added to a solution in orderidentify any barium cation present in the solution?

A. CrO42'

B. so42-

C. C2Oa2'

D. co32-

5 8. What reaction took place when nitric acid was added tothe precipitate formed from mixing Ba2+ and COtz-,

A. Ba2CO3(s) + 2 HNO3(aq) -+2 BaNO3(aq) + CO2(g) + H2O(1)

B. BaCO3(s) + 2 HNO3(aq) -+Ba(NO3)2(aq) + COz(e) + H2O(1)

C . Ba2CO3(s) + HNO3(aq) -+BaHCO3(aq) + BaNO3(aq)

D. BaCO3(s) + 2 HNO3(aq) -+Ba(NO3)2(aq) + CO2(s) + Hz(e) + OzG)

5 9. Why is it NOT possible to use hydrochloric acid insteadof nitric acid in experiment II?

A . Chloride anion will interfere with the reactions.B. Nitrate anion will interfere with the reactions.C . Hydrochloric acid is too weak to react.

D. Hydrochloric acid is too strong to use.

6 0. If an unknown mixture was treated with K2SO4 and a

precipitate formed, what can be concluded?

A. The unknown contained no calcium or magnesiumcations.

B. The unknown contained a barium cation.C. The unknown contained a strontium cation.D . The unknown contained either a strontium cation, a

barium cation. or both.

tfr 1. Which of the following anions does form a whiteprecipitate with silver cation but does NOT react withnitric acid?

A. SulfateB. CarbonateC. ChlorideD. Iodide

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62. The information from Table I was used to derive thefollowing flow chart:

Ba2*. sr2*, ca2*. Mg2*

lv7Al

BaV(s)1L )L )L

Sr-'. Ca- , Mg-

)+ )-+SrW1s.1 Ca- , Mg'

lxorY/ \

CaXG) Mg'*

VMgzzG)

Which of the following choices represents anINCORRECT anion-toletter correlation?

A. V = COtz'B. W = SO42-

C. X = CzO+2-

D. Z=OH-

ofCaY(s)

Passage X (Questions 63 - 69)

Water can be analyzed for various inorganic componentsusing a series of qualitative tests for metals in their variousoxidation states. This process is referred to as qualitativeanalysis. Many of the tests involve the formation of an

insoluble salt. The formation of a precipitate results fromthe addition of an anion to an aqueous cation solution. Someanions may complex with more than one metal, so multipletests are necessary. Molar solubility is different for each salt,so occasionally an anion may precipitate with one cationover another and the second metal can be "masked". For thisreason a schematic of the series of tests is evaluated.Insoluble to the eye is defined as less than one mgdissociating into one mL of solution. By looking at thecombination of precipitates formed, the cations can benarrowed down until one is chosen with reasonable accuracy.

A researcher analyzed an aqueous sample believed tocontain roughly equal concentrations of Ag+, gu2+,7n2+,

and Na+. Table 1 lists solubility data useful in developing a

schematic for the tests. An ideal schematic sequentiallyidentifies one cation at a time, leaving the remaining cationsin solution.

Compound Ksp Molar solubilityAgzS 1.6 " 19-49 y3 3.4 x 10-17 MZnS 2.5 * 19-22 YP 1.6 x 10-11 MCaS 1.1 * 16-11 142 3.5 x 0-6 MCaCO3 8.7 x l0-9 M2 9.3 x o-5 MAg2CO3 3.1 " 19-12 tr43 1.2 x o-41,tZnCO3 2.6 ,. 16-10 142 1.4 x o-5 vAgCl 1.6 * 16-10 y2 1.2 x l0-5 MZnCl2 1.1 x 10-4 M3 3.1 x l0-2 M

Table 1

The values listed in Table 1 are all determined in distilledwater at 25"C. h is occasionally necessary to conduct tests atan elevated temperature, in order to carry out qualitativeanalysis and the separation of cations. Sodium salts are notlisted in the table, because sodium salts are infinitely solublewith nearly all anions, compared to the other metals listed inthe data. To identify sodium cation, a flame test is oftenapplied. By applying a flame to a small sample of thesolution, electrons in sodium are excited. When they relaxback to their ground state, photons are emitted. In the case

of sodium, the light emitted is orange in color.

63. When 0.1 gram ZICO3 is added to 100 mL of water at25'C, which of the following statements is true?

A . It dissolves completely.B. It dissolves almost completely, with only a small

portion not dissolving.C. It dissolves slightly, with most of the ZnCO3

remaining not dissolving.D . None of the sample dissolves.

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64. It is best to visually distinguish a sample tube_acontaining Znz+ from a separate sample tube

containing Ag* by the addition of:

A. 0.10 M Na2S.

B. 0.10 M NaCl.C. 0.10 M NaNO3.D. 0.10 M Na2CO3.

65. In which solution is Ag2S MOST soluble?

A. 0.10 M Ag+(aq)B. 0.01 M Ag+(aq)C. 0.10lr4 S2-(aq)

D. 0.01 M S2-(aq)

66. What is the molar solubility of ZICO3 in 0.010lfZnCl2?

A. 2.0 x 10-12 MB. 2.0 x 10-10 MC. 2.0 x 10-8 MD, 1.4 x 10-8 M

67. Which of the following salts is MOST solublewater?

A. CaCO3B. Ag2CO3C. ZICO3D. AgCl

68. Which of the following values accurately depictssolubility product of Car(PO+)2 if the molarof Ca3@Oa)2 is represented by x?

A. 18 x3

B. 27 x4C. 54 x5

D. 108 x5

69. To remove a cation from solution, the solutionflowed through an ion exchange column where iprecipitate the cation while releasing a morecation. Which of the following filters will NOT

A. Ag+(aq) through a column containing NaCI(slB. Zn2+laql through a column containing CaS(slC. Ca2+(aq) through a column containingD. Ca2+1aq1 through a column containing

Fassage Xt (euestions 70 - 75)

Calcium salts are generally insoluble in water at standard::rperature. Biologically, this is beneficial in that the Ca2+"n_is soluble enough to be transported through aqueous,.edium, yet insoluble enough to be a major component of:-:uctural features in the human body, such as bones. Listed- Table I are some Ksp values at2l.C for various calcium;..1t s :

Calcium Satt Ksp (@ 27'C)CaF2 4.9 * 16-11 y3CaS04 6.t x 10-6 u2CaC03 9.t x t0-9 M2Ca(OH)2 4.3 x 10-6 M3Ca3(POfi2 1.3 * 19-32 y5

Table IThe solubility products of the given salts can be

: , npared only when the anions bonded to calcium carry their-ne.negative charge. For instance, the solubility products: - CaF2 and Ca(OH)2 can be directly compared to determine:"::ir relative solubilities. To enhance the solubility of a:r-cium salt, an anion that complexes calcium can be added: solution. The anion competes for calcium in a complex:,qrilibrium. A complex equilibrium is defined as the:--upling of at least two equilibrium reactions where the::rduct of one reaction is the reactant in another equilibrium::rction. Reaction 1 and Reaction 2 combine to form a;,,nplex equilibrium involving calcium cation.

Ca(OH)Z(s) + HZO(I) E- Ca2+1aq1+ 2 OH-(aq)

Reaction 1

Ca2+1aq1+ Na2CO3(s) T.L CaCO3(s) + 2 Na+(aq)

Reaction 2

A change in calcium cation concentration affect both:rirtions. If the Reaction 2 shifts to the right, then Reaction- rs forced to shift to the right in order to maintain itss-.rilibrium.

-{.1 . What is observed when sodium carbonate is added to asaturated calcium hydroxide solution with undissolvedcalcium hydroxide on the bottom ofthe flask?

A. The pH of the solution will increase more than irwould increase had it been added to an unsaturatedsolution of calcium hydroxide.

B. The pH of the solution will increase less than itwould increase had it been added to an unsaturatedsolution of calcium hydroxide.

C . The pH of the solution will decrease more than itwould decrease had it been added to an unsaturatedsolution of calcium hydroxide.

D. The pH of the solution will decrease less than itwould decrease had it been added to an unsaturatedsolution of calcium hvdroxide.

Copyright O by The Berkeley Review@ 213 GO ON TO THE NEXT PAGE

BufTerpH --+

77. The maximum lcu2*l in apH = 14 solution would bewhich of the following values?

A. 4.3 x 108 M Ca2+B. 4.3 , 1g-2 y1 gu2+C , 4.3 * 16-6 14 6u2+D, 4.3 x 10-10 M Ca2+

72. Enough CaCl2 is added to water so that not all of itdissolves, and thus some CaCl2 remains as a solid onthe bottom of the flask. The [Cl-] is then measured.Addition of which of the following ro the solurion willincrease the chloride ion concentration (iCl-l)?A. WaterB. Silver nitrateC. Calcium fluorideD. Sodium phosphate

73. Which of the following graphs depicrs the log lCa2+1versus the buffered pH of an aqueous buffer solution ascalcium hydroxide is dissolved into different solutions?

7 4. If addition of salt to water makes the water warmer oncethe salt has dissolved, which of the following are truefor solvation of that salt?

A . AH is positive, AS is negative.B. AH is negative, AS is positive.C. AH and AS are borh positive.D. AH and AS are both negative.

75. Which of the following, when added to water at 300 K.will yield the highest LCa2+l?

A. CaF2B. CaSO4C. CaCO3D. Ca(oH)2

BufferpH --->

BufferpH ->

BufferpH _+

Passage Xll (Questions 76 - 82)

Smog is a collection of several gas pollutants. Mostprevalent in the conglomeration of impurities are the sulfurand nitrogen oxides. The brown color so familiar in smoggyskies is due to the presence of nitrogen dioxide (NO2). This,along with nitric oxide (NO), make up the majority of theairborne nitrogen oxides. The nitrogen oxides can be inter-converted by a series ofequilibrium reactions.

2NO(e)+Oz(g)5*2NO2(g)Reaction 1

2 NOz(e) -q== NzO+(e)

Reaction 2

NOz(e) + Oz(e): NO(g) + O3(g)

Reaction 3

Reactions I and 2 are both exothermic while theReaction 3 is endothermic. Reaction 3 requires the additionof light to transpire. As the day progresses, sunlight carriesout Reaction 3, resulting in the highest ozone concentrationin the early afternoon. Figure 1 depicts the concentration ofpollutants during a hypothetical day:

cooz

Time

oI(ooI@

oo6

IIn.-iII

oo+

II

o3

o9c!

Noz

Ooci

-

NO

Figure 1

The amount of the nitrogen oxides in the air can bereduced by bubbling air through an aqueous transition metalhalide solution such as MnCl2(aq). Nitrogen dioxide (NO2)reacts with the water to form nitric acid (HNO:) and nitrousacid (HNO2), both of undergo deprotonation and bind themanganese, forming a complex ion. Reactions 4 and 5 makeup a complex equilibrium, as do Reactions 4 and 6.

2 NOzG) + HzO(t) : HNO2(aq) + HNO3(aq)

Reaction 4

MnCl2 + 2 NO2-(aq)

-Mn(NOz)z+2Cl'(aq)Reaction 5

MnCl2 + 2 NO3-(aq) + Mn(NOiz + 2 Cl- (aq)

Reaction 6

When reactions are added to produce an overall reaction,the equilibrium constants for the reactions are multiplied toobtain the equilibrium constant for the overall reaction. Thisis true for any type of equilibrium reaction.

II

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7 6. Which of the reactions has a negative entropy change?

A. Reaction 2B. Reaction 3

C. Reaction 5D. Reaction 6

7 7. Which of the following will NOT increase the [N2Oa]?

A . Decreasing the temperature.B. Increasing the pressure.

C . Addition of water vapor to the air.D. Addition of NO2 to the air.

78. If the following reaction is carried out in a closedreaction vessel, what will be observed for theof the system if the temperature is doubled?

A(e) + B(e) =-

C(e) + D(g) + E(g)

AH = -1l2kJlmole

A. The pressure will remain the same.B. The pressure will increase by less than lO}Vo.C . The pressure will exactly double.D . The pressure will increase by more than l00%o.

7 9. What can be concluded about the change in enthalpyReaction 2?

A. Because a bond is broken, theendothermic in the forward direction.

B. Because a bond is broken, theexothermic in the forward direction.

C. Because a bond is formed, theendothermic in the forward direction.

D. Because a bond is formed, theexothermic in the forward direction.

reactior

reaction

reactim

80. If Reaction 1 is carried out in a closed pistonwith an external pressure of 1 atm, what occurs0.1 atm of NO gas is added to the system?

A . The volume decreases bv more than ten

B. The volume decreases by less than ten perceril.C . The volume increases by less than tenD . The volume increases by more than ten

time

8 2. Which of the following graphs accurarely depicrs theNO2 concentration in a flask containing reaction 2before and after it has been bubbled through MnCl2(aq)?

81. Which will shift reaction 4 to the left?

A . Addition of sodium hydroxide to the solutionB . Addition of manganese(Il) chloride to the solutionC. Removal of nitrate from the solutionD. Removal of water from the solution

B.

Passage Xlll (Questions 83 - 89)

Physiological response is correlated with climate. Thiscan be proven by observing the shortness of breath and/orheadaches experienced by mountain climbers and skiers whenthey rapidly change altitudes. These symptoms are associatedwith a disorder referred to as hypoxia, a deficiency in theamount of oxygen that reaches body tissue. Hypoxia is atemporary disorder that in some cases can be fatal in the shortterm. After time, the disorder will disappear because thebody becomes acclimated to the new environment.

The cause of hypoxia is a result of the decrease inoxygen in the environment caused by the increase in altitude.The partial pressure of oxygen gas at sea level in most partsof the world is just over 0.20 atm. At an altitude of onemile above sea level, the partial pressure of oxygen gas isroughly 0. 16 atm. Over time, the body adjusts to the loweroxygen content by increasing the production of hemoglobin(Hb), the molecule that binds oxygen. Reaction 1 expressesthe binding of oxygen by hemoglobin.

Hb(aq) + 4O2@e + Hb(O2)a(aq)

Reaction 1

The equilibrium expression for Reaction I is:

1q = [Hb(ozl+l

tHbltozlaFigure 1

Hemoglobin is composed of four separate polypeptidestrands, each capable of binding a ferrous (Fe2+) cation, heldtogether as one molecule. By producing more hemoglobin,the body can shift the equilibrium to the right (forwarddirection). This results in more oxygen diffusing across thecell membrane. Once in the cell, myoglobin transports oneoxygen molecule from the interior of the cell membrane tothe mitochondria.

In acclimation, the amount of myoglobin does notchange. Acclimation takes a period of time that ranges fioma week or two to months. For the 1968 Mexico CityOlympics, several athletes trained in high altitude toapproximate the environment of Mexico City (elevation7500 feet above sea level). For those athletes that trainedcloser to sea level, many did not perform well in the"thinner" air. Conversely, many current runners train in themountains for sea-level races so that their blood can provideextra oxygen to starving muscle cells.

8 3 . Once hemoglobin transfers oxygen to the cellmembrane where it is absorbed, how many myoglobinmolecules should be present per hemoglobin?

A. 1

B. 2

c. 4

D. 8

A.

time

time

- rpyright @ by The Berkeley Review@ 215 GO ON TO THE NEXT PAGE

84. Which of the following changes will NOT increase theamount (in moles) of free oxygen in the body?

A. Increased ventilation, resulting in an increasedamount of air transferred through the lungs.

B . An increase in the production of HbC . Increasing the partial pressure of CO consumed in a

normal breath.D . Increasing the total amount of blood.

8 5. What will be observed when a long-time mountainresident travels back to sea level?

A. They will experience hypoxia due to the lowerpartial pressure of oxygen at sea level.

B. They will experience hypoxia due to the higherpartial pressure of oxygen at sea level.

C . They will experience an increase in vitality due tothe lower partial pressure ofoxygen at sea level.

D. They will experience an increase in vitality due tothe higher partial pressure of oxygen at sea level.

8 6. In the presence of carbon monoxide, hemoglobinpreferentially binds CO over 02 in a ratio close io 200: 1. Over time, what occurs in the blood of someonemoving to a carbon monoxide rich environment?

A. The increased CO reduces rhe amount of Hb thatbinds oxygen, so the Hb(Oz)+ decreases. Tocompensate, the bgdy produces more Hb andincreases its rate ofrpspiration.

B. The increased CO reduces the amount of Hb thatbinds oxygen. sor' the Hb(O)a increases. Tocompensate, the i body produces more Hb andincreases its rate dfrespiration.

C. The increased CO reduces the amount of Hb thatbinds oxygen, so the Hb(O)a decreases. Tocompensate, the body produces less Hb anddecreases its rate ofrespiration.

D. The increased CO reduces the amount of Hb thatbinds oxygen, so the Hb(OZ)+ increases. Tocompensate, the body produces less Hb anddecreases its rate of respiration.

8 7 . How does hypoxia affect the amount of myoglobin?

A. Myoglobin increases initially and then decreases.B. Myoglobin increases initially and then remain at

higher concentration.C . Myoglobin decreases initially and then remain at

lower concentration.D. Myoglobin remains constant.

Copyright @ by The Berkeley Review@ 216, GO ON TO THE NEXT

8 8. What activity will most likely result in hypoxia?A. Repelling down a mountainB. SCUBA divingC. Water skiing on the oceanD. Snow skiing

8 9 . Why should an athlete choose to train at high altitudes?

A. At higher altitudes, the partial pressure of oxygenis greater, thus the athlete's body will increase itshemoglobin count.

B. At higher altitudes, the partial pressure of oxygeris less, thus the athlete's body will increase irrhemoglobin count.

C. At higher altitudes, the partial pressure of oxygeris greater, thus the athlete's body will decrease ibhemoglobin count.

D. At higher altitudes, the partial pressure of oxygeris less, thus the athlete's body will decreasehemoglobin count.

Passage XIV (Questions 90 - 96)

The equilibrium of a gas-phase reaction can vary withfie dimensions of the container. This is observed when a:eaction has a different number of gas molecules on the twosides of the reacrion equilibrium. Although the equilibrium;onstant remains constant when pressure, volume, and moles.ire altered, the ratio of product to reactant can vary. Thefollowing reaction is a typical reaction which shows a

'jependence on the dimensions of the container:

NzO+(e) --! 2 NO2(g)

Reaction 1

The reaction may be carried out in a glass flask, where:ne volume of the container remains constant while the.nternal pressure may vary. The reaction may also be carried:irt in an enclosed piston, where the internal pressure remains:onstant while the volume of the container may vary. LeChAtelier's principle predicts that when the internal pressurerses, Reaction I will shift to the left to alleviate the increase:m pressure. Le Chdtelier's principle also predicts that when:e volume of the container is increased, Reaction 1 will;hift to the right to fill the volume. The equilibrium:onstant for Reaction I at 25"C is 4.i2 x 10-3 atm.

Independent of the starting conditions and presence or.rbsence of an inert gas, the ratio of the square of the partial:ressure of nitrogen dioxide to the partial pressure of:rritrogen tetraoxide is constant, given that the temperature.tJes not change. The ratio of nitrogen dioxide to dinitrogen-atraoxide increases as the temperature is increased.

I tl . The reaction as written is:

A. endothermic with AS > 0.B. endothermic with AS < 0.C. exothermic with AS > 0.D. exothermic with AS < 0.

I 1. The reaction will shift ro the right with all of thefollowing changes EXCEpT:

A . addition of N2Oa(g).B. an increase in volume at constant pressure.C . a decrease in pressure at constant volume.D. addition of He gas to the system at constant

volume.

I L Which of the following accurately shows K"q in termsof AG?

A. Keq=-ln RTAGB. Keq =

ln RT

IQq = eo%r

K* = e-a%r

AG

C.

D.

Cmyright @ by The Berkeley Review@ 2t7

D.

GO ON TO THE NBXT PAGE

93. What is NOT true for a reaction with an equilibriumconstant of L0 x 105?

A . There is mostly product present at equilibrium.B. When starting with all reactants, the shift to reach

equilibrium is insi gnificant.C. The K"O for the reverse reaction is 1.0 x l0-5D. The same reaction when carried out with a catalyst

will have the same equilibrium constant.

9 4. If the value of K"q at a certain temperature is greaterthan one, what will be observed after radiolabeled15NzO+ is added to an equilibrium mixture of NO2 andN2O4 in a rigid container?

A . The amount of radio-labeled NO2 increases, whilethe amount of radio-labeled N2O4 decreases.

B . The amount of radio-labeled NO2 decreases, whilethe amount of radio-labeled N2Oa increases.

C . The amount of radio-labeled NO2 increases, whilethe amount of radio-labeled N2O4 remainsconstant.

D . The amount of radio-labeled NO2 decreases, whilethe amount of radio-labeled N2O4 remainsconstant.

95. As the handle of a piston container filled with anequilibrium mixture of NO2 and N2O4 is lifted, whatoccurs?

PNonl. IhePNzoo

II. The piston cools down.

m. The mole percent of N2O4 increases.

A. I onlyB. II onlyC. I and II onlyD. II and III only

9 6. The Ks, for the reaction as written is:

A. K- = P*o,

"Y PNroo

B.

c.

(Pruo")2A€o =-' PNzo+

K^^ = PNzo+

\ r PNO,

K* = PNzo+,

(pNoz).

Questions 97 - 100 are NOT based

on a descriptive Passage.

97 . If the molar solubility of an MX type of salt is defined

as y, then the solubilitY Product is:

A. yB. 2v

c. v2D.4y2

98. MgF2(s) would be most soluble in which of the

following solutions?

A. 0.10 M CaF2(aq)

B. 0.10 M NaF(aq)

C. 0.10 M NaCl(aq)D. 0.10 M MgCl2(aq)

9 9. If the forward rate constant for a one-step reaction is

four times the reverse rate constant, then which of the

following is true?

O. Keg = 0-0625

U. KeQ = 0.25

a. Kee = 4.00O. KeQ=16.00

100. The amount of a salt that dissociates into water is

ALWAYS increased by which of the following

changes?

A . Increasing the temPerature'

B. Decreasing the temperature'

C . Increasing the amount of water'

D . Decreasing the amount of water'

Copyright O bY The Berkeley Review@ 2ra THAT'S

r.e 2.D 3.B6.A 7.8 8.D

11. A 12. B 13. D16. C 17. B 18. B2t. B 22. B 23. D26, C 27. A 28. C31. C 32. D 33. C36. A 37. A 38. D41. B 42. A 43. A46. B 41. C 48. A51. c 52. B 53. A56. C 5',7. A 58. B61. C 62. A 63. C

66. C 67. B 68. D'71. C '72. D '73. C16. A '7',7. C 78. B81. D 82. A 83. C86. A 87. D 88. D91. D 92. D 93. B

4.D 5. C

9.A l0.B14. C 15. B19. B 20. B24. B 25. D29. D 30. A34. A 35. A39. A 40. D44. A 45. C

49. A 50. B54. D 55. D59. A 60. D64. B 65. D69. C 70. A'14. B ',75. D79. D 80. c84. B 85. D89. B 90. A94. A, 95. C

96.B e7.c elJ 99. C 100. c

Equilibrium Passage Answers

Choice B is correct' This question involves first determining the value of the equilibrium constant (K".) at

'!32:^?I!,lT:,::t""T.t# the partial l':i"T d 90? at 70i"C. Converting rozlc ,',to Kelvin yields tf! il".According to Table 1., at37S 6,IGg = 9 atm-l. The solution is as follows:

K"q =9 - (pcg')2 =(Pc9{ =

PcgrY =g ... (pcozY = 36 ...pcoz = 6 atm.,choice B(Pco)2(Poz) e)2G) 4 '"\-'

Choice D is correct. Adding Co2 (a product) to an equilibrium mixture forces the reaction to the left. \A/hen thereaction shifts to the left, reactants (oz(c) and Co(g)i increase. This eliminates choices A and B. Because Ko.changes only with temperature, and the temperaiure did not change in this case, choice C is invahdl tyqdefault, (the elimination of three wrong choicei), choice D is the correct answer.

Choice B is correct' Because the value of K"o increases as- the temperature increases, products increase upon theaddition of heat to the reaction. This means that heat lies on the reactant side of the reaction and thus thereaction is endothermic as written. The AH is positive for an endothermic reaction. pick B and feel warm.

Choice D is correct. The equilibrium constant is just that, a constant. It is specific for a given temperature, so itvaries only with a change in, temperature. The equilibrium may shift with changes in either the pressure,volume or concentration, but the value of K remains constant. This makes choice D the best answer.

Choice C is correct. The following chart summarizes the partial pressures during the course of the reaction:Reaction: 2Co(g) oz(g)

-

2Co2G)Initially: 0

Shif t: +2x0 1.00

+X _t- _2x

Equilibrium: 2x x I -2xThe total pressure of the system is the sum of the partial pressures_of each component at any given time duringthe reaction' The question here asks for the total pressure at equilibrium, ,o tir" equilibrium"partial pressuresmust be added. To solve this question, use the

"qr'ruiio.,,Ptotul = PCOZ + PCO + POz = (1 -2x) +2x+ x= l- + x.

The value of x can be no larger than 0.5 atm., because you can lose no more CO2(g) than you start out with (1.00atrn')' The value of x must be somewhere between 0 and 0.5, because some of the carbon dioxide is lost, but notal] of it' This makes the following relationship 1.5 > Ptotul > 1 true, which is choice C, your best choice.

Choice A is correct. Compressing the reaction vessel results in reduced volume and increased pressure. Upondecreasing the container volume, prior to any shifting of the reaction, the partial pressure and concentration ofeach gas increases, but the mole ratio is the same until the reaction shilts. This is why no answer choicescontained the term concentration. Compressing the container decreases the free area in which the gases canexist' The more crowded environment favors the formation of CO2, because carbon dioxide lies on the lesscrowded side of the reaction. This means that moles of Co2 increase, moles of CO decrease, and moles of g)2decrease' This eliminates choice B. When moles of Co decrease, the mole fraction of Co decreases, so choice Cis eliminated. The temperature did not change, so the equilibrium constant remains the same, so choice D iseliminated' The best answer is choice A. As i note of interest, all of the partial pressures increase, but carbondioxide increases by the greatest amount. The shift to re-establish equiiibrirrrn never completely offsets thestress, so the result is that Co and 02 have slight increases in partiai pressure overall, *i1it" co2 shows asignificant increase in partial pressure.

Choice B is correct. Addition of Ca(oH)2(aq) to an equilibrium mixture of Reaction 1 serves to remove Coz(g)from the reaction mixture- To compensate ior lost io21g;, Reaction 1 proceeds in a way so as to regenerateCozG), but it is not completely regenerated. This eliminaies choice A. ro produce CozG), the reactiin mustshift right, which consumes both o2(g) and Co(g). This results in a decrease of both 0219 ind Co(g). The bestanswer is choice B. Total moles of gas decrease, so the total pressure of the system decreases.

r-;right @ by The Berkeley Review@ 219 Section III Detaited Explanations

8.

9.

Choice D is correct. Kst for the reaction is determined by plugging the values from Table 2 into the equilibriurexpression. Upon doing so, the value is found tobe5.44 atm., makirg choice D the best answer.

r.,,= (PHnr)2 - e-72)2 - e.7z)(2.22) = 2.72x2.72 = 2.22x2= s.44atm.' PH2 7.36 7.36 7.36

Choice A is correct. Addition of the gaseous HBr (a product in Reaction 1) to the equilibrium mixture pusLosthe reaction to the reactant side (right), and thus increnses both the partial pressure and mole percent of H1. -:you are of sound mind and body, then you should pick choice A. As long as the temperature remains consta:,:the concentration of Br2[) (or any other pure liquid) remains constant. The concentration of a liquid is measure;as density, which changes with temperature. This eliminates choices C and D, so don't pick them.

Choice B is correct. The greatest amount of H2(g) forms from the reaction that generates the greatest baiireaction or smallest forward reaction. Choices A and C are eliminated, because less H2(g) is present initral;than rn choice B, therefore less H2(g) exists at equilibrium. In addition, the difference between choices A and f,is the amount of Br2(l), and liquids do not affect the equilibrium distribution, so both yield the same amount ::HZ(g). There is only one correct answer per question, so identical answer choices should both be eliminated. -'

choice D went one hundred percent in the reverse direction, then 0.50 atm. of H2(g) would be generated. Thrvalue is haif of the amount of H2(g) in choice B, so choice D is eliminated. To generate an equivalent amount :rH2(g) at equilibrium as starthg with 1.00 atm. H2(g) and excess Br2(1), requires starting with 2.00 atm. HBr(g,.

Choice A is correct. An increase in temperature results in the addition of heat to the equilibrium mixtur;Because the mole percent of product (HBrf increases, heat must be viewed as a reactant in the reaction.

Heat + H2(B) + Br2(l) + 2HB(g)

When heat is viewed as a reactant, the reaction is defined as being endothermic, so choice A is the corre:lanswer. An endothermic reaction has AH value that is a positiue number. Choice C is eliminated, because lh:pressure changed with both the influx of heat and the shift of the reaction, so it cannot be an isobaric (consta:ii,pressure) process. Choice D is eliminated, because the passage states that Reaction 1 is an equilibrium mixtw-

12. Choice B is correct. The definition of equilibrium is a state where the forward and back reactions have equrate. Choices A and D are equivalent, so they are both eliminated. The reaction lacks Br2(I), and desp:liquids not affecting equilibrium, keep in mind that the system is not in equilibrium. Liquids can be a limitrrreagent. Because there is no Br2(1) present, the reaction cannot move in the forward direction, even if it wanteito. This makes choice B the best answer. At this point, we are uncertain it is correct, but it is the best ansrrchoice. To determine the direction in which a reaction proceeds, it is necessary to compare the value of Lreaction quotient (Qr*) to the equilibrium constant (K"q). The two possible scenarios are: Qr* ) K"q, in whiicase the numerator is too large in the equilibrium quotient (products are in excess of reactants), making Lreaction shift to the left (reactant side) to reach equilibrium. The second scenario is where Q.* < K"q, in wh;case the denominator is too large in the equilibrium quotient (reactants are in excess of products), making treaction shift to the right (product side) to reach equilibrium. We solve for Q as follows:

q,^ = (Pu.e'.)2 =: = t .,'r?r) ... e,* < K*qpH2 1

Because Qt* < K"q, Reaction 1 should shift right (to the product side). The problem here is that there is nopresent, so there can be no reaction. This confirms that Br2 is the limiting reagent. To quickly determinedirection of a reaction, put K"O and Qr.* alphabetically and turn the greater than or less than sign into an(i.e., the K"q > Q.* becomes K"q -t Q11, meaning the reaction proceeds forward to reach equilibrium).

13. Choice D is correct. All reagents are present, so a reaction is possible. To answer this question, solve for Qr*

e,* =(PHfuf =+rryI... Keg < e* by a smal amountPH2 1.5 1.36

Because Keg < Qt*, the reaction proceeds to the left (reactant side). This decreases HBr(g) and increases H2so choices A and B are eliminated. The shift is small, because Q and K are close, so choice D is the best anslr-

mnftil

10.

17.

J]IL {

J

ih*Im.

t*ft:

f "15

lllu:rxqT,l

Ip:

frrw:ullT''ri'*ti

:r*siiu,ul;

irl.UtreIjr'.{ATs

4[J

Copvright O by The Berkeley Review@ 22o Section III Detailed Expl

14.

15.

Choice C is correct. If the flask is initially charged with only HBr(g), then the graph of H2(g) over time startsat 0, eliminating choice A. Choice D is a iitration curve, so if is an incorrect unswe, choice.

'"fhe graphs depict

the progression of a reaction. Reactions proceed with a gradual tapering off (choice C), not an iirstantaneousshut-off (choice B). Pick C, because H2(g) increases most rapidly initially, then levels off until equilibrium.

Choice B is correct. "...NOT affect the partial pressure of H2(g)?" Addition of NaOH(aq) removes HBr andindirectly shifts the Reaction 1 equilibrium to the right, through a complex equilibrium. Addition of H2(g) andHBr(g) directly affect the equilibrium, shifting the equilibrium to the opposite side of the compound aaald. Ifthe reaction is already at equilibrium, then neither a solid nor a liquid (such as Br2(l) in thiJ case) affects theequilibrium. For this reason, choice B is a particularly swell answer choice. A solid or liquid can only affectthe net results (equilibrium conditions) if the system is not at equilibrium and the compound. being added (thesolid or liquid) is the limiting reagent. When added in, the system can then achieve equilibrium. -

Choice C is correct. According to the two reactions, only PC13 and NO react with C12, so PCl5 and NOCI are notoxidized by CI2. This eliminates answer choices B and D. By looking at the oxidation states of the reactantsand products, PCl3 (P = +3), PCl5 (P - +5), NO (N = +2), and NOCI (N = +3), it shows that the forwarddirection in both Reaction 1 and Reaction 2 as drawn are oxidation half-reactions for the non-metal compounds.The reactants get oxidi zed by chlorine in the forward reaction as written. Comparing the two values of Ksqshows that reaction 2 lies further to the right (product side) than does Reaction 1. Thus, Reaction 2 favors theformation of products more so than Reaction 1. This implies that NO reacts with Cl2 more readily than pCl3reacts with Cl2. The greater the reaction with Cl2 means more oxidation by C12. Choose C to feel good.

Choice B is correct. Initially, there is twice as much PCl3 as Cl2, and their sum pressure is 1.50 atm. \A/hen thereaction shifts in the forward direction, total pressure decreases, so choice D is eliminated. According to themole ratio, the initial partial pressure of phosphorus trichloride is 1.0 atm. and the initial partial pressure ofchlorine gas is 0.5 atm. The chart below summarizes the partial pressures during the course of the reiction.

Reaction: PCl3(g) ClzG) E4 PC15(g)

0Initially: 1.0 0.5

Shift: -x -x

16.

Equilibrium: 1- x 0.5-x--+ +x

X

The total Pressure of the system is a sum of the partial pressures of each component at any given time duringthe reaction The question here asks for the total pressure at equilibrium, so the equilibrium partial pressuresmust be added. To solve this question, use the formula I Pl = Ptotul. In this case the following holds true:

Ptotul = PpCls +PClz + Ppcls = (1 - x) + (0.5 - x) * x= 1.5 - x.

If everything reacted to the limiting reagent capacity, then x is 0.5. But the reaction does not go to completion,so the value of x falls between 0 and 0.5. This is because some, but not all, of the reactant is lost. Choice A iseliminated, because 1.5 > P16121 > 1. To solve this question, you must estimate the magnitude of x. The value ofx is greater than 0.25, because over half shifts to products to reach an equilibrium constant of 1.95. This makeschoice B the best answer. To solve it more precisely, assume P1s1n1 at equilibrium is 1.25 (the borderlinebetween answer choices B and C). If P1614 at equilibrium is 1.25 atm., then x = 0.25 atm., and at equilibrium,PPCIa = 0.75 atm', Pclz = 0.25 atm., and Ppglu = 0.25 atm. The following Q would result:

,-r-- - Ppcts = 0.25 - 1 - 1.33 ... Ko^ ) O.", so more must shift over.-',. - (r'pcl3;(pcl2) - (o'zsxo ,5) - u?s

This means that x is between 0.25 and 0.50 atm., and thus P1s1; at equilibrium is between 1.00 and 1.25.

Choice B is correct. The reaction listed in the question is the reverse reaction of Reaction 2. When a reaction isreversed, the equilibrium expression is inverted, because reactants and products are inter-converted. Theresulting equilibrium constant is the inverse of the original value. The inverse of something times 104 issomething times 10-5, making choice B the best answer. Choice A is an incorrect answer that could have beenreached if you inadvertently decided the values must multiply to 1.0 x 10-14 (a common value in acid-basechemistry). Choice C was the result of taking the square root, which is wrong in this case. Choice D iseLiminated because the forward and back reactions are different.

iight @ by The Berkeley Review@ 22t Section III Detailed Explanations

T9, Choice B is correct. Welcome to the wonderful world of math. The following chart summarizes the changes Lr.

partial pressures as the system goes to equilibrium. You must solve for the x value. Keep in mind that theanswer you seek is actually 2x,but you must first solve for x before you solve for 2x.

Reaction: 2 NO(g) CIZ(g) :Initially: 0 0

Shift: +2x +x --+Equilibrium: 2x x

From here, it is a matter of plugging the equilibrium values into the equilibrium expression.

r_= (PNocrF -G.0-2x)2 = 72 - 1 -2.7x704

(prvo)2(pcrr) 1z*;21*) (2x)2(x) 4*3

The easiest thing to do here is to take the reciprocal of each side and then solve for x, keeping in mind that i|,:answer choices are 2x.

4x3 = 7 =4.9x10-5 = 49x10-6.'. x3= 12.25x10-6.'. ,=!tZmrtO-Z2.1x 704

Thecuberootof 12.25 liesbetween2and3(because23=8and 33=27),solet's ca\lit2.? PNO= 2(2.?x1.0-2)=='xL0-2, which is choice B. The math is not so bad, if you approximate.

Choice B is correct. Immediately, choices A and C are eliminated, because there is no reaction possible, rti.*rthe reaction has only one reactant (and lacks C12). From here, it comes down to interpreting the magnituce :rK"O values. The reaction that forms the greatest amount of Cl2 is the reaction with the Kgq that most fa',-;rsformation of reactants (because C12 is a reactant in Reaction 1 and Reaction 2). Reaction 1, with the srr,izK"O, most readily favors the formation of reactants out of the two reactions. Because of the squared t€rrn '"9need to look more closely in most cases, but the K"nvalues are so drastically different for the two reach::sthat you are safe ignoring the squared terms in the Kgq of Reaction 2. The larger K"n is ursssislsd -,r',il

Reaction 2, so choice D is out and the best answer is choice B.

Choice B is correct. This is a rehash of an ongoing theme in this passage. Addition of Cl2 shifts both Rea:- rqr

1 and Reaction 2 in the forward direction. The question here entails the magnitude of the shift. Reachc: ;,

with a K"O value of approximately 1, shifts the equilibrium to the product side to use roughly half of th.e -r:gas that was added. Adding Cl2 to reaction 2, with a KuO that is extremely large, will shift the equiiibriu: nn

the product side to use nearly all of the Cl2 gas that was ailded. Choice B is the best answer.

Choice B is correct. To determine the direction of a shift requires solving for the equilibrium quotient (Q).

2 NOCI(g)

1.0 atm.-2x

7 -2x

20.

21.

li

c::1t

fl,;

flT

22.

- Plctr o.4o - 1Qrx =,_---]j|' -2.5 ... K"q < Q.*, so the reaction shifts left.(PpcrrIPcrr) (0.40X0.40) 0.40

When K > Q, the Q is too small, so the products must increase and the reactants must decrease (K -> Q . ,

forces the reaction to move to the left (reactant side). You really should pick choice C. Choice D is elirnr:because a catalyst affects only the kinetics (rate), and not the thermodynamics of a reaction. A ca::-simply lowers the activation energy, but has no effect on the heat of reaction (AH), the free energy -(AG) for the reaction, or the equilibrium constant (K) for the reaction.

This is an experiment that is used to determine the equilibrium constant for the gas phase reaction listed .:passage. In interpreting the apparatus in Figure 1, you must note the two reactants are being reacted or-:'.valve is opened between the gas line and the flask. The reactants are mixed in a ratio of 4}l2 to 1CS-they just-so-happen to react in a ratio of 4: 1". This means that the reactants always exist in a 4 :7 rat:-:the products always exist in a 2 : I ratio. The reactants are never completely depleted, thus the total p:(Pr,it - 2x) of the systern can never reach a value as low as 0.60 atm. Overall, the number of molecules of ln'decreasing as the reaction proceeds, which accounts for the decrease in pressure experienced by the syste:approaches equilibrium. The final total pressure is not given on the graph, but a AP is a negative valurpassages seems hard, but it is really testing your organization skills.

cflto

Copyright @ by The Berkeley Review@ 222 Section III Detailed

23.

24.

Choice D is correct' The purpose of the experiment is to determine the equilibrium constant (K) for Reaction 1.The value of the equilibrium constant varies with the temperature, so the temperature must be held constant toaccurately evaluate the value of the equilibrium constint. This means that the best answer is choice D.Hopefully the other answer choices did not seem that tempting.

Choice B is correct. Ugon heating a closed, isochoric system, the internal pressure increases according to theideal gas law (PV = nRT). This in essence is Charlei' law, eliminating choice A, which has Charles' lawlisted incorrectly' Because the pressure decreases rather than increases,*another variable must be changing.The system is in a flask, so the volume can't change. This means that moles must be decreasing, causing thepressure to decrease:

- T!" only way for moles to change in a closed system, is for a chemical reaction to takeplace' Reaction 1 shifted from the reactant side (five gis molecules) to the product side (three gas molecules).

This means that the reaction proceeds in the forward direction to consume ihe heat added to th"e system. Thismeans that the reaction as written must be endothermic, making choice B correct.

Choice D is correct. Tit question requires setting up the equilibrium relationship that transpires during thecourse of the reaction. The reaction goes as followsr

/.J.

Reaction: CSz(g) 4HzG)

Initially: 0.20 0.80Shift: -x -4xEquilibrium: 0.2-x 0.8-4x --.+

CH+(g) 2H2S(g)

00+x +2xx2x

At any point, including equilibrium, the total pressure of the system is found according to Equation 1.

Ptotut = PCSz + Pg, + Pgpq + PUzS = (0.2 - x) + (0.8 - 4x) + x + 2x = 7.0 _ 2xThe initial pressure of the system was 1.0 atmospheres when the valve was first opened to start the reaction.The pressure of the system at equilibrium is 1 - 2x atm., therefore the change in inteinal pressure from the startof the reaction until equilibrium is 2x. This means that the partial pr"rr.ri" at equilibrium due to H2S (whichis also 2x) equals the change in internal pressure during th" ,"u.iiorr. The purtiut pressure due to CH4 atequilibrium is one-half of the change in internal pressure.

-This makes choice p ihe best answer.

Choice C is correct. This question requires that you deduce that the value of x can never exceed 0.2. This upperlimit of 0'2 is due to the initial pt"tt,rt" of CS2 being only 0.2 atm. You can lose no more CS2 than you hiv -available as a reactant. Because of this, no matter what value for x is used (any value between o and'0.2), 0.g -4x will always be greater than 0.2 -x. This means that Pg, is always greater thin pgg, because 4x must alwaysbe less than 0'8. The best answer is choice C. If the reaction goes completely in the forward direction, thenproducts will exceed reactants and Pgg, < PcH+, eliminating choice A. becar-,se there is no CH4 or H2S in theinitial reaction mixture, the only CH4 or H2S present during the reaction or at equilibrium, is the result of thereaction proceeding in the forward direction. For every otr" CH4 that forms, two molecules of H2S form, so pplrgis always greater (not less) than Pggn, eliminating choice B. If the reaction barely proceeds in the forwarddirection, then the reactants will exceed the products and Pg, > Puzs. This eliminates choice D.

Choice A is correct. The equilibrium constant is the pressure of the products divided by the pressure of thereactants. Answer choices B and D (which both have reactants aivided by products) are eliminated. Thecoefficients from the balanced equation must be included as exponents in the

"qritlbri.r* expression, so choice Ais the correct choice. All of the reactants_and products are gur"r, thus they utl U"torlg in the K"n expression.Keep in mind that solids and pure liquids are not includeJ in the equilibrium exprJssion. tn-e'equihuriumconstant for this reaction is referred to as Ko, because it is a gas system and the values are partial pr"Jr.rr"r.

Choice C is correct. The total pressure of the system at equilibrium is a sum of the partial pressure of eachcomponent gas at equilibrium. The total pressure of the system is found as follows:

Ptoral = PCS2 + Pg, + Pggn + PHzS = (0.2- x)+ (0.8 - 4x) + x+2x=1..0 -2xChoices A and D are eliminated, because the total pressure is less than 1.0 atm. The final pressure of the;,vstem at equilibrium is 7 - 2x. The value of x cannot be greater than 0.20 atm because the cS2 gas is thelimiting reagent and only 0.20 atm. of CS2 is initially pr"r".i. Because the value of 2x is subtracted, the finalDressure drops, but it cannot drop to a value less than 0.6 atmospheres. The best answer is choice C.

:ight @ by The Berkeley Review@ 223 Section III Detailed Explanations

29. Choice D is correct. The equilibrium constant (a numerical value), changes with the temperature, but it dcrernot change with volume, pressure, moles of reactants, or moles of products. Choice A involves adding hydrogegas (a reactant). The reaction shifts to the right to reestablish equilibrium, but the equilibrium consta:r:remains constant. This is to say that the equilibrium position may change, but the equilibrium constant doetnot. Choice B involves removing all four gases. To determine which escapes most readily involves molecu}:mass and effusion rates, but that is irrelevant in this question. Losing both products and reactants causes thereaction to shift to reestablish equilibrium, but the equilibrium constant remains constant. Choice B *eliminated. Choice C involves adding hydrogen sulfide gas (a product). The reaction shifts to the letl ::reestablish equilibrium, but the equilibrium constant will remain constant. Choice D involves increasing '-:utemperature. The equilibrium position and the equilibrium constant both change with a change in temperatu:eThe value of the equilibrium constant can only be changed by temperature.

Choice A is correct. Given that the initial partial pressures of H2O and C12O are both 0.5 atmospheres, a:ufl,

both are reactants, it is possible initially for the reaction to proceed in the forward direction. As such. :einitial forward rate cannot start at zero. This eliminates choices C and D. There is no HOCI present iruhr{,therefore the reverse reaction rate must start at zero. This reaffirms that choices C and D are invalid. A. :mereaction proceeds to equilibrium, the reactant pressures decrease, thus the forward reaction slows gradu:Jl,rm'until equilibrium is reached. Equilibrium occurs when the rate forward is equal to the rate reverse. Beca-.

the forward rate diminishes with time, the best answer is choice A.

31. Choice C is correct. Reactions that can shift when the pressure of the system changes are reactions that Ian unequal number of gas molecules on the reactant and product sides of the reaction. In Reaction 1, therethree reactant gas molecules and two product gas molecules, so Reaction 1 shifts in the forward direction rrtotal pressure increases on the equilibrium system. In Reaction 2, there are two reactant gas molecules andproduct gas molecules, so Reaction 2 cannot shift in either the forward or reverse direction if the total presincreases. In Reaction 3, there are two reactant gas molecules and one product gas molecule, so Reaction 3 si

in the forward direction if the total pressure increases on the equilibrium system. This means that iReaction 1 and Reaction 3 shift when the total pressure of the system changes. The best answer is choice C

Choice D is correct. The equilibrium constant of a reaction remains constant unless the temperature ctsystem is changed. Changing the volume of the system does not change the equilibrium constant (althoue:reaction may shift and the equilibrium ratios change). This eliminates choices A and B, which show th;in its equation form. As the volume of the system increases, the reaction shifts to the side of the reactiormore molecules (in this case the reactant side), thus the ratio of HCOCI (a product) to HCI (a rea;decreases as the reaction reestablishes equilibrium following a change in volume. The best answer is cholce

Choice C is correct. Sulfur dioxide and oxygen are both reactants, according to Reaction 1. When mixin=atm. SO2 with 0.50 atrr.. 02, Reaction 1 proceeds in the forward direction to esiablish equilibrium. The rof K"O for this reaction (582), as listed in the passage, is well above one. At equilibrium, there is signii;tmore SO3 than both SO2 and ()2. This eliminates choices B and D. The SO2 should start higher than Sebut it diminishes twice as fast as 02, thus the SO2 line should drop below the 02 line at equilibrium. U

completely depleted (alI 0.75 atm. are consumed), 0.375 atm. of 02 would be consumed, leaving 0.125 a[*This is because SO2 is the limiting reagent, if the reaction goes to completion. This makes choice C '$rlanswer.

Choice A is correct. The equilibrium constant for reaction 2 is 8.67 x 10-3 at 500'C. Choice A is the rereaction from Reaction 2, so the equilibrium constant is the reciprocal of 8.61 x 10-3 at 500"C. This leai

30.

32.

JJ.

{

:

t{u

34.

value with 102, not 10-3 in it. This makes choice A the mismatch of K"O and reaction. Choice B is the :

Reaction 2 except the values have been cut in half. The result is that the K"n as written should be theroot of 8.61 x fO-3. fhis leads to a value with 10-2 in it, therefore choice B sLems reasonable and thusThe equilibrium constant for Reaction 3 is 3.26 x 10-6 at 500"C. Choice C is the reverse reaction from

and valid. The only mismatch is found with choice A

where all the values have been cut in half. The equilibrium constant as written should be the square roiltreciprocal of 3.26 x 10-6 at 500'C. This leads to a value with 102 in it, so choice C seems reasonable a.r,jChoice D is Reaction 3, where all the values have been doubled. The equilibrium constant as written sl

the square of 3.26 x 10-6 at 500'C. This leads to a value with 10-11 or 10-12 in it, so choice D seems re:

(sil:

effi

M

Affi

Copyright @ by The Berkeley Review@ 224 Section III Detailed Ex

J5.

36.

Choice A is correct. The conversion from the pressure unit of torr to the pressure unit of atmospheres involvesdividing the value in torr by 760 (the conversion factor from torr to atm.). Because the Ko value is given interms of torr-1, the value must be multiplied by 760 in order to convert it into a value in terms'of atm-l. The bestanswer is choice A.

Choice A is correct. An endothermic reaction is set up as follows:

heat + reactants q*products.The cooling of the reaction can be treated as the removal of heat. When heat is removed from an endothermicreaction, because the heat is acting like a reactant, the reaction shifts to the left (reactant side) to regeneratethe lost heat. This shift causes an increase in reactants and a decrease in products. The product to reactantratio therefore decreases. Choices B, C, and D all account for the same shift, so the three choices can beeliminated. The best choice is thus choice A.

The interesting feature of this reaction is that the ratio of the reagents is 1 : 1 : 1 : 1, so the total pressure of thesystem never changes. The reactants always decrease by the same amount, and the products always increase bythe same amount as the reaction proceeds from the initial conditions to the equilibrium conditions. The purposeof this experiment is to prove that equilibrium is pathway independent, implying that whether the reactionstarts with all reactants or all products, it will reach the exactly the same equilibrium.

Choice A is correct. Increasing the volume of the reaction vessel has no effect on the equilibrium, because thereare equal moles of gas molecules on both sides of the reaction. Adding moles of either reactant or product willshift the equilibrium to the opposite side. Because the reaction is exothermic, it will shift forward with adecrease in temperature and reverse with an increase in temperafure. Thus, the best answer is choice A.

Choice D is correct. The equilibrium constant is less than 1.00, so the reactants are more abundant than theproducts. This means at equilibrium, both CO2 and H2 have concentrations less than half of their initialconcentration (1.00 M). The only answer that shows this is choice D. Note that the final concentrations are thesame as in trial I. Regardless of the starting concentrations, the equilibrium concentrations will be the same ifthe total pressure and the temperature are the same.

Choice A is correct. When reaction vessel III was cooled from 1000"C to 500'C, the partial pressure of carbondioxide gas (CO2) increased. This increase in partial pressure for carbon dioxide indicates that the reactionshifted to the right (product side). The reaction must therefore be exothermic, because as heat is removed fromthe system, the reaction shifts to the product side to produce heat. Any reaction that releases heat in theforward direction is exothermic. This means that as heat is added to the reaction, the reaction shifts to theleft (to the reactant side). This results in a decrease in the partial pressure of CO2 and a decrease in theequilibrium constant . The correct answer is choice A.

Choice D is correct. A change in pressure will never change the equilibrium constant. This eliminates choices Aand B. Because there is the same number of gas molecules on both sides of the equation, a change in pressure hasno effect on the equilibrium. This means that the equilibrium does not shift so the best choice is D.

Choice B is correct. Because there is an equal number of gas molecules on both sides of the reaction (twomolecules on each side), the total pressure of the system will not change with shifts in the reaction. Thiseliminates choices A and C. Because there is no water present initially, the reaction must shift left to reachequilibrium. This results in an increase in both water and carbon monoxide, therefore both slopes shouldincrease equally. This is shown only in choice B.

Choice A is correct. V\4ren the equilibrium constant (K"q) is greater than 1, the concentration of products isgreater than the concentration of reactants. As such, choices B and C are incorrect statements, and can beeliminated. For an exothermic reaction, the equilibrium of the system shifts to the right (product side) as thetemperature of the system decreases. This means that the products increase and the reactants decrease as thetemperature of the system decreases. Both choices B and D are eliminated because of this, leaving only choiceA still standing. If the products out number the reactants, then the equilibrium constant is greater than 1.0, sothe best answer is choice A.

5/.

38.

39.

pyright @ by The Berkeley Review@ 225 Section III Detailed Explanations

43. Choice A is correct. The molar solubility values of the sulfate salts are greater than the molar solubilitl-values of the carbonate salts, as shown by the values in Table 1. This makes statement I valid. The addition ofcomplexing agents helps to dissolve cations into aqueous solution by forming ligand bonds to the metal. Themost common example is hemoglobin, which serves to make iron dication and trication more water soluble.This makes statement II valid. Given that both statements I and II are true, choices B and C are botheliminated. Hydrochloric acid is a strong acid, therefore chloride anion is not affected by the presence ofhydronium ion. As a result, chloride salts show no pH dependence. Sulfide is a weak base, therefore sulfideshows a pH dependence. This makes statement III invalid, so the best answer is choice A.

Choice A is correct. The most accurate value for KrO is found when the most accurate measurement of the molarsolubility is used in the calculation. The most accurate measurement of the molar solubility is found with ttlesalt having the greatest molar solubility. As a general rule, the larger the value, the less significant the errorr

in its measurement. The highest molar solubility is found with Ag2SO4, at 7.3 x 70-z M. The best answer fuchoice A. This question is just a different way of asking which compound has the greatest molar solubility.

Choice C is correct. The anion chosen to remove Zn2+ selectively is the anion that forms the least soluble sahwith zinc dication, rather than with silver cation or strontium dication. According to molar solubility valuesthe least soluble chloride salt is AgCl, so chloride anion is not a good choice. According to molar solubilitryvalues, the least soluble carbonate salt is ZnCOg, so carbonate anion is a possible choice. According to molmsolubility values, the least soluble hydroxide salt is Zn(OH)2, by a substantial margin, thus hydroxide anicmis the best choice so far. According to molar solubility values, the least soluble sulfate salt is ZnSO4 so su-l-fa@

anion is a possible choice. The greatest difference between zinc and the next most soluble cation comes n-idhhydroxide anion. The best answer is thus choice C.

46.

47.

Choice B is correct. Silver chloride is the only chloride salt that is highly insoluble. A precipitatewhen chloride is added to solution indicates that silver cation must be present. This makes choice B theanswer. Choices A and D cannot be concluded, because no precipitate is expected between chloride andzinc or strontium. This means that zinc or strontium may or may not be present.

Choice C is correct. When the cation mixture is treated with chloride anion, the least soluble chloride sa,lt

silver chloride, therefore chloride will remove the Ag+ from solution. When the solution with theremaining cations is treated with sulfide dianion, the least soluble sulfide salt is lead sulfide, thersulfide will remove the Pb2+ from solution. When the solution with the two remaining cations is treatedhydroxide anion, the least soluble hydroxide salt is zinc hydroxide, therefore hydroxide will remove thefrom solution. The cation that will remain in solution is strontium (Srz+), thus the best answer is choice C.

48. Choice A is correct. An acid can protonate a basic salt in solution (which in some cases serves to form thesoluble ionic form of the product). The acid serves to reduce the hydroxide concentration which allowsthe basic salt to dissociate. This is a complex equilibrium that develops as a result. A complex equilit'usually enhances the solubility in aqueous solution. Pick A to feel perky.

49. Choice A is correct. Because the solution has OH- present already, the solubility of AgOH will be hinderedthe common ion effect where hydroxide is the common ion. AgOH will be most soluble in the solution withleast hydroxide anion present. The least hydroxide is found in the solution with the most H+ presenlsolution consequently has the lowest pH. Pick A to feel spunky.

50. Choice B is correct. Because all three salts dissociate into one cation and one anion, the compound rtgreatest molar solubility also has the greatest value for its solubility product (Ksp). The MZ salt fo

heaviest and has the least mass dissociate into solution, so MZ has the lowest moles of salt in solution and

has the lowest solubility product. This eliminates choices C and D. Deciding between MX and MYmore work. More mass of MY dissociates that MX, but MX has a lower molecular mass. The moles of ND(

over 100 while the moles of MY equal 9 over 120. This means that there are 0.07 moles MX and 0.075

MY is present in a smaller volume of solution than MX, so MY has a greater molarity and solubili['than MX at 25"C. The correct answer is choice B.

44.

45.

Copyright @ by The Berkeley Review@ 226, Section III Detailed

51. Choice C is correct. Because the solubility of a salt increases with temperature, answer choices B and D areeliminated. It can also be read from the third column in Table 1 that more MX and MY dissociate into water ata temperature greater than 25'C than what dissociates at 25.0". This also eliminates choices B and D. Themolarity of MX at 31.6'C and the molarity of MY at27.4"C are found as follows:

7g Qo

Mrtxo31.6.c=100g'mole-1 =0.070mo1 = 7 M Mrrrryozz.+.a-120g'mole-1 =0.075mo1 = 7.5M0.0262L 0.0262L 2.62 0.02s3 L 0.0253 L 2.53

There are more moles of MY than MX in a smaller volume of solution, therefore saturated MY at 27.4"C is moreconcentrated than saturated MX at 31.6'C. The best answer is choice C.

Choice B is correct. Exactly 7.0 grams of MX dissociates into 28.9 mL of water at 25.0'C according to Table 1.

The mass that will dissolve into 25.0 mL is proportionally less. The volume ratio is just under ninety percent, soroughly ninety percent of 7.0 grams MX will dissociate. The best answer is choice B.

Choice A is correct. As with any question like this, first consider units. liters must be in the denominator, sochoices C and D are eliminated. At 25"C,9.0 grams of MY requires that enough water be added to reach 26.7 mLof solution when fully dissociated. This makes choice A the best answer. To determine the molarity of thesolution, the 9.0 grams must first be converted into moles, by dividing 9.0 grams by the molecular mass of MY(120 grams per mole). This value is divided by the volume of the solution as measured in liters (0.0261 L). Thevalue of 0.0253 preys upon the possibility you read the wrong column. The best answer is choice A.

Choice D is correct. As the temperature rises, the volume of the solution increases according to Table 1.

Liquids, as a general rule, expand with increasing temperature. As the volume increases, both the molarityand the density of a solution decrease. The best answer is therefore choice D.

Choice D is correct. Density is defined as mass per volume. The densities of all of the solutions are greaterthan pure water (1.00), so choices A and B are eliminated. It is easy to forget to consider the mass of the waterwhen determining the density. At 25'C there are 7 .0 grams of MX and 25.0 grams of H2O in 28.9 mL of solution,9.0 grams of MY and 25.0 grams of H2O :rl26.1mi, of solution, and 4.0 grams of MZ and 25.0 grams of H2O n28.1.mL of solution. The densest solution (greatest mass in the least volume) is found with the saturated MYsolution. This makes choice D the correct choice.

Choice C is correct. The solution with the highest boiling point is the solution with the highest molality ofimpurities. Molality is defined as moles solute per kg solvent. MX has a lower molecular mass than MY, thus1.0 grams of MX generates a greater moles of solute than 1.0 grams of MY. Because the density of waterdecreases as the temperature of water increases, the mass of 10 mL water is greater at the lower temperature.This means that the largest molality is found with 10 mL water at 50'C, because the mass solvent is least. Thebest answer is therefore choice C.

Choice A is correct. An anion can be used to identify a cation when it reacts differently with that cation thanthe other cations. Of the five anions used in table I, all five form a precipitate with barium cation. Thismeans that to identify barium (distinguish barium from the other cations) the anion chosen must precipitateexclusively with barium. Only chromate CrO42- forms a precipitate with barium and no other cation. Todistinguish barium cation from the others, chromate should be added to solution. If a precipitate forms, thenbarium was present in the solution. If no precipitate forms, then barium was not present in the solution. Thebest answer is choice A.

Choice B is correct. Nitric acid protonated carbonate anion to form carbonic acid (H2CO3), which subsequentlydecomposed to form water and carbon dioxide. Because the carbonate anion carries a -2 charge, two equivalentsof nitric acid are required. The two nitrate anions and barium cation are spectator ions. The best answer istherefore choice B. This can also be inferred from the passage, where it is stated that nitric acid is added toreact with carbonate anion to form carbon dioxide gas. The barium cation carries a +2 charge and carbonateanion carries a -2 charge, therefore the two must be in a one-to-one ratio in the salt. This eliminates choices Aand C.

-)2.

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r";right @ by The Berkeley Review@ 227 Section III Detailed Explanations

59.

60.

61.

62.

Choice A is correct. Hydrochloric acid is not used, because it would introduce the chloride anion, whichaccording to table II, forms a precipitate with silver cation. This means that chloride interferes, makingchoice A the correct choice.

Choice D is correct. The salt K2SOa delivers the anion SO42- (sulfate) to solution. From table I, sulfateprecipitates with both barium and strontium cations. The conclusion reached upon observing the formation of a

precipitate should be that either strontium or barium is present in solution. It cannot be determined whether ornot calcium or magnesium cations are present because they do not form any detectable precipitate with sulfate.The best answer is choice D.

Choice C is correct. Looking at the third reactivity column of table II shows that a white precipitate formsupon the addition of silver cation with both carbonate and chloride anions. Addition of nitric acid dissolvesthe carbonate away, but it does not react with the chloride. The best choice is the chloride anion, choice C.

Choice A is correct. The first anion added precipitates with only barium. This according to the chart ischromate (CrOa2-\. Answer choice A shows this to be carbonate (CO32-), making choice A an incorrect choice

Just to be certain, anion W precipitates with strontium but not calcium and magnesium. According to the chart,

this is sulfate (SO4Z-; so choice B is valid. Anion X precipitates with calcium but not magnesium. According toTable 1, it is either carbonate (COrz-, or oxalate (CzO+z-), so choice C (oxalate) is valid. Anion Z precipitateswith magnesium. According to the chart, this is hydroxide (OH-), so choice D (hydroxide) is valid.

The focus of this passage is on the solubility of different cations in the presence of anions. Solubility is one

the more favored topics on the MCAT so far, so be certain to make sense of this passage. The key factremember is that when comparing solubility, look at the molar solubility, not the Ksp value.

63. Choice C is correct. This question boils down to whether Qsp or KrO is larger. When Q > K there are too ma

products, in which case the salt cannot completely dissociate. \zVhen Q < K, there is room for more products'which case the salt completely dissociates. When Q = K, it is exactly saturated. In this question, it iseasier to decide whether the molarity of ZnCOg in the event it fully dissociates exceeds the molarconveniently listed in Table 1. The molarity if ZnCOg fully dissociates is calculated as follows:

0.1 grams

64.

0.1 mole = 0.8 mole=0.0008mo1e+0.0008mole =0.008M125 1000 0.10 Lt25.4l#anf.sl/mole0.008 M > 0.000014 by a large amount. This means that very little of the ZnCO3 dissociates. On this qyou must go even further. It will barely dissociate according to the numbers which implies choice C. Do

is expected of you: be kind to strangers, brush after every meal, and pick choice C.

Choice B is correct. To distinguish one thing from another visually, you must be able to see a difference inbehaviors. The behavior here is the formatLn of a precipitate. Tb visually distinguish a tube wltt. Zn2+

a tube with Ag+, an anion that precipitates with only one of them must be used. The passage defines the

for seeing a precipitate as one mg per mL solution. Assuming the average salt to have a molecular mass

L00 grams pet *o1", means that the cut-off for visual delectability is roughly 0.010 M. You must locabanion that shows a greater molar solubility than 0.01 M for one cation, but less for the other. Sulfides show

biggest ratio of numerical values, but in a practical sense, neither dissolves to a detectable extent to r-distinguish. The numbers are different in a multiplicative manner, but they are close in an additiveChoice A is eliminated. With the chloride salts ([Ag+] = 1..2x 10-5, while lZnz+t = 3.1 x 10-2), there is awhere the silver chloride precipitates while the zinc chloride dissociates. The zinc concentratidetectable to the eye. In other words, we can see the quantity of salt that dissociates. Choose B, and feel

Choice D is correct. This question addresses the common ion effect from a conceptual perspective. Silver sis most soluble in a solution with the least amount of common ion. This eliminates choices A and Cquestion is whether or not silver cation or sulfur dianion more affects the calculation. In this case/ udeciding which leads to a greater x value, Krp = ( 91]1(x) gt Krp = (2x)2(0.01). in a solution with 0-01 \[

65.

the sohibility is 104 x Ksp. In a solution with'0.01 M S2-, the sdlubility is 5 x square root of KrO. Beca

KrO for Ag2S is so small, ihe sq.rare root of KrO is a larger number than KrO. The best answer is choice D.

Copyright @ by The Berkeley Review@ 224 Section III Detailed Ex

66. Choice C is correct. This question is purely mathematical. The key factpresent in solution initially. The set-up and solution are as follows:

Reaction: ZnCO3(s)

Initially: excess

Shift: - x

Equilibrium: whocares?

is recognizing that there is 0.01 M Cf

+ co32-

0

+ zn2+1aq)

0.01.#+x

0.01 + x

+X

X

n/.

68.

59.

Choice D is correct.is needed:

pn.

Ksp=[zt+]tcoe2-l ...Kp =(o.ot + x)1*;

Uponignoringx: Kro =2.0x10-10=(O.Of +x)1x; = 0.01(x) .'. x=2.0xt0-8M

The correct answer is choice C.

Choice B is correct. The most soluble salt is the salt with the highest molar solubility (not necessarily thehighest solubility product). This is a "read-the-chart" type of question. According to data in Table 1, thelargest molar solubility of the choices is found with Ag2CO g (1,.2 x 10-4 M). The trick here is picking thecorrect column in the table to reference (molar solubility rather than KrO). To be more than you can be, pick B.

To calculate the solubility product for the calcium phosphate salt, the solubility reaction

Ca3(POa)2 -q* 3Ca2+1aq; +ZPOa3-(aq)

Kro = [Ca2+]s1e9ne1z = 1ax)3(2x)2 = (27 xr1ro^ *2) = 108 x5

The best answer is choice D.

Choice C is correct. An ion exchange column involves competition by two cations for an anion bound to thecolumn. The cation of the less soluble salt will precipitate with the anion, which is bound to the column. Thetwo cations to consider are the one in solution and the one originally bound to the column. The columns works ifthe cation in solution is less soluble with the anion in the column than the cation originally coupled with theanion in the column. Silver chloride (AgCl) has a lower molar solubility than sodium chloride (NaCl),therefore silver cation precipitates with chlorine anion preferentially over sodium cation. Choice A is a validstatement. Zinc sulfide (ZnS) has a lower molar solubility than calcium sulfide (CaS), therefore zinc cationprecipitates with sulfide anion preferentially over calcium cation. Choice B is a valid statement. Calciumcarbonate (CaCO3) has a higher molar solubility than zinc carbonate (ZnCOg), therefore calcium cation doesnot precipitate with carbonate anion preferentially over zinc cation. The column of choice C will not work,thus choice C is the best answer. Calcium sulfide (CaS) has a lower molar solubility than sodium sulfide(Na2S), therefore calcium cation precipitates with sulfide anion preferentially over sodium cation. Eventhough the two salts have different stoichiometry, they can be compared directly through their molarsolubility values. Choice D is a valid statement.

Choice A is correct. Because there is solid calcium hydroxide at the base of the flask, it can safely be assumedthat the solution is saturated. As sodium carbonate is added to the flask, calcium carbonate will begin toprecipitate. This reduces the calcium ion concentration, causing more of the calcium hydroxide solid can beginto dissociate in order to regenerate calcium ion in solution. As calcium hydroxide dissociates into solution, thehydroxide concentration increases. As the hydroxide ion increases, the solution's pH increases, eliminatingchoices C and D. Had the solution not been saturated, there would not be any solid calcium hydroxide presentto dissociate into solution. This means that the pH increases more in the solution with solid calcium hydroxidepresent than the solution with no calcium hydroxide present. The best choice is therefore answer A.

Choice C is correct. The maximum calcium ion concentration at pH = L4 can be determined using the solubilityproduct of calcium hydroxide (Krp = [Ca2+][OH-]2). Because the pH is 14, the pOH is 0, so the hydroxide anionconcentration is 1.00 M. This means that when plugging into the solubility product to determine the calcium ionconcentration, it can be determined that the calcium ion concentration equals the numerical value of thesolubility product (Krp = [Ca2+]). The solubility product is 4.3 x 10-6 M3. The best choice is C, 4.3 x 10-6 M.

:nvright @ by The Berkeley Review@ 229 Section III Detailed Dxplanations

Ina Choice D is correct. In order to increase the chloride concentration in solution, an anion must be added thatprecipitates calcium cation out from solution. This allows the calcium chloride solid present on the bottom oithe flask to dissociate into solution to regenerate the lost calcium cation. As the calcium chloride dissociatesthe chloride anion concentration increases. Choice C should be eliminated immediately, because the solubilitrof calcium chloride is reduced by the common ion effect. Choice B is eliminated, because the silver cation n'il-precipitate chloride anion from solution. This is one of the solubility rules you may want to know. Silverhalides are insoluble in water. The answer is either choice A or D. The sodium phosphate adds phosphateanion into solution, which then precipitates out of solution as calcium phosphate and thus decreases the

calcium cation concentration. The solid calcium chloride dissociates to counteract the decrease in calciur.concentration and in doing so increases the chloride ion concentration. The best answer is choice D.

Choice C is correct. It is important to realize that the solution is buffered, so the calcium hydroxide tha:dissociates does not drastically affect the pH. The buffer absorbs the hydroxide that is released as calcir.r:-hydroxide dissociates. The amount of calcium hydroxide that dissociates is controlled by the pH of the

solution. At a high pH, there is a large amount of hydroxide present in solution, so only a small amount icalcium hydroxide dissociates. As the pH decreases, the hydroxide anion concentration decreases, resulting =an increasing calcium ion concentration. The overall effect is that as the pH of the buffer increases, calci r-rhydroxide is less soluble due to the common ion effect. This results in an inverse relationship, so choices A ar,lB are eliminated.

The question may also be addressed from a complex equilibrium perspective. As the buffer pH decreases, '.:uP

amount of hydronium in solution increases. Increasing the hydronium concentration forces the second reaction:mthe forward direction. This reduces the hydroxide concentration. To compensate, the first reaction shrnm

forward to regenerate hydroxide. In doing so, calcium cation is released into solution. This confirms that lwcalcium ion concentration increases as the pH decreases.

Ca(OH)2(s) + H2O(l) --- Ca2+1aq; + 2 OH-(aq)

H3O+(aq) + OH-(aq).$2HzO(I)Ca(OH)2(s) + 2H3O+(aq): Ca2+1aq; + aH2O(l)

The net result is that as the [HSO*] increases, the [Ca2+] increases. As the calcium cation concentraincreases, the log of the calcium cation concentration increases, therefore the log of the calcium :

concentration increases as the pH decreases. Because both the x-axis and y-axis are based on a log scale

relationship is linear, but not necessarily with a slope of one. The slope is negative (due to the negative sigrdetermining pH). The best answer is choice C.

74. Choice B is correct. If the dissociating of a salt into water increases the temperature of the water, *dissociation (the solvation process) releases heat. The release of heat makes the solvation process e

which makes the enthalpy change (AH) a negative number. This eliminates choices A and C. The entr;increases when a salt dissociates into solution, because the ions have more freedom to randomize withinsolution than within the lattice. The change in entropy (AS) for dissociation is therefore positive. The cosanswer is choice B.

Choice D is correct. This question requires viewing Table 1. The solubility of calcium fluoride and ca,

hydroxide are directly comparable using the solubility product, because fluoride and hydroxide carry the

anionic charge. The solubility of calcium sulfate and calcium carbonate are directly comparable usin::solubility product, because sulfate and carbonate carry the same anionic charge. Choices A and C

73.

75.

eliminated due to their lower solubility product values. Because calcium hydroxide forms thlgg isrLs ',cft

dissociating into solution, the solubitity product is 4x3, where x is the calcium ion concentration 1[CaZ-] . 1

value of " i[Cu2*]) is some number times 10-2 for Ca(OH)2. For calcium sulfate, the solubility product :swhere x is the calcium ion concentration ([Ca2+]). The value of x ([Ca2+]) is some value times 10-3 for cdsulfate. The value is greater for calcium hydroxide, so the best answer is calcium hydroxide, choice D.

Copyright @ by The Berkeley Review@ 230 Section III Detailed Exp

'7.

-3.

This is a difficult passage because as one reaction shifts, it affects the other reactions. It is easiest to view thereactions as independent, and observe how specific reagents are changing during each step. For instance, theshift of reaction L will affect the partial pressure of NO2 which will in turn affect reactions 2,3, ar.d 4. Theresult is a complex equilibrium.

Choice A is correct. A negative change in entropy results when the system becomes more ordered. The systembecomes more ordered when it decreases in volume or changes phase to a more ordered phase. Reaction 2 goesfrom two gas molecules to one gas molecule, therefore in has lost in entropy. Reaction 3 goes from two gas

molecules to two gas molecules, therefore entropy has not changed. Reaction 5 goes from three ions in solution tothree ions in solution, therefore entropy has not changed. Reaction 6 also goes from three ions in solution tothree ions in solution, therefore entropy has not changed. The best answer is therefore choice A.

Choice C is correct. Decreasing the temperature shifts reaction 2 in the forward direction, because it isexothermic. This results in an increase in [N2O4]. Increasing the pressure shifts reaction 2 in the forwarddirection. This results in an increase in [N2Oa]. Water vapor in the air reacts with the NO2 in the air(according to reaction 4), and thus reduces the amount of NO2. Reaction 2 shifts in the reverse direction to re-establish equilibrium. This decreases (NOT increases) the [N2Oa]. Choice C is the correct choice. Addition ofNO2 shifts reaction 2 in the forward direction which increases the [N2Oa].

Choice B is correct. According to the ideal gas law, when the temperature of a gas system is doubled, thepressure doubles, as long as the volume of the system and moles of gas remain the same. The reaction vessel is aclosed steel container, so the volume of the system cannot change. The problem here is that the moles of gas

change with the increase in temperature. Because the reaction is exothermic, an increase in temperaturepushes the reaction in the reverse direction. This increases the reactants and decreases the products. There aremore products than reactants in the balanced equation, so a shift in the reverse direction results in fewer molesof gas in the system. The increase in pressure due to the ideal gas law is not be as great as expected, due to thedecrease due to the shift in equilibrium. The ideal gas law predicts that change in temperature doubles thepressure, while the shift in reaction (reduction in moles of gas) predicts that the pressure decreases. Thechange in moles is a smaller factor than the increase temperature, because at the least, the moles of gas wouldbe two-thirds of their original value. The moles cannot be cut in half according to the balanced equation.Therefore, the overall change in pressure is a little less than double. This is best described as choice B.

Choice D is correct. A bond must be formed in reaction 2, because the two NO2 molecules are combining to formone N2O4 molecule. This eliminates choices A and B. Bond formation is an exothermic process, thereforechoice D is the correct answer.

ff,1. Choice C is correct. The total pressure of reaction 1 at equilibrium is 1.00 atmospheres, because the internalpressure can equilibrate with the external pressure in a piston system. The pressures equilibrate by having thepiston plunger either rise up (increasing the volume) or drop down (decreasing the volume). The addition of0.10 atmospheres of NO gas initially increases the total pressure of the system to 1.1,0 atmospheres, before thesystem can equilibrate. The piston rises to accommodate the change in pressure (and equilibrate the internaland external pressures), thus increasing the volume of the piston. The addition of NO gas displaces thereaction from equilibrium. The reaction is no longer at equilibrium, so it reacts to re-establish equilibrium.Excess reactant is present, so the reaction shifts in the forward direction (io use up the excess reactant). There

are three moles of reactants and only two moles of products, so the number of moles decreases, causing the

volume to decrease slightly. The overall result is a slight increase in volume, making choice C correct. If therewas no shift in the reaction, the volume would increase by exactly ten percent, therefore the increase in volumemust be less than ten percent.

tl,. Choice D is correct. Addition of sodium hydroxide to solution deprotonates HNO2 (HNO3 is a strong acid andhas already fully dissociated), and thus shifts reaction 4 to the product side to re-establish equilibrium.Addition of manganese(Il) chloride (MnCl2) to solution removes both NO2- and NO3- from solution throughcomplexing of the ligands. To re-establish equilibrium in reaction 4, the reaction must shift right to make moreNO2- and NOg-. Removal of nitrate (NOS-) from solution results in a shift in the product direction to re-

establish equilibrium. Removal of water (a reactant) results in a shift in the reverse direction (left) to re-

establish equilibrium. Choice D is the best answer available to you.

-9.

-:p;zright @ by The Berkeley Review@ 231 Section III Detailed Explanations

82.

83.

84.

Choice A is correct. At equilibrium, the NO2 concentration is constant. \A/hen bubbled through water, it rea;m

to form nitrous acid (HNO2) and nitric acid (HNO3), which can both deprotonate and form ligand bonds to tne

manganese cation. This means that the NOZ(g) is depleted when it is bubbled through an aqueous solutio:- trtuttgutr"r" chloride. Graphs C and D are eliminated, because they show that the concentration of \C!increises with the addition of aqueous manganese chloride solution. The NO2(g) is gradually regeneratec bu

both the reverse reaction of reaction 2 and the forward reaction of reaction 1. This means NO2 is regeneraed(and thus increases), but not to the level it was initially at. This is best shown in graph A.

Choice C is correct. The passage states that hemoglobin binds four oxygen molecules while myoglobrn brubonly one oxygen molecule. To transfer all of the oxygen, there must be four myoglobin molecules per hemoglo'hm

This should be co- on knowledge from biology. Pick choice C, and start the passage off on the right foot.

Choice B is correct. Increasing the rate of respiration increases the uptake of oxygen. This results in an incl"eam

in free oxygen, which lessens-slightly by the shift in equilibrium to the oxygenated hemoglobin. Overall" :iue

amount oi fiu" oxygen increases, so choice A is eliminated. An increase in the amount of hemoglobin results r rr

shift in the equilibrium in the forward direction. This reduces the amount of free oxygen. Choice B is thers;mthe best answer. An increase in CO consumed results in more bor.rnd sites on the iron of hemoglobin, so less or-r:

can bind. Less oxygen bound results in more free oxygen. Choice C is eliminated. More blood results in

moles of all components in the equilibrium including oxygen. Choice D is thus eliminated. Select choice B.

85. Choice D is correct. A long-time mountain resident has more hemoglobin in their blood than a long-time

level resident. When the long-time mountain resident descends to sea level where there is a higher abund

of oxygen, they experience increased vitality due to the increased partial pressure of oxygen gas. Pick D.

86. Choice A is correct. Because CO binds the iron of hemoglobin preferentially over 02, any CO present in ti'r,e

bind iron cation, and thus reduce the amount of 02 that can bind. To compensate, respiration increases, into increase the amount of air consumed. Over time, the body produces more Hb (this is a result of acclima

The best answer is choice A.

Choice D is correct. Because hypoxia does not directly affect the cellular uptake rate of oxygen/ onc€

absorbed from the lungs, myoglobin (present in cells) is not affected directly by hypoxia. Without considi

any other factors, the myoglobin should remain constant. The best answer is choice D.

Choice D is correct. Hypoxia results from a drastic decrease in the amount of oxygen present in the air.

decrease is associated with an increase in elevation. Repelling down a mountain and scuba diving both inr

increases in the oxygen present. In scuba tanks, the gas may be mixed with helium (an inert gas) to comP€

for the greater u*o.ttrt of air consumed per breath (due to the increased pressure under water). If the tanks

not paiiaily filled with helium, too mulh nitrogen and oxygen would enter the body. Of the last two ch;

only snow skiing involves a high elevation, so choice D is the best answer,

Choice B is correct. According to the passage, there is less oxygen present at higher elevations. By train::qhigher elevations, where the aii has less moles of oxygen, the body acclimates by producing more Hb One

atilete returns to a lower elevation, the increased amount of FIb remains for a short time, before the bod--

re-acclimate. For a short period of time, an athlete can increase their oxygen carrying capacity (ano

increase their cellular metabolism). The best answer is choice B.

I

87.

88.

89.

Copyright @ by The Berkeley Review@ 232 Section III Detailed ExP

This passage mimics a passage that appeared on the MCAT years ago. The reaction is a typical example of gasphase equilibrium, and is found in most every general chemistry text book. The reaction is often .tred todemonstrate Le ChAtelier's principle. In this passage, the reaction is chosen because it has an unequal numberof reactant molecules as product molecules. This means that changes in the condition of the system (volume andpressure) can shift the equilibrium reaction, but the equilibrium constant remains the same, as long as atemperature change does not accompany the volume or pressure change. The passage also points out that aninert gas does not disturb an equilibrium, as long as the container is rigid (which a glass container is assumed tobe). You may wish to note that if inert gas is added to an expandable container, it can disrupt the equilibrium,because the partial pressures of the component gases are changed.

Choice A is correct. Because the reaction involves the breaking of a bond, it must be an endothermic reaction.This is confirmed in the last sentence of the passage, which states that the ratio product to reactant increasesas the temperature increases. Because there are two molecules formed from just one, the AS for the reaction aswritten is greater than zero (positive). The best answer is therefore choice A.

Choice D is correct. Addition of reactant (NZO+) shifts the reaction to the right (product side), thus choice Ais valid. An increase in volume disrupts equilibrium, and results in the reaction shifting from the side with onemolecule to the side with two molecules. This results in a shift to the right, so choice B is valid. A decrease inpressure disrupts equilibrium, and results in the reaction shifting from the side with fewer molecules (one) tothe side with more molecules (two). This results in a shift to the right, thus choice C is valid. Choices B and Cshould both have been eliminated, because they are the same answer and cannot both be correct. When addinghelium gas (an inert gas) to the reaction at constant volume, the partial pressures of N2O4 and NO2 do notchange. Because the partial pressures do not change, the reaction is not displaced from equilibrium, thus itdoes not shift in either direction. This makes choice D the correct answer.

Choice D is correct. The equation relating AG and K"n is AG = -RT * Keg. To isolate Ksq, both sides of theequation are first divided by -RT. To eliminate the natural log function, the two sides of the equation must beexponents of e. This makes choice D the correct answer. The derivation is shown below.

AG = -RT lnK-^ .'. Ir.K-^ = - AG -Ac/

RT "&q=e /nr

Choice B is correct. When the equilibrium constant has 105 associated with it, it is said to be large. As such,the reaction distribution at equilibrium is almost exclusively products, making choice A valid. If the reactionstarts as mostly reactants, it shifts nearly one hundred percent to form products, making the shift significant.Choice B is NOT true. The equilibrium constant for the reverse reaction is the reciproial of the equitibriumconstant for the forward reaction. This makes choice C valid. A catalyst increases the reaction rate (in boththe forward and the reverse directions), but it does not affect the equilibrium constant. This makes choice Dvalid. The best answer is choice B.

Choice A is correct. Because the equilibrium is dynamic, the forward and reverse reactions are continuallytranspiring. When the equilibrium is disturbed by the addition of N2O4 [abeled or not), the equilibrium islost and the reaction must undergo a net shi.ft forward to compensate for the excess reactant. The reversereaction continues as well, but not to the degree of the forward reaction. The result is that the amount of N2O4decreases and the amount of NO2 increases. The radiolabeled nitrogen will eventually be evenly distributedbetween the products and reactants, once equilibrium has been re-established. This makes choice A the bestanswer. This is referred to as scrambling of the label.

Choice C is correct. Because there are more molecules on the product side than the reactant side, the reactionshifts to products as the volume of the piston increases. This makes statement I a valid statement. Thereaction is shifting in the forward direction, which is endothermic, therefore heat is absorbed by the reaction.The result is that the temperature decreases. This makes statement II a valid statement. Do not mistakenlythink of PV = nRT, because the volume changes and the pressure changes, the temperafure was not changed tocause the volume or pressure change. As the reaction shifts to the right, NZOa decreases and the amount ofNO2 increases, causing the mole fraction of N2O4 to decrease. This makes statement III an invalid statement.The best answer is choice C, both statements I and II.

qiright @ by The Berkeley Review@ 233 Section III Detailed Explanations

96. Choice B is correct. In this case, the equilibrium constant is the product squared (because of the stoichicoefficient) divided by the reactant. This makes choice B the best answer.

97. Choice C is correct. \Mhen an MX salt dissociates into water, it forms M+ cation and X- anion. The soluproduct (KrO) is equal to tM*ltx-l = y2. You really have no altemative but to pick C.

Choice C is correct. This is a question involving the common ion effect. Because F- is present in solutionchoices A and B, and Mgz+ is present in choice D, all of the choices except C are eliminated due to the commion effect. It is only in choice C that the compound does not have a common ion (either Mg2+ or F-) present.

Choice C is correct. Equilibrium is the state in which the forward reaction rate equals the reverse reactirate. For a one-step reaction at equilibrium, k1[R] = kr[P]. K"O is defined as products over reactants, whichmanipulated as follows:

98.

99.

Given kflRl = k,[Pland tqo =fil,x"q =*Because the forward rate is four times the reverse rate,4k, can be substituted for kp and thus K"n is 4k,by kr, which is 4. Choose with dignity; choose C.

100. Choice C is correct. Choice D is eliminated, because less solvent reduces the amount of a salt thatdissociate into solution. Because a solvation reaction can be either exothermic or endothermic, the effect onsystem by a change in temperature varies, and is thus unpredictable. This eliminates choices A and B. Incases/ an increase in temperature results in an increase in the amount of salt that dissociates into solutiorLthere are some exceptions. Only choice C, increasing the solvent, always increases the amount of a saltdissociates into solution. As more solvent is added, more compound can dissociate. Be sure that you rthat this question is asking about the amount (in mass or moles) of salt dissociated, not the concentration (solubility).

Copyright @ by The Berkeley Review@ 234 Section III Detailed Ex

,$eCtion IVAcids and Bases

by Todd tsennett

Strong Acid:pH = -log [HXl

Strong Base:

pOF{=-1ogIMOHI

Wbak Acid: ,

pH,= |nr"- { r"g tHAl

WeakBase:

Buffer:

' pH=pKa-ug JSIHA I

or

' PH = PKa -log rnoleq Coaiugate Basq

moles Conjqqate Acid

Terminologya) Fundamental Definitions ,.

b) Water-based Acid.Base Chemistry:i. Acid Dissociationii Base llydrolysis ' , ,l

c) Determination of Reagent StrenEtri. Strong Acidsii. Weak Acids .

iii. Very Weak AcidS I l

iv. Strength and the pK Scalev. StrongBsesvi. Weak Basesvii Very Weak Bases

Types of Acids and Bas€o ,

a) Haloacidsb) Oxyacidsc) Metal Oiides and Metat fiydroxidesd) Organic Acidse) Polyprotic Acids

Calculating ptla;t Determining pH ' '" :

b) LogiReviewc)'pf1forStrongReagen[5''''.d) pH for Weak Reagents

Co4iugate faiis , ,

a) Typicat Conjugate Pairsb) Relationship:of pKa and pKb

. c\ Ilenderson.flasselbalch,Hquation

'IfutrIEKrcLEYl)6.E-y.1.fiv'

Speciahzing in MCATI Preparation

Acids & Bases Section Goals

av Know the definitions for an acid and some common examDles of acids.An acid can be defined as either a proton donor, an electron-pair acceptor, or a comHgO+ when added to water. Typical examples include acetic acid (H:CCOzg),1

pound that yieldshydrochloric acidH3O+ when added to water. Typlcal examples include acetic acid (F-I3CCO2H),

(HCl), and sulfuric acid (H2SOa). You are expected to recognize the common acids.

av Know the definitions for a base and some common examDles of bases.

av

A base can be defined as either a proton acceptor, an electron-pair donor, or a compound that yieldsOH- when added to water. Typical examples include sodium hydroxide (NaOH), ammonia (NHE),and potassium tert-butoxide ((CHg)SCOK). You are expected to recognize the common bases.

Be able to calculate the pIl of agueous solutions of base or acid.The pH of a solution is defined as the negative log of the hydronium ion concentration in the solution.You must be able to determine the pH of the solution, knowing the concentration and strength ofthe.species. The pH of a weak acid is greater than the pH of a stiong acid when the two are inlqualmolar concentratrons.

Understand what is meant by the strength of a reagent.The strength of a reagent is the measure of its degree of dissociation in water. A compound that fullydissociates in water ls said to be strong, while a reagent that only partially dissociates in water issaid to be weak. You should be able to determine the relative strengths of acid from chemical featuresand its pH in the aqueous solution.

Know how conjugate pairs and buffers work.A buffer is formed when a weak acid and its conjugate base are combined in an aqueous solution.An equilibrium exists between the two species, sb as long as both are present in solution, thehydronjum ion concentration will remain fairly constant, and the pH will also remain constant. Thiseffect is known as "buffering." You must understand buffers and how pH is determined using theHenderson-Hasselbalch equation.

Recognize and specific types of compounds.You should understand why metal oxides and metal hydroxides are basic. You should understandwhy non-metal oxides axd non-metal hydroxides are aiidic. Be familiar with typical examples suchas jcid rain and soil pH. Recognize that the conversion from a non-metal oxide (Lewis aiid form)to a non-metal hydroxide (Bronsted-Lowry form) involves hydration of the acid. The strength ofthe reagent is not affected by this conversion.

Understand the terms associated with polvprotic acids.Polyprotic acids have multiple pKa and pKb values. You must understand the conceptual andmailiematical relationshlps <jt tn'e variable's to one another. You must be familiar with t6rms suchas "normaility" and "equivalents." Know the typical examples of polyprotic acids and their formulae.

"3

'3

o?

General Chemistry Acids and Bases Introduction

Acid-and-base chemistry is a foundation for understanding organic chemistry,biochemistry, and physiology. As such, we shall address it from severalperspectives, observing what effect the varying of a solvent has on the nature of areaction. Water and other protic solvents solvate charged species, so acid-and-base chemistry can be monitored by the gain and loss of charge, associatedspecifically with the gain and loss of an ionized hydrogen atom (H+). The terms"protonated" and "deprotonated" are derived from the ionized hydrogen atom,rvhich is just a proton. In a lipid environment, charges cannot be stabilized, so

acid-and-base chemistry is typically considered from an electron-transferperspective, where bonds are broken and formed. The different definitions ofacids and bases are the result of differences in solvent, not chemical reactivity.

Once the definitions are established, reactivity will be considered. There are twocommon misconceptions to clear up, if we are to understand acid and basechemistry better. The first common misconception is that the pH scale has fixedlimits. The second common misconception is that weak acids automatically havestrong conjugate bases. Let us address both of these before anything else.

Acidity is typically measured in terms of [H+] on the pH scale. The pH value of asolution is determined by taking the negative log of the hydroniumconcentration. Because neutral water has a hydronium concentration of 10-/ M,due to autoionization of watet, neutral water has a pH = 7.0. Acidic solutionshave pH values less than 7.0, while basic solutions have pH values greater than7.0. The only limits to the value of pH are associated with concentration andsolubility, so the pH scale is limitless in theory. In practice, there are limits.

The pH scale does not range from 0 to 1-4. The pH of a compound can be negatiae

for highly concentrated strong acids and greater than 14 for highly concentrated

strongbases (e.g.,10 M HCI has pH = -1, and L0 MNaOHhas pH = 15).

The relative strength of an acid can be determined from the relative strength ofits conjugate base. This relation is reversible: The relative strength of a base can

also be determined from the relative strength of its conjugate acid. Using therelationship between pKu and pK6 for conjugate pairs in water at 25'C, either pKl-alue can be determined from knowing the other. For instance, a weak acid witha pKu of 7 has a conjugate base with a PKU of 7. Note that the conjugate base ofihe weak acid is a weak base. This is not a typo! It is the relative reactivities thatare compared, not the absolute reactivities. If acid HA is stronger than acid HB,then base A- is weaker than base B-. This can be demonstrated by comparingReaction 4.1 with Reaction 4.2:

HA(aq) + H2O(1)

PKa=7

HB(aq) + H2O(l)PKa=9

=+ H3O+(aq) + A-(uq)

PKu=7Reaction 4.L

.- H3O+(aq) + B-(uq)

PKU=5Reaction 4.2

The lower pK3 value indicates that HA is a stronger acid than HB while the

larger pK6 value indicates that A- is a weaker base than B-.

The stronger the acid, the weaker its conjugate base. lt is not necessarily the case

that weak acids haae strong conjugate bases. A prime example is a weak acid with

PKa = 7. The pK6 of the conjugate base is 7, which is not strong.

Copyright @ by The Berkeley Review 237 Exclusive MCAT Preparation

General Chemistry Acids and Bases Terminology

TerminologyFundamental DefinitionsA perfect place to start a discussion of acids and bases is with the definitions ofan acid, a base, and strength. There are three definitions proposed for both acidsand bases, although each considers an acid as the opposite of a base. The firstdefinition is the Arrhenius definition, which states that an acid yields Hgo*when added to water, while a base yields oH- when added to water. An acidicaqueous solution therefore has a higher hydronium concentration thanhydroxide. This definition is useful when doing calculations of pH and pKu. Thesecond definition is the Brsnsted-Lowry definition, which states that an acid is aproton (H+) donor and that a base is a proton (H+) acceptor. This definition isbased on what occurs in a protic solvent, where reactions are viewed as proton-transfer reactions. The third definition is the Lewis definition, which states thatan acid is an electron-pair acceptor while a base is an electron-pair donor. Thisdefinition is based on what occurs in an aprotic solvent, and is typically used inorganic chemistry. For this section, o.rly thu Arrhenius aeiinition will beemphasized. Table 4.1 lists the different definitions of acids and bases.

Term Definition ExampleArrhenius acid yields HgO* when added to H2O aq: [H3O+]> [OH-]Arrhenius base yields OH- when added to H2O aq: [OH'] > [HeO*]

Bransted-Lowry acid Proton donor HX in protic solventBronsted-Lowry base Proton acceptor KOH in protic solvent

Lewis acid Electron pair acceptor BF3 in aprotic solventLewis base Electron pair donor NH3 in aprotic solvent

Table 4.1

Water-based Acid-Base ChemistryFor our purposes, acid and base chemistry is to be considered as reactions tha:only occur in aqueous solution. Therefore, to understand water-based acidit'.'and basicity, it is vital to understand the properties of water. water -,amphoteric, meaning that it may act as either an acid or a base. Neutral wate:naturally dissociates into hydronium and hydroxide, according to Reaction 4.3.

2H2O(t): H3O+(aq) + OH-(aq)

Reaction 4.3

Only a small fraction of water dissociates into solution. Water at 25"C, in r.:absence of an acid or a base, dissociates enough to generate a solution with 1--M H3O+(aq) and t0-7 M OFf(aq). This meanslhat neutral water has both a Lit-.hydronium and a little hydroxide. A neutral aqueous solution is defined as c:iiin which [HSO*] = [OH-], and therefore pH = pOH. If the solution is at 25"11

then the pH and poH are both equal to 7.0. For an acidic aqueous solutic:[HsO+] > [OH-], pH < pOH, and the pH of the solution is less than 7.0. Equa-,for a basic aqueous solution, [OH-] > [H:O*], pH > pOH, and the pH of r,:solution is greater than 7.0. There are several perspectives according to whir,;solution may be deemed acidic or basic, and regardless of the reasoning behr:,:each one, you must know how all of these different perspectives relate to c:eanother. Any acidic solution has a pH of less than 7.0, hydronium in it be,:-smore abundant than hydroxide, and it can turn blue litmus paper red.

4

i{rCopyright @ by The Berkeley Review 23a The Berkeley Relienn

General Chemistry Acids and Bases Terminology

Acid DissociationWhen an acid is mixed with water, it is said to dissociate. This is the reason weuse the term acid dissociation, expressed quantitatively with an acid dissociationconstant (K3). The acid dissociation constant is nothing more than the

equilibrium constant for the dissociation reaction of an acid in water. Each

molecule of an acid when added to water, dissociates to form hydronium ion(HsO*) and its conjugate base (expressed generically as A-) upon reaction withone water molecule. Reaction 4.1 is an acid dissociation reaction in water.

HA(aq) + H2O(l) : H3O+(aq) + A(aq)

Reaction 4.1

Equation 4.1 is the equilibrium expression (used to solve for the equilibriumconstant) for the acid dissociation reaction shown in Reaction 4.1.

K- _[H3O+][A] (4.1)" [HA]

Equations 4.2 and 4.3 are the equations for converting between PKa and K".

PKa=-logK3

Ka = 10-PKu

As the relative strength of an acid increases, its Ka increases and its pK3

decreases. This means that stronger acids have higher K6 values and lower pKn

values. The K3 and pK6 of an acid depend on the strength of the acid, but nof its

concentration.

Base HydrolysisWhen a base is mixed with water, it is said to undergo hydrolysis. This is thereason we use the term base hydrolysls, expressed quantitatively with a base

hydrolysis constant (K6). The base hydrolysis constant is nothing more than the

equilibrium constant for the hydrolysis reaction of a base in water. Eachmolecule of a base when added to water, hydrolyzes one water molecule to formhydroxide ion (oH-) and its conjugate acid (expressed generically as HA).Reaction 4.4 is a base hydrolysis reaction in water.

A-(aq) + H2O(l) -- HA(aq) + HA(aq)

Reaction 4.4

Equation 4.4 is the equilibrium expression (used to solve for the equilibrium:onstant) for the base hydrolysis rpaction shown in Reaction 4.4'

r. _ [HA][OH-] @.4)r'b = LA-l

Equations 4.5 and 4.6 arc the equations for converting between PKI and K5.

pKb=-1ogK6

K6 = 1g-PK6

,\s the relative strength of a base increases, its K6 increases and its pK6 decreases.

-his means that stronger bases have higher K6 values and lower pK6 values. As

=een with the acids, the K6 and pK6 of a base depend on the strength of the base,':'tt

not its concentration.

(4.2)

(4.3)

(4.s)

@.6)

-opyright @by The Berkeley Review 239 Exclusive MCAT PreParation

General Chemistry Acids and Bases Terminology

Tying everything together for a conjugate set, we observe that as the acid getsstronger, its conjugate base gets weaker. The overall correlation is shown inFigure 4-1.

Overall RelationshipAs acid strength t, Ku 'l., pKa I, conjugate base strength J, K6 J, pKU T

Figure 4-1

Determination of Reagent StrengthThe strength of a reagent is determined strictly from the Ku (or pKu) and the K6(or pK6). Strength is a measure of the completeness of a reaction in water. Termsthat are sometimes used instead of "dissociation" are ionization and electrolyticnsture. The stronger the acid, the more electrolytic it is, because it conductselectricity better due to the greater number of ions in solution. For water toconduct electricity, there must be ions in solution to transfer the electron charge.

Because the equilibrium expressions are comparable, the values can be correlatedfor a conjugate pair. At 25'c, Equation 4.7 describes the relationship betweenpKu and pK5.

pKa (HA) + pKU (A) = 14 (4.7)

The strength of a reagent is measured by its ability to carry out a reaction inwater. The stronger an acid, the more readily it dissociates into water. Thestronger the base, the more readily it undergoes hydrolysis when mixed withwater. Be careful not to confuse the concentration of a reactant with its strength.A highly concentrated weak acid may have a lower pH than a strong acid in lowconcentration.

Example 4.1what can be said of the pK6 associated with the conjugate base of the moreelectrolytic acid of a pair of acids?

A. The pK6 associated with the conjugate base of the more electrolytic acid isgreater than the pK5 associated with the conjugate base of the lesselectrolytic acid.

B. The pK6 associated with the conjugate base of the more electrolytic acid issmaller than the pK6 associated with the conjugate base of the lesselectrolytic acid.

C. The pK6 associated with the conjugate base of the more electrolytic acid isequal to the pK6 associated with the conjugate base of the less electrolyticacid.

D. The pK6 values associated with the conjugate bases of two electrolytic acidscannot be compared.

SolutionThe acid that is more electrolytic is the acid that forms more ions, and thus isbetter able to conduct electricity. The stronger acid dissociates more, and indoing so, produces a greater ion concentration. The more electrolytic acid is thestronger acid, and according to Figure 4-10, the stronger acid has a conjugatebase with a higher pK6 (associated with the weaker conjugate base.) This makeschoice A the best answer.

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General Chemistry Acids and Bases Terminology

Strong AcidsStrong acid.s are acids that dissociate fully when mixed with water. That is to say

that irong acids ionize completely into hydronium and a conjugate base when

added to water. Reactants "o.rr"tf "ompletely

into products, so the Ku value of a

strong acid is very large. It is safe to assume that an acid that does not have a

measirable Ka (i.;., it has a Ku that is too large to measure, making it greater than

1.0) is a strong acid. Reaction 4.5 shows the dissociation of a strong acid, HX.

HX(aq) + HzO(l)

-

H3O+(aq) + X(aq)

Reaction 4.5

The equilibrium constant for the acid dissociation reaction shown in Reaction 4.5

has a Lrge numerator and a minuscule denominator. As such, Ku is significantly

greater tian 1.0, and pKu is negative for HX. This is summarized in Figure 4-2.

r" = [Hr-O*Ir] >> 1 ... PKa < o-

tHX]

Figure 4-2

It happens that all of the strong acids can be classified as either haloacids or

o*yuiidr. Some typical strong acids are listed in Table 4'2'

Acid Name PKa

HCI Hydrochloric acid -7

HBr Hydrobromic acid -7

HI Hydroiodic acid -9

H2SOa Sulfuric acid -9 (PKar)

HN03 Nitric acid a

HCIOa Perchloric acid -10

Figure 4-3

Table 4.2

Weak AcidsWeak acids are acids that only dissociate partially when dissolved into water'

They do not fully ionize into conjugate base and hydronium ion in water.

Reactants are converted partially into products, so dividing products by reactants

indicates a small value for Ku (a value that is less than 1.0). Any acid with Ku less

than 1".0 is a weak acid. Reaction 4.1 shows the dissociation of a weak acid'

HA(aq) + H2O(l) .- H3O+(aq) + A(aq)

Reaction 4.1

The equilibrium constant for the acid dissociation reaction shown in Reaction 4'1

is less than 1.0. However, because our definitions must apply to biochemistry as

well as general chemistry, weak acids are classified as weak or very weak. For

instance] aspartic acid is iaid to have an acidic side chain, while the side chain of

leucine is considered to be neutral. We will define any acid with a K2 less than

1g-1a (and thus a pKu greater than 14) as a very weak acid, meaning that the pKn

range for a weak icid is from 0 to 14. This is summarized in Figure 4-3.

K" -[H3o+][A].where 1 > Ka > 10-14 ... 0 <pKu < 14" tHA]

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General Chemistry Acids and Bases Terminology I

Weak acids are members of a number of different classifications, includingoxyacids of low oxidation state, one haloacid, carboxylic acids, alky-lammoniums, and phenols. Table 4.3 lists some typical weak acids and theircorresponding pKu values.

I

tt

t

t-

t_

,JIL

fi,

D"

$rIr

I

l

Huc

il[Mi']r

,4"

tc.nn

Acid Name PKaCI3CCO2H Trichloroacetic acid 0.64

Cl2HCCO2H Dichloroacetic acid 7.27

H2SQ3 Sulfurous acid 1.82 (pKu1)

HCIO2 Chlorous acid r.90

CIH2CCO2H Chloroacetic acid 2.82

HF Hydrofluoric acid 3.15

HNO2 Nitrous acid 3.41

HC02H Formic acid J./4

H3CC02H Acetic acid 4.74

2,4-(H1C)2C6H3NH3+ 2,4- dimethylanilinium 5.08

4-H2NC6HaNHg* 4-aminoanilinium 6.18

H2CO3 Carbonic acid 6.36 (pK61)

4-O2NC5HaOH 4-nitrophenol 7.75

HCIO Hypochlorous acid 7.46

HBrO Hypobromous acid 8.72

NH,r+ Ammonium 9.26

HCN Hydrogen cyanide 9.32

HIO Hypoiodous acid 10.66

Table 4.3

Very Weak AcidsVery weak acids dissociate less than water. The Ku value of a very weak acid bless than 10-14, because products divided by reactants is less than 1.0 * 10-1i(K*). Given that water is thought to be neutral, an acid with a Ku less than 10-1{(and pKq greater than 14) is a very weak acid. This is summarized in Figure 4-4.

." ==rr9!I4; where Ka < 10-14 ... pKu > 14

Figure 4-4

Strength and the pK ScaleThe same rules that apply to acids also apply to bases, except that hydrolysl.rather than dissociation, is considered, and K6 replaces Ku. As the pK value for ;compound decreases, its strength increases (this is true for both acids and basesAs a rule, the stronger the acid, the weaker its conjugate base. strong acids har-"very weak conjugate bases, and strong bases have very weak conjugate acid-*The odd sounding relationship is that weak acids have weak conjugate bases. Itmay seem peculiar, but the conjugate base of a weak acid is most often a wea_libase. You have seen this relationship before with buffers, although it is unlike.-*that it was emphasized in your general chemistry courses.

,rl;a

S@u

fruidtibm

rl{nilui'

Sumr

iltE

'mmm

Copyright @ by The Berkeley Review 242 The Berkeley Revieu

General Chemistry Acids and Bases Terminology

Table 4.4 defines the strength of an acid and a base in relative terms.

Compound NumericalData

Conjugate NumericalData

Strong acid

(Fully dissociates in H2O)e.g., HCI (PKa = -7)

Ku>>1

PKa<0

Very weak base

(Hydrolyzes less than H2O)

e.g., Cl- (PKU = 21)

K6 < 19-14

pK6 > 14

Weak acid(Partly dissociates in H2O)

e.g., RCO2H (PKa = 3-5)

10-14<Ku<1

0<pKu<14

Very weak base

(Partly hydrolyzes in H2O)

e.g., RCO2- (pKU = 9-11)

10-14<Ku<1

0<pK6<14

Very weak acid(Dissociates less than H2O)

e.g., CH4 (PKa = 49)

Ku < 19-14

pKz > 14

Strong base

(Fully hydrolyzes in H2O)

e.g., CH3- (PKa = -35)

Kb>>1

PKU<o

Table 4.4

The strength of a reagent is measured by its ability to react in water. The morereadily an acid dissociates into water, the stronger it is. The more readily a base

undergoes hydrolysis, the stronger it is. Do not confuse the concentration of areactant with its strength. Both affect pH. Within conjugate pairs, to calculatethe strength of one reagent from the strength of its conjugate, use Equation 4.7'

On the exam, one of the required skitls will be determining what the question isasking. In terms of acid and base chemistry, many of the questions will be morecomplicated versions of the basic question of "which acid is stronger?"

Example 4.2\Atrhich of the following acids would yield the highest pH in water?

-A; 0.10 M HCIB. 0.i0 M HCloa"e. o.1o M HBrp. )o.ro M HCo2H

SolutionThe highest pH in water results from the lowest hydronium concentration. Thisis associated with the weakest acid, which in this case is carboxylic acid, choice

D. Choices A, B, and C are all strong acids that fully dissociate in water.

Example 4.3Which of the following is the BEST choice to titrate 0.10 M H3CNH2(aq)?

A. O.5O M KOHB. 0.05 M NH3C. 0.10 M HBrD. 0.10 M HCO2H

SolutionTo ensure complete reaction, the titrant must always be a strong reagent. Thiseliminates choice B (a weak base) and choice D (a weak acid). In this case, the

solution being titrated is a weak base solution (H3CNH2 is a weak base), so

strong acid is added. Choices A is a strong base and choice C is a strong acid.

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General Chemistry Acids and Bases Terminology G

Example 4.4\A/hich of the following acids dissociates to the GREATEST extent when added towater?

A. HNO3B. HNO2C. H3PO4D. H2CO3

SolutionThe most complete dissociation is associated with the strongest acid. Thestrongest acid in this case is nitric acid (HNo3), choice A. These questions show1.fu-.of the -Ty ways relative acidity can be compared. In addition todissociation, acidities can als.o-be computed by pKu values (lower is more acidic),Ku values (larger is more acidic), pH varues (to-"t is more acidic for equal acijconcentrations), electrolytic strength (more electrolytic is more acidic), andreactivity with bases (stronger acids react with weaker'bases).

Strong Basesstrong bases, like strong acids, react completely when they are added to water.strong bases fully hydrolyze water, so they completely ionize when dissolvedinto water. Reactants are completely conveited inio products; thus, the K6 valueof a strong base is much greater than 1.0, because products divided by re"actantsis very large. Any base that does not have a meazurable K6 (or has a 16 that ismuch greater than 1.0) is a strong base. Reaction 4.6 shows tne hyarotysls of thestrong base, KOH:

MOH(s) + H2O(l)

-- M+(aq) + OH-(aq)

Reaction 4.6

The equilibrium constant for the base hydrolysis reaction shown in Reaction 4.6has a large numerator and a minuscuie denominator. Its value, therefore, issignificantly greater than 1.0. This results in a negative pKu value for a strongbase. This is summarized in Figure 4-5.

"o =tYlrl?Tr >> 1 .'. pKu < o

tMoHl

Figure 4-5

It happens that all of the strong bases are either hydrides, hydroxides, alkoxides,amides, or carbides. some typical strong bases are listed in Table 4.5. No pK6values are listed, because all of these bases when added to water will formhydroxide.

m!tffi,e

W'qfimr

!ilm- :

'Meuu

aml

#m'ffi

/,fl*Imffi

& ud/il

mrn

lheIIffilesilFlg:

-t.Hxril-rffi

--4

Base Name

KH Potassium HydrideNaOH Sodium Hydroxide

KOCH3 Potassium MethoxideNaNH2 Sodium Amide

Li(CH2)3CH3 Butyl Lithium

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Table 4.5

The Berkeley Review CoF

General Chemistry Acids and Bases Terminology

Weak BasesWeak bases, like weak acids, partialiy react when added to water. Weak basespartially hydrolyze water, so they only partially ionize when dissolved intowater. Reactants are only partia\ converted into products, so dividing productsby reactants is placing a small number over a large number, leading to a smallvalue for K6 (ess than one). As with the weak acids, because water is consideredto be neutral, we will define any base with a K6 less than 10-14 (and thus pK6greater than 14) to be a base that is too weak to consider. This means that thepK6 range for a weak base is from 0 to 14. It should be safe to assume that anybase on the test for which you are given a K6 value (or pK6 value), is likely to bea weak base. A base with less than 100% hydrolysis is a weak base. Reaction 4.7shows the hydrolysis of a weak base, A-.

A-(aq) + H2O(l) : HA(aq) + OH-(aq)

Reaction 4.7

The equilibrium constant for the base hydrolysis reaction shown in Reaction 4.7has a small numerator and a larger denominator, making it less than 1.0. Thisresults in a small positive pK6 value for a weak base. This is summarized inFigure 4-6.

Kb = tH4l-tqH l' where 10-14 < Kr < 1 .'. 74 >pKu > otA-l

Figure 4-6

There are many different types of weak bases. some typical weak bases are theconjugate bases of the acids listed in Table 4.3. In addition, other common weakbases include the carboxylates (RCo2-), the alkyl amines (RNH2), bicarbonate(HCO3-), carbonate (COe2-), phosphate (PO+3-), and phenoxides (C6H5O-).

VeryWeak Basesvery weak bases do not undergo any significant (or detectable) hydrolysis inwater. The K6 value for a very weak base is less than L0-14, because the productsdivided by reactants is less than 1.0 x 10-14 (K-). Given that water is consideredto be neutral, a base with a K6 less than 10-14 (and pK6 greater than 14) is a veryweak base. This is summarized in Figure 4-7.

xo = tH4ll9H-1. where K6 < 10-14 ... pK6 > L4

IA-I

Figure 4-7

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General Chemistry Acids and Bases Terminology

Example 4.5Which of the following compounds when added to water does NOT yield asolution where hydroxide ion is in greater concentration than hydronium ion?

A. Na2CO3B. HCO2NaC. HNO3D. C6H5ONa

SolutionA base is a compound that hydrolyzes water to generate hydroxide, and thu-"form a solution where hydroxide is more concentrated than hydronium. Thisquestion is in essence asking, "Which compound is NOT a base?" Choice C, nitricacid, is NOT a base, making choice C the answer.

Example 4.6lzVhich of the following bases has the STRONGEST conjugate acid?

A. O.5O M KOHB. 0.50 M LiN(CH3)2C.0.50MNaHD. 0.50 M NaHCO3

SolutionThe concentration does not affect the strength. No matter what concentration :rs

added to solution, the value of K6 is constant. This means that this questior:simply addresses the strength of the compound. The strongest conjugate acid isassociated with the weakest base. In this question, choice A is a metal hydroxrdechoice B is a metal amide, and choice C is a metal hydride. Choices A, B, and Care strong bases. The only weak base is the bicarbonate base, choice D.

Example 4.7\Alhich of the following is the BEST choice to titrate 25 rnL of 0.10 M HCO2H(ac ?

A. O.1O M KOHB. 0.05 M H3CNH2C. 0.10 M HBrD. 0.05 M H3CCO2H

SolutionTo ensure complete reaction, the titrant must be a strong reagent. This eliminaMchoice B (a weak base) and choice D (a weak acid). In this case, the soluhcmbeing titrated is a weak acid solution (HCO2H is a weak acid), so strongmust be added. Choices C is a strong acid, so the only strong base is choice A,KOH.

Copyright @ by The Berkeley Review 246 The Berkeley

General Chemistry Acids and Bases Types ofAcids and Bases

iilffiriffiffi iiii iiiiffi#$es

HaloacidsThe haloacids are the series of HX acids, whose X represents a halogen. Theacidity of haloacids is predictable based on the size of the halide ion. The largerthe halide, the more acidic the haloacid. This means that HI > HBr > HCI > HF.This trend in acidity is attributed to the increased stability of the conjugate baseas it increases in size. The negative charge is more diffuse on the larger anion.The more diffuse, the less basic the anion, and thus the more acidic its conjugateacid. A second, and perhaps easier, way to view the acidity of haloacids involvesestimating their bond strengths. The longer the bond, the weaker the bond, as ageneral rule in chemistry. As you descend the halogen column in the periodictable, halogen size increases. As halogen size increases, the bond length of an H-X bond must also increase. As the bond length increases, the bond strengthdecreases, and the H+ can be removed more readily. This approach works eventhough bond dissociation energies are determined from homolytic bondbreaking, as opposed to the heterolytic cleavage associated with acid-basechemistry. Ions (H+ and A) are the result of heterolytic bond breaking, but thecorrelation between bond strength and acidity still holds for haloacids. Figure 4-8 summarizes the effect of halide size on acidity.

H_FH- CI

H- Br

Descending a column:Halogen size increases.'. Bond length increases.'. Bond strength decresases.'. Dissociation increases.'. Acidity increases

Figure 4-8

Example 4.8\ /hich of the following conclusions can be made concerning the relative aciditiesof haloacids?

A. Acid strength increases with increasing electron affinity of the halide.B. Acid strength increases with increasing electronegativity of the halide.C. Acid strength increases with increasing ionic radius of the halide.D. Acid strength increases with decreasing isotopic abundance of the halide.

SolutionAs you descend a column in the periodic table (such as the halogen column),acidity increases due to the increase in atomic size. The anion formed upondeprotonation is more stable as it increases in size, so acidity increases as thecolumn is descended. Electron affinity and electronegativity decrease as thecolumn is descended, and they have no bearing on acidity. Isotopic abundanceand atomic mass do not affect acidity, either, so choice C is the best answer.

An important fact to recall about haloacids is that they are all strong acids, exceptfor hydrofluoric acid (HF). HF has a pKa of roughly 3.3, so it does not fullydissociate when added to water. As we have observed in studying other topicsrelated to the properties of this chemical family, this weaker acidity is attributedto the smaller atomic radius of fluorine relative to the other halogens.

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General Chemistry Acids and Bases Types of Acids and Bases

Example 4.9Which of the following acids has the largest Ku value?

A. HFB. HClC. HBrD. H2S

SolutionThe largest Ku value is associated with the strongest acid. Because Br is thelargest anion of the halogens listed, it dissociates from the proton most readilr..This makes HBr the most acidic, so you should pick choice C. Choices D (H2Sfcan be eliminated, because S is of roughly comparable size with, but lesselectronegative than, Cl. This means that HCI is a stronger acid than H2S.

Within a period (row) of the periodic table, it is electronegativity that dictates thestrength of an acid, not atomic radius. A prime example of this idea is therelationship between ammonia (H-NH2), water (H-OH), and hydrofluoricacid (H-F). The strongest acid of the three compounds is the hydrofluoric acid,because fluorine is more electronegative than both nitrogen and oxygen. Theatomic size does not change that noticeably between N, O, and F, because they al1

have the same valence level (n = 2). The periodic trend that most changes is theelectronegativity.

Oxyacids (Non-metal Hydroxides and Non-metal Oxides)Oxyacids are also an acid type to be familiar with. They differ from haloacids, r,that the hydrogen dissociates from an oxygen rather than a halide. In an oxyacidthe acidic hydrogen is bonded to an oxygen, which in turn is bonded to a centra-atom (which in some cases can be a halogen). The simplest rule is that the moreoxygen atoms there are bonded to the central atom, the more the oxygen atomswithdraw electron density from the central atom, and thus the more acidic theoxyacid. This can be summarized as the resonance effect (which is typicall',classified as an organic chemistry concept). Unlike the haloacids, where size isimportant, in oxyacids, the electronegativity of the central atom is mos:important. As a general rule, for every additional oxygen on the central atomthe pKa of the acid will drop by approximately 5 pK3 units. Table 4.6lists son.etypical oxyacids.

Oxyacid Name Oxyacid Name

HN02 Nitrous acid HNO3 Nitric acid

H2S03 Sulfurous acid H2SOa Sulfuric acid

H3PO3 Phosphorous acid H3POa Phosphoric acid

H2CO3 Carbonic acid HO2CCO2H Oxalic acid

HCIO Hypochlorous acid HCI02 Chlorous acid

HCIO3 Chloric acid HCIOa Perchloric acid

HBrO Hypobromous acid HBrO2 Bromous acid

HBrO3 Bromic acid HBrOa Perbromic acid

Hio Hypoiodous acid HiO2 Iodous acid

HIOg Iodic acid HIOa Periodic acid

Copyright @ by The Berkeley Review 24A

Table 4.5

The Berkeley Reviev

General Chemistry Acids and Bases Types of Acids and Bases

\rvhen comparing the relative strength of oxyacids, the number of excess oxygenatoms and the electronegativity of the centrai atom must both be considered. Forcarbonic acid, there is an exce-ss oxygen count of one. What is meant by excessoxygen count is the number of oxygen atoms exceeding the number or hyaroge.atoms. The excess oxygen count often turns out to be the number or o*y[u.,atoms double-bonded to central atom. The more n-bonds to oxyge' fror'in"central atom, the more resonance withdrawal from the central atom, and thus the3o11aci!5 the compound. This explains why H2so4 is a stronger acid thanH2so3' when two compounds hav-e the same excess oxygen count, the nextfactor to consider is the electronegativity of the centril"atom. The moreelectronegative the central atom, the more it withdraws electron density from theacidic proton, increasing the acidity. This is in essence the inductive effect.sulfur is more electronegative than carbon, so H2so3 is a stronger acid thanlzCOg' As you leamed in organic chemistry, the resonance effect ii greater thanthe inductive effect.

Example 4.10\Mhich of the following acids has the LARGEST pKu value?A. HCIO2B. HIO2C. HC1O3D. HIO3

SolutionThe largest pKu value is associated with the weakest oxyacid, which is associatedwith the acid having the smallest number of excess oxygens and the leastelectronegative central atom. Choices C and D are elimin-ated, because thtcontain two excess oxygens each. Choice B is better than choice A, because theiboth have one excess oxygen, but iodine is less electronegative than chlorine.This question could have easily asked for the strongest acfr. It is important torealize that most questions will be asking for eithJr the strongest acid or theweakest acid among the answer choices.

"The wording may be in reference to

conjugate bases, electrolytic nature, ionizability, Ku ialues, pKn values, pHvalues, or re-activity. The secret is to do enough pru"ti." questioni so that ytuencounter all the different possibilities at least ifew times before your test.

As a point of interest, oxyacids result from the hydration of non-metal oxides.This is the cause of acid rain, where most often nitrogen oxides and sulfur oxides(Lewis acids) react with moisture in the air to forri Bronsted-Lowry acids. Algweling it pH for rainfall may also be observed in environments rich in carbondioxide (which hydrates to become carbonic acid). Reaction 4.g shows thehydration of carbon dioxide, while Reaction 4.9 shows the hydration of sulfurtrioxide.

Coz(s) + H2o(t) H2CO3(aq)

non-metal oxide oxyacid

Reaction 4.8

SOz(g) + H2O(t) -+ H2SO3(aq)

non-metal oxide oxyacid

Reaction 4.9

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General Chemistry Acids and Bases Types ofAcids and Bases

Non-metal oxides act as Lewis acids (electron-pair acceptors), while theirhydrated counterparts (non-metal hydroxides, commonly known as oxyacids) actas Brsnsted-Lowry acids. Both SO3 and H2SOa react with hydroxide to formHSO4-, so they are equivalent in terms of strength. The difference is that SO3 ishydrated to become H2SOa. By a similar reaction, sulfur dioxide (SO2) formsH2SO3 (sulfurous acid) and nitrogen dioxide (NOZ) forms both HNO2 (nitrousacid) and HNO3 (nitric acid). These are major components of acid rain. Acidrain is a combination of rain and air-bome pollutants, such as sulfur oxides andnitrogen oxides. It is often treated with steam to convert from the Lewis acidform (non-metal oxide) into the Bronsted-Lowry form (non-metal hydroxide) andthen neutralized with calcium oxide.

Metal Hydroxides and Metal OxidesJust as non-metal oxides are Lewis acids and non-metal hydroxides are Bronsted-Lowry acids, metal oxides are Lewis bases and metal hydroxides are Bronsted-Lowry bases. You should be familiar with these general classifications. Metaioxides are basic and will form metal hydroxides when treated with water. Aprime example is calcium oxide (CaO), which forms calcium hydroxide(Ca(OH)f when hydrated. Reaction 4.10 shows the Lewis acid-base reaction of ametal oxide and a non-metal oxide, while Reaction 4.11 shows the Brsnsted-Lowry acid-base reaction of a metal hydroxide and a non-metal hydroxide. Thereactants in Reaction 4.11. are the hydrated form of the reactants in Reaction 4.10.

SOs(g) + CaO(s) CaSO+(s)

neutral saltnon-metal oxideLewis acid

metal oxideLewis base

Reaction 4.L0

Organic AcidsLet us consider three types of organic acids: carboxylic acids, phenols, and alk' -

ammonium salts. It is important that you recognize these functional groups ai.:know their pKa ranges. For carboxylic acids and alkyl ammoniums, you shou*:know their organic pKu range and their range in amino acids. In carboxylic ac-and phenols, the proton comes off of an oxygen that is involved in resonance. I:an alkyl ammonium cation, the proton comes off of a nitrogen. Figure 4-9 sho''r,:a generic carboxylic acid and its pKu range, a generic phenol and its pKn rar.t=and a generic alkyl ammonium cation and its pKx range. The AA in parenthes*.refers to amino acid terminals, while the R designation refers to an alkyl grouF.

H2SOa(aq) +

non-metal hvdroxideBran-Lowiv acid

:o:

,.A,,"Carboxylic acid

pKulp; = 3-5

pKu144y = 2-3

Ca(OH)2(aq) $ Ca2+1aq;

metal hydroxide cationBron-Lowrv base

Reaction 4.11

:olrPhenol

PKa = 9.5-10.5

Figure 4-9

+SOa2-(aq) + 2H2C"

anion

Alkyl ammonium cado:pKa(R) =9-11

pKa(AA) = 9-70

Iti:s;:llni

vll

r;m]t

ff,og

HH\,S

tO-/,'\RH

Copyright O by The Berkeley Review 250 The Berkeley Keliam

General Chemistry Acids and Bases Types of Acids and Bases

Polyprotic AcidsPolyprotic acids are acids which yield multiple equivalents of hydronium(Hgo+) when treated with a base. The three most common examples fromgeneral chemistry are carbonic acid (H2Co3), sulfuric acid (H2so4), andphosphoric acid (H3Po4). In addition to these three acids, there are also commonamino acids that qualify as either diprotic (no active proton on the side chain) ortriprotic (an active proton on the side chain). Concentrations of polyprotic acidsare often given in terms of normatity (N). Normality is defined- is moles ofequivalents per liter solution. This is to say that a 1.0 molar diprotic acid solutionwould be listed as "2.0 normal," because there are two equivalents of acid.

Example 4.11\Alhich of the following acid solutions is 3.0 N?A. 1.00 M alanineB. 1.50 M carbonic acidC. 1.50 M phosphoric acidD. 7.25 M sulfuric acid

SolutionNormality (N) is found by multiplying the morarity of an acid by the number ofprotons per molecule. In choice A, alanine (H3NCH(CH3)Co2H) is diprotic, sothe normality is 1.00 x 2, which equals 2.0 N. Choice A is eliminated. ln choice B,carbonic acid (H2Co3) is diprotic, so the normality is 1.50 x 2, which equals 3.0N. Choice B is the correct answer. In choice C, phosphoric acid 6jro4) istriprotic, so the normality is 1.50 x 3, which equals +.5 N.

-cnoice C is eliminated.

ln choice D, sulfuric acid (H2so4) is diprotic, so the normality is 7.25 x 2, whichequals 2.5 N. Choice D is eliminated.

Polyprotic acids have multiple pKu values, one for each dissociable proton. Bydefinition, the first proton removed is more acidic than the second or," ,"*or"d,rc pKal is always lower than pKu2. some difficulty may arise when you considerthe,pKu y3ll"_r of a polyprotic acid and its conjugate base. For a diprotic acidsuch as H2Co3, the first proton removed correiponds to the ru.or,d protongained by the conjugate base. H2co3 and HCo3- are a conjugate pair wiih pruland pK62 summing to 14. The futl dissociation of carbonii acid is shown inReaction 4.1.2 and Reaction 4.13.

H2CO3(aq) + H2O(1) -PK"t -PKuz

Reaction 4.12

pK-"l__::.-

-

PKtr

Reaction 4.13

H3O+(aq) + HCO3-(aq)

HCO3-(aq) + H2O(l) H3O+(aq) + CO32-(aq)

Determining the amount of base needed to neutralize a polyprotic acid is atypical question from general chemistry. while the MCAT doeJ not emphasizesolving mathematical problems, understanding the setup is still important,because understanding equivalents can help to determine th" pH of mixtures.

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General Chemistry Acids and Bases Types of Acids and Bases

) a.

Example 4.L2How many mL of 0.40 MH2SOa?

.i ,,.,

i.

A. 62.5mLB. 80.0 mLC. 100.0 mLP,,) t25.0 mL

NaOH are required to neutralize 100 mL 0.25 M

',,. I... | ... ''1 :: i l<lL::': ':' 1'' '

i . --.9'

SolutionIn order to neutralize an acid, an equal mole quantity of hydroxide must be

added to the hydronium source. The general relationship is shown in Equation4.8.

moles oH- = moles H* .'. (MoH-)ffoH-) = (M11+)(Vg+)

Substituting into Equation 4.8 yields the following results:

(4.8)

(MoH-)ffos-) = (Mg+)(VH*)(o'ao $ffeH-) = 2 x (o'25 M)(1oo mL)

voH- = zx (0.2s iralx 100 mL =(o.so )x 100 mL = 125 mL\0.+O Vt/ \0.+o /

Choice D is the best answer. Choice A would have been the result of yourcalculation, if you had forgotten to multiply the molarity by the number oiequivalents, a common mistake with these types of questions.

Example 4.13How many milliliters of 0.60 M HCI are required to neutralize 3.0 grams CaCOrl

A. 50 mLB. L00 mLC. 200 mLD. 300 mL

SolutionAccording to the balanced equation, two molecules of HCI are required for eve:rone molecule of CaCO3. The balanced equation is shown below:

CaCO3(s) + 2HCt(aq)

-

CaCl2(aq) + CO2(g) + H2O(l)

The mathematical setup is:

moles OH- = 2 x moles COg2- = (Mg+)(Vg+)

z* 3'08 =(0.60M)(Vs+)

1oo g/_ot"

0.05 moles = (0.50 M)(Vg+) .'. Vg+ = 0.10 L = 100 mL

Choice B is the best answer. Choice A would have been the result of vc'=calculation, if you had forgotten to multiply the molarity by the number :,r

equivalents, a common mistake with these types of questions.

Copyright @ by The Berkeley Review 252 The Berkeley Reliw

General Chemistry Acids and Bases Types of Acids and Bases

Example 4.14citric acid (C6H6o7) has three dissociable protons with pKu varues of 3.r4, 4.79,and 5.20. \Arhich of the following solutions would have the lowest pH values?-Y 50 mL 0.10 M citric acid(aq) + 25 mL 0.20 M NaOH(aq)B. 25 mL 0.10 M citric acid(aq) + 50 mL 0.10 M NaoHiaqjg,<5 mL 0.20 M citric acidlaql + 75 mL0.20 M NaoHiaqi

_ D. 50 mL 0.20 M citric acid(aq) + 50 mL 0.10 M NaoHiaqj

SolutionFor questions involving mixtr res, it is important to think of the reagents in termsof equivalents. The lowest pH belongs io the solution that is most acidic. Themost acidic solution is the solutiott *heru the fewest equivalents of base relativeto citric acid have been added. In choice A, the NaoH(aq) solution is 1.5 timesthe volume and twice the concentration of the citric acid'solution, so there arethree equivalents of NaoH(aq). In choice B, the NaoH(aq) solution is double thevolume of the citric acid solution and of equal concentration, so there are twoequivalents of NaoH(aq). In choice C, the NaoH(aq) solutionis three times thevolume of the citric acid solution and of eqral concentration, so there are threeequivalents of NaoH(aq). In choice D, the NaoH(aq) solution is of equar vorumeand half the concentration of the citric acid solution, so there is only one-half ofan equivalent of NaoH(aq). The fewest equivalents are found in choice D, sochoice D is the best answer.

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General Chemistry Acids and Bases Calculating pfl

ffiruuxfitiffi iiiiHffi

Determining pHA solution's acidity is measured in terms of hydronium concentration ([H3O+])using the pH scale. The pH value of a solution is determined using Equation 4.9.

pH = -1og [HeO+] (4.e)

By manipulating Equation 4.9 so as to isolate hydronium concentration, Equation4.10 can be derived.

lHsO+l = 10-PH (4.10)

Because neutral water has an HgO* concentration of 1,0-7 M, due to theautoionization of water, neutral water has a pH of 7.0. Acidic solutions have pHvalues less than 7.0, while basic solutions have pH values greater than 7.0. Thereare no limits to the pH scale other than those imposed by the strength anCconcentration of the acid or base in solution.

On the test, you can almost assume that they will have questions about both pHand pOH for an aqueous solution. Just as Equation 4.9 defines pH, Equation 4.11

defines pOH.

pOH = -log [OH-] (4.11r

This means that in order to calculate pH or pOH, it is necessary to determine theHgO* concentration or the OH- concentration. Once this is accomplished, it i:simply log math (negative logs, actually). There are no calculators allowed o:this test. The volume of a solution does not matter in determinhg pH; only t|.;concentration is important. Equation 4.72 may be used to interconvert betu,'eer'pH and pOH in an aqueous solution at 25'C.

pH+pOH=14 (4.1:

Example 4.15What is the pH of a solution where the hydroxide concentration is 106 tir:.,e=

greater than the hydronium concentration?

A.4B.6c. 10

D. 13

SolutionBecause hydroxide is in greater concentration than hydronium, the solutior'. :-.

basic, so the pH must be greater than 7. This eliminates choices A and B. Ti*trick here is to use the log scale correctly. Because the concentrations differ l-,' ;factor of 106, the pH and pOU differ by log 706, which is 6. If the pH is 10, r.,urthe pOH is 4, which are different by 6. It the pH is 13, then the pOH is 1, anC -edifference is 12. This means that the best answer is choice C. This is a tnrlrquestion where understanding the conversion between the log scale a: iconcentration scale is pertinent. Most students choose D, because they add : rrthe neutral pH of 7. It is important to recognize why choice D is a tr1::rumistake.

IT

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F--t--I

l.-

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General Chemistry Acids and Bases Calculating pH

Log ReviewBecause pH is based on logs, it is helpful to do a quick review of logs. When youmultiply numbers, you add their logs. when you divide numbers, you subtracttheir logs. You can solve for any log given that log 2 = 0.3 and log 3 = 0.48.Drawn in Table 4.7 arc examples of how to solve for a log value.

Table 4.7

Log calculations should be carried out only to the level of approximation. Do notforget that log of 10x = x, so knowing the power of 10 that some quantity is canbe a useful hint when selecting the correct log value. Often, only one answerchoice relates to the correct power of 10, which saves a great deal of time spent incalculating. AIl sorts of shortcuts and tricks will be presented during thefollowing sample problems. Before moving on in this section, however, make aconcerted effort to work through Table 4.7 and understand thoroughly how toestimate log values.

Determining pH for Strong ReagentsCalculating the pH for strong acid and strong base solutions is based on theconcept of full dissociation. The calculation of the pH for a strong reagentfollows an easy pattern. For a strong acid, Equation 4.9 is employed, where theconcentration of the strong acid ([HX]) is substituted for the concentration ofhydronium ([HSO+]). For a strong base, Equation 4.11 is employed, where theconcentration of the strong base ([MOH]) is substituted for the concentration ofhydroxide (tOH-l). Calculating pH and pOH from concentrations is most easilydone when the concentration is written in scientific notation. This is becausetaking the log of a number in scientific notation is more convenient in terms ofbookkeeping.

Number Mathematical Calculation Log Value2 G ven value 0.301

J C ven value 0.477

4 Log 4 = Log (2x 2) = leg 2 + 1og2 = .301, + .301 = .602 0.602

5 Log 5 = Log (10 - 2) = log 10 - Iog 2 = 1..00 - .301. = .699 0.699

6 Log 6 = Log (3 x 2) = 1s*2 + 7og,2 = .307 + .307 = .602 0.778

LogT is approximated as being closer to log 8 than log 6 0.845 +.018 Log 8 = Log(2x2x2) = 3 (log 2) = 3(.301) =.903 0.903

9 Log9 = Log (3 x 3) = lsg 3 + log 3 = .477 + .477 = .954 0.954

0.33 Log0.33 = Log (1 *3) = log 1 - log 3 = 0 - .477 -0.477

0.50 Log 0.50 = Log (7 - 2) = log 1 -log2= 0 - .301 -0.301

7.20 Log 1,2 = Log (6 * 5) = log 6 - 1og, 5 =,778 - .699 = .079 0.079

1.25 Log7.25 = Log (5 * 4) = log 5 - log 4 = .699 - .602 = .097 0.097

1.33 Log 1.33 = Log (4 * 3) = log 4 -log3 = .602 - .477 = .725 0.125

t.40 Log 1.40 =Log,(7 *5) = log 7 -log5 =.845 -.699 =.746 0.746

1.50 Log 1.50 = Log (3 - 2) = log3 - log 2 = .477 - .301. = .176 0.176

1.60 Log 1.60 = Log (8-5) = 1og8 -1og5 =.903 - .699 = .204 0.204

1,.67 Lo91,.67 = Log (5 + 3) = log 5 - log 3 = ..699 - .477 = 222 0.222

1.75 Log7.75 =Log(7 -4) = leg 7 -Iog,4=.845 - .602= .243 0.243

1.80 Log 1.80 = Log (9 - 5) = log 9 - log 5 = .954 - .699 = .255 0.255

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General Chemistry Acids and Bases Calculating pll

Example 4.16\Atrhat is the pH of 0.0020 M HCl(aq)?

A. 2.00B. 2.70e. 3.00D. 7.00

SolutionDetermining an exact numerical value involves calculations. The pH for anr'solution is defined as pH = - log [H3O+], so for a strong acid, the pH is - log [HX].It is best to use scientific notation for the concentration.

pH=-log(2x10-3)pH = - (log2 +log 10-3; - - log 2 - log 10-3

pH = - log2- C3) = 3 -log2pH=3 -0.3=2.7

The correct answer is choice B. Choice D should have been eliminated, becausethe solution is acidic, so pH is less than 7.00. Choices A and C could also har-e

been eliminated, if you noted that the log values ended in ".00". For a log to be a

whole number, the concentration must be a power of ten. The fact that theconcentration was .002 tells us that the log could not be a whole number. Fromthis example, you should derive a shortcut for use in the future. You might takenotice thai the- log of 2 x 10-3 is equal to 3 - log 2, so why not remember this aniskip a few steps in the future. Use the relationship: - log (z x 10-Y) = y - log z.

This shortcut applies to all negative log calculations, including the conversicr.from Ku to pKu. For instance, the pKs for a weak acid with K6 equal to 4.L x 1Cr-f

is 6 - log 4.1. This value can be estimated to be greater than 5.0 (which is equal. t.6 - log 10), but less than 5.5 (which is equal to 6 - log 3). A range of 5.0 to : :should be good enough to choose the correct answer from four choices. \A'irexact numerical questions on the MCAT, your goal should be to narrow th.e

answer choice range enough so that three wrong answers may be eliminated.

Example 4.17\zVhat is the pH of 100 mL of 0.030 M HBr(aq)?

I

T

I

A. 1.30;9.-':1.r,c. 2.30D. 7.00

-:w \j2

/-!

\ ..r-7"

,l

IXm,l

'trtn{

]HI]

1t;fuf

trq]lSolution

The volume of the solution does not affect the pH, unless another solutior. sadded. Using the shortcut, the pH is found as follows:

pH = - 1og (3 x 70-2) = 2 - log3pH=2-0.48=7.52

The correct answer is choice B. Choice D should have been eliminated, becau-'*

the solution is acidic, so pH is less than 7.00. Choices A and C could also ha-''*

been eliminated, if you noted that the log values ended in ".30". This n'o::-rlcome from logarithmic insights.

Copyright @ by The Berkeley Review 256 The Berkeley Rerim

General Chemistry Acids and Bases Calculating pH

Example 4.18What is the pOH of 0.050 M KOH(aq)?

d..'.i.goB. 7.70c. 72.30D. 72.70

t.

SolutionThe pOH of a basic solution is found in a manner similar to getting the pH of anacidic solution. Using the shortcut, the pOH is found as follows:

pOH = - log (5 " t0-2) = 2 - log 5

pOH=2-0.7=1.30The correct answer is choice A. Choices C and D should have been eliminated,because the solution is basic, so pH is greater than 7.00, and therefore pOH is lessthan 7.00. The pH of the solution can be found using pH = 14 - pOH.

Example 4.19lVhat is the pH of 200 mL of 0.00391 M KOH(aq)?

2.41.

2.6777.3911.59

SolutionThis question would seem to be quite difficult at first glance; but if you follow therules, it is easy. Because KOH is a strong base, it will fully dissociate whenadded to water. Plugging values into our shortcut method yields the following:

pOH = - log (3.91 x 10-3) = 3 - log 3.91

3 - log 10 < 3- log3,91 < 3 - log 3 .'. 2 <pOH <2.5

If 2 < pOH < 2.5, then 12 > pH > 11.5

The correct answer is choice D. Choices A and B should have been eliminated,because the solution is basic, so pH is greater than 7.00. Choice C is eliminated,because it does not fit into the range for the correct number. Some of you mayhave chosen to approximate 3.91 as 4, and solved accordingly. This method isfine, too.

You should be able to determine pH or pOH for strong compounds in less thanfifteen seconds. While the MCAT does not offer up many calculation questions,if you are fortunate enough to get a pH calculation question, you should finish itquickly, and carry the time you save over to more difficult questions.

A.B.C.D)

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General Chemistry Acids and Bases Conjugate Pairs

Determining pH for Weak ReagentsBecause weak acids and weak bases do not fully dissociate, it is necessary to usethe Ku and K6 values to determine their H3O+(aq) and OH-(aq) concentrations.K6 and K6 are equilibrium constants for the respective compounds in water.Equation 4.7 shows the relationship between the pKu and pK6 for an acid and itsconjugate base. Equation 4.13 is based on the relationship of equilibriumconstants to K* (the dissociation constant for water).

K3xK6=19-14 (4.13)

According to Reaction 4.1, weak acids dissociate into hydronium and conjugatebase when added to water. Equal parts of conjugate base and hydronium ionform. To determine the [H3O+], the dissociation reaction and the dissociationconstant (K6) must be employed. This too, like the strong acids, is a systematicprocess to master, in the course of which we shall discover another shortcut. Tounderstand the process, consider some concentration of a weak acid (HA) with a

pKu between 2 and 12. The setup for the reaction is shown in Figure 4-10.

Figure 4-10

Substituting values into the acid dissociation expression leads to Equation 4.14"which can be applied if the concentration is greater than the Ku, and if pKu fallsbetween 2 and 72. Equation 4.14 does not generate precise answers, but it doesgive a very close approximation.

lHgO+llHso+l -[Hgo*]2IHA]

Given: [A-] = fi3o+1, then 11u = [A-l[Hgo*] -.

tHA]

Reaction: HA(aq) HzO(l)

Initially: [HA]nit excess

Shift: -x -xEquilibrium: [HA]ini1-x irrelevant

negligible 0

-------+ +X +XXX

IHA]

=

Ii!

I{l

fiulL

uffi

l@

W

0.

ifleI

"" = t119ilt ...[Hso*]2 = Ka x tHAl ...IH:o*j = {ru

^ [He]" tHA]Plugging this value into Equation 4.9 for IHSO+], yields Equation 4.14

pH = -log {t(" . tF Al

Equation 4.74 can be further manipulated to generate Equation 4.15 (the s

equation), which will save you time, once it is understood and mastered.

pH = -log1/K" " tF Af = - logy'Ka + (- log fiHati

pH = -log(Ku;1/z *(-tog IHAIl/r) = -ttorKu -Llog [HA]

pH = +log

Ku - lr"rtHAl =lpK" -

I log [HA]

pH =lpK" -llog [HA]

To ensure that Equation 4.15 makes conceptual sense, let's consider the pHweak acid solution. If you add more acid, the pH should decrease. AccordiEquation 4.15, increasing [HA] lowers the pH. Stronger acids have :

dissociation, so they should form solutions of a lower pH. Equation 4.15 sthis by including the pKu term. Getting the pH of a weak acid should be eagi.-

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General Chemistry Acids and Bases Calculating pll

Example 4.20ri\hat is the pH of 1.00 M HF with pKa = 3.32?

A. 0.50B. 1,.66

c. 2.00D. 6.64

SolutionThis question is made easy by applying Equation 4.15. Two requirements forEquation 4.15 to work are that: L) the weak acid concentration must be greaterihan the K6 and 2) the pK2 must lie between 2 and 72. Both of these criteria are:ret, so Equation 4.15 may be applied.

pH =lpK" -Ilog[HA]=l1a.Sz;- 1log 1.0= 1.66-l(O) = f.Oe-2--2"22"2The correct answer is choice B. Because [HA] is 1.0 M, the pH is half of the pKu.U the acid concentration is 0.10 M, then the pH is half of the pKa + 0.5. Thisrreans that the pH of a weak acid can be estimated quickly.

Example 4.21i\hat is the pH of 0.Q7562 M HCO2H with a pKu of 3.642?

.\. 2.259B. 2.383c. 2.759D. 2.883

-\."., i -;1

SolutionThe pH can be estimated from one-half of the pKu. If the acid concentration were1.00 M, then the pH would be 1.821 (half of the pK6). If the acid concentration.,r-ere 0.10 M, then the pH would be 2.321. (half of the pKu + 0.5). If the acid

-oncentration were 0.010 M, then the pH would be 2.821. (half of the pKu + 1.0).

The concentration falls between 0.L0 M and 0.001 M, so the pH must fall between1,321 and 2.821. This eliminates choices A and D. Because the concentration isrust less than 0.10 M, the pH should be slightly higher than2.321-, making choiceB the best answer.

To date, the MCAT has not presented numbers this difficult, so given that theserumbers now can be handled without difficulty, any weak acid pH calculation:an be made easy. The answer to a question like this one in a multiple-choiceexam can be approximated quickly and simply using this technique. The answer

-hoices may be considered as other expressions of.

A. 1.827 +0.438B. 2.327 + 0.062C. 2.327 + 0.438D. 2.827 + 0.062

Equation 4.76 is the equivalent equation for the pOH of a weak base solution. Its derived in the same fashion as Equation 4.15, but K6 replaces Ku, [A-] replaces

[HA], and base hydrolysis is considered instead of acid dissociation.

pOH =l pKo - I log [A-]'2-2 (4.'t6)

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General Chemistry Acids and Bases Calculating pfl

A.B.

,e}D.

Example4.22\A/hat is the pH of 0.20M sodium propionate, if it has Kb = 7.2x 10-10?

Less than 3Between 3 andTBetween 7 andL1Greater than 11

SolutionThe solution is basic, so the pH is greater than 7.0, eliminating choices A and B.The weak base concentration is greater than K6, and pK6 falls between2 and 12,so Equation 4.1.6 can be applied. The pK6 for the base is 10 - log 7.2, which isslightly more than 9. A good estimate is 9.2.The pOH can be estimated from one-half of the pKg. If the base concentrationwere 1.00 M, then the pOH would be 4.6 (half of the pK6). If the baseconcentration were 0.10 M, then the pOH would be 5.1 (half of the pK6 + 0.5).The concentration is 0.20 M, so the pOH lies between 4.6 and 5.1. This meansthat pH lies between 8.9 and 9.4, making choice C the best answer.

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Copyright @ by The Berkeley Review 26() The Berkeley Revia

General Chemistry Acids and Bases Coqiugate rairs

rc$ffi iii

A conjugate pair consists of an acid and a base that exchange one proton; one

gains a proton and the other member of the pair loses that proton. An acid whendeprotonated forms its conjugate base, and a base when protonated forms itsconjugate acid. Reaction 4.1 and Reaction 4.4 show that HA and A- are a genericconjugate pair. The charge difference between the acid and the base in a

conjugate pair is +L.

Example 4.23The conjugate base of HCO3- is which of the following?

A. CO32-B. H2CO3C. COzD. HCO2-

SolutionA conjugate base is formed when an acid loses one proton. This eliminateschoice B, which happens to be the conjugate acid of HCO3-. The conjugate base

of HCO3- is CO32-, so choice A is the best answer. The dissociation reaction ofbicarbonate into hydronium and carbonate is shown below:

HCO3-(aq)+H2O(t):$H3O+(aq)+CO32-(aq)acid conjugatebase

Example 4.24lAtrhich of the following pairs of compounds is NOT a conjugate pair?

A. NH3/NHa+B. HzCOa/HCOe-C. HzSOg/HSOg-D. Po43-/nzPo+-

SolutionIn choice A, NH4+ is formed when a proton is added to NH3, so NH4+ and NH3differ by one proton. This makes them a conjugate pair, and it eliminates choiceA. In choice B, H2CO3 is formed when a proton is added to HCO3-, so H2CO3and HCO3- differ by one proton. This makes them a conjugate pair, and iteliminates choice B. In choice C, H2SO3 is formed when a proton is added toHSO3-, so H2SO3 and HSO3- differ by one proton. This makes them a conj^ugate

pair, and it eliminates choice C. Choice D is the best answer because, PO4r- andH2PO4- differ by two protons, so they do not constitute a coniugate pair.

Recognizing conjugate pairs should be effortless. Knowing how to apply theconcept of a conjugate pair to solving problems is the more useful ability.

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General Chemistry Acids and Bases Conjugate Pairs

Typical Conjugate PairsTypical conjugate pairs either are observed in biological examples or are commonlaboratory buffer systems. There are many weak acids and conjugate bases, so itis a good idea for you to recall classes of compounds, rather than specificcompounds. common conjugate pairs where both components are weak includethe carboxylic acids/carboxylates (RCo2H/RCo2-), the alkyl ammoniums/alkr1amines (RNH3+ /RNH2), and phenols /phenoxides (C6H5OH/COHSO-) Specificpairs that are likely to appear on the MCAT with great frequency are carbonicacid / bicarbonate (H2Co3 /HCo3-) and phosphoric acidldihydrogen phosphate(H1POa/H2PO+-), due to their presence in physiological systems. It is a goodidea to know the acid-base properties of compounds that are common inphysiology.

The distribution within a conjugate pair is dictated by the pH of the solution,The conjugate pair favors the conjugate acid form in the presence of hydroniurrn-The conjugate pair favors the conjugate base form in the presence of hydroxide-The exact distribution is determined by the relationship between pH of thesolution and the pKu of the conjugate acid. Figure 4-11 shows this relationship.

If pH > pKu, the solution is basic relative to the compound, so it is deprotonated..

If pH < pKu, the solution is acidic relative to the compound, so it is protonated"

Figure 4-11

The pH refers to the surrounding solution (environment) in which thecompounds exist. The pKa refers to the conjugate acid that exists in solution-The compound responds to the pH of the solution. Because of the importance osrelating pH to pKu, common pKu ranges should be known. Some common pK*ranges are shown in Figure 4-9. Carboxylic acids have pKu values of 2 to 5. Inamino acids, the carboxyl terminal has a pKu between 2 and 3. Because of tlwlow pKz value relative to physiological pH, the physiological form of carboxr-i-rcacids is the deprotonated form. Ammonium and alkyl ammoniums have pxevalues of 9 to 11. In amino acids, the amino terminal has a pKu between 9 and 11"Because of the high pKu value relative to physiological pH, the physiologica,ilform of amines is the protonated form. Figure 4-12 summarizes some corunonphysiological compounds and their natural states:

AcidRCO2H (PKa = 3-5)

RNH3+ (PKa = 9-10)physiological form, because 7.4 < 9

H2POa- (PKaz=7.2)

H2CO3 (PKar = 6.4)

Conjugate Base

RCO2- (pKu = 9-11)physiological form, becauseT.4 > 5

RNH2 (pKu = 4-5)

HPO42- (pKtz = 6.8)physiological form, becauseT.4 > 7 -

HCO3- @Kt 2=7.6)physiological form, because 7.4 > 6.1

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Figure 4-12

The Berkeley

General Chemistry Acids and Bases Conjugate Pafos

Relationship of pKu and pK6Because the members of a coniugate pair exchange one proton, their respective

equilibrium constants are related. An acid when deprotonated forms itsconjugate base, and a base when protonated forms its conjugate acid. Equation

4.7 shows the relationship of the pKu for the acid and the pK6 for its conjugate

base. This relationship is useful for interconverting between pK values.

Howevet, when polyprolic acids are involved, it gets a little more complicated.

The relationship is emphasized in Reaction 4.12 and Reaction 4.13. Also keep inmind that Equation 4.7 applies only at 25"C in water.

Example 4.25Whatls the pKu for ammonia, given that the pK6 for ammonia is 4'7?

A. 4.7B. 7.0c. 9.3D. 33

,LtPku' '"3*

..'r,-.t l'{,r -'

,{.iIr"{,

SolutionIt is not 9.3! The compound with a pIG of.9.3 is ammonium (NH4+), the

conjugate acid of ammonia. Ammonia is a weaker acid than ammonium, so ithas a pKu greater than 9.3 The best answer is choice D,33, indicating thatu*moniu is such a weak acid that when added to water, there is no detectable

dissociation. The mistake of choosing 9.3 is easy to make, one that most students

make routinely. But the equation pKa (HA) + pKb (A) = 14 is for conjugate pairs,

not for the same compound.

t ,.,

- rq

,t r?.

1l jl

The test-writers are more likely to apply this concept to carbonic acid (H2CO3),

which has two dissociable protons, and thus has both pKul and pKa2' The

correct relationships between pKu and pK6 for the two respective conjugate pairs

are pK61 + PKUZ = 1+ and PKa2 + PKbl = 14' Also note that pK21 is always less

than pKaZ, because the first proton is more acidic than the second one, bydefinition.

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General Chemistry Acids and Bases Conjugate Pairs

Henderson-Hasselbalch EquationThe pH of a solution comprise of both components in a weak conjugate pair can

be determined using the Henderson-Hasselbalch equation, which is shownbelow as Equation 4.17. For the Henderson-Hasselbalch equation to hold true,both the acid and its conjugate base must be present in appreciable concentrationin solution.

(4.17)oH = pK" *,on[Conjugate baselr I * "[Conjugateacid]

The equation shows that as the [conjugate base] increases, the pH of the bufferincreases. It also shows that as the [conjugate acid] incfeases, the pH of the

buffer decreases. Equation 4.77 is derived from the Ku equation.

*"=ffiF.rHeo*r =""(ffi)

taking -log of everything yields: -1og I H3o+] = -logKa - bg(tl1:f )"\ta-l I

-rog B{3o+l =-rosKa r"r(ffi)= pH =pKa -r"r(ffi) =pKu +t"*(#)

Equation 4.17 can be used with either concentration units or mole quantities forHA and A-. This means that Equation 4.17 can be rewritten as Equation 4.18.

pH =pKu+1ogmoles Conjugate Base (4.18)moles Conjugate Acid

According to Equation 4.17, tlre addition of water to a conjugate pair mixture has

no effect on the pH of the solution. The compounds are diluted, and thus less

concentrated, but the pH remains the same.

Exarnple 4.26\Atrhich of the following solutions has the GREATEST pH?

A. L0 mL 0.10 M NH3 with 15 mL 0.10 M NH+*/B-75 mL 0.10 M NH3 with 10 mL 0.10 M NH4+-C. 10 mL 0.10 M HCO2Na with 15 mL 0.10 M HCO2H.'D. 15 mL 0.10 M HCOiNa with 10 mL 0.10 M HCO2H

SolutionThe pH of a conjugate mixture can be determined using Equation 4.18. To make

the pH high, the pKu must be high, and the mixture must be rich in terms of

conjlgate base. The pKu of ammonium is greater than the pKu of a carboxylicacid (in this case, formic acid), so choices C and D are eliminated. Choice A ha-'

more acid than conjugate base, so the pH is less than the pKu for the acid' Choice

B has more base than conjugate acid, so the pH is greater than the pKu for the

acid. This means that choice B has the greatest pH.

Copyright @ by The Berkeley Review 26,4 The Berkeley Reviel

#ffi rusiiffi ii'i#e$#1'$5#fi#iii,iffi #$H$eC

I. Acid and Base Definitions

II. Dissociation and Colligative Properties

III. Oxyacids

IV. Acidity of Thiols and Alcohols

V. Organic Acids

VI. Electron-Withdrawing Effect and Acidity

VII. Weak Acid pH Equation

VIII. Aspirin and Antacids

IX. Household Acids and Bases

X. Stomach Acid and pH

XI. Tooth Decay and pH

XII. Acid Rain and Scrubbers

XIII. Amino Acids pK2 Values

Questions Not Based on a Descriptive Passage

Acids and Bases Scoring Scale

Raw Score MCAT Score

84 - loo l5-1566-85 to-1247 -65 7 -934-46 4-6t-55 t-3

(r -7)

(B - 15)

(r4 - 2r)

(22 - 28)

(2e - 54)

(42 - 48)

(35 - 4r)

(4e - 55)

(56 - 62)

(63 - 70)

(7r - 76)

(77 - 84)

(85 - el)

(e2 - lOO)

Passage I (Questions 1 - 7)

A base can be defined in three different ways, dependingon its solvent. The definitions are:

1. The Arrhenius definition states that a base vieldsOH-(aq) when added to water.

2. The Br6nsted-Lowry definition states that a base is aproton acceptor.

3. The Lewis definition states that a base is an electronpair donor.

The same base can fit the description of all threedefinitions. The perfect example of a base that fits the threedefinitions is ammonia (NH:), which can donate its lone pairof electrons to accept a proton from water, to yield an

aqueous hydroxide anion (OH-(aq)).

Although a base may simultaneously fit the threedefinitions, each definition has its own unique application.For the calculation of the pH of water-based solutions, theArrhenius definition is the most applicable. The Brgnsted-Lowry definition is applicable in a more general sense,

because it accounts for acid-base chemistry that takes place ina protic solvent other than water. The Lewis definition ismost commonly applied to organic chemistry where bases

donate their lone pairs to empty p-orbitals. A Lewis base can

react in any solvent, including aprotic ones.

A Lewis base also may be referred to as a nucleophile.The electrons of the nucleophile are donated to the partiallypositive site on an electrophile (which also may be a protonin addition to a partially positive carbon). The partial or fullcationic charge is often the result of an excess of protons onthe electrophile. The term "nucleophile" is derived from thestrong affinity of the substances for positive charge (thecharge of protons in an atomic nucleus). All three definitionscan account for the nucleophilic nature of a base.

1. If the [OH-] of a

pH will be:

pH = 8.0 solution is tripled, the new

,.! - \'l-l ' {'rt (j'

\".*.,}Y.

C.ir--

^ I.+.-l

r :i */

^-t+ 'I

i lcl. t-.i tl<- \2.

i<-* i-,

*,f,

t. - |-\-

.a ') i;

2. Which of the following bases would have theLARGEST pK6 value?

A . A base that undergoe s 20Vo hydrolysis in waier

.'p. Abase that undergoes 15% hydrolysis in water

--€ . A base that undergoes 107o hydrolysis in waterD. 'A base that undergoes 57o hydrolysis in water

Copyright @ by The Berkeley Review@ 267 GO ON TO THE NEXT PAGE

t-'"

-t.L). r

HCIOHBrO2

)ri2co3HNO2 tt

\.2

t, t, 1.-;t'>, - --

\._iC r a) -, ii . r-) ;'...

t*

t .l t."-n. ri , / t''F I

r"EbI

,\'/-.V

,f -, .J. *(-.,.rr(r c-*)

Let i* '1"tis l"o g' e-c* q (

'')

6 . Which of the following definitions does NOT describean acid?

--4. Aelectron-pair acceptorB. A proton donor

,-e ._A compound that produces hydronium ion in water

D.,. rA nucleophilic molecule

i

As the strength of an acid increases, which of the

following does NOT happen?

--k: The acid becomes more electrolytic.B. The acid dissociates more.

. C . The acid has a lower pKu.

D. 'The acid has a lower Ku.

V Which ol the lollowing is an Arrhenius base?

A.B.

(C'-n.

I r.:,i rr i lr' i'

c'*'l

4. Which of the following would be the BEST choice toneutralize 25 mL 0.10 M HCIO3?

L. 25 mL 0.10 M NaOH(aq)B. 250 mL 0.010 M NH:(aq)

.9. 25 mL 0.10 M HCI(aqtD. 250 mL 0.010 M HCO2Htaq)

!

' r;" t \ L ' ':' 6l' \'\ ; /--r | {'i r'r r"

.-,.-L,)i.,'i)

As the conjugate acid for a base gets stronger, the base:

A. exhibits a decreasing pK6 value.

B. can react with weaker acids.

C . acquires a higher pOH value in water.D . requires more moles of acid to be neutralized.

Passage ll (Questions 8 - 13)

The strength of an acid in water is defined by its abilityto dissociate into hydronium and conjugate base. Strongacids generate more ions in solution. As the number of ionicimpurities in the water increases, so does the solution'selectrical conductivity. Current does not readily pass throughdistilled water, so the degree of an acid's dissociation can beestimated by a solution's ability to conduct electricity.

To determine the correlation between acid strength andelectrical conductivity, a researcher places the two ends of anopen circuit into a container of water so that the solutionbecomes part of a closed circuit. Two volumetric tubes arepoised above the solution, allowing for an exact quantity ofsolution to be added to the circuit solution. Figure 1 showsthe apparatus used in the experiment.

Tube A Tube B

12V

Figure IIn different trials, 25 mL aliquots of acid are added, and

the current is measured at different points in the wire. Anaverage current was recorded. Any deviation in current atdifferent sites can be attributed to errors in measurement, as

the circuit is a single loop, so current should be uniformthroughout. Table I lists the average current in the wire andthe contents ofeach tube in six separate trials.

Trial Tube A Tube B Current1 0.10 M HCI Nothins 5.94 amns

2 0.10 M HCIO Nothing 0.42 amps

-t 0.10 M HCI 0.10 M KOH 6.03 amps

4 0.10 M HCIO 0.10 M KOH 5.98 amps

5 0.10 M KOH Nothing 5.89 amps

6 0.10 M KOAc Nothing 5.92 amps

Table IThe voltage of the battery is constant for the duration of

the experiment.

8. What current would be expected, if 25.0 mL of 0.10 MHF were added to the aqueous solution?

A. 0.02 ampsB. 0.48 ampsC. 5.42 ampsD. 6.17 amps

Copyright @ by The Berkeley Review@ 264 GO ON TO THE NEXT P

9. How can it be explained that there is no differencebetween the current readings in Trial 5 and Trial 6?

A. Both KOH and KOAc are strong bases.B. Both KOH and KOAc are weak bases.

C . Neither compound dissociates into ions in water.D . The number of ions in solution does not depend rrm

the base strength.

I 0 . In Trial 2, the water is acting as:

A. aBrgnsted-Lowry base.

B. an Arrhenius acid.C. a Lewis acid.D. an amphoteric species.

1 I . Which of the following is NOT true?

A. The solution in Trial I has a higher boiling p'nmthan the solution inTrial 2.

B. The solution in Trial I has a higher freezing p',rmthan the solution inTrial 2.

C . The solution in Trial I has a higher ostr"nrcpressure than the solution in Trial 2.

D. The solution in Trial 1 is more electrolytic thansolution in Trial 2.

12. All of the following would show a current of rou_eirS

amps when added to the solution EXCEPT:

A. 0.10 M HNO3.B. 0.10 M H2SOa.

C. 0.10 M NaOH.D. 0.10 M NH3.

13. What might be the reason for the similaritiesthe results in Trial 3 and Trial 4?

A complete reaction transpires only in Trial l.A complete reaction transpires only in Trial 4After reaction, whether it is a strong acid w

strong base or a weak acid with a strong ba-ie-

same concentration of spectator ions remains.Only anions conduct electricity, and bothhave the same amount of anions after reaction

o

il&

A.B.C.

D.

Passage lll (Questions 14 - 21)

The term oxyacid is coined from the oxygen, the key

component of this type of acid. Oxyacids contain a non-

metal atom (highly electronegative) bonded to oxygens, and

the acidic hydrogen is bonded to an oxygen. A typicalexample is nitric acid (HNO3), where nitrogen is the

electronegative atom that is bonded to the oxygens, one ofwhich is also bonded to a hydrogen. Some oxyacids ofinterest are those containing halides, sulfur, phosphorus, and

nitrogen.

The primary rule for predicting the strength of an oxyacid

is that the greater the number of oxygens attached to the

central atom, the more acidic the compound. For example,

HCIOa is more acidic than HCIO3. The secondary rule forpredicting the strength of oxyacids is that it increases as the

electronegativity of the central atom increases when the

number of oxygens is equal between two oxyacids. Table I

shows the trend for the halide acids:

Acid Formula K" value PKa

Perchloric HCIOa 1.4 x 10e -8.8

Perbromic HBrO4 2.6 x l}s -4.6

Periodic HIOa I .-5 x 102 1.8

Chloric HClO3 8.9 x 10-1 0.1

Chlorous HCI02 1.3 x 0-t 1.9

HyDochlorous HOCI 5.4 x g-8 I --'\

Hypobromous HOBr 2.3 x g-e 8.6

Hvooiodous HOI 1.7 x 0-11 10.8

Table 1

Table I shows that the more dominant of the two causes

of strengthened acidity is the increase in oxygens attached to

the central atom of the acid. This exceeds the effect ofchanging the halide central atom. A rough approximation is

that each additional oxygen will lower the pKu of the acid by

approximately 5. Fluorine cannot expand its octet toaccommodate multiple oxygens, so it is not among the

elements that form halogen-based oxyacids.

1 4. The acidity of halide-containing oxyacids increases

directly with which of the following?

A . The increasing electronegativity of the halide

B . The increasing bond length of the H-O bond

C . The increasing size ofthe halide

D . The increasing bond angle of H-O-X

15. Which sequence accurately describes the relative strength

of oxyacids?

A. HIOa > HCIOa > HCIO3 > HBrO3-B. HCIOa > HBrO3 > HIO4 > HIO3C. HIOa > HCIO4 > HBrO3 > HC1O3

D. HCIOa > HCIO3 > HBrO3 > HIO2

Copyright @ by The Berkeley Review@ 269 GO ON TO THE NEXT PAGE

t 6-: If HCIO has a pKu =7.26, then the pH of a 0.10 MHCIO solution is which of the following?

A. 3.63;. 1.1c . 7.26D. 8.26

t_::

i :.'. \

17 . If pKnl for a diprotic acid is 7.8, then the BEST choice

for pKu2 for its conjugate base, formed when the acid

loses a proton, is which of the following?

A. 3.9B. 6.2c. 7.8D. 12.9

18. If the dissociation of an acid is exothermic, then(assuming that entropy is negligible):

A . Ku should increase as the temperature increases.

. '8-. Ka should remain constant as temperatirre increases.

C ; Ka should decrease as the temperature increases.

D . Ku never changes with varying temperature'

19. Which of the following values MOST accuratelydescribes the pKu for HBrO2'based on Table 1?

A. 0.63 I.B. 1.44c. t.gzD - 2.85 i,

20. Which of the following acids would show the

GREATEST dissociation in water?

A. H3POaB. HNO2C. HIO2D. HCIOa

21. What is the pH for a 125-mL sample of 0.10 MHClOa, given that HCIOa is a strong acid?

" .,, t\;-t{"-'

i " . t, tj

[li'

A. -1 i

B.0.1 \'C- tll-'. I |-''

Passage lV (Questions 22 - 28)

The acidity of thiols (RSH) is observed to be greater thanthe acidity of alcohols (ROH). This is determined bycomparing the relative pKu values for corresponding alkylgroups attached to thiols and alcohols. Table I lists pKuvalues for various alcohols and thiols with comparable alkylsubstituents. From these values, it is possible to determinethe relative acidities of the two classes of compounds.

Alcohol PKa Thiol PKaH3COH t6.1 H3CSH 10.3

H3CCH2OH t6.3 H3CCH2SH 10.6

(H3C)2CHOH tt.o (H3C)2CHSH I 1.0

(H:c):coH 17.8 (H3C)3CSH 11.4

Table IThe difference in acidity between alcohols and thiols is

attributed to the polarizability of their respecrive conjugatebases. The larger the anion (more correctly, the atomcarrying the negative charge in the conjugate base), the morediffuse the electrons will be, and thus the more polarizablethe electron cloud ofthe anion. The result is that the electroncloud is spread over more area, increasing the stability of theanion (conjugate base).

As the stability of the conjugate base increases, thebasicity ofthe conjugate base decreases. The final correlationis that as the basicity of the conjugate base decreases, theacidity of the conjugate acid increases. This leads ultimatelyto the conclusion that as the size of the atom to which theacidic hydrogen is attached increases, the acidity of thecompound increases. This can also be correlated to the bondlength of the bond between the acidic hydrogen and the atom.The longer the bond, the weaker that bond will be.

22. Which of the following compounds would be MOSTacidic?

A. H3CSCH3B. H3COCH3C. H3CNHCH3D. H3CCH2CH3

23. From the table of pKu values, what conclusion can bedrawn about the role of alkyl groups?

A . Alkyl groups are electron-withdrawing and decreaseacidity.

B. Alkyl groups are electron-donating and decreaseacidity.

C . Alkyl groups are electron-withdrawing and increaseacidity.

D. Alkyl groups are electron-donating and increaseacidity.

Copyright O by The Berkeley Review@ 27o. GO ON TO THE NE\T

24. Which of the following halogen-containing acids is theSTRONGEST acid?

A. HFB. HCIC. HBrD. HI

25. HCI is considerably more acidic than H2S, because tL.relative acidities of the compounds formed by:

A . atoms in the same row of the periodic table deper:on the electronegativity of the non-hydrogen arc:in the compound.

B. atoms in the same column of the periodic tab,:depend on the electronegativity of the non-hydror::atom in the compound.

C . atoms in the same row of the periodic table deper,:on the size of the non-hydrogen atom in t::compound.

D. atoms in the same column of the periodic ta: :depend on the size of the non-hydrogen atom iri _:{,:

compound.

2 6. How does ethanol compare with its correspondine e::,thiol?

A . Ethanol has a greater value of Ku.B. Ethanol dissociates more completely in water.C. H3CCH2O- is a stronger base than H3CCH1S-D. Ethanol yields a greater [H:O+].

2 7. When an atom in question is not directly attachec :: :thm,

acidic hydrogen, then the acidity of that com:,-,mdlcorrelates to the electronegativity of that atom. ::,: ms

size. This is known as the inductive effect. Ac: ..to the inductive effect, which of these acid: :

STRONGEST?

A. H3CCH2CO2HB. H3CCCI2CO2HC. H3CCSCO2HD. H3CCI2CO2H

2 8 . Which sequence accurately describes the relatir; s

of these acids?

A. H3CCH2SH > (H3C)3CSH > H3CCO1HB. (H3C)3CSH > H3CCH2SH > H3CCO1HC. H3CCO2H > H3CCH2SH > (H3C)3CSHD. H3CCO2H > (H3C)3CSH > H3CCH2SH

Pi

Ct(R

8rithr

:iilfuc ,U

nfu ru

rrfidji rrrnr

I

*J

Passage V (Questions 29 - 34)

Organic acids are weak acids with positive pKu values.

Common types of organic acids include carboxylic acids(RCOzH) and phenols (C6H5OH). Electron-withdrawing

groups along the backbone of an acid increase its acidity, and

thus lower its pKu. Chloro, fluoro, and nitro groups are

among the common electron-withdrawing groups. Table 1

lists some organic acids with their respective pKu values.

They are all monoprotic organic acids.

Table 1

When a weak acid dissolves into water, it partiallydissociates according to its Ka. Equation l, can be applied to

calculate the pH of an aqueous solution of a weak acid.

PH=|Pr'-|log[HA]

Equation L

Equation 1 is derived from the equilibrium expression for

the dissociation of a weak acid into water. It applies only ifthe acid concentration exceeds K6 by 100 fold and if the pKu

of the weak acid is between 2 and 12'

log 2 = 0.3 log 3 = 0.48

2 9 . What is the pH for an acid solutionof 0.50MandaKa-8x 10-8? ,

l-

A':OcPH<3.0 -, 7.'8. 3.0 < pH 3 3.5c:-'3.5<pH<z.o

.D. pH>7.0

with a concentrationl,.t: 1 -.-)

l,'ri:''

Structure Formula PKa

oil

ctrcl \oHCl3CCO2H 0.64

oo,*{Ho" p-O2NC6H4CO2H 3.40

O-{"" C6H5CO2H 4.21

oil

HrC/t- O"H3CCO2H 4.14

ozN OH p-O2NC6HaOH 7.18

OH C6H5OH 10.01

Copyright @ by The Berkeley Review@ 27t GO ON TO THE NEXT PAGE

30. What is the mass percent of carbon in pheno-.' (CoHsOH)?

-_i _s -,7 . _A. 66.7Vo 1 ':* v loo = .u'i'ii'on i' r{':-ilLc. 76.670

D.B2.4vo 3:r**,':o-,.lou -i. . .*

F- --t" I,.n ''i''

3 1 . Which of the following acids would yield the lowest pH

value once completely neutralized by strong base?

A. Acetic acid (H3CCO2H) r\1 '. '" I ' '

..'8. Trichloroacetic acid (CI3CCO2H)-- C. p-Nitro benzoic acid (O2NC6H+COzH)

D. Benzoic acid (C6H5CO2H)

3 2. When comparing equal molar aqueous solutions of acids

from Table l, what is TRUE?

A. 1.0 M acetic acid has a higher boiling point than

1.0 M p-nitro benzoic acid.

B . 1.0 M tricloroacetic acid has a higher boilirlg point' than 1.0 M p-nitrophenol (p-OzNCOH+QH).

.4'. 1.0 M tricloroacetic acid has a higher freezing point

than 1.0 M benzoic acid..D . 1.0 M acetic acid has a higher freezing point than

l.oMphenol. 1...tr

3 3. Which of the followingconjugate base?

.--A-;lPhenol (COHSOH)

D. p-Nitrophenol (p-O2NC6HaOH)

C. Acetic acid (H3CCO2H)

D. Trichloroacetic acid (ClrCCOzH)

lL'. *n"n 0.10 moles of an unknown weak acid HA are- -; dissolved into 100 mL of H2O, the pH for the solution\ -i, 3.7. What is the concentration of its conjugate base

(A-)?

A. tA-l >2x10-4B. tA-l =2 x 10-4

C . tA-l <2 x 10-4

'D?The Ka of the weak acid must be known to

acids has the $TRONGEST

Passage Vl (Questions 35 - 41)

The strength of an organic acid varies directly with thestrength of any electron-withdrawing groups attached to itsorganic backbone. For instance, the acidity of acetic acid canbe increased by adding an electron-withdrawing group to themethyl carbon. The acidity ofacetic acid can be decreased byadding an electron donating group to the methyl carbon. Theeffect on the strength of the acid varies with the location andelectronegativity of the electron-withdrawing group and ismost substantial when the functional group withdraws by* av of resonance. The following set of phenols demonstratesthe effect ofthese substituents on acidity:

Organic Acid -p&

1.2

8.4

G* r 0.0

t0.4

11.2

Figure 1

Increased acidity can be attributed to the withdrawal ofelectron density (through either the inductive effect orresonance) from the O-H bond of phenol. The loss ofelectron density weakens the bond, causing it to break in a

heterolytic fashion more easily. This makes the compoundmore acidic. Resonance withdrawal is most pronounced whenthe substituent is in the ortho or para position, so thestrongest acids are those with electron-withdrawing groupsattached to the phenyl ring at the ortho or para positions.Resonance is substantially weaker from the meta position.The inductive effect is weaker than resonance from anyposition.

3 5. The LOWEST pH value would be associated with:

A. 0.10 M O2NC6HaOH (p-nitrophenol).B. 0.50 M O2NC6HaOH (p-nitrophenol).C. 0.10 M C6H5OH (phenol).D. 0.50 M C6H5OH (phenol).

Copyright @ by The Berkeley Review@ 272 GO ON TO THE NEXT

36. A 100 mL 1.0 M phenol solution has a pH of:

A. 10.0.B. 7.0.c. 5.0.D. 1.0.

3 7. According to the pKu data in Figure 1, which compoun:is the STRONGEST electron-withdrawing group?

A. O2N-B. H3CO-C. H3C-D. H3CCO-

3 8. The conjugate base of p-nitrophenol has a pK6 of:

A. 10.4.B. 1.2.c. 6.8.D.3.6.

3 9 . How many grams of methoxy phenol must be adde; m,

100 mL of pure water to reach a pH of 6.1 ?

A. 12.4 gramsB. 2.48 gramsC. 1.24 gramsD. 0.62 grams

4 0 . What is the Ku associated with p-methyl phenol?

A. 4.0 * 10-10B. 2.5 x 10-10C. 4.0 x 10-11

D, 2.5 x 10-11

s

a:

41. What is the pH of0.15 M phenol with

A. 10.0B. 1.0c. 5.8D. 5.5

a solution made by mixing i10 mL water?

LD

E"

fi.pF

"FL4

l1

t3.

--,r-'

Passage Vll (Questions 42 - 48)

A strong acid fully dissociates when dissolved in water,and the pH of the resulting solution can be found by takingthe negative log of the strong acid concentration:

pH = Jog [strong acid]

Equation 1

For a weak acid, the determination of the pH is not as

simple, because the acid does not dissociate completely whenadded to water. To find the pH of a weak acid in water, the

[HfO+] must first be determined from the weak acid

equilibrium:

HA(aq)- H+(aq)+A-(aq)

lHAlinitt"r rc-7 0Initial:

Dissociation: - x

Equilibrium: [HA]in1si.1-x

+x +xx+10-7 x

10-7 is assumed to be negligible compared to x, andx is assumed to be negligible compared to [HA]616u1

The [H3O+],can be determined from the Ku value. The

following is a derivation of Equation 2, used to determine thepH for an aqueous solution of a weak acid.

Ka = [H3o+][A-] - [H:o*]2 ... Ka x [HA] = [H:o*]2tHAl FrAl

[H:O+]={Ku*[HA]' pH = -log [H:O*] = -log {Ku * [tt,q]

-log {ru * 1He1 = -loe{G - log{iffi= - ltogKu - llog tHAl = -1pK2 - llog tHAl2222

... pH = lpKu - log IHAI22Equation 2

Equation 2 works only when the value of Ks is less than

lHAl, and when pK6 is between 2 and 12.

42, For which of the following acid solutions does Equatiort

2 NOT work?

A.B.

@D.

HCIOHzCOrHIH5C6CO2H

43. What assumption is inherent in the derivation ofEquation 2?

Ai,'[HA] < [H:O+],8. [H3O+] = [A-lY. tHal = te-lD. [HA] < [A-]

Copyright @ by The Berkeley Review@ 273 GO ON TO THE NEXT PAGE

t

,44. ll the pH ol a 0.10 M weak solution is

the concentration of the conjugate base?

-. A. 0.r0 Me B. 2.0 x 1o-2 tvt(f) z.o x ro-3 M

D. 2.0 x 10-4 M', f A"

*r.F c-" / 91- )' '.' \ 11

;.,;;'i''-raiIr

,\ -. - l a1 r{ -l

P,. J_. a.

t 11 : i .-)rr

-i

seda'r5 S'-

_.^ -r'-/x2.70, what is'

;

45. Which of the following equations can be used todetermine the pOH of an aqueous weak base solution?

7<1pOu=loglOH-l -"t V '4'"-'\ '"ni14'

B. pOH =lpK6 - \og tA-l221C? pou= lpK6 - -Lloe tA-lv22

P. poH = t+ - (Jpru - rroe FrAl)

46, If a weak acid is titrated with enough strong base so that

[A-] > [HA], then for the resulting compound:

A. [H3O+] < K2.B. pH < pKa.

-{ luzo+l > tA-].D. [HA] < [HrO+].

t*ul

kt.L--*'lr

".r-,-- t{<

.],-' ..1

r ,\tr

An-- Wttl"tr acid yields the GREATEST conjugate base

& concentration when 0.010 moles HA are added toenough water to make the final volume 100 mL?

-,A. HFB. HNO3

C. HCND;. tt3CCO2U

48. A series of0.10 M weak acid solutions, each containinga different acid are compared. What is true for the acidin the series that has the GREATEST Ku value?

A. Its aqueous solution has the lowest conjugate base

concentration.

E'-_Itr aqueous solution has the lowest pOH.

G. -Liluqueous solution has the highest [H:O+].D . Its aqueous solution has the highest ratio of HA to

A-.

Passage Vlll (Questions 49 - 55)

When the pH of the stomach drops to a value less than2.0, the excess acidity may be reduced by consuming anantacid. The role of the antacid is to neutralize the HCI inthe gastric fluid. Some common bases found in antacids areCaCO3, MgCO3, Mg(OH)2, and NaHCO3. Calciumcarbonate, magnesium carbonate, and magnesium hydroxideneutralize acid in a two-to-one ratio. Sodium bicarbonateneutralizes only in a one-to-one ratio.

The gas buildup experienced after consuming an antacidis due to carbon dioxide gas, which forms in the stomachwhen carbonic acid (formed upon the full protonation ofcarbonate) decomposes into water and carbon dioxide. Theprincipal antacid ingredients of some common, commerciallyavailable products are listed in Table 1.

Aspirin (acetylsalicylic acid) can irritate the stomachlining. In an acidic environment, aspirin does notdeprotonate, so the molecule remains neutral and thus lipidsoluble. As a result, aspirin is able to work its way into themembrane of the stomach lining and then into hydrophilicpores within the membrane. It deprotonates inside thehydrophilic pore, causing ion levels to build up in thepocket, until osmotic prgssure forces water into the interiorregion of the membrarie, where parietal cells swell andrupture. The net effect is to eat away the stomach lining.Figure 1 shows the structure of aspirin and its ionizationequilibrium.

+H+o-

Figure I49. At what pH is aspirin MOST soluble in water?

A. 1.5

B. 3.4c.1.4D. 9.2

5 0. How many acidic sites are there in acetylsalicylic acid?

A. 1

8.2c. 3

D.4

Copyright @ by The Berkeley Review@

Yo+

o

ct13Yo?

o

cll3

Alka-Seltzer Aspirin, NaHCO:, citric acidBufferin Aspirin, MgCO3, aluminum glycinateMilk of Magnesia Mg(OH)2Rolaids Dihvdroxvaluminum sodium carbonateTums Calcium carbonate

Table 1

51. Which antacid gives the MOST neutralizing strengthper gram?

A. Magnesium carbonate (84.3 grams per mole)B. Calcium carbonate (100 grams per mole)C. Magnesium hydroxide (58.3 grams per mole)D . Sodium bicarbonate (83.9 grams per mole)

5 2. Which of the followingMOST damage to stomac

A. o

@Hg

@Hs

5 3 . How many acidic protons can be neutralized pmmolecule of dihyroxyaluminum sodium carbonam(Al(oH)2NaCo3)?

A. I8.2c. 3

D.4

The K"O for the ionization reactionacid in water is:

A . much greater than 1.00.

B . barely greater than 1.00.

C . barely less than 1.00.

D. much less than 1.00.

of acetylsali

5 5. Which of the following mixtures results in a

solution?

A. 1 equiv. acetylsalicylic acid + I equiv. HCIB. 2 equiv. acetylsalicylic acid + I equiv. HCIC. 1 equiv. acetylsalicylate + 1 equiv. HCID. 2 equiv. acetylsalicylate + i equiv. HCI

compounds would cause the

h lining?B.O

.A,Aocu.u--.D. O

o+o

4o

C. Od

54.

274 GO ON TO THE NEXT P

Passage lX (Questions 56 - 62)

Many common products used every day have acidic andbasic properties. An example is the antacid taken by manyheartburn sufferers to neutralize excess stomach acid (HCl).Antacids often contain hydroxide anion, carbonate anion, orboth in their salt forms. The contents of a normallyfunctioning stomach can reach a pH as low as 1.7 (highlyacidic) and in extreme cases can reach a pH as low as 1.0.

Because the pH scale is a log scale, a decrease of 0.7 pH unitsrepresents a hydronium ion (H3O+) concentration that is500Vo greater. Equation I is used to determine the pH of a

solution.

pH = -log [H:O+]

Equation 1

Table I lists some common household acids.

Acids Formula Product

Acetic acid H3CCO2H Vinegar

Ascorbic acid C6HgO6 Vitamin C

Hypochlorous acid HC1O Bleach

Acetyl salicylate H3C2OC6HaCO2H AspirinSulfuric acid H2SOa Battery acid

Table ITable2lists some common household bases.

Bases Formula Product

Ammonia NH: Windex

Sodium Bicarbonate NaHCO3 Baking Soda

Sodium Hydroxide NaOH DranoSodium Lauryl sulfate HrC(CHr) Na

Table 2

Bases when added to neutral water raise the pH of thesolution to a value greater than 7.0. Acids when added toneutral water lower the pH of the solution to a value less

than 7.0. A pH of 7.0 is considered to be neutral, becausepure (distilled) water has an [H3O+] = 1.0 x 10-7.

Acid in solution may be represented as either H3O+ or

H+, depending on the solvent. Both representations are

equivalent ways ofdescribing an acid.

t5F. A hydroxide anion is formed when water loses a proton

t (H+) to another water (known as autoionization). If the

concentration of dissociated hydroxide anion (OH-) indistilled water is 1.0 x 10-6 M, the pH of the solutionmust be: i,

A. 1.0, becadse [H:O*] = [OH-]..8. 7.0, becalrse water always has a pH of 7.0.

.-C._6.0, because lHfO*l = [OH-].

'D,. 8.0, becAuse the solution is rich in base (OH-).

.t{.Copyright @ by The Berkeley Review@ 27s GO ON TO THE NEXT PAGE

-,[57. What is the hydronium ion concentration ([H3O+]) of

an aqueous solution with a pH value of 1.7?

.*. l.'7 M H3O+

{B: r2.0 x 10-2 M H3O+g1 1.0 x t0-7 M H3O+D'. 5.0 x 10-13 M H3O+

"*. tf l0 mL of an aqueous solution of a strong acid with'r, pH = 2.0 were mixed with 100 mL of pure water, then| '/ the linal pH value would be: n {""*. to w L 1. | , 'rL

A. less than 2.0. s"l**;: ;'"L:il''10""'o r'Q' {'un"'t " *t'" ' 'i * r"''{ v

'. C. exactlY3.0. ;''' i''"i"" '"\ " n. D. greaterthan 3.0. t-1 1.. . t t:1{{ > ----

'r. ?.i '' +'- ""i'-'1 " :' r r ..-",' J & ",..-J !

5 9 . How many grams of CaCO3 are needed to neutralize 50

mL of stomach acid at pH = 2.0 completely, if thefollowing equation represents the neutralizationreaction?

CaCO3(aq) + 2 H+(aq) -> Caz+1aq7 + CO2(g) + H2O(1)

A. 25 mgB. 50 mgC. 100 mgD. 1.0 grams

6 0. Which of the following household products would NOTundergo an acid-base reaction with Windex?

. A. Vinegm."B1 Aspirin

{ 9, -Dtuno

-D:'Bleach

61. Which of the following when added to an aqueous

solution at pH = 6.0 would NOT raise the pH of the

solution?

,. '1{. Distilled water. B. Shampoo

...e1 xnacidD. -Vitamin C

i,_,_--

62, How many milliliters of 0.20 M H3O+ are required to

neutralize 1.68 grams of baking soda completely?

NaHCOI(aq) + H+(aq) -> Na+(aq) + COz(g) + HzO(l)

A. l0 mLB. 25 mLC. 50 mLD. 100 mL

Passage X (Questions 63 - 70)

The average human being produces between two andthree liters of gastric fluid in the normal day. Gastric fluid ishighly acidic (due to the presence of hydrochloric acid). It issecreted by the mucous membrane of the stomach lining toaid in the digestion of food. The average pH of this fluid isaround 1.5. The acid concentration necessary for this pH is0.030 M, if the acid is a strong acid. The acid in gastric fluidis strong enough and concentrated enough to dissolve(oxidize) metals with a positive oxidation potential.

The lining of the stomach includes parietal cells that are

tightly fused to form junctions in the stomach wall. Thesecells have a cell membrane that is permeable to neutralmolecules (such as water), but not to ions (such as Na+ and

Cl-). Hydronium ion, responsible for the acidity of thestomach, is a byproduct of metabolism. Carbon dioxide, a

byproduct of metabolism, is a non-metal oxide that convertsto an oxyacid when combined with water. The hydration ofcarbon dioxide (CO2) to carbonic acid (H2CO3) is shown inReaction 1. Carbonic acid is a weak acid that partiallydissociates into bicarbonate anion and a proton, as shown inReaction 2.

COz(g) + H2O(l) € H2CO3(aq)

Reaction 1

H2CO3(aq) .- H+(aq)+ HCO3-(aq)

Reaction 2

The proton, along with a chloride anion, is carried across

the membrane into the stomach via active transport.Enzymes assist the migration of both the proton and chlorideanion from the blood plasma into the interior of the stomach.These ions remain in the stomach until removed, because ofthe impermeability of the cell membrane to ions.

Eating stimulates the production of the enzymeresponsible for active transport and thus the release of acidinto the stomach to hydrolyze food molecules. Some protonsare absorbed by the mucous lining of the stomach wall,resulting in small localized hemorrhages. About 30 millioncells are destroyed per hour in normal stomach activity. As a

result, the stomach's entire lining is regenerated roughly onceevery 72 hours. Excess acid production increases thishemorrhaging and, in the worst case scenario, an ulcerdevelops.

6 3. What can be concluded about the following reaction?

HZCO:(aq) + NaCl(aq) a NaHCOg(aq) + HCl(aq)

A . AG > 0; requires active transport in vivo.B. AG < 0; requires active transport in vivo.C . AG > 0; does not require active transport in vivo.D . AG < 0; does not require active transport in vivo.

Copyright O by The Berkeley Review@ 276 GO ON TO THE NEXT PT

64. Which solution has the LOWEST pH?

A. Carbonated waterB. Distilled waterC . Salt waterD. Lime water (CaO(aq))

65. Which of the following metals can be oxidized b,'"

gastric fluid?

A. CopperB. GoldC . SilverD. Zinc

66. Which of the following reactions is NOT catalyzed b-'

acid?

A. Hydrolysis of disaccharides

B . Hydrolysis of polypeptidesC . Hydrolysis of esters

D . Hydrolysis of alkanes

67. Which will NOT reduce the hydronium ion (H3O-concentration in the stomach?

A . The consumption of waterB . The consumption of baking soda (NaHCO3)

C . The consumption of aluminum metalD . The consumption of solid food

6 8. What is the pH of 0.030 M HCI(aq) solution?

A. 0.7B. 1.5

c. 3.0D. 7.0

6 9 . Which of the following beverages when consumed c,:':*

NOT promote the decay of the stomach lining?

A . Orange juice (citric acid)B. Lemonade (citric acid)C. Milk (lactose)

D. Coca-Cola (phosphoric acid)

7 0. Which of the following has the LOWEST pH?

A. 0.03 M HCI(aq)B. 0.01 M HCI(aq)C. 0.03 M H2CO3(aq)D. 0.01 M H2CO3(aq)

T

{

Passage Xl (Questions 71 - 76)

As a general rule, the solubility for basic salts (conjugate

bases of weak acids) increases as the pH of the solutiondecreases. This is to say that basic salts are more soluble inacidic solution than in neutral solution. The calcium salts

such as calcium hydroxide (found in cement), calciumcarbonate (marble), and hydroxyapatite (tooth enamel) are

perfect examples. Hydroxyapatite is Ca5@O+):OH but can

also be written as 3Ca3(PO4)2' Ca(OH)2. Cavities result

when the hydroxyapatite dissolves away. When small

crevices are formed in teeth, acid-producing bacteria build up

and further dissolve away the worn region in the enamel. The

acid that dissolves the enamel is formed when aldoses

(aldehyde sugars, such as glucose) are oxidized to their

respective aldonic acid compounds. To reduce the dissolving

of the enamel, toothpaste and mouthwash containing fluoride

are recommended. The fluoride will substitute for hydroxidein the salt to form the less soluble fluorapatite Ca5(POa)3F(also known as 3Ca3(PO12' CaF). This lengthens the

lifetime of tooth enamel.

The solubility product (Ksp) for calcium fluoride (CaF2)

is 3.4 x 10-11 M3, for pure calcium phosphate (Ca:(PO+)z)

is 1.4 x 16-26 tr4s, and for calcium hydroxide (Ca(OH)2) free

of calcium oxide is 2.3 x 10-8 M3. The lower solubility ofcalcium fluoride compared to calcium hydroxide is mimicked

in the reduced solubility of fluorapatite compared tohydroxyapatite.

The following chart lists the molar solubility of calcium

hydroxide at varying pH along with the gram solubility per

100 mL solution at varying pH. The temperature was held

constant at25'C for all values:

pH Molar solubility Gram solubility

1 2.3 x 10I8 1.7 x 101e

2 2.3 x lO16 Ll x 1017

3 2.3 x l}ra 1.7 x 1015

1 2.3 X 106 1.1 x 107

il 2.3 x lO-2 1.7 x 10-l

t2 2.3 x l}-a 1.7 x 10-3

Table L

Table I shows that the effect of pH on the solubility ofbasic salts is drastic. This is because pH is measured on a

1og scale, while solubilities are measured on a linear scale'

71. As sodium fluoride (NaF) is added to an aqueous

solution with undissolved hydroxyapatite in it, what

happens to the solution?

A. The pH of the solution increases.

B. The pH ofthe solution decreases.

C. The pH of the solution remains constant at a value

greater than 7.0.

D. The pH of the solution remains constant at a value

less than 7.0.

Copyright @ by The Berkeley Review@ 277 GO ON TO THE NEXT PAGE

7 2. Calctum hydroxide is BEST described as which of the

following?

A. A strong acidB. A strong base

C. A weakacidD. Aweakbase

7 3, If the solubility of enamel in a 1.00 M solution of acid

HA is greater than the solubility of enamel in a 1.00 Msolution of acid HB, which of the followingconclusions can be drawn concerning to the two acids?

A. HA when dissolved into 100 mL pure water has a

higher pH than HB.B. The conjugate base of HA is a stronger base than

the conjugate base of HB.

C . HA has a higher Ka value than HB.

D. HA has a higher PKa value than HB.

74. Which of the following changes will ALWAYSincrease the amount of hydroxyapatite that dissolves

into solution?

A. Adding a strong acid to the solution

B. Adding a strong base to the solution

C . Increasing the pH of the solutionD. Lowering the amount of water in the solution by

evaporation

75. An example of an aldonic acid is drawn below. Withwhich of the answer selections listed should the aldonic

acid share a similar pKa value?

A. Acetic acid (H3CCO2H)

B. Hydrochloric acid (HCl)C. Nitric acid (HNO3)

D. Sodium carbonate (Na2CO3)

7 6. ln which of the following solutions does tooth enamel

dissolve the FASTEST?

A. 100 mL 0.10 M HClB. 100 mL 0.0010 M HCIC. 100 mL 0.10 M HF (Ka = 6.8 x 10-a)

D ' loo mL o'oolo M HF (Ka = 6'8 x 1o-a)

l

i

Passage Xll (Questions 77 - 84)

Normal rainfall has a pH between 5.6 and 6.0, dependingon the amount of carbon dioxide in the air. Excessiveamounts of atmospheric carbon dioxide lower the pH of rain.However, even with high carbon dioxide concentration, thepH of rain does not drop much below 5.4. While rain rich incarbon dioxide is more acidic than normal rain. it is notconsidered tobe acid rain.

Acid rain is attributed to non-metal oxides present in theair, like sulfur trioxide, and the nitrogen oxides. These non-metal oxides combine with atmospheric moisture to formnon-metal hydroxides, known as oryacids. Reaction 1 showsthe hydration of sulfur trioxide, converting it from a non-metal oxide into a non-metal hydroxide.

SO:(g) + H2O(g): H2SOa(g)

Reaction 1

Sulfur oxides are common by-products produced duringthe combustion of coal. A scrubber (a chamber at the base ofan exhaust stack filled with high pressure steam and calciumoxide) can help neutralize the non-metal oxides in theindustrial emissions of coal-burning factories. As theexhaust passes through the chamber, steam hydrates the non-metal oxide and converts it from an aqueous Br/nsted-Lowryacid into an airborne Lewis acid. Then as this non-metalhydroxide solution flows across the calcium oxide in thescrubber, it is neutralized by being converted into calciumsulfate and water. This is shown in Reaction 2.

H2SO4(aq) + CaO(s): CaSO4(s) + H2O(t)

Reaction 2

If the large quantities of acidic industrial waste generatedby burning fbssil fuel are not treated in this manner, they canproduce rain with a pH as low as 4.0. Acid rain is known todamage bodies of water by changing the bacteria population.Acid rain damages plants by changing soil pH. Changes inpH affect the structure ofplant enzymes needed for the uptakeand transport of nutrients. The enzymes become ineffectiveat lower pH because of their altered structures.

7 7. What chemical reaction is indicated in Reaction 1?

A . The neutralization of sulfur trioxideB. The acidification of sulfur trioxideC . The conversion of a Br@nsted-Lowry

Lewis acid.D. The conversion of a Lewis acid into

Lowry acid.

7 8 . A non-metal oxide can be characterized as;

A. amphoteric.B . an Arrhenius base.C. aBronsted-Lowry acid.D. a Lewis acid.

Copyright O by The Berkeley Review@

acid into a

a Br@nsted-

79. Which of the following compounds would creare :solution with the LOWEST pH when added to warer?

A. SO2B. CO2C. N2O5

D. MgO

What is the pH of 0.10 M HOCI(aq), which has a pK.of 1.46?

A . 1.00B. 4.23c. 6.46D. 1.46

Which of the following could be a component of ac:rrain?

A. Na2OB. NzC. NO2D. Hz

A11 of the following qualifications affect the acidity c,an oxyacid EXCEPT:

A . the oxidation state of the central atom.B . the electronegativity of the central atom.C . the size of the central atom.D . the number of n-bonds to the central atom.

All of these statements about H2S04, H2S03, an:their constituents are true EXCEPT:

A. SOr2- has a lower pK61 than SO42-.B. 1.0 M H2SO4(aq) has a lower pH than 1.0 \l

H25O3(aq).C. H2SOa has a higher Ku1 than H2SO3.D. H2SO3 is more electrolytic than H2SO,1.

Which sequence correctly lists the four acids shou:beiow according to their INCREASING pK. value?

PKal H2SOa < PKa HNO3 < PKal HzSO: < PK'HN02

PKaHNO3 < PKal H2SOa < PKaHNO2 < pK''H2SOj

PKal H2SO3 < PKa HNO2 < PKal H2SOa < PK,HNO:

PKa HNOz < PKal H2SO3 < PKa HNOj < PKa.H2SOa

80.

81.

82.

83.

84.

A.

B.

C.

27a

D.

GO ON TO THE NEXT PAGE

Passage Xlll (Questions 85 - 91)

Amino acids can be classified as either diprotic ortriprotic acids, depending on their side chain. A genericamino acid is shown below in Figure 1:

oil

HrN+1 -zc\c' o-

12HRFigure 1

The chemical structure in Figure I represents an aminoacid in a pH - 7 aqueous environment. At a pH less than thepKu of the carboxyl terminal (pKat ), the carboxylate group is

protonated and exists in its neutral carboxylic acid form.Similarly, at a pH greater than the pKu of the amino terminal(pK62 or pKa3), the ammonium group is deprotonated and

exists in its neutral amine form.

If the R group (side chain) of an amino acid exhibitsacid-base properties, then the amino acid is triprotic. Theside chain may be deprotonated at pH = 7, depending on thenature of the functional group. The carboxyl terminal pKulies between 1.8 and 2.6, and the amino terminal pKu lies

between 8.8 and 10.6 for all twenty common amino acids.The pKa for the protonated form and the pK5 for the

deprotonated form add up to a value of 14 at25"C.

PKa @rotonated form) + PKb (deprotonated form) = 14

Equation 1

The pK2 values for common side chains range from 3.9

to 13.2. Table 1 lists the R group and pK2 values for seven

common triprotic amino acids.

Amino acid Side chain group pKa

Aspartic acid -CH2CO2H 3.88

Glutamic acid -CH2CH2CO2H 4.32

Histidine -CH2C=CH-N+H=CH-NH- 6.05

Cysteine -CH2SH 8.36

Tyrosine -CH2C6HaOH 10.07

Lysine -CH2CH2CH2CH2NH3+ 10.80

nine -(CHz)rNHC(NH2)=lrt[12+ 13.21

Table IPhysiological pH is considered to be 7.4, although

gastric fluids and the fluid contained in lysosomes have

considerably lower pH. At physiological pH, a diproticamino acid exists predominantly in its zwitterion form.

8 5. What is the relationship between the pKu and pK6 forlysine?

A. PKat + pK61 = l{B. PKat + pK62= lQC. PKaZ + PK62 = l{D. pKaZ + pK63 = l{

Copyright @ by The Berkeley Review@

8 6. At physiological pH, which of the following aminoacids exists predominantly as a cation?

A. ArginineB. Glutamic acid

C. GlycineD. Histidine

87. What explains the lower pKa for the side chain

associated with cysteine than the one associated withserine (R = -CHZOH)?

A. Oxygen is less electronegative than sulfur.

B. Sulfur is less electronegative than oxygen.

C . Oxygen has a greater atomic radius than sulfur.

D . Sulfur has a greater atomic radius than oxygen.

8 8. The carbon chain associated with aspartic acid is shorterthan the one associated with glutamic acid, whichresults in a:

A . stronger inductive effect, making aspartic acid's side

chain more acidic.

B. weaker inductive effect, making aspartic acid's side

chain more acidic.

C . stronger inductive effect, making aspartic acid's side

chain less acidic.

D. weaker inductive effect, making aspartic acid's side

chain less acidic.

8 9 . At pH = 7, what are the applicable values for histidine?

A. PKat - 1.81; PKa2= 6.05;PK63 = 4.85

B. PKa: = 9.15 pKb2 = 7.95; pK61 = 4.85

C . pKtZ = 7.95; PKa2 = 6.05; pK6 = 9.15

D. PKa: = 9.15; pKb2 = 1.95; pKU: = 12.19

90. What is the normality of 0.50 M glutamic acid inwater?

A . 0.50 N H3N+CH(CH2CH2CO2H)CO2H(aq)

B. 1.00 N H3N+CH(CH2CH2CO2H)CO2H(aq)

C. 1.50 N H3N+CH(CH2CH2CO2H)CO2H(aq)

D. 2.00 N H3N+CH(CH2CH2COZH)COZH(aq)

91. All of the following amino acids have a neutral side

chain at pH = 7 EXCEPT:

A. cysteine.

B. histidine.C. lysine.D. tyrosine.

279 GO ON TO THE NEXT PAGE

ot

Questions 92 - 100 are NOT basedon a descriptive passage.

Which of the following solutionspH?

A. 0.01 M HCIO (pKa = 7.5)B. 0.05 M HCO2H (pKa = 3.6)C. 0.10 M HF (pKu = 3.1)D. 1.00 M HCN (pKa = 9.1)

has the LOWEST

93. Acid rain can possibly contain all of the followingcompounds EXCEPT:

A. oxidized non-metals.B. hydrated non-metal oxides.C . electron pair acceptors.D . metal hydroxides.

9 4. Which of the following conjugate pair-mixtures has theLOWEST pH?

l0 mL 0.10 M HF(aq) with 15 mL 0.10 M KF(aq)l5 mL 0.10 M HF(aq) with 10 mL 0.10 M KF(aq)10 mL 0.10 M NH3(aq) with 15 mL 0.10 MNH++(aq)

D. 15 mL 0.10 M NH3(aq) with 10 mL 0.10 MNH4+(aq)

95. Which of the following relationships applies tophosphoric acid?

A. pKat + pK61 = l{B. PKaZ + pK62 = l{C. pKa: + pK63 = l{D. pKat + pKrj = l{

96. What is the normality of a solution made by mixing50.00 mL of 1.5 M H3POa with 50.00 mL of purewater?

A. 3.00 N H3POaB. 2.25 N H3PO4C. 1.50 N H3POaD. 0.75 N H3PO4

A.B.C.

Copyright @ by The Berkeley Review@ THAT'S ENOUGH CHEM FOR

1.C6.D

11. B16. B21. C26. C31. B36. C41. D46. A51. C56. C61. D66. D7t. A16. A81. C86. A91. C96. B

2.D1.D

12. D17. D22. C21. B32. B31. A42. C47. B52. C5',7. B62. D61. D72. B11. D82. C87. D92. C97. B

3.C8.8

13. C18. C23. B28. C33. A38. C43. B48. C53. D58. D63. A68. B13. c78. D83. D88. A93. D98. C

4.A 5.C9.D 10.A

14. A 15. D19. D 20. D24. D 25. A29. C 30. C34. B 35. B39. C 40. C

44. C 45. C49. D 50. A54. D 55. D59. A 60. C

64. A 65. D69. C 70. A74. A 75. A19. C 80. B84. A 85. C

89. D 90. C94. B 95. B99. B 100. C

97. A 0. 10 M solution of unknown material has a pH4.62. It can BEST be described as a:

A. strong acid.B. weak acid,

C. strong base.D. weak base.

98. The HIGHEST pH is observed in which of thefollowing solutions?

A. 0.10 M HNO2(aq)B. 0.10 M HNO3(aq)C. 0.10 M NaNO2(aq)D. 0.10 M NaNO3(aq)

9 9. What is the pH of 0.050 M H3CCH2CO2H? (pKa fcrH3CCH2CO2H is 4.89)

A . 1.30B . 3.10c. 4.89D. 9.78

100. Which pH is INCORRECT for the correspondingsolution ?

A. 0.10 M HBr(aq) has pH = 1.00B . 0.10 M HCO2H(aq) has pH = 2.32

C. 0.10 M NaOAc(aq) has pH = 5.13D. 0.10 M KOH(aq) has pH = 13.00

E$"

1.

Acids and Bases Passage Answers

Choice C is correct. A quick look says that the pH should increase when the hydroxide ion concentrationincreases, which eliminates choices A and B. The pH should not increase by a full 3 units. An increase by 3

units would mean 1000 times the base concentration. This implies choice C is the most intuitive choice. Thisproblem can be solved longhand as well. The question deals with [OH-] and pH, so some form of conversion mustoccur. First, pH must be converted to pOH, using the relationship pOH = 14 - pH (14 - 8 = 6.0). Then, pOH must

be converted to [OH-], usi_ng the relationship [OH-] = tO-poH_1[OH-] = 10-6). Tripling the amount of OH- yieldsa concentration of 3 x 10-6 M. The new pOH is - log (3 x 10-o;, which equals 6 - log 3, which is approximately5.5. The new pH is equal to 74 - 5.5 = 8.5, so choice C is in fact correct.

Choice D is correct. The larger the value of pK6, the weaker the base. The weaker the base, the less itundergoes hydrolysis to form hydroxide anion when added to water. This makes choice D the correct choice.

Choice C is correct. An Arrhenius base is defined as a base that yields OH-(aq) upon addition to water. Of theanswer selections, only choice C is basic. Choice C must be correct.

Choice A is correct. HCIO3 is a weak acid that can be neutralized using a base of some sort. This eliminatesboth choice C and choice D. Choice B is a weak base, so it cannot fully react, eliminating choice B. Pick choiceA. For titration, 25 mL of strong base is ideal, because there are 25 mL of equimolar acid present. Although wehaven't reviewed it yet, titration is the quantitative addition of an equal mole quantity of a reactant.

Choice C is correct. As the conjugate acid gets stronger, the conjugate base gets weaker. The weaker theconjugate base, the higher its pK6 value, so choice A is eliminated. The weaker the conjugate base, the strongerthe acid that is required to react with it, so choice B is eliminated. The weaker the conjugate base, the lowerthe hydroxide anion concentration, so the higher the pOH and the lower the pH. This makes choice C correct.The weaker base requires the same moles of acid as a stronger base to be neutralized, as long as they are in equalconcentration. The strength of the acid required for reaction is different for the two bases, but the moles ofhydroxide are the same from equimolar bases, regardless of their strengths.

Choice D is correct. The passage provided the three definitions for bases. The definitions for acids is theopposite of that for bases. tnir rr,uuns that the Lewis definition of an acid is an electron-pair acceptor. ChoiceA is thus valid. The Brsnsted-Lowry definition of an acid is a proton donor, making choice B valid. TheArrhenius definition defines an acid as a compound that produces hydronium ion upon addition to water. Thismakes choice C valid. Choice D must therefore be the best answer. A Lewis base, not a Lewis acid, is definedas a nucleophilic molecule.

Choice D is correct. By definition, as the strength of an acid increases, it dissociates to a greater extent whenmixed with water. This eliminates choice B. By dissociating more, the acid produces more ions, making thesolution more electrolytic. Choice A is eliminated. A stronger acid produces more hydronium ion, so Ku shouldincrease and pKu should decrease. This both eliminates choice C and makes choice D correct.

Choice B is correct. Hydrofluoric acid, HF, is a weak acid, so it partially dissociates in water. The currentwould be minimal. HC1O is also a weak acid, so an HF solution has conductivity similar to an HC1O solution.The best answer is choice B. Pick choice B, and be a happy camper.

Choice D is correct. Both KOH and KOAc are salts. As such, they will both completely dissociate into water,so the ion concentration is the same in both solutions. This eliminates choice C. KOH is a strong base, whileKOAc is a weak base, so choices A and B are also eliminated. The best answer is choice D,

10. Choice A is correct. HCIO is a weak acid, so water must be acting as a base by removing a proton fromhypochlorous acid. This eliminates choices B and C. Water can be an amphoteric species, meaning it can act as

either an acid or a base, but in this reaction it acts only as a base. Choice D is eliminated. By gaining a protonfrom HCIO, water is acting as a proton acceptor, making it a Brsnsted-Lowry base. Choice A is the best answer.

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Choice B is correct. The solution in Trial 1 shows a greater current than the solution in Trial 2, so the solution inTrial 1 is more electrolytic than the solution in Trial 2. Electrolytic is defined as the ability of a solution toconduct electricity. This eliminates choice D. HCI is a strong acid, while HCIO is a weak acid. As such, moreions are present in solution during Trial 1 than during Tfial 2. This explains the greater conductivity in Trial 1

than Trial 2. Other colligative properties include boiling point elevation, freezing point depression, andosmotic pressure. Because the solution in Trial t has more ionic impurities (and thus more total impurities), thesolution in Trial t has a higher boiling point, lower freezing point, and greater osmotic pressure. This makeschoice B the correct answer.

Choice D is correct. As seen in the experiment, full dissociation leads to a current of approximately 6.0 amps.Nitric acid is a strong acid, so choice A should generate a current of roughly 5.0 amps. Sulfuric acid has onestrong proton, so choice B should generate a current of roughly 6.0 amps. Sodium hydroxide is a highly solublesalt, so choice C should generate a current of roughly 6.0 amps. Ammonia is a weak base with no charge, scaddition of ammonia to water forms only a few ions. The current cannot reach 6.0 amps, so choice D is the bes:answer.

Choice C is correct. A strong base is mixed with a strong acid in Trial 3. A complete reaction occurs, leavirrsbehind potassium cation (K+) and chloride anion (Cl-). A strong base is mixed with a weak acid in Trial4. -1.

complete reaction occurs, leaving behind potassium cation (K+) and hypochlorite anion (ClO-). A completereaction occurs in both Trial 3 and Trial 4, because in each case a strong base has been added. This eliminateschoices A and B. Because equal amounts of strong base (KOH) are added to equal mole quantities of acid in bot}'Trial 3 and Trial 4, both reactions go to completion, generating the same number of ions in solution. This favor,choice C. Choice D is invalid, because cations are also in equal concentration, and cations conduct electrons.

Choice A is correct. From the relative acid strengths derived from the pK. values, HCIOa is more acidic thrHBrO4, which in turn is more acidic than HIO4. This same trend is seen with HOCI, HOBr, and HOI. Becau-.eCl is more electronegative than Br which is more electronegative than I, the trend indicates that tl-..electronegativity of the halide dictates the relative acidity of an oxyacid. This fact was also given in tL=passage. Pick answer A, and feel what it is to be correct. Choice C applies to haloacids, not oxyacids.

Choice D is correct. To determine the relative strength of the oxyacids, both the number of oxygens and ti.;electronegativity of the halide must be considered. HCIOa is a stronger acid than HIO4, because chlorine *.more electronegative than iodine, so choices A and C are both eliminated. Choice B is eliminated, becaus:HIOa is a stronger acid than HBrO3, due to the extra resonance structure associated with the additior.:-oxygen. The best answer is choice D, since it follows both the decline in number of oxygens trend and r:,=decreasing electronegativity trend.

Choice B is correct. For a weak acid solution with [HA]1r.,itial > Ka, use the shortcut equation to determine --pH. The shortcut equation, pH = lrpKu - llog [HA], applies if the pK6 lies between 2 and, 1,2. The math is ::follows:

pH = l-pKu - llog [HA] =7-g.ZAy- llog (0.10) = 3.ffi - l(-t) = 3.63+ 0.5 = 4.132' 2- 2 2- 2

The correct answer is choice B.

Choice D is correct. The pKu2 value must be larger than the pKul value, because the first proton is more aci:uthan the second proton (by definition), and the stronger the acid, the lower its pKu value. Given that pKu- s7.8, the value of pKu2 must be a number larger than 7.8. The only answer larger than 7.8 ts L2.9, so pick chc:,*D, and feel satisfied that you did. The pK6 (not pKu) for the conjugate is 6.2, but the question asks for pKu2.

Choice C is correct. Since the dissociation reaction is exothermic, heat is given off when forming the produ;;.Thus when heat is added to the system, it acts as an inhibitor of product formation according to Le ChAteLe: iprinciple. This means that products decrease and reactants increase, as the system is heated. The r*,(equilibrium constant) is a measure of the products over the reactants, so the value of Ku decreases with :-*addition of heat to the system. The best answer is choice C. Choice D should be eliminated immediately.

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Choice D is correct. We know HBrO2 is less acidic than HCIO2, since both acids have the same number ofoxygen atoms, and chlorine is more electronegative than bromine. Given that HBrO2 is the weaker acid, thepKu for HBrO2 must be greater than 1.9 (the pKu of HCIO2), so the only answer possible is choice D.

Choice D is correct. Dissociation refers to the breaking of the bond between the acidic proton and the conjugatebase, so that the conjugate base and a hydronium ion are formed. The greatest dissociation is associated withthe strongest acid. This question is asking for the strongest acid. The strongest acid is the acid with thegreatest number of excess oxygens and the most electronegative central atom. This makes choice D the bestanswer. This answer could also have been determined by recalling the six strong acids listed in the text. Onlychoice D is one of these strong acids. A strong acid fully dissociates, while weak acids partially dissociate.The amount of dissociation can be determined from the Ku and vice versa. For instance, a 1.0 M weak acid with10% dissociation results in 0.90 M HA undissociated acid, and the formation of 0.10 M A- and 0.10 M H3O+. Theequilibrium constant for the dissociation reaction (Ka) is shown below:

6" = [HeO"][Al _ (0.1)(0.1)

=0.01 =0.0111" tHAl 0.e 0.e

Choice C is correct. The volume of the solution does not affect the pH; only the concentration and strengthaffect the pH. The first step is to identify the type of acid. Since HC1O4 is a strong acid, it fully dissociates,so the major source of protons in the aqueous solution is from the dissociation of HClOa.

pH = - 1og [H3O+] - -log [HCIO+] = - log(0.10) = - (-1) = 1

This best answer is choice C.

22, Choice C is correct. The most acidic compound is the compound that most readily loses H+. In choices A, B andD, the hydrogens are bonded to carbon. In choice C, there is a hydrogen that is bonded to nitrogen. The sizedifference between atoms is significant only when the atoms are in different rows of the periodic table.Nitrogen and carbon are in the same row of the periodic table, so they are comparable in size. When atoms arein the same row of the periodic table, the most important factor to consider when looking at acidity iselectronegativity. Nitrogen is more electronegative than carbon, making a hydrogen on nitrogen more acidicthan a hydrogen on carbon, so choice C is the most acidic. It is a common mistake to not notice that thehydrogens are all on carbons in choices A and B. Be careful not to make mistakes like this.

23. Choice B is correct. It can be observed from the data in Table 1 that as the number of methyl groups increases,the pKu value increases. An increase in pKu is indicative of decreased acidity. Acids are defined as electron-accepting, so electron-withdrawing groups increase acidity, while electron-donating groups decrease acidity.This makes choice B correct.

l1' Choice D is correct. In haloacids (binary compounds), the acidic proton is directly bonded to a halogen, sorelative acidity can be discerned from the features of the halide. Halogens are in the same column of theperiodic table, so the important factor when considering bonding is the size of the halide. Because iodine is thelargest of the halogens, the HI bond is the weakest, so HI is the strongest haloacid. Pick choice D, and smile tothe world.

15' Choice A is correct. Sulfur and chlorine are adjacent to one another in the same row of the periodic table, sochoices B and D are eliminated. Because chlorine is smaller than sulfur, the increasing size does not correlatewith acidity. This eliminates choice C. Chlorine is more electronegative than S, so Cl draws electrons from Hmore than S does. This makes HCI a stronger acid that H2S. The electronegativity predicts the acidity bestfor atoms in the same row, so pick answer choice A to get this one correct.

16. Choice C is correct. Table 1 shows that thiols have lower pKu values than their corresponding alcohols, soethyl thiol is more acidic than ethanol. Because ethanol is a weaker acid (has a lower Ku value than ethylthiol), ethanol dissociates less than ethyl thiol when added to water, and ethanol yields a lower [H3O+] thanethyl thiol. These three facts eliminate choices A, B, and D. Because ethanol is a weaker acid than ethylthiol, the conjugate base of ethanol (CH3CH2O-) is a stronger base than the conjugate base of ethyl thiol(CH3CH2S-), making choice C the correct answer.

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Choice B is correct. The most electronegative atom that differs from choice to choice is chlorine. This highdegree of electron withdrawal due to chlorine results in an increase in acidity. This implies that becausechlorine is more electronegative than the other atoms, it withdraws electron density the most and thusincreases the acidity to the greatest extent. The answer of answer choices is choice B.

Choice C is correct. Of the answer choices, the carboxylic acid is the most acidic, because of the resonance

associated with the carbonyl bond. This eliminates choices A and B. From this point, data in Table 1 must be

analyzed. Table 1 shows that as the substitution of the thiol decreases, the acidity increases, so the priman-thiol is more acidic than the tertiary thiol. This makes choice C the best answer.

Choice C is correct. The solution is acidic, so the pH is less than7.0, eliminating choice D. For a weak acic

with pKu between 2 and 72 in an aqueous solution with [HA]6itial ) Ka, use the shortcut equation to determine

the pH. If Ku is 8 x L0-8, then pKu is 8 - log 8 = 7 - 0.9 = 7.1.. The [HA] is 0.5 M, and log 0.5 = - log 2 = -0.30.

RH=|RKu -llog[HA] =L1z.t!-l"S(0.50) =r.55 - 11-o.s) =3.55 + 0.75=3.7

The value falls in the range of choice C, so the best answer is choice C.

Choice C is correct. Mass percent is defined as the mass of one component atom relative to the total mass of tLe

compound. In this case, we are interested in carbon within phenol. Units cancel, so we can use atomic masses,

mass yo carbon - mass of carbon - 72g,C x100% + 72%= 72 <ZZ<ZZ=g1"h100 94 90mass of molecule 94 g C6H5OH

The mass percent of carbon is between 72'h and 80%. Choices A and B are too small, while choice D is too largtOnly choice C,76.6"/o, falls into the range, so choice C is the best answer'

Choice B is correct. When an acid is titrated to its equivalence point, it has been completely converted into r:.conjugate base. The lowest pH value after neutralization is found with the weakest conjugate base, assumi-lthat the concentrations are all equal. The weakest conjugate base corresponds with the strongest acid (that r,'

the acid with the lowest pKu). This is because the stronger the acid, the weaker its conjugate base. Using f =

data from Table 1, the acid with the lowest pKu value is choice B, trichloroacetic acid. Pick choice B please"

Choice B is correct. Increasing the number of impurities dissolved into an aqueous solution raises the boi-lr:point and lowers the freezing point of a solution. The total impurity concentration depends on the solu:*

concentration, the number of particles it dissociates into, and the degree to which it dissociates. All of ---*

choices are monoprotic weak acids of equal concentration (1.0 M), so it depends only on the dissociation of ':*acid (acid strength). Acetic acid is weaker than p-nitro benzoic acid, so choice A is eliminated. Tricloroace-racid is stronger than p-nitrophenol, so choice B is correct. Tricloroacetic acid is stronger than benzoic acid --:

choice C is eliminated. Acetic acid is stronger than phenol, so choice D is eliminated.

Choice A is correct. The acid with the strongest conjugate base is the weakest acid. The highest pK6 vak; ''+

associated with the weakest acid. Referring to the data in Table 1, the acid with the highest pKa is phe:.---

Choice A is the correct answer, since the highest pKu value indicates the weakest acid.

Choice B is correct. This question appears to be difficult at first glance. Flowever, if you realize that whe: m

acid dissociates, equal parts of hydronium and conjugate based are formed, then you know that [H3O+] = l-i-All that is required is converting from pH to [H3O+], using [HaO*] = 10-PH. The pH of the solution is 3'7, sc :u[HeO*] is 10-3'7= 190.3x 10-4. E".urrr"log 2=0.3,itistruethat100'3 =2. This*"uttthat100'3x 10-4 = 2x'-+ffre HgO+l is2x10-4 M, and [HSO+] = [A-], so [A-] =2x10-4. The correct answer is choice B.

35. Choice B is correct. The lowest pH is found in the solution with the greatest amount of hydronium. lrtehydronium concentration is affected by the strength and concentration of the acid. The most acidic solu:lmresults from the strongest acid in highest concentration. In this question, p-nitrophenol has a lower pKu :mphenol, so choices C and D are eliminated. Choice B has the higher concentration of p-nitrophenol, so choi:: B

is the best answer. Pick choice B for best results and greatest test satisfaction.

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36. Choice C is correct. The solution is acidic, so the pH is less than 7.0, eliminating choices A and B. For a weakacid with pKu between 2 and 12 in an aqueous solution with [HA]initial > Ka, use the shortcut equation todetermine the pH. The pKu of phenol is L0, and the [HA] is 1.0 M. The log of 1.0 is zero, so the pH is simplyhalf of the pKa. Half of ten is five, so the pH is 5.0. Choice C is the best answer.

Choice A is correct. A stronger electron-withdrawing group on the aromatic ring of the phenol makes thephenol more acidic, thus decreasing its pK" value. As a result, a substituted phenol with an electron-withdrawing $oup has a lower pKu than phenol. From the acids listed in Table 1, the acid with the lowestpKu is p-nitrophenol. The electron-withdrawing group on p-nitrophenol is O2N-, so choice A is terrific.

Choice C is correct. For a monoprotic acid and its conjugate base, pKa + pKb = 14. From Table 1, the pKu valuefor p-nitrophenol is 7.2. This means that pK6 for the conjugate base (p-nitrophenoxide) is 14 - 7.2 = 6.8. Thecorrect answer is choice C.

Choice C is correct. This problem is a twist on the normal approach (i.e., it's harder than typical problems).

This problem is most easily solved using the shortcut equation, pH = Ipfu - hog [HA], to solve for [HA].2- 2-pH=

fRKa -qogtHAl ;.6.L= L3t.z1-|l"flHAl= 5.6-LrogtHAl .'.0.5=-hog[HA],solog[HA1=-1

log [HA] = -1 .'. [HA] = 10-1 M = 0.10 M

The concentration of the acid is 0.10 M, so the next step of the solution is to determine the moles of weak acidneeded to make 100 mL of a 0.L0 M solution. To make 100 mL 0.1 M H3COC5H4OH, 0.01 moles of the acid mustbe added to enough water to make 100 mL aqueous solution. The final step is to convert 0.010 moles into grams.The molecular mass of H3COC5HaOH is 724 grams per mole, so 0.01 moles is 7.24 grams H3COC5HaOH. Thecorrect answer is choice C.

Choice C is correct. To convert from pKu to Ku, tlle relationship is.Ku = 10-PKa. The pKu of p-methylphenol is10.4 (as given in Table 1). The K, is therefore 16-10'a = 160'6 *-10-11, when expanded into scientific hotation.Given log2= 0.3 and log4 = lo,g2+log2= 0.3 + 0.3 = 0.6. Accordingto the anti logrelationship,100'6 = 4. Thismakes the value of Ku - 4 x 10-1I. Pick choice C and be a success story in acids and bases.

Choice D is correct. The addition of water dilutes the aqueous phenol solution, but it does not react with thephenol. Using the relationship, M1V1 = M2Y2, the new molarity is found to be 0.10 M. As is becoming the

norm, this problem is most easily solved using the shortcut equation, pH = lrpKu - qog tHAl.

pH =lpXu - hog [HA] =l(rO.O) - ltog (0.10) = S.0 - l(-1) = 5.0 + 0.5 = 5.52' 2- 2 2- 2

The correct answer is choice D. Because the solution is acidic, choices A and B should have been immediatelyruled out.

Choice C is correct. The shortcut equation, as presented in the passage, works only for weak acids. It does NOTwork with a strong acid. The only strong acid in the choices is hydroiodic acid (HI), which you should know isa strong acid, because HI is stronger than HCl, a common strong acid. Choice B (carbonic acid) and choice D (acarboxylic acid) should immediately be recognized as weak acids. Pick choice C for the greatest success here.

Choice B is correct. In the first step of the derivation, [H3O+] is substituted for [A-], as shown in the conversionfrom [H3O+][A-] to [H3O+12. The basic assumption is that when the acid dissociates, equal amounts of H3O+and A- are formed in solution, and the little (10-l M) H3O+ formed from water is insignificant. This makeschoice B the best answer.

Choice C is correct. Regardless of the molarity of the weak acid, when the pH of the solution is 2.70, the[HsO+] in solution is 1,0-2J M. This means that the [HaO+] is equal to something " fO-s (where "something" is avalue less than ten). Because [HaO*] = [A-] when a weak acid dissociates, it should be concluded that [A-] =something x 10-3, which makes choice C the best answer.

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45. Choice C is correct. Starting from the K6 relationship for the hydrolysis of a weak base in water, the

equilibrium expression can be manipulated to yield the relationship pOH - -log {G " tAl This relationshipcan be manipulated in the same way that the pH relationship was manipulated in the passage to yield thesame type of equation as the shortcut equation, except that it's for a weak base. The logic is that the pOHdepends on the base, hence the formula contains K6 rather than Ku, and [A-] rather than [HA]. This makeschoice C the best answer. Choice A is applied in general to all basic solutions, but it is not the best answer.Choice D may sound tempting, but that is how the pOH for a weak acid solution is found, not for a weak basesolution.

Choice A is correct. According to the Henderson-Hasselbalch equation, when [A-] > [HA], the pH is greater

than the pKu, because the log of the [A-] to [HA] ratio is a positive value. This eliminates choice B. If pH isgreater than the pKu, then by definition -log [HSO*] is greater than the -1og Ku, hence iog Ku is greater thanthe log [HsO*] (when both sides of an inequality are multiplied by -1, the inequality sign must be reversed).Because log Ku is greater than the log [H3O+], Ku is greater than the [H3O+], making choice A the best answer.Choice C cannot be true, because as pH goes up, [A-] increases as [H3O+] decreases, making [A-] > [HaO+] at allpoints of a reaction after the initial mixing of the weak acid into water. This eliminates choice C. Choice Dcan be true only if [HA] = 0, which is physically impossible with the equilibrium constant favoring HA. Itshould always be true that [HA] > [HaO*], which eliminates choice D. The best answer is choice A.

Choice B is correct. The greatest conjugate base concentration is found with acid that dissociates to the greatesiextent. The acid that dissociates most completely is the strongest acid, which describes choice B (HNO3), theonly strong acid among the choices. This question is just another way of asking, "Which acid is stronger?"

Choice C is correct. As the value of Ku increases, the strength of the acid is increasing. As the acid getsstronger, the degree of dissociation when added to water increases. This causes the conjugate base concentrationto increase, which eliminates choice A. The amount of H3O+ increases, which causes the pH to go down ancthus the pOH to increase. This eliminates choice B and confirms that choice C is the best answer. Because thtacid dissociates more, the amount of HA decreases and the amount of A- increases, causing the ratio of HA to A-to decrease. This eliminates choice D.

Choice D is correct. Aspirin (acetylsalicylic acid) when added to water dissociates into the anionic conjuga;=base and a proton, just like all other carboxylic acids. The dissociation obeys the laws of equilibrium, resultir.:in an equilibrium constant known as Ka. Because it obeys the laws of equilibrium, the anion form (more wate:-soluble form) is most abundant when the H+ (proton) is removed from solution. The lowest concentration ::protons ([HSO*]) is found at high pH. The highest pH results in the greatest water solubility for the aspir--.because it exists in its anionic form. This, in essence, is the common ion effect. The best answer is thus choice D

Choice A is correct. Acetylsalicylic acid has only one hydrogen on the carboxylic acid functional group tha: ,'acidic. All of the other hydrogens are bonded to carbon, which does not make them acidic. The correct ans-,\:::for the number of acidic protons is therefore only one. Choice A is the best answer.

Choice C is correct. The first three choices can all neutralize two equivalents of acid, while choice D, sod:-:nbicarbonate, can neutralize only one equivalent of acid. This means that the most neutralizing strength :e'gram is found with the compound having the lightest molecular mass of the compounds capable of neutralu:5two equivalents of acid. Magnesium hydroxide has the lowest molecular mass of the first three answer cho:r=*The correct answer is therefore choice C.

Choice C is correct. The molecule that can cause damage to stomach lining is the acidic molecule that ior-:Et:once within the membrane pocket of parietal cells. The only acidic compound of the choices is benzoic a:ri.answer choice C. It is the most similar in structure to aspirin, so it is the best answer.

Choice D is correct. Dihydroxyaluminum sodium carbonate is composed of the dihydroxyaluminum .;:rlmr(which loses two hydroxide substituents when hydrolyzed), sodium cation (with no basic properties il'urd,

carbonate dianion (which can neutralize two acidic protons). The net result is that the two hydroxides arr; TrnE'

carbonate can neutralize four equivalents of protons, making choice D correct. You might also consider r:carries a +3 charge and Na carries a +1 charge, so the sum charge of the bases must be -4.

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Choice D is correct. Acetylsalicylic acid is a carboxylic acid, and carboxylic acids are weak acids. Weak acidshave K" values that are less than one. The pKu for a carboxylic acid ranges from 2.0 (as seen with the carboxylterminal of amino acids) to 5.0 (as seen with regular alkyl chains). This results in a range for Ku of 1O-2 to 10j5,which yields a Ku that is much less than 1.00. This vague question is best answered by choice D.

Choice D is correct. A buffer forms when the conjugate acid and base of a conjugate pair are present in roughlyequal molar ratio. Choices A and B should immediately be eliminated, because the mixtures are made up oftwo acids, and not of an acid and base pair. Choices C and D involve a strong acid and weak base, so the correctchoice involves partial titration of the weak base. It is in choice D that an equal molar ratio of the weak baseand its conjugate acid are present. The one equivalent of HCI will convert one of the two equivalents ofacetylsalicylate into acetylsalicylic acid, while one equivalent remains as acetylsalicylate. The equalquantity of the two components of the conjugate pair results in a buffer (with a pH equal to the pKu of the weakacid). The best answer is choice D.

56. Choice C is correct. The first word in the question is "If", so keep that in mind. The mantra that you chantabout water having a pH of 7 does not apply in this speculative question. If the only hydroxide ion in solutionis formed when one molecule of water loses a proton to another molecule of water, then the hydroxide anionconcentration [OH-] must equal the hydronium cation concentration ([H3O+]). This is true within distilledwater. This means that if [OH-] = 1.0 x 10-6 M, then [HSO*] must also equal 1.0 x 10-6 M. The negative log of 1.0x 10-6 is 6.0, so the pH of the water must be 6.0. The best answer is choiie C. For this hypothetiLl solutiin, pH= 6 and pOH = 6, which means that pH + pOH = \2, not 14. This may bother your sense of what is right in iheworld of acid and base chemistry, but keep in mind that pH + pOH = 14 applies only at 25'C. At highertemperatures, there is more autoionization, so more hydronium and hydroxide are generated. At 37'C forinstance, pH of distilled water is 6.8, so pH + pOH = 73.6.

Choice B is correct. The hydronium ion concentration in solution can be.found from the pH of the solution.Because the pH is defined as pH - -1og [$eO+J, the hydronium ion concentration can be found by manipulatingthe relationship to yield [H3O+] = 10-P_H. The hydronium ion concentration in a pH = 7.7 solution is i0-1.7 M-,which is therefore roughly equal to 10-2 M. The best answer selection is choice B.

Choice D is correct. The volume increases from 10 mL to 110 mL when 100 mL of distilled water is added to thesolution, so the concentration of the acid must decrease by a factor of 11. The log of 10 is equal to 1, so the log 11is slightly greater than 1. This implies that the pH increases by just a little more than 1.0, because the [H3O+]goes down by a factor of 11. This makes the final pH greater than 3.0. The best answer is greater than 3.0,which is choice D. This is true only of a strong acid solution. If the solution were an aqueous weak acid, the pHincrease would not be as significant as with the strong acid.

Choice A is correct. The hydronium ion concentration in a pH = 2.0 solution is 1.0 x 10-2 M. There are 50 mLpresent, so the moles of H+ are found by multiplying 0.010 M by 0.050 L = 5.0 x 10-4 moles H+. Only half themoles of CaCO3 are necessary to neutralize the H+, because it is a 2 : 1 ratio of H+ to CaCO3 in the balancedequation. This means that 2.5 x 10-4 moles CaCO3 are required to neutralize all of the hydronium ion (HSO*).2.5 x 10-4 moles when multiplied by 100 grams p"r tt oi" yields 2.5 x 70 2 grams Ca-O3, which equals 25milligrams. Choice A is the best possible answer you can choose in a situation like this.

Choice C is correct. Windex contains ammonia, which according to Table 2 is a base. An acid can react with abase, while a base cannot react with another base. The question is thus asking: "Which of the followingcompounds is not an acid?" Vinegar, aspirin, and bleach are all listed as acids in Table 1, so they all can reactwith ammonia (Windex). The best answer is choice C, Drano, because Drano is a base and not an acid.

Choice D is correct. Raising the pH of a solution requires reducing (or diluting) the concentration of hydroniumion (H3O+). Adding distilled water reduces the hydronium ion concentration by diluting the solution. Thismakes choice A invalid. Shampoo and antacid are bases, which when added to solution, react with the H3O+ion and thus reduce the hydronium ion concentration and raise the pH of the solution. Choices B and C aretherefore invalid. Adding an acid lowers -- not raises : the pH of the solution, so choice D (vitamin C), anacid, is the only answer choice that will NOT increase the pH of the solution when added. Pick choice D witha smile.

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Copyright @ by The Berkeley Review@ 287 Section IV Detailed Explanations

62. Choice D is correct. 1.68 grams of baking soda is equal to 0.020 moles NaHCO3 as determined by dividing the1.68 grams by 84 grams per mole. The equation for the reaction shows a 1:1 ratio between baking soda and the

hydronium ion for full neutralization, so 0.020 moles of H3O+ are required to reach full neutralization. The

molarity of the acid solution when multiplied by the volume of the acid solution must equal0.020 moles HgO*.The molarity of the acid solution is given as 0.20 M, so the volume of the acid solution must be 0.10 liters, whichis 100 mL. Selecting answer choice D in a situation like this, is the best thing to do.

63. Choice A is correct. The formation of a strong acid from a weak acid and neutral salt is unfavorable.Hydrochloric acid is stronger than carbonic acid, so the reaction as written has AG > 0. To carry this o:ut ittaiuo, it must be coupled with some other favorable reaction. As stated in the passage, this process is active

transport. The correct answer is therefore choice A.

Choice A is correct. The lowest pH is associated with the most acidic solution. Both salt water and distilledwater are neutral (have a pH of 7.0), so choices B and C are eliminated. Carbon dioxide (a non-metal oxide)

when dissolved into water gets hydrated to form carbonic acid (H2CO3, a non-metal hydroxide). Calciumoxide (a metal oxide) when dissolved into water gets hydrated to form calcium hydroxide (Ca(OH)2, a metaihydroxide). Non-metal hydroxides are acidic and metal hydroxides are basic, so the best answer is choice A.

Choice D is correct. Acid is capable of oxidizing any metal with a low ionization energy. ln this question, onlr'one answer is correct, so it is better to ask yourself which metal is most easily oxidized (or least stable't.

Relying on either trivial knowledge or experience, you know that copper, gold, and silver are relatively air-

stable, as evident by their commercial uses in wiring and currency. All three tarnish over enough time. Zincmetal is most readily oxidized (more so than hydrogen gas even). The best answer is choice D. If you are

interested in smugglin g zinc coins across the border, it is recommended that you refrain from ingesting them to do

so. If you don't know about the relative reactivity of these metals, you just have to take a guess and move on.

Choice D is correct. This is really a misplaced organic chemistry question. But then agakt, the passage is on a

physiology topic, emphasizing that acidity and basicity are applicable to many areas of study. Alkanescannot be hydrolyzed whether an acid is present in solution or not. Disaccharides, polypeptides, and esters ar€

all broken down in the stomach by acid-catalyzed hydrolysis. The best answer is choice D.

Choice D is correct. Consuming water dilutes all solutes (including HgO* in an acidic solution), thus tl'ehydronium ion concentration is reduced by consuming water. Baking soda can consume some of the hydronium ic:r

by way of an acid-base reaction. Aluminum metal can be oxidized by hydronium ion, resulting in the formatia:oi aluminum trication and hydrogen gas. Therefore, choices A, B, and C are all eliminated' Consumption c:

solid food induces the release of hydrochloric acid into the stomach. The correct answer is choice D.

Choice B is correct. This answer is taken straight from the first paragraph of the passage. 0.030 M hydronirr.has a pH of 1.5. However, in the event you skipped reading the passage, you can solve for the pH b-;

determining the negative log of the hydronium concentration (pH = - log [H3O+]). The pH of a 0.030 M HCsolution is --log (3 ;1 1O-Z; = 2 -log3. The value is less than 2, but greater than 1. The correct answer is choice B.

Choice C is correct. Stomach lining is deteriorated by reacting with acid. All of the answer choices exce:(

milk contain acid, so they all promote the decay of the stomach lining. The best answer is choice C. Ti:s

question is asking for you to identify which compound is not an acid, so look for th choice that is differe:-:"

Lactose is a disaccharide composed of galactose and glucose.

Choice A is correct. The lowest pH results from the presence of the strongest acid in the highest mc:,rconcentration. Hydrochloric acid ii the strongest acid (stronger than carbonic acid), and 0'030 M is a hig!*concentration than 0.010 M. The correct answer is therefore choice A.

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7'1., Choice A is correct. The first question is whether the pH of the solution changes. Because fluorapatite is rmsoluble than hydroxyapatite, hydroxide anion is released as fluoride anion exchanges for the hydroxide ihydroxyapatite. When hydroxide anion is released, the pH of the solution increases. Although the ph :rgreater thin 7.0, choice C is incorrect, because the pH does not remain constant. The best choice is A.

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Copyright O by The Berkeley Review@ 2aa Section IV Detailed ExPlanatioa,

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/3.

Choice B is correct. Calcium hydroxide when added to water yields hydroxide anion (OH-), making it basic.This eliminates choices A and C. The question is whether it is a strong or weak base. These questiois on theYCAT are tough, because who knows what the test-writer wants? Beiause calcium hydroxide does not fullydissolve into water at high enough concentrations, it can be considered a weak base. However, because it reactswith weak acids, and it does fully dissociate at low concentrations, it must be a strong base. The best answerbeyond any questioning and rational argument you can present is choice B (unless, of course, you wish to chooseD)' On some questions, like this one, there are two reasonable answers, but the writer of the question wants youto pick just one best answer.

Choice C is correct. Enamel is more soluble in the more acidic solution. Because enamel is more soluble in theHA solution than the HB solution, the HA solution must be more acidic than the HB solution. Both the HAand HB solutions are of equal concentration, so it must be that HA is a stronger acid than HB. This means thatif HA is dissolved into water, it has a lower pH than when HB is dissolved into water. The conjugate base ofHB is a stronger base than the conjugate base of HA. HA has a higher Ku and a lower pfu thai HB. Thiseliminates choices A, B, and D. Choice C is the best choice.

Choice A is correct. The hydroxyapatite dissolves most readily in acidic medium, so the addition of acidincreases its solubility. Thus, the amount of hydroxyapatite that dissolves into solution increases as acid isadded to solution. Addition of base increases the pH of the solution, meaning that choices B and C are the sameanswer. The MCAT presents questions like this on occasion. When they hive two identical answers wordeddifferently, eliminate them both. Decreasing the solvent reduces the amount of salt that has dissolved intosolution. The best answer is choice A.

Choice A is correct. Sodium carbonate (choice D)should be eliminated immediately, because it is a base. Ofthe three remaining choices, both hydrochloric acid (choice B) and nitric acid (choice C) are strong acids. Thealdonic acid drawn is not a strong acid, so choices B and C are eliminated. The structural similarit"y because ofthe CO2H grouP on both acetic acid and gluconic acid (the aldonic acid drawn) makes answer A the best choice.

Choice A is correct. Tooth enamel is a basic substance, so it dissolves fastest in an acidic solution. Table 1 showsthat the solubility increases as pH decreases. This means that the most acidic solution (solution with thelowest pH) dissolves tooth enamel the fastest. The lowest pH results from the strongest acid in the highestconcentration. The highest concentration is found in choices A and C, so choices B and D are eliminated. Thestronger of the two acids is HCl, leading us to conclude that the 0.10 M HCI solution dissolves the tooth enamelthe fastest. The best choice is answer A.

Choice D is correct. In Reaction 1, sulfur trioxide is hydrated. The addition of water (hydration) does notchange the acidic or basic nature of a compound. This eliminates choices A and B. As stated in the passage, thehydration of a non-metal oxide (a Lewis acid) converts the compound into a non-metal hydroxide (a Broirsted-Lowry acid). This tells us that choice C is wrong and makes choice D the best answer.

Choice D is correct. A non-metal oxide acts exclusively as an acidic (and not a base), so it is neither amphotericnor an Arrhenius base. This eliminates choices A and B. A non-metal oxide has no protons, so it cannot be aBrsnsted-Lowry acid. This eliminates choice C. By having n-bonds, a non-metal oxide is a good electrophile,which makes it a Lewis acid. The best answer is choice D.

Choice C is correct. The most acidic compound when combined with water creates the solution with the lowestpH, although the compound's concentration has an effect on the pH of the solution as weII. Metal oxid.es arebasic, so choice D (MgO) should be eliminated. The three acids formed when choices A, B, and C are added towater are H2SO3 (from SO2 and H2O), H2CO3 (from CO2 and H2O), and HNO3 (from N2O5 and H2O),respectively. If you knew that nitric acid is stronger than either sulfurous acid or carbonic acid, then you couldhave picked choice C. However, if you didn't know that tidbit of trivia, then you could have determ-ined theirrelative strength from their oxidation states and from the electronegativities of their central atoms. Sulfurhas a +4 oxidation state in SO2 (and H2SO3), carbon has a +4 oxidation state in CO2 (and H2CO3), andnitrogen has a +5 oxidation state in NzOS (and HNO3). Nitrogen is the most electronegative and has thehighest oxidation state. The nitrogen-based oxyacid has the greatest propensity to gain an electron pair,making HNO3 the most acidic. Pick choice C.

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Copyright O by The Berkeley Review@ 289 Section IV Detailed Explanations

80. Choice B is correct. The solution is acidic, so the pH is less than 7.0, eliminating choice D- For a rveak acid

with pKn between 2 and, 12 in an aqueous solution with [HA]6itia1 ) Ka, use the shortcut equation to determine

the pH. The pKu is7.46 and the [HA] is 0.10 M.

pH = l, pKu - ltog [HA] = L1z .+e7 - l"r (0. 10) = 9.7 s - \-1) = 3.73 + 0.50 = 4.23

ZL

The best answer is choice B. Choice A can be eliminated, because HCIO is a weak acid. A 0.10 M strong acid

has a pH of 1.0. An equimolar weak acid, because it dissociates less than a strong acid, would have a pH

greatei than 1.0. Choice C can be eliminated, because the pH is less than the pKu.

Choice C is correct. This is a freebie question, if you read the passage carefully. Non-metal oxides are found in

acid rain, particularly the oxides of sutfur and nitrogen. Choice A is eliminated, because it is a metal oxide'

making it is basic, not acidic. Choice B should be eliminated, because nitrogen is present in the air all the time

and can be found in all rain, not just acid rain. Choice D is eliminated, because it is neither acidic nor veffabundant in our atmosphere. This means that the correct answer is choice C'

Choice C is correct. As the oxidation state of the central atom in an oxyacid increases, the compound becomes

more acidic. This is because the number of n-bonds between oxygen atoms and the central atom increases. This

increase in n-bonds results in more electron withdrawal (by way of resonance) from the central atom, making ilelectron-poor. The compound therefore becomes more acidic. This means that both the oxidation state and the

number of n-bonds to the central atom affect the acidity of the oxyacid, so choices A and D are eliminated' As

the electronegativity of the central atom in an oxyacid increases, the compound becomes more acidic. This i5

because of the inductive effect. The increase in electron withdrawal from the central atom, due to the inductire

effect, makes the central atom electron-poor. This means that the electronegativity of the central atom affects

the acidity of the oxyacid, so choice B is eliminated. The size of the central atom has no bearing on the aciditr

of the oxyacid, because the acidic proton is not directly bonded to the central atom. This makes choice C *re

best answer.

Choice D is correct. Because H2SOa has more oxygen atoms than H2SO3, it is more acidic. The more acidir

compound has a lower pKu, afrigher Ku, a weakei ionjugate base (with a higher pK5), and generates a lo$-er

pH i,1 water. This makes choices A, B, and C valid stitements. The fact that H2SOa is more acidic thmr

H2SO3 means that H2SOa dissociates more when added to water, resulting in a higher ion concentration, and'

tf,us a'nigner cond,rc-tivity. H2SO4 is more electrolytic (has higher conductivity) than H2SO3, so choice D fo

an invalid statement, which makes it the correct answer.

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Choice A is correct. The stronger the acid, the lower its pK6 value, so the lowest pKu is found with I

strongest acid. Having more oxygen atoms on an oxyacid increases its acidity, so HNO2 and H2SO3 cannot

the niost acidic compJunds. This eliminates choices C and D. Sulfuric acid is stronger than nitric- acid, whi

you can determine fiom their oxidation states (S is +6 is H2SO4, while N is +5 in HNO3). This makes choice

the best answer.

Choice C is correct. Lysine is a triprotic acid. The first proton released (from the carboxyl

corresponds to the third proton gained by the conjugate base. This means that from a conjugate pair p

the pti"-to-pK6 relationship is: iKul + pkue = r+. itris is not one of the answer choices, so we next consider

,""o.alrotonln lysine. ti'," ,".ond protorireleased (from the amino terminal) corresponds to the second prc

gained. This means that from a conjugate pair perspective, the pKu-to-pK6 relationship is: pKs2 + 9KVZ---

This makes choice C the correct answer] buito finish our discussion of the correlation between pKu and pKb'

must consider the third proton. The third proton released (from the side chain) corresponds to the first proto

gained. This means that from a conjugate pair perspective, the pKu-to-pK6 relationship is: pK63 + pK61 = ld"

Choice A is correct. At physiological pH, the carboxyl terminal is deprotonated (pH > pK.u (carboxyllerrr

so it carries a negative "nutg".

"At physiological pH, the amino terminal is protonated (pH < PKa (

terminal)), so it cairies a positi-ve charge.--In order for the amino acid to be a cation, the side chain must carrl

positive charge. Arginine has a sid"e chain pKu of 13,'21'., so pH'PKq: Thilmel:s,1111,'n:j^t:::i.T;;;;;;;rJ,lJit ."rti"r u fositive charge. choice-A is the best unr*"i.. Choice B is eliminatedll"-..11t",^t:.*

anion. Choice C is eliminated, becausJ the side chain cannot carry a charge. Choice D is eliminated, becan

histidine is d.eprotonated (pH > PKu (side chain)), so it carries no charge'

Copyright @ by The Berkeley Review@ 29o. Section IV Detailed Ex

87. Choice D is correct. The difference between the side chains of serine and cysteine is that serine has an alcoholgroup (O-H), while cysteine has a thiol group (S-H). This means that the difference in acidity lies in thedifference between oxygen and sulfur. Choices A and C are eliminated immediately, because we know fromtheir relative positions in the periodic table that oxygen has a smaller atomic radius and is moreelectronegative than sulfur. When comparing the acidity of protons bonded to different atoms that occupy thesame column of the periodic table, we find that the most significant factor influencing the acidity is atomicradius. The best answer is choice D.

Choice A is correct. Based on the pKu values in Table 1, aspartic acid (with a side chain pKu of 3.88) is moreacidic than glutamic acid (with a side chain pKu of 4.32). This eliminates choice C and D. A shorter chainresults in the electron-withdrawing NHg* group's being closer to the side chain carboxylic acid group, whichincreases the acidity of the carboxylic acid group. This makes choice A the best answer.

Choice D is correct. According to Table 1 in the passage, the pK" values for histidine are: pKul between 1.8 and2.6, pKa2 = 6.05, and pKu3 between 8.8 and 10.6. At a pH of 7.0, the carboxyl terminal and the side chain aredeprotonated, while the amino terminal is protonated. This is because pKa3 > pH > pKaZ > pKa1. Since thecarboxyl terminal is deprotonated, we consider pK53 rather than pKu1. Since the side chain is deprotonated,we consider pK62 rather than pKn2. And since the amino terminal is protonated, we consider pKu3 rather thanpK61. This means that the pK values of interest are pK"3, pK62, and pK63, making choice D the best answer.

Choice C is correct. Normality is defined as the moles of equivalents per liter. Because glutamic acid istriprotic, it yields three acidic protons per molecule. For each mole of glutamic acid, three moles of acidicprotons can be generated. The normality in this case is the molarity multiplied by a factor of three (N = M x 3).The molarity is 0.50, so the normality is 1.50. The best answer is choice C.

Choice C is correct. Cysteine is neutral when protonated. The side chain pKu for cysteine is 8.36, so at pH = 7.0,

pH < pKu. Under such conditions, the side chain of cysteine is protonated, and thus neutral. Choice A iseliminated. Histidine is neutral when deprotonated. The side chain pKu for histidine is 6.05, so at pH = 7.0,

pH > pKu. Under such conditions, the side chain of histidine is deprotonated, and thus neutral. Choice B iseliminated. Tyrosine is neutral when protonated. The side chain pKu for tyrosine is \0.07, so at pH = 7.0, pH <pKu. Under such conditions, the side chain of tyrosine is protonated, and thus neutral. Choice D is eliminated.Lysine is neutral when deprotonated (as is the case with the three basic amino acids). The side chain pKu forlysine is 10.80, so at pH = 7 .0, pH < pKa. Under such conditions, the side chain of lysine is protonated, and thuscationic. Choice C is the best answer, because the side chain is charged.

Choice C is correct. The lowest pH is associated with the solution with the greatest hydronium concentration.The greatest hydronium concentration depends on the concentration and the strength of the acid. To lower pH,the concentration of an acid may be increased, or a stronger acid, with a lower pKu value, may be employed.This means that the correct answer is a combination of lowest pKu and greatest concentration. Choice A getseliminated for having the lowest concentration and a high pKu value. Choice B gets eliminated, because it has

a lower concentration and greater pKu than choice C. The pH in choice C is half of the pKu + 0.5, which is 1.65

+ 0.5 = 2.15. The pH in choice D is half of the pKu, which is 4.55. Choice C has the lowest pH.

Choice D is correct. Acid rain is contains airborne Lewis acids. Non-metal oxides such as sulfur oxides andnitrogen oxides make up most of the acid rain we study. This means that choice A is an example of acid rain.Once these non-metal oxides react with moisture in the air, they become hydrated, so acid rain does containhydrated non-metal oxides (also known as non-metal hydroxides). A Lewis acid is an electron-pair acceptor,and non-metal oxides qualify in this category. Choice C is a valid description of an acid rain component. Metalhydroxides are basic, so they are not found in acid rain. Choice D is the best answer.

Choice B is correct. The pH of a conjugate pair is found using the Henderson-Hasselbalch equation. The lowestpH is attributed to the conjugate pair having the acid with the lowest pKu and the mixture that most favorsconjugate acid. The pKu of ammonium is 9.26 (although you really just need to know it's between 9 and 11),

while the pKu of hydrofluoric acid is 3.17. This eliminates choices C and D. Choice B has more acid thanbase, while choice A has more base than acid. The best answer is choice B.

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Copyright @ by The Berkeley Review@ 29t Section IV Detailed Explanations

95. Choice B is correct. For a conjugate pair at 25'C, the pKu of the acid when added to the pK6 of its conjugatebase is equal to 14. This question boils down to which of choices represents a conjugate apair. Choice D can beeliminated immediately, because both values are pKu values, and if they happen to sum to 14, it's purelr-coincidental. Phosphoric acid is triprotic, so there are three pKu values. The first proton oTf (pK21)corresponds to the third proton back on (pKUe), so pK^1 + pKOg = 14. The second proton off (pKu2) corresponds tothe second proton back on (pKfd, so pKu2 + pKUz = 14, making choice B the correct answer. The third proton off(pKas) corresponds to the first proton back on (pKUr), so pKu3 + pK61 = l{.

H3Poa(aq) + H2o(t) # H3o+(aq) + HzPo+-(aq)

First proton offlThird proton on .'. pK61 + pK63 = l{

H3O+(aq) HPOa2-(aq)

# H3o+(aq) + Po43-(aq)

Third proton offlFirst proton o4 ;. pKu3 + pKbl = 14

Choice B is correct. 1.5 M H3POa = 4.5 N H3PO4, because there are 3 equivalents of H+ per H3PO4. The

normality is found by multiplying the molarity by the number of equivalents. Doubling the volume by addingpure water cuts the concentration (measured in either normality or molarity) in half. This makes normality one

half of 4.5 N, so choice 8,2.25 N, is the best answer.

Choice B is correct. Because the pH is less than seven, the solution must contain an acid rather than a base.

This eliminates choices C and D. If the acid were a,strong acid, it would fully dissociate in water. This wouldlead to a hydronium concentration of 0.10 M, which would make the pH = 1. Given that the pH is greater thart1.0, the acid does not fully dissociate, so it is a weak acid. This eliminates choice A and makes choice B the

best answer.

Choice C is correct. The highest pH is found in the solution with the highest concentration of hydroxide.Hydroxide concentration depends on the compound and its concentration. Choice A is a weak acid, so the pH is

low. This eliminates choice A. Choice B is a strong acid, so pH is very low. This eliminates choice B. ChoiceC is the conjugate base of a weak acid, so it is a weak base. Choice C could be the correct answer. Choice D isthe conjugate base of a strong acid, so it is a very weak base, meaning it has no significant impact on the pH otthe aqueous solution. Because the pH of choice D is7.0, choice D is eliminated. The best answer is choice C.

Choice B is correct. The solution is acidic, so the pH is less than 7.0, eliminating choice D. The acid is weak, so

it partially dissociates in water. If the acid were strong, it would fully dissociate, leading to a hydroniurnconcentration of 0.050 M and thus a pH of 1.30. Carboxylic acids are weak acids, so the pH is higher than 1.3

This eliminates choice A. The pH is less than the pKu, so choice C is eliminated. The only choice remaining is

choice B. If you choose to do so, you can solve for the pH exactly. For a weak acid with pKu between 2 and 12 inan aqueous solution with [HA]6itial > Ka, use the shortcut equation to determine the pH. The pKu is 4.89. The

[HA] is 0.05 M, and log0.05 = - log20 = - (log 10 + log3) = - (1 + 0.3) = -1.3.

pH = lpK, - llog [HA] = 1 1+.AO; - ltog (0.050) = 2.45 - lf-t.Sl = 2.45 + 0.65 = 3.10'2'-222-2

The value confirms that choice B is the best answer.

100. Choice C is correct. Hydrobromic acid (HBr) is a strong acid, so a 0.10 M solution has a pH of 1.00. Choice A 1s

valid and thus eliminated. Formic acid (HCO2H) is a weak acid, so it does not fully dissociate. It has a pHgreater than 1.0 but less than 7.0. Choice B could be valid, so it is unlikely. To eliminate it with certainty, rr-e

need to do a calculation. We shall hold off on calculating until we have evaluated the other answers. Sodiurnacetate (NaOAc) is a weak base, so it does not fully hydrolyze. It has a pH greater than 7.0 but less than 13-f .

Choice C is invalid, so it is the best answer. Potassium hydroxide (KOH) is a strong base, so a 0.10 M solutionhas a pH of 13.00. Choice D is valid and thus eliminated.

PKaz -H2POa-(aq) + H2O(1)

Second proton offl

HPOa2-(aq) + H2O(l)

-pKb, n3L/'(aq) + IlrLJ4'

Second proton on .'. pKu2 +pI<b2=14

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Copyright @ by The Berkeley Review@ 292 Section IV Detailed Explanations

Section VBuffers

a) Buffer Compositiotr . , .

b) pH Range for Buffers :

c) Buffer Reciped) Physiological Buffers

.

Titration Curvesa) Quantitative Keactions,b) Fundamental Curve Shapesc) Plotting a Titration Curved) Concentration Dffectse) Strength Effectsf) Polyprotic Acids

Indicatorsa) Composition and Functionb) Detecting the Equivalence Pointc) f,stimating Solution pll

Buffers andTitration

by Todd Bennett

o-5

Equivalents O,lO M KOH added

Specializing in MCAT Preparation

By knowing the relative mole quantities of what has been mixed in solution, a pH value can beestimated from the position on d titration curve corresponding to the mjxture. Tlis requires beingable to identify the Shape of a titration curve based on ihe com'ponents in the mixture.

Understand how polyprotic acids affect the plt of a solution.Polyprotic acids have multiple

"q.tiuul"Know that amino acids aie a subset of-polyprotic acids. 'Be able to determine the pH at middleequivalents points by applying the eqriatibh for pl (averaging the two respective'pKs values).

uffers & Titration Section GoalsKnow how to a te the of a solution usi a titration cury€.

Know the all titration curves.

Know the role of an indicator in titration and solution

Know how and buffers work.

and be able to titration curves for titration.

On titration curves associated with strong reagents, the pH at equivalence is always equal to 7.0.On titration curves associated with a weal a-cid orweakbase, titiated by,g strong r'eage'nt, tne pHat the half-titrated point is equal to the pK2 of the weak acid. fhe ig ul

"qiiuuftn.; ;;;'b;approximated by averaging the pKa of the weak acid and the pH of the titrant.

An indicator is used to make the endpoint of the titration visible. An indicator is a species thai hasa different color for the conjugate acid and conjugate base. l\4ost indicators ut" o.gurii. .ornpo.rtraiwith a_great degree of -conjugation, and the colbr"is caused by a transition from th"e ru level tb the n*wlrn a grear oegree or conlugatlon/ and tne color ls caused bv a tlevel. Indicators can also be used to predict the pH of a soluiion.

A buffer is formed when a weak acid and its conjugate base are added to the aqueout roi"ti"ttBecause there exists an equilibrium between the lwg Jpecies, as long as both are prdsent in solution,the hydronium_ion concentration will remain fairly'constant, the"refore the pil will also remaintne nyclromum-lon concentratlon wrll remarn tatrly constant, therefore the pH will also remainconstant. The effect is know-n as."buffering." You mu"st understand buffers andhow the pH is foundfrom the Henderson-Hasselbalch equatioh.

I

'v You must be able to identify the titrant and the species being titrated when you look at the titrationcurve. Features to note are the weak acid lip, the'strong acid-sigmoidal shap'e, the ascent (associatedwith the titration of an acid by a base) or di:scent (asso"ciated #lth tne titraiion of a base bv an acid)of the_ curve, and the numberbf inflection points (indicative of whether the compound is folyproti6or polybasic).

T{ll'

m:'J![

urmr

mfil

General Chemistry Buffers and fitration Introduction

A buffer is a solution where pH remains relatively constant after the addition ofeither strong acid or strong base. The pH may vary slightly, but for all intentsand purposes, it does not change significantly. Buffers play a major role inphysiology and biochemistry, so understanding how they work is critical. Theycan be made in one of two ways. The first method involves combining a

conjugate pair in roughly equal mole portions. The second method involvespartially titrating a weak acid with roughly half of an equivalent of strong base,

or by partially titrating a weak conjugate base with roughly half of an equivalentof strong acid. The pH of a buffered solution is determined using theHenderson-Hasselbalch equation. The data associated with buffers are generallyeasy to work with in a conceptual sense. Titration curves have buffering regions,so understanding buffers can help you to understand titration better.

Neutralization is the mixing of equal mole portions of an acid with a base,regardless of their concentrations and strengths. To neutralize an acid, an equalmole quantity of base must be added to solution. To neutralize a base, an equalmole quantity of acid must be added to solution. A neutralized solution hasmoles HgO* equal to moles OH-. Addition of a base to an acid (or acid to a base)yields water and a salt upon neutralization. This is shown in Reaction 5.1:

HX(aq) + MOH(I) ===- HzO(i) + MX(aq)

Reaction 5.1

Depending on the strength of the acid and base, the pH at the neutralizationpoint (also referred to as the endpoint and the equivalence point in titration)varies. Neutralization does not mean to make the pH of the solution equal to7. When the base is stronger than the acid, the neutralized solution is slightlybasic, so the pH is greater than 7.0. When the base is weaker than the acid, theneutralized solution is slightly acidic, so the pH is lower than 7.0. When the base

and acid are equally strong, the neutralized solution is neutral, so the pH is equalto7.0. The three possible combinations are summarized below.

HX(aq)+MOH(I)-H2O(l) +MX(aq)

Strong acid + Strongbase

Strong acid + Weak base

Weak acid + Strong base

PHat equivalence = 7'0

PHat equiva]".,." < 7'0

pHat equival"r,." ) 7.0

The pH is not alzaays 7 at the equioalence point of a titration. For a weak acid

titrated by a strong bnse, the equiaalence point is the point at which it is completely

conaerted into its conjugate base. The conjugate base will yield a pH greater than 7

so the pH is greater thsn 7 at the equiaalence point.

The last section in this chapter involves detecting the equivalence point. Visualindicators that change color upon changing pH are a typical component of anygeneral chemistry curriculum. They are often highly conjugate organicmolecules. The color associated with an indicator is a reflective color, resultingfrom the absorbance of a photon accompanied by the excitation of an electronfrom a n-bonding molecular orbital to a ru-antibonding molecular orbital. Thetransition energy changes when the compound gets deprotonated, so the energyof the photon absorbed, and ultimately the color of light reflected also changes.Hence, any pH-sensitive chromophore is an indicator.

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General Chemistry Buffers and Titration Buffers

Buffers '

Buffer CompositionBuffers are solutions that resist drastic changes in pH. Buffers are made of aroughly equal mole mixture of a weak acid and its weak conjugate base in anaqueous solution. Both the acid and the base of the conjugate pair must be weakin order to form a buffer solution. This is so that the equilibrium between thetwo species can be controlled by the environment. With approximately equalmolar quantities of conjugate acid and conjugate base in solution, the solution isresistant to pH change caused by the addition of either strong acid or strong baseto solution. When both of the species in the conjugate pair are weak, the buffercan equilibrate in both the forward and reverse directions of the reaction toabsorb any hydronium or hydroxide that may be added to solution. Addition ofa strong acid to solution converts the weak base into its conjugate acid. This haslittle to no effect on the pH. Likewise, addition of a strong base to the solutionconverts the weak acid into its conjugate base and has little to no effect on thepH.

To emphasize the need for roughly equal portions, the following experiment maybe studied. In this study, three mixtures of acid and conjugate base aregenerated. In the first system, weak acid is in extreme excess relative to itsconjugate base. In the second system, the two species are in roughly equalconcentrations. In the third system, weak base is in extreme excess relative to itsconjugate acid. The results are shown in Figure 5-1.

System I: Mix 999 pafts HA with l part A-. To this mixture, add 1 part OH-.

Initiallv: [A-] = t = after addition of 1 oart ott-' [A-] - z' [HA] 999 ' [HA] 998

Because Ka = [H+]* ++, and K3 is a constant, when -[A J- aorUtes, [H+]tHAl IHAImust be cut in half, and the pH changes. System I is NOT a buffer.

System II: Mix 500 parts HA with 500 part A-. To this mixture, add 1 part OH-.

Initially: [A-] =soo -+afteraddition of l part ott-' [A-l -sor' tHAl s00 [HAl 4ee

Because K6 = [H+] " g+, and K3 is a constant, when 1{ 6ur"1y changes,tHAl -

[HA][H+] is constant, and the pH doesn't change. System II is a buffer.

System III: Mix 2 parts HA with 998 part A-. To this mixture , add 7 part OH-.

tnitially: lA-l =sga =afteradditionof l part o"-' IA-] -ggg' [HA] 2 ' IHAI 1

Because Ka = [H+]* el, and K3 is a constant, when i{l aorutes, [H+]tHAl [HA]

must be cut in half, and the pH changes. System IIi is NOT a buffer.

Figure 5-1

The conclusion is that the pH remains constant only if the weak acid and itsconjugate base are in roughly equal concentration. The addition of strong base orstrong acid shifts the ratio of weak acid (HA) to its conjugate base (A-), but thepH does not change, if the A- : HA ratio is close to 1.0.

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General Chemistry Buffers and Titration Buffers

pH Range for BuffersThe experiment in Figure 1 explains why the weak acid and weak conjugate base

must be present in roughly equal parts. If they are not relatively close in molequantity, then the system does not act as a buffer. According to convention, theratio can not exceed 10 : 1. Substituting 10 : 1 and 1 : 10 into the Henderson-Hasselbalch equation shows us that the range of a buffer is the pKa of the weakacid + 1. The pH of a buffer solution obeys the Henderson-Hasselbalch equation,which is shown in Figure 5-2.

pH = pKu + togJ4 I .'.pH = pKa + ro* Yolslg*gqt" uu:g

" [H,ql r I

Moles conjugate acid

Lowest pH =pKa + logl- =pK3 - I Highest pH =pKa + Iog10 =pK6 + 1

.'. PH range = PKu + 1

Figure 5-2

The derivation of the Henderson-Hasselbalch equation from the Ka equation ison page 264. The Henderson-Hasselbalch equation shows that as [conjugatebase] increases, buffer pH increases. The Henderson-Hasselbalch equation alsoshows that as [conjugate acid] increases, buffer pH decreases. It also offersquantifiable verification of the concept that when pH is greater than the pKu, thesolution is rich in conjugate base.

Example 5.1If water is added to a buffer solution with pH = 3.96, what happens to the pH?

A. The pH increases slightly.B. The pH decreases slightly.C. The pH remains the same.D. If the pH is greater than 7, then it decreases. If the pH is less than7, then it

increases.

SolutionAddition of water to a buffer equally dilutes the concentration of the weak acidand its weak conjugate base. This means that the mole ratio of the weak base tothe weak acid does not change upon the addition of water. According to theHenderson-Hasselbalch equation, the pH of the solution does not change because

pK3 is constant and the fraction has not changed. The result is that the pH of abuffer does not change when it is diluted. This is why the Henderson-Hasselbalch equation can be written as moles A- over moles HA, as well as [A-]over [HA].

Knowing the buffer range is important when making a buffer, which isaccomplished in two steps. First, an acid within the range must be chosen. ThepKn of the acid should be as close as possible to the desired pH. Second, a

mixture containing both weak acid (HA) and its weak conjugate base (A-) shouldbe formed so that the two species are in roughly equal concentration. Select anacid for the buffer whose pKa value lies within the +1 range of the desired pH.For buffering at pH values between 2 and 5, carboxylic acids are typical. Forbuffering at pH values between 8 and 11, amines are typical.

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General Chemistry Buffers and Titration Buffers

Buffer RecipesBuffers can be made either by mixing the conjugate pair together, or by partiallytitrating either component in a conjugate pair of weak reagents. The partialtitration method can be either to half-titrate the weak acid with strong base, or tohalf-titrate the weak conjugate base with strong acid. First, a weak acid must bechosen with a pKz value close to the desired pH. Buffers can be mixed by any ofthe methods shown in Figure 5-3.

Weak acid + the salt of the conjugate base in roughly equal moleproportions (e.g., HCO2H with HCO2Na)

Weak base + the salt of the conjugate acid in roughly equal moleproportions (e.g., NH3 with NHaCI)

Weak acid and roughly half of an equivalent of strong base (e.g.,HOAc with half equivalent KOH)

4. Weak base and roughly half of an equivalent of strong acid (e.g.,H3CNH2 with half equivalent HCI)

Figure 5-3

When using either of the last two methods, titration takes place until the desiredpH is achieved within the buffering region. The buffering region (where the pHdoes not change appreciably) is found in the middle area of the titration curve,between the starting point and the equivalence point. Buffering occurs only withthe titration of a weak reagent by a strong reagent. A strong acid combined withits conjugate base or a strong base combined with its conjugate acid do notproduce a buffer, so strong acid and strong base titration curves have nobuffering region.

1.

2.

J.

Example 5.2Which of the following solutions results inHA has apKu of 4.7?

A. HA with one-half equivalent of A-B. A- with one equivalent of HAC. HA with one-third equivalent of OH-D- :A- with one-third equivalent of H3O+

a buffer with a pH of 5.0, given that

' *-l'<;-"Fr-

,\

SolutionThe pH is greater than the pKu, so the solution must be rich in the deprotonatecspecies. In choice A, [HA] > [A-], so the pH is less than pKu (4.7), meaning tha:choice A can be eliminated. In choice B, [HA] = [A-], so the pH equals pKu (4.7,

meaning that choice B can be eliminated. [n choice C, one-third of an equivaleniof A- forms from the reaction, and two-thirds of an equivalent of HA is left over,In choice C, [HA] > [A-], so the pH is less than pKu (4.7), meaning that choice Ccan be eliminated. In choice D, one-third of an equivalent of HA forms from thereaction, and two-thirds of an equivalent of A- is left over. In choice D, [HA] <

[A-], so the pH is greater than pKu (4.7). Choice D is the best option for thosewho like to pick the correct answer. To solve for the exact numerical value, theHenderson-Hasselbalch equation must be employed, The values differ by 0.3

and the antilog of 0.3 is 2, so the correct answer must form a 2 '. 7 ratio a-conjugate base to acid.

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General Chemistry Buffers and Titration Buffers

Example 5.3

\Vhich of the following combinations, when mixed in the correct ratio, produce a

buffer?

xiNuou/Nuct :,., t!B. NaNO3/HNO3C Hct/xot-l ' \'D.:NHg/HCl , ':'

SolutionChoice A is a strong base and a neutral salt, which does not make a buffer.Choice B is a neutral salt and a strong acid, which does not make a buffer.Choice C is a strong acid and a strong base, which also does not make a buffer.By elimination, choice D is correct. To be a buffer, the weak base (NH3) must be

half-titrated by the stlong acid (HCl). Ammonia and hydrochloric acid make abuffer when mixed in the correct ratio (2 : 1)'

Example 5.4\\4rich of the following does NOT form a buffer when added to NaHCO3(s)?

A. NaOH(aq)B. HCI(aq)C. H2CO3(aq)D. H2o(l)

SolutionAdding half of an equivalent of NaOH to a sodium bicarbonate solution converts

half oflhe HCO3- to its conjugate base CO32-. The mixture of the two forms a

buffer, so choice A is correct. Adding half of an equivalent of HCI to a sodiumbicarbonate solution converts half of the HCO3- to its conjugate acid H2CO3. The

mixture of the two forms a buffer, so choice B is correct. Adding an equivalent ofH2CO3 to a sodium bicarbonate solution results in equal portions of the HCO3-and its conjugate acid H2CO3. The mixture of the two forms abuffer, so choice C

is correct. Adding water does not make a buffer, because water is amphoteric, so

it does not convert sodium bicarbonate to either its conjugate acid or conjugate

base. To be a buffer, there must be both a weak acid and its weak conjugate base

present in solution at the same time. This means that choice D is correct.

Example 5.5If the iatio of base to acid in a conjugate pair is 3:1, and the weak acid has d Ka =1.0 x 10-5, what can be said about the pH of the buffer solution?

pH<5pH=S ( 1:" t';5<pH<7PF{>7

SolutionIf the ratio of base to acid were 1:1, the pH would equal the pKu. Becauss the

base is in excess, the pH is greater than the pKa. The pKu is -1og 1.0 x 10-5 = 5.

This eliminates choiies A and B. According to the Henderson-Hasselbalchequation, the pH is log 3 greater than the pKu. Log 10 = 1, and log 10 is greater

than log 3, so pH (= pKu + log 3) < 6. The best choice is C'

A"B.

iCt''s.

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General Chemistry Buffers and Titration Buffers

.lu{-\.-'r. -

,'{ d-- -.' ,-)r'\

r it"'e'?'-

Example 5.6If 1.0 moles of a weak acid in 1.0 liters of water are treated with 0.4 moles of

2lrengrbase, what is the pH of the solution? (Ku for the weak acid is 2.0 x 10-4)

(a, -pfi <e.zB. pH=3.7 f'+-'C. 3.7 <pH <7D. pH>-7 ,:''i

-

Solution

't(l

If the ratio of base to acid,were 1:1, the pH would equal the pKu. The pKu is - log2.0 x L0-4 = 4 - 1og2 = 4 - 0.2 = 3.7,as hinted at by ttre unr*", cfioices. To be half-titrated, it would require 0.5 moles o{ strong base. At the half-titration point, the

tH = pKu. with only 0.4 moles of base, the halfway point is not yet reached andthere is excess weak acid relative to conjugate base (0.6 moles to 0.4 moles).According to the Henderson-Hasselbalch equation, the pH is less than the pKu.The best answer is choice A. The pH is less than pKu, -hl.h is 3.7. Be aware thatthis question could about the [H+] as well as the pH of the solution. If a solutionis half-titrated, then the K6 = [H+]. This can belricky, but the answer is foundfrom the Henderson-Hasselbalch equation. pH = pKu + log Q.4, so pH < pKa.

Example 5.7A buffer made by mixing 100 mL of 0.5 M HoAc (Ka = 1.g * t0-5) with 25 mL of1.0 M KoH has a pH approximately equal to which of the following values?

7

A. 0.2&> +,2c. 7.0D. 9.3

Solution

.: i'-'-1 /-"'

vThere are 0.05 moles or.Hoa" present and 0.025 moles of KoH present. Thismeans that exactly half of an equivalent of strong base has bebn added to a weakacid, converting half of the original weak acid to its conjugate base. Half of theoriginal weak acid remains unreacted. This means that pFl= pKu. The value forpKu is solved for as follows:

pKa = -log (1.8 x t0-s) = -log 1.8 - log 10-s = -log 1.8 - (-5) = -log 1.8 + 5

Log 1.8 is less than 1, so the pH is greater than 4 and less than 5. Choice B is best,

Example 5.8A buffered solution initially has a pH of 8.31. when five drops of 12 M HCI areadded to a 500-mL beaker fitled with this buffered solution, what would beexpected for the final,pH value?A. 3.31 "

,, ;t.\

.. -61 e.zo , j, L .

"18. g.31 i:'

E. 8.36

SolutionAdding a little acid decreases the pH slightly. Choice A shows a lower pH valuebut a drastically lower pH value. The change is small, so the answer is ihoice B.

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General Chemistry Buffers and Titration Buffers

Physiological BuffersPhysiological pH is considered tobe7.4 (although venous and arterial pHvary,and gastric fluids are highly acidic). The pKu value for physiological acidsshould be considered relative to this value in order to determine the structure ofthe compound in aiao. When the pH of the environment is greater than the pKuof the species, it exists predominantly in its deprotonated state. Equally, whenthe pH of the environment is less than the pKu of the species, it existspredominantly in its protonated state.

A prime example of physiological buffering involves carbonic acid, which is usedto buffer the blood of biological systems. Arterial blood is oxygen rich (and thuscarbon dioxide-poor), so it has a slightly higher pH than venous blood which iscarbon dioxide-rich. Carbon dioxide when added to water undergoes thecomplex equilibrium shown as Reaction 5.2.

COz(g) + H2O(l) + H2CO3(aq) : HCO3-(aq) + H+(aq)

Reaction 5.2

This equilibrium regulates blood pH, so any conditions (or disorders) that affectcarbon dioxide levels in physiological systems also affect the pH and buffering ofblood. For instance, emphysema hinders the uptake of oxygen from the lung andthe release of carbon dioxide into the lung. The blood can compensate for thereduced uptake of oxygen by increasing heart rate and producing more redblood cells. But the release of carbon dioxide is not as easily adjusted for. Theconsequence is that carbon dioxide levels in the blood increase, causing the bloodto be more acidic than normal. This condition is known as respiratory acidosis.This can inhibit the binding of oxygen, as is demonstrated by the Bohr effect.Using the same logic, it can be concluded thal respiratory alkalosis results from theexcessive loss of carbon dioxide.

Gastric fluids are highly acidic (rich in HCI), so the loss of gastric fluids results inthe loss of acid from the body, producing a condition known as metabolic alknlosis.Loss of HCI can accompany vomiting or the pumping of the stomach. When thestomach is pumped, the acidic solution that is removed must be replenished toreduce the risk of metabolic alkalosis. As a point of interest, food poisoning anddrug overdose patients are given a solution of charcoal and water to drink, whichabsorbs the organic toxin. The porous carbon matrix binds organic moleculesbetter than the water, especially compounds with n-bonds.

Waste products exit the body in a chemically neutralized state (and believe methat if they didn't, you'd notice), so the lower end of the GI track must be basic.Losing lower intestinal fluids results in the loss of basic metabolites, in particular,HCO3- (bicarbonate). Diarrhea (which can be caused by Olestra consumption)results in the loss of basic metabolites, producing a condition known as metabolicacidosis. The intertwining of acid-base chemistry and physiology is a perennialfavorite on the MCAT. Figure 5-4 summarizes the pH-related disorders of thebody.

Retention of CO2: Blood pH .L .'. Respiratory acidosis

Loss of CO2: Blood pH t .'. Respiratory alkalosis

Loss of HCO3-: Blood pH J .'. Metabolic acidosis

Loss of H3O+: Blood pH t .'. Metabolic alkalosis

Figure 5-4

a t-{ = {'ts* n} o P*l---{--/

,-[ t ?E^ ""'J**'*'c4

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General Chemistry Buffers and Titration Titration Curves

ffif#affi$fir,i$fiffip$Quantitative Reactions (Titration Theory)Acid-base titration is the quantitative addition of a titrant acid to a base insolution, or of a titrant base to an acid in solution. What is meant by quantitativeaddition is that the volumes of acid and base are measured precisely. Thetitration is said to reach its equivalence point when the moles of acid are equal tothe moles of titrant base. The volumes must be known precisely in order todetermine the relative concentrations, using the relationship that moles are equalto the product of volume and molarity. we shall consider two versions oftitration. The first case is the titration of a strong reagent by a strong reagent,such as strong acid by strong base or strong base by strong acid. The second caseis the titration of a weak reagent by a strong reagent, such as weak acid by strongbase or weak base by strong acid. There is no titration of a weak reagent with aweak reagent, because the two weak reagents do not react with one another.

We shall first consider what is in solution at different points during the titration.The two species can be mixed in three different ways: excess of one, excess of theother, or in equal portions. Let us address each of these scenarios, along with theinitial point of titration. When less than one equivalent of the titrant is added tothe original reagent, this mixture is found in a region on the curve before theequivalence point. When exactly one equivalent of the titrant is added to theoriginal reagent, this is the equivalence point. When more than one equivalent ofthe titrant is added to the original reagent, this mixture is found irr a region onthe curve beyond the equivalence point. Table 5.1 shows the different conditionsand pH calculations along a strong-by-strong titration curve.

Strong Acid titrated by a Strong Base

Point on Curve Species in Solution pH Range pH CalculationInitial Pure strong acid pH 7 pH=-log[HX]ir1i112iBefore equivalence Leftover HgO* pH <7 pH=-log[H3O+]u*""",At equivalence H2O & neutral salt p}{=7 pH=TatequivalencePast equivalence Leftover OH- pH>7 pOH=-log[OH-]"*."r,

Table 5.1

When a weak reagent is titrated by a strong titrant, it's different than the strong-by-strong titration. The difference when dealing with a weak reagent is that a

weak conjugate is formed as the product, so there is an equilibrium betweenproducts and reactants. The pH calculations must consider all species in solutior'that affect the pH. Table 5.2 shows the different conditions and pH calculation-.along a weak-by-strong titration curve.

Weak Acid titrated by a Strong Base

Point on Curve Species in Solution pH Range pH Calculation

Initial Pure weak acid pH <7 pH =1pKu -llog [H-r- )- ') -

Before equivalence HA and A- (Buffer) PKa+1 pH =pKa+togJal

At equivalence A- (diluted) pF{>7 pOH = f_ pfu - llog t-r I

Past equivalence Leftover OH- p}{>7 pOH=-1og[OH-]"r.g..

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Table 5.2

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General Chemistry Buffers and Titration Titration Curves

Fundamental Curve ShapesThe mixtures listed in Table 5.1 can be studied quantitatively with the help ofgraphs. Strong acid and strong base titration curves have distinct features.Figure 5-5 shows the titration of a strong acid by a strong base. We shall call thisCase 1a.

mL strong base added

Case 1a: Strong acid titrated by a strong base

Figure 5-5

Figure 5-6 shows the titration of a strong base by a strong acid. We shall call thisCase 1b.

j Note that strong curves sharean equivalence point atpH = 7.

Equivalence point (pH = 7)

moles OH- moles H\66"6

mL strong acid added

Case 1b: Strong base titrated by a strong acid

Figure 5-6

In Cases 1a and 1b, the equivalence point is at pH = 7.0, because the neutralizedproduct is a neutral salt formed from the reaction of a strong acid with a strongbase. The graph, although not extended far enough to tell, is symmetric aboutthe equivalence point. The shape of the curve is referred to as sigmoidal. Thereagent when initially added to water fully dissociates (or in the case 1b,hydrolyzes with a base), giving the highest concentration at first. Over thecourse of the titration, the concentration is reduced. This means that the pHgradually increases during the entire titration, with rapid pH change near theequivalence point. Because pH is measured on a log scale, the graph assumes thecharacteristic sigmoidal shape. The reasoning behind this is that as youapproach pF{=7, each change of 1.0 in the pH requires ten times less titrant. Forinstance, in going from pH = 1 to pH = 2, the hydronium concentration goes from0.10 M to 0.01 M, a change of 0.09 M H3O+. In going from pH = 2 to pH = 3, thehydronium concentration goes from 0.01 M to 0.001 M, a change of 0.009 MHgO*. The change in hydronium concentration is ten times less when the pH isone unit closer to 7.0.

Equivalence point (pH = 7)

moles HXinit = moles OH addea

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General Chemistry Buffers and Titration Titration Curves

The mixtures listed in Table 5.2 can be studied quantitatively with the help ofgraphs. Weak acid and weak base titration curves have d.istinct features. Case2a, shown in Figure 5-7, shows the titration of a weak acid by a strong base.

Because [HA]= [AlEquivalence point (pH > 7)

PH = PKu

Lip-o-weakness

mL strong base added

Case 2a: Weak acid titrated by a strong base

Figure 5-7

case 2b, shown in Figure 5-8, shows the titration of a weak base by a strong acid.

Lip-o-weakness

Because [A

PH = PKuEquivalence point (pH < 7)

Note that strong curves share half-equivaience point at pH = pKo.

mL strong acid added

Case 2b: Weak base titrated by a strong acid

Figure 5-8

Titration curves exhibit an initial cusp when the reagent being titrated is weak.This is referred to as a lip-o-ueakness, and may be used to distinguish the natureof the reagent from its titration curve. The lip-o-weakness is due to the fact thatthe equilibrium between weak acid and conjugate base heavily favors one of thetwo species, so the pH changes significantly. This can be seen the experiment inFigure 5-1. The pH at half-equivalence point is always equal to the pKu, whichcan be inferred from the Henderson-Hasselbalch equation.

In Case 2a, the equivalence pH is above 7.0, because the neutralized product is aweak base (the conjugate base of the weak acid). The weaker the acid titrated.the stronger the conjugate base formed at equivalence, and consequently thehigher the pH at the equivalence point. The pH of the conjugate base depends onboth the concentration and the K6 of the base at equivalence.

In Case 2b, the equivalence pH is below 7.0, because the neutralized product rs aweak acid (the conjugate acid of the weak base). The weaker the base titrated.the stronger the conjugate acid formed at equivalence, and consequently thelower the pH at the equivalence point. The pH of the conjugate acid depends c::both the concentration and the Ku of the acid at equivalence.

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General Chemistry Buffers and Titration Titration Curves

Example 5.9All of the following are feature of a titration curve in which both reactants arestrong EXCEPT:

A. an equivalence pH of 7.0.B. a sigmoidal graph shape.C. the most rapid pH change near pH = 7.

D. a large change in pH as the first few drops of titrant are added.

SolutionThis question tests your knowledge about strong-by-strong titration curves.When both reagents are strong, the neutralize each other and leave behind a pH= 7 aqueous salt solution. This makes choice A valid. Because the reagent fullydissociates (in the case of a strong acid) or hydrolyzes (in the case of a strongbase), the highest concentrations are initially observed. The result is a slowchange i. pH until just before equivalence. This causes the curve to be sigmoidaland makes choice B valid. The pH change is always rapid near the equivalencepoint, so for strong-by-strong curve, the pH change near pH = 7 is the mostdrastic. Choice C is valid. The pH of the titration mixture is nearly constant atthe start of the titration. This is to say that a strong-by-strong curve is "lip-free"in the beginning. Choice D is invalid, and is thus the best answer.

Example 5.10All of the following are features of a titration curve in which one reactant is weakand the titrant is strong EXCEPT:

A. a half-equivalence pH equal to pKu.B. a sinusoidal graph shape.C. a pH at equivalence that is not equal to 7.

D. a large change in pH as the first few drops of titrant are added.

SolutionThis question tests your knowledge about weak-by-strong titration curves.When a weak reagent is half-titrated by a strong titrant, half of the originalspecies in converted to its conjugate. This leaves half of the original reagentunreacted in solution. Because the two components of the conjugate pair are inequal concentration, the pH of the solution is equal to the pKu. This makeschoice A a valid statement and eliminates it. A sinusoidal graph implies a sinewave, which is not observed with weak-by-strong titration curves. This makeschoice B invalid, and thus the correct choice. When a weak acid is fullyneutralized by a strong base, it forms its conjugate base. This results in a solutionwith pH greater than 7. The equivalence pH is not equal to 7. When a weak baseis fully neutralized by a strong acid, it forms its conjugate acid. This results in a

solution with pH less than 7. The equivalence pH is not equal to 7. The pH atequivalence for a weak-by-strong titration curve is not Z so choice C is valid andthus eliminated. Early in a weak-by-strong titration curve, the equilibriumbetween conjugates favors the one initially in solution. As a titrant is added, theequilibrium shifts drastically, resulting in a big change in the hydroniumconcentration. A big change in the hydronium concentration causes a significantchange in the pH of the solution. This makes choice D a valid statement andeliminates it.

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General Qhemistry Buffers and Titration Titration Curves

To understand titration curves better, Table 5.3 shows how the pH is calculatedat different points along a titration. The equations used to calculate pH are all inTable 5.2. The titration represents the titration of a hypothetical weak acid with apKu of 5.0 and an initial concentration of 0.10 M. The weak acid is titrated by0.10 M KoH(aq). The pH represents the solution pH after the titrant base hasreacted with the weak acid in solution. The value points in Table 5.3 are plottedin Figure 5-9, and by connecting the dots (yes, that's right, studying for theMCAT involves a little connect-the-dots), a titration curve can be generated.

Table 5.3

Initially (at 0.00 mL added), the pH is found using the shortcut equation. Beforeequivalence (from 1.00 mL added to 49.00 mL added), the pH is found using a

modified version of the Henderson-Hasselbalch equation, where the moles OH-added are substituted for moles A- (given that the OH- converts to A-) and 50 -mL OH- is substituted for moles HA, because that describes the leftover HA. -{:equivalence (at 50.00 mL added), the pH is approximated by averaging pH of thetitrant base and the pKu of the acid. The approximation is off by 0.15, so the plicolumn shows the actual pH at equivalence. After equivalence (beyond 50.00 n.Ladded), the pOH is found by taking the negative log mL OH- - 50 (for what r.reacted)over the total volume (50 + mL OH-).

mL KOH pH calculation pH

0.00 On=of,^ -llog [HA] =7-+rog (0.10) =2.5 +0.5 = 3.0 3.00

1.00 pH =pKu + log-rnL-QH- = 5.0 + togJ-L = 5.0 - log49- -50 - mLOH- "49 o 3.31

2.00 pH =pKu + log-rnl-Qtl-- = 5.0 + log2.0 = 5.0 - log24-50 - mT.OH- "4R n 3.62

5.00 pH =pKu + log--ml-Q,H- = 5.0 + log*9 = 5.0 - log 9-50 - ml.OH- "45 o 4.05

10.00 pH =pKu * tog#*fio- = 5.0 + tosfl* = 5.0 - los 4 4.40

12.50 pH =pKa + log--!0L-QH- = 5.0 + logg* = 5.0 - log 3-50 - ml-OH- "i7 q 4.52

25.00 pH =pKa + tog--![L-QH- = 5.0 + [og25 = 5.0 + Iog 1- "50 - mLOH- "ts 5.00

37.50 pH =pKu + log--InL,QH- = 5.0 + 1"g4+ = 5.0 + log 3- 50 - mT.OH- "12 \ 5.48

40.00 pH =pKa + log---mLQH- = 5.0 + log4!.0 = 5.0 + log 4"50 - ml,oH- "1 o n 5.60

45.00 pH =pKu + log-- nnL-QH- = 5.0 + 1og450 = 5.0 + log 9* "50-mLOH- "so 5.95

48.00 pH =pKu + log--rnlOH- = 5.0 + trST* = 5.0 + log24 6.38

49.00 pH =pKu + log--mL-QH- = 5.0 + log49-} = 5.0 + 1og49- "50-mLoH- "to 6.69

50.00 o11= PKa +PHtitrant =5.0 + 13 =18')2) 8.85

51.00 pOH = -1og [OH-]"*"ur, = -log(0.10 x-L)- pH = 14 - pOH 11.00

55.00 pOH = -log [OH-]",,."r, = -1og(0.10 x-L)= pH = 14 - pOH 11.68

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General Chemistry Buffers and Titration fitration Curves

Plotting a Titration CurveThe points from Table 5.3 are represented in Figure 5-9 as circles. The curves isdrawn to fit the circles. The shape of the curvJfor a weak acid is distinct. Thebuffer region is not perfectly flat, showing that pH changes slightly in the bufferregion. The pH values at 1.00 mL and 2.00 mL are not eiuci, because theHenderson-Hasselbalch equation does not hold as well outside the pKu + 1range. Nevertheless, the values are close enough to generate a reisonabletitration curve.

pH12

11.68

3.62

3.31

3.00

20 25 30

mL 0.10 M KOH added

Figure 5-9

The titration curve should become familiar with enough examples. What makescurves useful is that they summafize a great deal of information. If you think oftitration curves in terms of equivalents and regions, you can extract a substantialamount of information from them. For instance, when 2r.27 mL of 0.1 M KoHhas been_ added, the pH is roughly 4.7 to 4.8. That range is small enough that aneducated guess can be made on a multiple-choice q.testio.r. We shall elmphasizeusing titration curves in lieu of calculating the pH, when it comes to buffers andother mixtures.

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General Chemistry Buffers and Titration Titration Curves

30 mL NaOH added

pKo-1<pH<pK pK"-<pH<pI{+1

Example 5.11What is the pH after 30 mL of 1.00 MHOAc? HOAc has a pKu = 4.74.

NaOH has been added to 100 mL 0.50 M

p lld -t ' *:. -" _.Klz.st rr

'f--=4'se ?'a-A{-492

D. 5.97

- 1, )'a t-\ '"

\ L(-o

SolutionBecause the strong base is twice as concentrated as the weak acid, only half thevolume of strong base (relative to the weak acid) is required to reach theequivalence point. This means that 50 mL of 1.00 M NaOH fully neutralizes the100 mL of 0.50 M HOAc. The halfway point of the titration is reached whenexactly 25 mL of 1.00 M NaOH has been added. At the halfway point, the pH ofthe solution equals the pKu of the weak acid. The additional strong base beyondthe 25 mL makes the pH of the solution slightly greater than the pKu of the acid,4.74. The best choice is answer C. The titration curve below shows a summary ofthe intuitive approach:

mL0.10MNaOH(aq) added

When 30 mL has been added, the mixture is just beyond the half-titrated point onthe titration curve (as shown by the arrow). This makes the pH fall into therange of pKu < pH < pKu + 1, according to the titration curve. According to theHenderson-Hasselbalch equation, the pH equals the pK3 + log (conjugate baseover acid). Past the half-titrated point, the concentration of the conjugate baseexceeds the concentration of the acid, so the ratio of conjugate base to acid isgreater than one. The log of a number greater than 1.0 is a positive value. Whena positive value is added to the pKu, the final value is greater than the pKu"confirming that pH > pKa. These questions should be answered quickly, usingeither a titration curve or the Henderson-Hasselbalch equation in a purelvconceptual manner.

5025

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General Chemistry Buffers and Tltration Titration Curves

Example 5.L2What is the pH after 70 mL of 0.20 M HCI has been added to 50 mL 0.60 MH3CNH2? H3CNH2 has a pK6 =3.42.A. 9.44B. 10.51c. 10.65D. 1.1..72

SolutionThe strong acid is one-third as concentrated as the weak base, so three times thevolume of HCI (relative to the H3CNH2) is needed to reach the equivalencepoint. This means that 150 mL of 0.20 M HCI fully neutralizes the 50 mL of 0.60M H3CNH2. The halfway point of the titration is reached when 75 mL of 0.20 MHCI is added. At the halfway point, the pH of the solution equals the pKu of theweak acid. Less than the 75 mL has been added, so the pH of the solution isslightly greater than the pKu of the conjugate acid, 10.58. The best choice isanswer C. The titration curve below shows how to estimate the value.

75mL 0.20 M HCI(aq) added

Example 5.13\Alhat is the pH of a solution made by mixing 10.0 mL 0.10 M HCO2H(aq) with 4.0mL 0.10 M KOH(aq)? The pKu for HCO2H is 3.64.

A. L.34B. 3.46c. 3.82D. 9.36

+, -,.I

SolutionThe best way to solve this question is to think in terms of equivalents. The weakacid and titrant strong base are of equal concentration, so 10 mL KOH is oneequivalent (the amount needed to reach the equivalence point.) If 5.0 mL areadded, then the acid is half-titrated, so pH = pKa. However, less than 5.0 mL hasbeen added, so pH is a little less than pKu. The pKu is 3.64, so the best answer ischoice B. Choice A is too much less than the pKu (more than 1,.0 is beyond the 10: L ratio, which would be when less than 1.0 mL of KOH had been added.)

pK"-<pHcpI{+1 pKu-1<pH<pK

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General Chemistry Buffers and Titration Titration Curyes

mL titrant strong base added

Concentration Effect on Titration Curve ShapeThe pH of a solution depends on the strength and concentration of the reagents,so strength and concentration also affect titration curves. The concentration ofthe reagents affects the dimensions of the titration curve, but not its basic shape.A strong-by-strong curve maintains the same fundamental shape (sigmoidal),but with varying concentrations, the curve may skew and elongate. Theequivalence point is always at pH = 7. A weak-by-strong curve also maintainsits same fundamental shape (i.e., has a lip-o-weakness), but with varyingconcentrations, the curve also skews and elongates. The half-equivalence point isalways at pH = pKu.

For a strong acid titrated by a strong base, as the concentrations of both reagentsincrease proportionally, the respective curves start lower and finish higher, butthey have the same distance in the x-direction (the mL axis). If the acidconcentration is increased but the base remains the same, then the curve startslower and stretches to a point farther from the origin along the x-axis for theequivalence point. If the base concentration is increased but the acid remains thesame, then the curve finishes higher and contracts to a point closer to the originalong the x-axis for the equivalence point. This is shown in Figure 5-10.

1.0 M KOH(aq)

0.1 M KOH(aq)

.01 M KOH(aq)

0.1 M HCI(aq) +

.01 M HCI(aq) +

Note that strong curves sharean equivalence point at pH = 7.

Figure 5-10

In each titration, the concentration of the strong acid is equal to the concentrationof the strong base, so the volume of base required is the same in each case. Thisis why the curves are similar in the x-direction. In the y-direction, the curvedepends on concentration. As the concentration lessens, the curve contracts withrespect to a line through pH = 7. Note that the initial pH is 0, 1., and 2

respectively for the three titration curves. The variation in the concentration ofbase also causes the ends of the curves to vary (in the region of excess titrant).But at equivalence, the pH is7,no matter what the concentrations are.

For a weak acid titrated by a strong base, as the concentrations of both reagentsincrease proportionally, the curve starts lower and finishes higher but advancesthe same distance in the x-direction (the mL axis) and has the same buffer region.If the acid concentration is increased but the base remains the same, then thecurve starts lower and stretches to a point farther from the origin along the x-aisfor its equivalence point. If the base concentration is increased but the acidremains the same, then the curve finishes higher and contracts to a point closer tothe origin along the x-axis for its equivalence point. This is shown in Figure 5-11.

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General Chemistry Buffers and Titration Tltration Curves

4--J.o M HoAc(aq) +

0.1 M HOAc(aq) +

.01 M HOAc(aq) +

1.0 M KOH(aq)

0.1MKOH(aq)

.01 M KOH(aq)

The equivalence points vary, because higherinitial HOAC concentration leads to a higher

OAc- concentration at the equivalence point.The approximate pH at equivalence is an

average of the pI( and pH of the titrant base.

As acid concentration lessens, initial pHincreases, and the size of the lip lessens.

Figure 5-11

In each titration, the concentration of the weak acid is equal to the concentrationof the strong base, so the volume of base required is the same in each case. Thisis why the curves are similar in the x-direction. In the y-direction, the curvedepends on concentration. As the concentration lessens, the curve contracts withrespect to a line through pH = pKu. The initial pH is I pKu,l pKa + 0.5, and !PKa + L.0 respectively for the three curves. The variation in the concentration ofbase also causes the ends of the curves to vary (in the region of excess titrant).But at half-equivalence, the pH is pKn, no matter what the concentrations are.

Strength Effect on Titration Curve ShapeThe examples in Figure 5-10 and Figure 5-11 show the effect of concentration ontitration curves. The shape of a titration curve is also a reflection of the strengthof the reagents. When the concentrations are uniform between titration curves,then the shape of the curve and location of key points give clues as to the natureof the acid being titrated. As the acid becomes weaker, there is a larger initial lip,a greater midpoint (where pH = pKu), and a greater equivalence point. This isshown in Figure 5-12. As acid strength increases, initial pH decreases, and thesize of the lip-o-weakness lessens.

Figure 5-12

Curves are nearly identicalthrough the buffer regioq

On each curve, because

tAl = tHL pH = plf

mL titrant strong base added

0.1MHq

mL 0.10 M KOH(aq) added

0.1MHOAc(aq) + o.l MKoH(9

0.1MHNOrt"o) *-0'@

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General Chemistry Buffers and Titration Titration Curves

Polyprotic Acid Titration CurvesPolyprotic acids are acids which yield multiple equivalents of hydronium ion(HSO*). Three common polyprotic acids with which every test-taker must befamiliar are carbonic acid (H2CO3), phosphoric acid (H3POa), and sulfuric acid(H2SOa). Polyprotic titration curves have multiple equivalence points, one foreach dissociable proton. But the protons are removed one at a time, so the curvesfor each proton are separate. Th"y should be treated as separate titration curvesthat happen to overlap on the same graph. Figure 5-13 shows the titration curvefor a typical diprotic acid, where both protons are weak. Examples that share thesame basic curve shape include carbonic acid, glycine, or a mixture of two weakmonoprotic acids in the same solution. Figure 5-14 shows the titration curve fora typical diprotic acid, where the first proton is strong and the second proton isweak. An example that shares the same basic curve shape is sulfuric acid. Figure5-15 shows the titration curve for a typical triprotic acid, where all three protonsare weak. Examples that share the same basic curve shape include phosphoricacid, citric acid, and glutamic acid.

Second equivalence point

First equivalence point

Weak first PH = PKar

protonmL strong base added

Figure 5-13

Second equivalence point

Figure 5-14

pH=pKThird equivalence point

Second equivalence point

pH=pK

First equivalence point

First equivalence point

mL strong base added

mL strong base added

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pH=pK

312

Figure 5-15

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General Chemistry Buffers and Titration Indicators

I ff. ruHffiIndicator Composition and FunctionAn indicator is an organic compound with extended conjugation. Its weak acid(protonated) form and conjugate weak base (deprotonated) form have twodistinct colors. The colors of both species are reflected colors. The energy statetransition that produces color involves the n-bonding and n-antibonding orbitalsand depends on the electron donating and withdrawing nature of substituents onthe n-system. The lone pair formed upon deprotonation affects this conjugation.When the pH of the solution is less than the pKu of the indicator (pHsotrtton <pKa (indicator;), the indicator exists predominantly in its protonated form (H-Ind > Ind-). If this is true, the solution assumes the hue of the protonated form ofthe indicator (H-Ind). when the pH of the solution exceeds the pKu of theindicator (pHsolution > pKa (indicator)), the indicator exists predominintly in itsdeprotonated form (Ind- > H-Ind). ir this is true, the solution assumes the hueof the deprotonated form of the indicator (Ind-). When the pH of the solution isequal to the pKu of the indicator (pHrolrrlion = pKu (indicator)) the indicator existsequally in its protonated and deprotonated forms itFl-tncit = [Ind-]). If this istrue, the solution assumes a hue that is the result of a mixture of the protonatedand deprotonated forms of the indicator. When the pH of the solutionis near thepKu of the indicator, the color varies with small changes i. pH. Reaction 5.3represents the dissociation of an indicator in water.

r i t,," .' H-Ind,..+ H+ + Ind- ar!'\:;i ''"" *'."

-y"uo* ';l;" , "'-l'-'Reaction 5.3

Table 5.4 shows the pH effects on the generic indicator in Reaction 5.3, alongwith a ratio of protonated-to-deprotonated species, and the color of the indicator.

Table 5.4

\Alhen the solution is bluish green, it can be concluded that the pH of the solutionis slightly (about 0.2 to 0.4 times) greater than the pKa of the indicator. Thismeans that the pH of the solution can be approximated from the color of thesolution. The color change range (and thus the useful range) of an indicator ispKa (Indicator) t 1. An indicator is generally used for one of two purposes. Thefirst is to detect the endpoint of a titration, and the second is to approximate thepH of a solution by observing the color of the indicator in the solution.

PHsolution Ratio of H-Ind to Ind- Mixture of colors to form solution colorpKu + 3.0 1 : 1000 1 yellow : 1000 blue .'. bluepK" + 2.0 1 :100 1 yellow : 100 blue .'. bluepKu + 1.0 1:10 1 yellow : 10 blue .'. greenish bluepKu + 0.7 1:5 1 yellow : 5 blue .'. blue-greenpKu + 0.3 1:2 1 yellow : 2 blue .'. bluish green

PKa 1:1 1 yellow: l blue .'. greenpKu - 0.3 2:7 2 yellow : 1 blue .'. yellowish greenpK^ - 0.7 5:1 5 yellow : 1 blue .'. yellow-greenpKu - 1.0 10:1 10 yellow : 1 blue .'. greenish greenpKu - 2.0 100:1 100 yellow with l blue .'. yellowpKu - 3.0 1000: 1 1000 yellow with l blue .'. yellow

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General Chemistry Buffers and fitration Indicators

Detecting the Equivalence Point (Using Indicators)Indicators make the equivalence point of a titration visible, either by convertingfrom an uncolored species to the colored conjugate or by converting from onecolored species to its conjugate, which has-a-differeni color. Ai indicatorchanges color at a pH near the equivalence point of the titration. The change incolor can thus be used to determine the equivalence point. Indicators are addedin,small quantities, so they do not interfere with the titration. The pH of thesolution thus dictates the color of the indicator. As the titration approachesequivalence, the pH changes rapidly, so the color changes rapidly. A colorchange indicates that equivalence has been reached, if the correct indicator ischosen.

The ideai scenario for an indicator is when the pH at equivalence equals the pKuof the indicator. The range of a color change for an indicator is centered on itspKa. For the titration of a weak acid with a strong base; the pH at theequivalence point may not be known exactly, so a best approximation of theequivalence pH must be made. When approximating the equivalence pH,consider that the pH at equivalence is greater than the pKu of the weak icidbeing titrated, but less than the pH of the strong base being added. The pH atequivalence lies roughly half way between the pKu of the acid and the pH bf thebase being added. A good approximation of the pH at equivalence is an averageof the pKu of the weak acid and the pH of the titrant strong base. The pKu of tfieindicator should be somewhere around (within + 1 unit) the average of the pKuof the acid being titrated and the pH of the strong base. For the titration of aweak base with a strong acid, the pH at the equivalence point is less than the pKuof the conjugate acid of the weak base being titrated. The pKu of the indicatorshould be somewhere around (within + 1 unit) of the average of the pKu of theconjugate acid of the weak base being titrated and the pH of the strong acid. Theactive range for an indicator is represented by the two titration curves in Figure5-16 and Figure 5-17.

Equivalencepoint

I mdi"u.o.

J resion

pH at equivalence point > 7.0

PH = PKtu.ia)

mL strong base added

Best scenario: PH"qr.ri'ulence = PKa(indicator) t 1

Best approximation: PKa(acid) + PH(titrant base)

= pKa(indicator) t 12

Choosing an indicator for the titration of a weak acid by a strong base

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Figure 5-16

The Berkeley Rel.ieu,

General Chemistry Buffers and Titration Indicators

PH = P41a"iay

pH at equivalence point < 2.0

Indicatorregion

mL strong acid added

Best scenario: PHequivalence = pKalindicator) I 1

Best approximation: PKa(coniugate acid; + PHltitrant acid)

-

=P(a(indicator) tlChoosing an indicator for the titration of a weak base by a strong acid

Figure 5-17

You should make note of the fact that the pH at equivalence for the titration of aweak acid by a strong base can be approximated quite .tor"iy uy averaging thepKu of the weak acid with the pH o? ?ne titrant stiong buse. 'This value will bewithin 0.5 of the actual value, meaning,that for the iitration of acetic acid bysodium hydroxide, the pKn of the iridicator should be greater than seven(somewhere around nine.) As the titration is carried out,

-the pH of solution

increases; thus, the indicator starts out in its protonated form irra "rr"r,tuuiybecome deprotonated. For the sample indicator in Reaction 5.2, the solutionwould go from yellow to green to blue.

For the titration of a weak base b-y a strong acid, the pH at equivalence can beapproximated by averaging the pKu of the ionjugate acid of thl weak base withthe pH of the titrant strong acid. This means that for the titration of ammonia byhydrochloric acid, the pKu of the indicator should be less than 2.0 (somewherearound 5.0.) As the titration is carried out, the pH of solution decreases; thus, theindicator starts out in its deprotonated form and eventually becomes protonated.For the sample indicator in Reaction 5.3, the solution #ould go frtm bl";-i;green to yellow.

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General Chemistry Buffers and Titration Indicators

,i-,-Eo ,1.-' i _ '-q-'

-,.tu '! J'

- -pL\(_. u

' !:";,-t{'': ''' k--\

- "t'l'\i r',"

,t . ?V^G,f !' ! ,,.. 1ci"l (-tt"'

Estimating Solution pH (using Indicators)For a solution of unknown pH, if the pH is within one unit of the pKu of theindicator, then the hue of the indicator can be used to estimate the pH of thesolution. Consider an indicator with a pKu of 6.83 that is used to test pool water.It is yellow when protonated and red when deprotonated. Table 5.5 can be usedto estimate the pH of the pool water.

PHsolution Ratio of H-Ind to Ind- Solution Color

pH > 7.83 1:10+ Red

PF{=7.83 1:10 Oransish redpF{=7.53 1:5 Red-orangepH=7.73 1:2 Reddish oransepH = 6.83 1:1 oranqepH = 6.53 2:1 Yellowish oransepH = 6.13 5:1 Yellow-orangepH = 5.83 10:1 Oraneish vellowpH < 5.83 10+: 1 Yellow

{.i*r':,*'i Table5-S

If the results of the pool water test is red, the water is basic; and because red isoutside the color change range, the exact pH cannot be approximated. If theresults of the pool water test is yellow, then someone has left some acid in it.Most people are aware that yellow pool water has some acid in it, and it shouldnot be swum in. The ideal color is orange with a slight hint of red. Indicators are

used in pH test kits for swimming pools and fish tanks. The kits usually containmore than one indicator, increasing the accuracy of the approximation. All of theindicators in such a test kit should have pKu values between 6.0 and 8.0, becausethe pH of the water should be around 7.0. Indicators are also used in pH teststicks, where a series of three or four indicators are on the stick. The pKu valuesof the indicators differ by roughly two units. This allows for a wider range of pHvalues from which one can correlate the color to the solution pH. For instance, apH stick with three indicators, with pKu values of 5.05, 6.98, and 9.11, has a rangeof roughly 4.05 to L0.11. This is because each indicator has a two-pH-unit range.

Example 5.14Given the following indicators on a pH stick, what is the pH of a solution thatyields X: red, Y: blue, andZ: red?

Indicator X: pK6 = 4'96; when deprotonated, it goes from yellow to red

Indicator Y: pKx = 7.01; when deprotonated, it goes from yellow to blue

Indicator Z: pKa = 8.98; when deprotonated, it goes from red to blue

A.58.6c.7D.8

SolutionBecause Indicator Y is blue, the pH must be at least one unit greater than 7.01

(the pKz of Lrdicator Y). Because Indicator Z is rcd, the pH must be at least one

unit less than 8.98 (the pK3 of Indicator Z). The only value greater than 7.01 and

less than 8.98 is 8, choice D.

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Buffersand

TitrationPassages

15 Passages

lOO Questions

Suggested Buffers and Titration Passage Schedule:I: After reading this section and attending lecture: Passages I, IV, VII, VIII & XI

Grade passages immediately after completion and log your mistakes.

II: Following Task I: Passages II, III, V, VI, IX, & XIV (41 questions in 55 minutes)Time yourself accurately, grade your answers, and review mistakes.

III: Review: Passages X, XII, XIII, & XVFocus on reviewing the concepts. Do not worry about timing

ffilll..,ffi 1,,. ;,...[...lirull ffi.*Speciahztng in MCAT Preparation

I. Buffer pH and Weak Acids

II. Buffer Composition

III. Buffer Chart and pKz Chart Passage

IV. Molecular Weight from Neutralization of an Organic

V. Conjugate Pair Titration Curve

VI. Titration Curves and Concentration Effects

VII. Titration Curves and Strength Effects

VIII. Normality and Neutralization

IX. Titration Curve of a Polyprotic Acid

X. Carbonate Titration Curve

XI. Indicator Selection

XII. Indicator Color and Solution pfl

XIII. Indicator Table

XIV. pH Sticks and Indicators

XV. Acidity and Electronic Influences

(r -7)

(8 - 14)

(15 - 21)

Acid (22 - 29)

(5O - 56)

(57 - 42)

(43 - 4e)

(5O - 56)

(57 - 63)

(64 - 70)

(7r - 76)

(77 - 85)

(84 - Be)

(eo - e6)

(97 - 100)

Buffers and Titration Scoring Scale

Raw Score MCAT Score

84 - IOO l5-1566 a3 10-t247 65 7 -9

54-46 4-6L-33 1-3

a

!l:l

Passage I (Questions 1 - 7)

An integral part of any biological study conducted invivo is the accurate simulation of body conditions with as

much precision as possible. As important as any factor tobiological systems is pH. To accommodate the need for a

constant and accurate pH, organisms of all types use buffersto maintain a relatively constant internal pH range. A bufferexists when there is both a weak acid and its weak conjugatebase present in solution in roughly equal molarconcentrations. It is important that both the acid and the base

be water-soluble and exhibit no side reactions. Organic acidsand their conjugate bases are best for this purpose.

The Henderson-Hasselbalch equation, Equation 1, is used

to calculate the value of the pH for a buffer.

pH = PK6 + tog E3!91lAcidl

Equation 1

In human blood, a buffer ofbicarbonate and carbonic acidexists. Inorganic phosphates also play a role in bufferingwithin the body. The buffer must have a pH that is relativelyclose to 7.4, known as biological pI1. This can be simulatedin the lab by starting with a weak acid whose pKu is as close

to 7.4 as possible. After the acid has been added to water, thesolution is titrated with strong base until a pH of 7.4 isobtained. This method ensures an accurate value. Buffers can

also be made by mixing equal molar portions of the weakacid and its conjugate base (the conjugate base may come inits salt form). The pH equals the pKu of the acid when abuffer is made with equal molar portions of acid and

conjugate base.

l. What is the pH of a solution made by adding 0.839grams NaHCO3(s) (MW = 83.9 grams/mole) to 100 mL0.10 M H2CO3(aq)?

PKal = 6.4 PKa2 = 10.8

A. 3.20B. 6.40c. 8.60D. 10.80

2. Which of the following pH values is the BEST choicefor the pH of a buffer initially at pH = '7.21 after HClhas been added?

A . 1.14B. ',7.21

c. 7.28D . 8.31

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\ .:,/.{3. What should be mixed to make a pH = 4.2 buffer?

PKalbenzoic acid) = 4.2

__l . lO grams C6H5CO2Na + l0 grams C6H5CO2H

,,,Y. 10 mL 0.10 M C6H5CO2H + 5 mL 0.10 M NaOH

C. 10 mL 0.10 M C6H5CO2H + 10 mL 0.10 MNaOH

D. 10 mL 0.10 M C6H5CO2H + 15 mL 0.10 M' NaOH

i,'. ,-.L'^ { i'r.r'r' . /- ).

,''i'i ' t-''

4. How many mL of 0.20 M NaOH must you add to 50mL of 0.10 M HF to produce a solution with a pH of3.3? (pKa = 3.3) ., .A. 10.0 mLB. 12.5 mL .., : .,,,:, iirC. 16.7 mLD. 25.0 mL

' \,/ irij Ji\.-

,I .

')'' ''' I5. What is the acetate anion concentration in a solution

made by mixing 20 mL 0.30 M HOAc with l0 mL 0.30MNaOH?

A. 0.07 M H3CCO2-B. 0.10 M H3CCO2-

C. 0.15 M H3CCO2-D. 0.20 M H3CCO2-

6 . Which of the following combinations produces a buffer?

.x(tomL0.25 M NH:(aq) +20mL0.25 M HCI(aq)

y. 20 mL 0.25 M NHaCI(aq) + l0 mL 0.25 M.'" HCI(aq)

9; l0 mL 0.25 M NH3(aq) + 20 mL 0.25 M NaOH(aq)

/6,,'20 mL 0.25 M NHaCI(aq) + 10 mL 0.25 M1*'

NaOH(aq)

7 , What is the pH of 100 mL of 0.10 M propanoic acid,

PKa(propanoic acid; = 5'0?

A.2.5 , "I---ri-l'-'.-\1-(_8.3.O \--c. 5.0 ,, r{ - ., .". ,i ''

D. 6.0 \ .l

Passage ll (Questions 8 - 14)

A buffer is composed of a weak acid and its conjugate

base. In order to maintain buffering, the ratio of the

conjugate pair must be less than 10:1, in favor of either

component. According to Equation l, when the two

components of the conjugate pair are roughly equal inconcentration, the pH of the buffer solution is approximatelythe pKa of the weak acid.

PH = PKa + log E3!91

Equation 1

A buffer is capable of consuming any acid or base that is

added to solution. The pH of an aqueous buffer solution

changes only slightly after an acid or base is added. This is

because there is an equilibrium between the conjugates.

Table 1 lists a series ofweak acids and conjugate bases, along

with the pKa values for each acid. The pKu value of the

weak acid and the acid-base ratio can be used to determine the

pH of a buffer from Equation l.

Weak Acid Conjugate Base pKu value

CIH2CCO2H CIH2CCO2K 2.82

HF NaF 3.t9

HCO2H HCO2Na 3.64

H3CCOCO2H H3CCOCO2Na 3.86

C6H5CO2H C6H5CO2K 4.19

C6H5NH3Cl C6H5NH2 4.62

H3CCO2H H3CCO2Na 4.',l4

C5H5NHCI C5H5N 5.16

4-NO2C6HaOH 4-NO2C6HaOK 1.15

HCIO KCIO 1.49

FIBlo KB10 8.61

NHaCI NH: 9.26

C6HaOH C6HaOK 10.01

H3CNH3CI H3CNH2 10.56

(H3C)2NH2Cl (H3C)2NH 10.78

Table 1

To make a buffer, an acid must be chosen that has a pKu

value within one unit of the target pH. The closer the pKu

value is to the pH, the better the buffer. Buffers are made

either by mixing the weak acid with its conjugate base, or by

partially titrating either the weak acid with strong base, or the

weak conjugate base with strong acid. When the pH exceeds

the pKu, there is more conjugate base present in solution.

8 . Which mixture does NOT produce a buffer?

A . H3CCO2H with 2 equivalents of H3CCO2K

B . NH3 with 2 equivalents of NH4CI

C . H2CO3 1.5 equivalents of KOH

D . H3CNH2 with 1.5 equivalents of HCI

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9 . What is the pH of a solution that contatnsweak acid and one part conjugate base?

A. pKa(weak acid) + 2

B. pKa(weak acid; + log 2

C . pKa(weak aciq - 2

D. pKa(weak acid; - log 2

10. At what pH would it be MOST difficult topH buffer?

A. 1.0

B. 3.0c. 5.0

D. 7.0

two parts

establish a

I 1 . What BEST describes an aqueous buffered solution?

A. An aqueous solution where the hydronium and

hydroxide concentrations are equal

An aqueous solution where the hydronium-to-hydroxide concentration ratio never exceeds 10:l or

is less than 1:10

An aqueous solution where the hydronium and

hydroxide concentrations are relatively constant

An aqueous solution where the hydronium-to-hydroxide concentration ratio is within one unit oithe pKs value fbr the weak acid

12. Addition of 1.00 mL of 0.10 M KOH(aq) to a solution

made by mixing 15.00 mL 0.10 M H3CCO2HIaq) with

10.00 mL 0.10 M H:CCOZNa(aq) results in:

A. a solution with a pH less than 3'14.

B. a solution with a pH between 3.14 and 4.14.

C. a solution with a pH between 4'74 and 5'14.

D. a solution with a pH greater than 5.14.

13. In which solution is there the LARGEST ratio o'

conjugate base to conjugate acid?

A. A solution of HCO2H and HCO2- with pH = 4'00

B. A solution of HC1O and ClO- with pH = 7'00

C . A solution of HBrO and BrO- with pH = 8.50

D . A solution of NH3 and NH4+ with pH = 9.50

14. What is observed after 5'00 mL HzO(l) has been adde:

to 50.00 mL of a buffered solution with pH initialll ':5.0?

A. The pH drops slightlY.

B. The pH remains constant.

C . The pH increases slightlY.

D . The pH increases drasticallY.

B.

C.

D.

Passage lll (Questions 15 - 21)

Buffers are aqueous solutions of weak acids and theirconjugate base. The pH of solution is dictated by theHenderson-Hasselbalch equation: pH = pKa + 1og base/acid.

This means that a buffer solution should be mixed in a

manner where the pKn of the acid is close to the pH desired.Table I lists the Ku and pKu values for some common

monoprotic weak acids:

Acid Ka value pKa value

H2NCONH3+ 6.6 x 10-l 0.18

HF 6.8 x 10-a 3.11

HCNO 3.5 x 10-4 3.49

HCO2H 1.'7 x 1O-+ 3.18

H3CCOCO2H 1.4 x l0-a 3.89

C6H5CO2H 6.5 x 10-5 4.r9

C6H5NH3+ 2.3 x 10-) 4.64

H3CCO2H 1.8 x 10-l 4.14

C5H5NH+ 7.1 x l0-o 5.16

HClO 3.5 x 10-6 1.49

B(OH): 5.9 x l0-10 9.22

NH4+ 5.6 x l0-10 9.26

HCN 4.9 x 10-ru 9.32

H3CNH3+ 2.2 x 10-11 r 0.66

(H3C)2NH2+ 1.9 x 10-r r0.11

Table 1

Polyprotic acids can also be used in making buffers. Adifficulty that arises with polyprotic acids involves the two ormore pKu values. For any given polyprotic acid, the

conjugate pair will buffer at the respective pKa for the acid ofthe conjugate pair. For instance, carbonate/bicarbonate willbuffer at a pH around 10.8 because pK62 of carbonic acid is

10.81. Carbonic acid/bicarbonate will buffer at a pH around6.4 because pKul of carbonic acid is 6.31 . Table 2 lists the

Ka values for some common polyprotic acids:

Table 2

As a point of interest, carbonates and phosphates are

believed to be the major contributors to buffering in humanblood. For years it was believed that carbonate played themajor role, but recent research indicates that phosphate mayplay a more significant role than carbonate in the overallbuffering. Within the kidneys, phosphates are known to playa significant role.

Acid Krl value K"2 value K^3 value

H2C2Oa 5.6 x 10-2 5.1 x 10-5

H25O3 1.3 x l0-2 6.3 x 10-8

H3POa 6.9 x 10-3 6.2 x l0-8 4.8 x 10-13

H2CO3 4.3 x lO-1 1.5 x 10-11

HzS 8.9 x 10-8 3.8 x l0-i3

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15. Combining all of the following results in a buffer withpH = 9.5 EXCEPT:

A . 1.8 equivalents NH3 with 1.0 equivalents NH4+.B. 1.6 equivalents NaCN with 1.0 equivalents HCN.C. 0.7 equivalents HCI with 1.0 equivalents NH3.

D . 0.65 equivalents NaOH with 1.0 equivalents HCN.

16. To make a buffer at pH = 10.83, which of the followingshould be mixed?

A. One-half equivalent of NaOHof H2CO3

B. One and one-half equivalentsequivalent of H2CO3

C. One and one-half equivalentsequivalent of H3PO4

D. Two and one-half equivalentsequivalent of H3PO4

with one equivalent

of NaOH with one

of NaOH with one

of NaOH with one

17. Which of the following solutions forms a buffer with apH greater than 7.0?

A. One and one-half equivalents of KOH with oneequivalent of H2C2O4

B . One-half equivalent of KOH with one equivalent ofH2CO3

C . One-half equivalent of KOH with one equivalent ofHCIO

D. One-half equivalent of HCI with one equivalent ofC5H5N

18. Which of the fbllowing mixtures would NOT result in a

solution with pH = 4.00?

A. Excess HCO2' mixed with HCO2H

B. Excess C6H5NH3+ mixed with C6H5NH2

C. Excess C6H5CO2- mixed with C6H5CO2H

D. Excess H3CCOCO2- mixed with H3CCOCO2H

19. Carbon dioxide when dissolved into the blood formscarbonic acid. What is observed in CO2-enriched blood?

A . The pH is less than 1.4.B. The tPO+3-l increases.

C . The [HCO:-] decreases.

D. The [HgO*] decreases.

20. In which of the following solutions is rhe conjugatebase in GREATER concentration than the acid?

A. HF(aq) with F-(aq) at pH = 3.00B. H2CO3(aq) with HCO3-(aq) at pH = 6.00C. HC2Oa-(aq) with C2O42-(aq) at pH = 4.00D. H2POa-(aq) with HPO42-(aq) at pH = 8.00

21. Biological pH is approximately 7.4. Which of thefollowing is NOT true about the concentration of bufferspecies at this pH?

A. [HCO:-] > [H2CO3]B. tHCO:-l > [CO:2-]C. tH2POa-] > FIPOoz-,D. [HPoa2-J t POo:-,

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Passage lV (Questions 22 - 29)

The formula weight of an acid can be determined bytitration, using a strong base of known concentration. Theprocess involves the titration of an exact mass of someunknown acid. Once enough base has been added to reachequivalence, the moles of base added are used to determine themoles of acid that were present in the solution initially. Themass of the acid divided by the moles of the acid gives theformula weight for the acid, not the molecular weight. Oftenboth weights are the same, but a difference arises if theunknown is a polyprotic acid. Only when the acid ismonoprotic is the formula weight equal to the molecularweight. Table I shows some phenols with their respectivepKu values. Al1 of the phenols are solids at roomtemperature and are monoprotic acids.

Table 1

If the molecular mass of an acid is known, the pH o: isolution can be calculated from its gram concentration .'isolution. To do this requires converting from grams ir..moles. Once the concentration is.known, the shoni*:equation, Equation 1, may be used to quickly calculate the iIofthe aqueous solution ofthe acid:

PH = | pr^ tlog

[HA]

Equation I

22. 1.0 gram of which of the above acids requires e\e;,r',30.0 mL of 0.20 M NaOH to reach equivalence?

A. Acetic acid (HjCCO2H) MW = 60B . Trichloroacetic acid (CI3CCO2H) MW = 151.-<

C. p-nitro benzoic acid (O2NC6HaCO2H) MW =-i-D. Benzoic acid (C6H5CO2H) MW = 122

Structure Formula PKa

p-O2NC6HaOH 7.2

Hscp-H3CCOC6HaOH 8.4

C6H5OH r0.0

OH p-H3CC6rH4OH 10.1

OH p-HTCOCaHsOH I 1.:

23. lf 1.0 gram of an unknown acid requires exactly 40.00mL of 0.25 M NaOH to reach the equivalence point,what is the formula weight for the unknown acid?

A. 50 grams per mole

8.. 80 grams per mole

Q., 1o0 grams per moleD . 125 grams per mole

U

. .:i *.-ls .- <i.t

2 4. Consider this graph:

pH

It;.i'I

I

i

{

mL of titrant

The graph represents the titration of:

--"k -, - H3CCO2H by NaOH..3. HCI by NaoH.

C AH, by H3CCo2H., D.-NH3 by HCl.(,-''

25. Which of the following indicators would be BEST forthe titration of p-nitrophenol(O2NC6HaOH) by NaOH?

A. Thymol blue (pH range of color change is 1.2 to' 2.8)B. Methyl red (pH range of color change is 4.6 to 5.8)

z-.e ..Bromthymol Blue (pH range of color change is 6.0\'-"'' to 7.6)

B( Phenolphthalein (pH range of color change is 8.0to 9.6)

, .. /1

i:, \:' ;

-''i' ':l11.!

2 6. Which of the following mixtures results in a bufferedsolution?

.x-to mL 0.25 M NaoH + 10 mL 0.25 M H3cco2Hprl20 mL 0.25 M NaOH + l0 mL 0.25 M H3CCO2H

! c. to mL 0.25 M NaoH + 2o mL0.25 M H3cco2H

-,.V. lO mL 0.25 M HCI + l0 mL 0.25 M H3CCO2H

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27 . TteBEST choice for a pH = 8.5 buffer would be whichAof the following?

.1 O2NC6HaOH with less than one full equivalent ofNaOH

,8. H}COC6HaOH with less than one full equivalent," of NaOH

,./. fuCC6H4OH with less than one full equivalent of- - NaOH

D. H3CCOC6HaOH with less than one full equivalent

r.., of NaOH

2 8. What is the formula weightoxalic acid (HOzCCOzH)?

,.. A. 45 grams per moleB:90 grams per mole*C . 135 grams per molefi. tSO grams per mole

weight) of

25.0 mL of an unknown acid when titrated by exactly30.0 mL of 0.100 M KOH(aq), requires seven drops of0. 100 M HCI(aq) to return to equivalence. What is the

concentration of the unknown acid?

A. 0.1217 M acid(aq)

B . 0.1 183 M acid(aq)

C.0.0849Macid(aq)D. 0.0817 M acid(aq)

(equivalentrtilL-li-rLi

2

,l

29.

Passage V (Questions 30 - 36)

Conjugate acid/base pairs are connected by a relationshipbetween pKu and pK6. Equation I shows the relationshipwithin a conjugate pair at 25"C in aqueous solution.

PKa (conjugate acid) + PKb (conjugate base) = 14

Equation 1

When titrating a weak acid or weak base, the portion ofthe curve following the initial drops of titrant up until justbefore the equivalence point is an equilibrium mixture of the

conjugate pair. As such, the titration curve of onecomponent in a conjugate pair have similarities to thetitration curve of the other component.

The titration curves for conjugate pairs are inverse graphs

that intersect at the half-titrated point. At this point, the pHof the solution equals the pKu of the conjugate acid. At thissame point, the pOH of the solution equals the pK5 of the

conjugate base. Figure 1 shows the titration of acetic acid(H3CCO2H) with strong base (NaOH) overlaid onto thetitration of sodium acetate (H3CCO2Na) with strong acid(HCl). In both titration curves, all species are in equal molarconcentrations. The pKu for carboxylic acids is generally

between 3 and 5.

12.5 25.0mL titrant solution added ---*

25 mL 0.10 M ECCO2Htitrated by 0.10 M NaOH

25 mL 0.10 M tLCCOzNatitrated by 0.10 M HCI

Figure I

Figure 2 shows the titration of methyl ammoniumchloride (CH3NH3CI) with strong base (NaOH) overlaid ontothe titration of methyl amine (CH3NH2) with strong acid(HCl). In both titrations, all species are in equal molarconcentrations. The pKu for amines is generally between 9and 11.

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12.5 25.0mL titrant solution added #

25 mL 0.10 M CF{3NH3CItitrated by 0.10 M NaOH

25 mL 0. l0 M CFhNH:titrated by 0.10 M HCI

Figure 2

Both a weak acid and weak conjugate base titration curr:of a conjugate pair show the same pH at the half-titratio:point (indicated by the empty circle on both graphs .

regardless of the initial concentration of the conjugal;species. The pH at the half-titration point in Figure 1 is les.

than the pH at the half-titration point in Figure 2. This r.because the pKu value of acetic acid is less than the pK,value of methyl ammonium cation. At this point, th:conjugate base concentration equals the acid concentratiortherefore, according to the Henderson-Hasselbalch equatioi-.pH = pKa.

30. By roughly how much do the two equivalents points :-the first graph differ?

Fewer than 2.0 pH unitsFewer than 4.0 pH units, but moreunlts

C. Fewer than 8.0 pH units, but moreunits

D. More than 8.0 pH units

31. The pH at equivalence is GREATEST for which ol ti..following titrations?

A. The titration of 0.10 M H3CCO2H by NaOHB. The titration of 0.10 M H3CCO2Na by HClC . The titration of 0.10 M CH3NH3CI by NaOHD. The titration of 0.10 M CHjNH2 by HCl

A.B. than 2.0 pH

than 4.0 pl-i

32. What is true in the titration of ammonia by

hydrochloric acid, when the pH of the solution is greaterthan the pKu for ammonium chloride?

A. [NH++] > [NH:]; Ka(ammonium chloride) > [H*]B. [NHa+] > [NH:]; Ka(ammonium chloride) < [H+]C. [NHa+] < [NH:]; Ka(ammonium chloride) > [H+]D. [NH++] < [NH:]; Ka(ammonium chloride) < [H*]

33. How does the pH at point a in Figure 1 compare to the

pH at point d in Figure l?

A. The pH at point a is more than 1.0 pH unit greater

than the pH at point d.

B. The pH at point a is greater than the pH at point d,

but the difference is less than one pH unit.The pH at point a is less than the pH at point d,

but the difference is less than one pH unit.The pH at point a is more than 1.0 pH unit lowerthan the pH at point d.

34. Given that H3CCO2H has a lower pKu value than

CH3NH3CI, which of the following statements is true?

A. H3CCO2H buffers at a higher pH value than

CH3NH3+.B. H3CCO2H has a conjugate base with a lower pK6

value than the conjugate base of CH3NH3+.

C. H3CCO2H dissociates less than CH3NH3+.D. H3CCO2H is a better electron pair acceptor than

CHrNH:+.

35. Given that H3CCO2H is a stronger acid than

CH3NH3+, which of the following statements is NOT

true?

A. H3CCO2H yields a lower pH value than

CH3NH3+ of equimolar concentration.

B. H3CCO2H has a conjugate base with a higher pKb

value than the conjugate base of CH3NH3+.C. H3CCO2H produces more conjugate base than

CH3NH3+ when added to water.

D. H3CCO2H is a worse proton donor than

CH3NH3+.

C.

D.

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36. If a similar experiment with identical concentrations and

volumes were conducted using hydrofluoric acid and

fluoride anion, what would be true, knowing that thepKu of hydrofluoric acid is lower than the pKu of acetic

acid?

A. The initial pH for the titration of HF would be

greater than the equivalence point for the titrationof H3CCO2Na.

B. When both HF and H3CCO2H are one-third titratedby equimolar NaOH, [F ] > [H:CCOZ-] and

Ka(acetic acid) < [H+1'

C . When both HF and H3CCO2H are one-third titrated

by equimolar NaOH, [HF] > [H3CCO2H] and

Ka(acetic acid) < [H+1'

D. When both HF and H3CCO2H are one-half titrated

by equimolar NaOH, Ka(acetic acid) < [H+] in the

HF titration.

Passage Vl (Questions 37 - 42)

Titration involves the quantitative addition of one reagentto another, where the concentration is known for only one ofthe species. Acids are often titrated by strong bases, so that

the concentration of the acid may be ascertained. Once the

equivalence point has been reached, the volume of titrant ismeasured. Using Equation 1, it is possible to solve for the

molarity of the acid, if the molarity of the titrant strong base

is known.

Mlacid)V(aci41 = M6ase;V6ase;

Equation 1

Besides the quantitative aspects, qualitative features ofthe acid may also be determined. The shape of the titrationcurve varies with the strength of the acid. For strong acids,

the shape is sigmoidal, with a nearly horizontal initialregion. For weak acids, the shape is not sigmoidal, with an

initial vertical ascent before leveling off into a horizontalbuffering region. As the degree ofthe initial ascent increases,

it can be observed that the acid being titrated is weaker. The

concentration of the acid also affects the titration curve. The

same fundamental shape is observed, but the pH values are

different.

A researcher conducts two experiments studying the

effect of acid concentration. In the first experiment, she

titrates the same strong acid, HCl, at three differentconcentrations, keeping the acid and titrant base in the same

concentration as each other. Figure I shows the threetitrations overlaid onto one graph. She finds that all three

titrations generate the same pH at their equivalence point,regardless of the initial concentration of strong acid.

t2.5 25.0

mL NaOH solution added --+

'' 25 mL 0.01 M HCI titrated by 0.01 M

25 mL 0.10 M HCl titrated by 0.10 M

-

25 mL 1.00 M HCI titrated by 1.00 M

Figure I

Copyright @ by The Berkeley Review@

NaOH

NaOH

NaOH

In the second experiment, the researcher titrates the same

weak acid, HOAc, at three different concentrations, keeping

the acid and titrant base in the same concentration as each

other. Figure 2 shows the three titrations overlaid onto one

graph. It is found that all three titrations generated the same

pH at their half-equivalence points.

12.5 25.0

mLNaOH solution added -----*

' 25 mL 0.01 M HOAc titrated by 0.01 M NaOH

25 mL 0.10 M HOAc titrated by 0.10 M NaOH

-

25 mL 1.00 M HOAc titrated by 1.00 M NaOH

Figure 2

Weak acid titration curves show the same equivalencrpoint, regardless of the initial concentration of weak acitiThe pH at the half-equivalence point in all three titratio:curves in Figure 2 is equal to the pKu for acetic acid. At th:half-equivalence point, the concentration of the conjugai:base equals the concentration ofthe acid; therefore, accordingto the Henderson-Hasselbalch equation, pH = pK2.

3 7. The initial pH is GREATEST in which titration?

A. The titration of 0.01 M H3CCO2H by NaOHB. The titration of 1.00 M H3CCO2H by NaOH

C . The titration of 0.01 M HCI by NaOHD. The titration of 1.00 M HCI by NaOH

3 8. What is true in the titration of acetic acid by sodiut:hydroxide when the pH of the solution is greater tha:the pKu for acetic acid?

A. [H3CCO2-] > [H3CCO2H]; Ka(acetic acidl > [H--B. [H3CCO2-]> [H3CCOzH]lKa(acetic acid) < [H-C. tH:CCO2-l < [H3CCO2H]iKa(acetic acid) > [H*-D. tH3CCO2-l < [H3CCO2H]i Ka(acetic acid) < [H-'

Titrations

326 GO ON TO THE NEXT PACE

39. The BEST explanation for the greater pH at the

equivalence point observed with the higher initial

concentration of weak acid can be attributed to:

A . the greater number of mL of base solution added to

reach the equivalence Point.the lower number of mL of base solution added to

reach the equivalence Point'the greater conjugate base concentration at the

equivalence Point.D. the lower coniugate base

equivalence Point.

B.

C.

concentration at the

4 0. How would the titration curves in Figure 1 be affected ifthe base concentrations were all doubled, while the acid

concentrations remained the same?

A. Equivalence would be achieved with half the

volume of titrant base, and the shape of the

titration curves would change.

B. Equivalence would be achieved with twice the

volume of titrant base' and the shape of the

titration curves would change.

Equivalence would be achieved with half the

volume of titrant base, and the shape of the

titration curves would not change.

Equivalence would be achieved with twice the

volume of titrant base, and the shape of the

titration curves would not change.

41. . Even though the NaOH concentration in the third trial

is 100 times greater than the NaOH concentration in the

first trial, the two graphs follow a similar slope' This

is BEST explained by which of the followingstatements?

A. The solution is a buffered solution, so the pH

change is minimal.B. The NaOH is a weak base and does not fully react

with the HCl.C. The pH is a log scale, so as the pH increases up to

7.0, the amount of base necessary to increase the

pH becomes less.

D. The pH changes only at the equivalence point'

C.

D.

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42. Which of the following ,eraphs represents the :esLrlts

that would be observed if the experiment described rr

the passage were carried out with ammonia (NH: t and

hydrochloric acid (HCIX

A.

A

I

pH

25 mL 0.10 M NH 1

- with 0. l0 M HCI

- 25 mL1.00MNH3

- 'rith 1.00 M HC1

12.5 25.QmL HCI solution added 4

25 mL 0.01 M NH iwith 0.01 M HCI

25 mL 0.10 M NH j

- with 0.10 M HCI

25 mL 1.00 M NH 3

with 1.00 M HCl

12.5 25.0mL HCI solution added +

25 mL 0.01 M NHwith 0.01 M HCI

25 mL 0.01 M NHwith 0.01 M HCl

D.

A

t

pH

25 mL 0.10 M NH 3

- with 0.10 M HCl

- 25mL1.00MNH3

- rvith 1.00 M HCI

25 mL 0.10 M NH 3

- with 0.10 M HCI

-25mL1.00MNH3- *ith 1.00 M HCI

12.5 25.0mL HCI solution added 4

25 mL 0.01 M NHwith 0.01 M HCI

t2.5 25.0mL HCl solution added =+

Passage Vll (Questions 43 - 49)

The four titrations curves shown in Figure 1, represent

the titration of three weak acids (HCN, HCIO, and HOAc)and the titration of the strong acid HCl. The equivalencepoint is represented by the dot at the vertical inflection pointof each curve. The respective conjugate base is shown nextto its equivalence point. At the start of each titration, 25 mLof 0.10 M acid are present. All of the acids are titrated by0.10 M KOH(aq).

12.5 25.0mL 0.10 M KOHaq) solution added +

Figure I

The initial and equivalence pH values were collected and

recorded for some of the titrations. Table 1 shows datacollected during the experiment.

Titration Initial pH Equivalence pH

HCl bv KOH 1.00 7.00

HOAc bv KOH 2.81 8.12

HCIO bv KOH 4.23

HCN bv KOH r 1.01

Table 1

The pKu for each weak acid can be found by measuring

the pH of the aqueous solution at the half-equivalence pointin its respective titration. For instance, the pKu for HCN is

9.32, which is the pH at the half-equivalence point of the

titration curve. The values are approximated, because the

points on the curve cannot be read that accurately. The pH

values are listed in Table I were recorded from a pH meter, so

they are considered to be reliable.

43. The greatest Ku value is found with which of the

following acids?

A. HCNB. HCIOC. HOAcD. HCI

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44. If pKu of HOAc is 4.14, the pH at the equivalence in

the titration of HCN is 11.01, and the initial pH in the

titration of HCIO is 4.23. then what else must be true?

A. Initial pH in the titration of HCN is 3.86; the pHat equivalence in the titration of HCIO is 9.61.

B. Initial pH in the titration of HCN is 5.16; the pHat equivalence in the titration of HCIO is 10.08.

C. Initial pH in the titration of HCN is 6.12; the pHat equivalence in the titration of HCIO is 11 .42.

D . Initial pH in the titration of HCN is 1.42; the pHat equivalence in the titration of HCIO is 10.34.

4 5. Which of the following statements are valid when

comparing the titration curve associated with a weak

acid by strong base titration to the titration curveassociated with a strong acid by strong base titration?

I. Strong acid titration curves have an initial drop inpH due to the dissociation of the protons, whileweak acid titration curves start with a plateau.

II. Weak acid titration curves have a buffer regionwhile strong acid titration curves do not.

m. All points before the equivalence point are less than

7 for both the titration of and weak acid and the

titration of a strong acid.

A. I onlyB. II onlyC . I and II onlyD. II and III only

Given that 0.10 M HIO has a pH greater than that o,

0.10 M HCN, we can conclude that:

A . the pKn of HIO is 8.61.

B. the pH at equivalence in the titration of HIO b',

KOH is 10.06.

C . the pH of 0.10 M HIO(aq) is 1.21.D . the difference between the pKu of HIO and the pH

at the equivalence point in the titration of 0.10 l'lHIO(aq) by 0.10 M KOH(aq) is less than 3 p!-'

units,

47. When 10 mL 0.10 M HCIO is mixed with 10 mL t-r -:M KCIO, the pH is 7.64. What is the pH after 30 :---

of water is added to raise the volume to 50 mL?

A. 7.51B. "/.64

c . 1.11D. 8.26

46.

48. Which of the following curves accurately represents

the titrations of 0.10 M NaCN(aq) by 0.10 M HCI(aq)

and 0.10 M NaCIO(aq) by 0.10 M HCl(aq)?

A.CN.

cl0

pH

1

mL 0.10 M HCI(aq) added 25.0

mL 0.10 M HCl(aq) added

c.CN-

cl0

pH

mL 0.10 M HCI(aq) added 25.0

D.cloCN-

pH

'7

mL 0.10 M HCI(aq) added

25.0

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25.0

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4 9. Which of these sequences relates the pK2 values of tft" -fKindicated acids in descending order?

A. PKa(ucN; > PKa@Clo; > PKa(HoAc) > PKa(HCl)B. PKa(uoac) > PKa(HClo) > PKa(HC$ > PKa(HCl)C. PKa(uctl >PKa@CN; > PKa@Clo) > pKa@oAc)

D. PKa(Hct; > PKa@oAc) > PKa(HClo) > PKa(HCN)

Passage Vlll (Questions 50 - 56)

A polyprotic acid is an acid that contains more than oneacidic proton. The second proton lost by the acid is never as

strongly acidic as the first proton lost by the acid. Thesecond proton can be removed using strong base, once thefirst proton has been completely removed from the acid.Concentrations of polyprotic acids are measured in terms ofnormality. Normality measures molar equivalence. Thenormality of an acid equals the molarity of the equivalentbase required to neutralize all of the acidic protons.

A researcher sets out to determine the effect of mixingsolutions together, by studying their pH before and aftermixing. Into a flask (Flask 1) he places exactly 25.0 mL of0.20 N H2SOa to be titrated by a solurion of NaOH ofunknown concentration. Into a second flask (Flask 2) heplaces exactly 40.0 mL of 0.30 N H3POa, also to be titratedby the same solution of NaOH. Both solutions are titrated insuccession to a visual endpoint, determined by the colorchange of an indicator. The quantity of base needed toachieve this is recorded accurately to the second decimal place.

50. If Flask I requires exactly 20.0 mL NaOH solution tobe neutralized, what must be the concentration of theNaOH solution?

A. 0.125 M NaOH(aq)B. 0.200 M NaOH(aq)C. 0.250 M NaOH(aq)D. 0.500 M NaOH(aq)

51 . Which of the following statements is true?

A . pKat is always larger than pKu2.B . pKaZ is always larger than pKu1.C . There is no rule for pKu2 or pKu1.D. pKaZ is greater than pKul only for the oxyacids.

52. A solution of 0.30 M H3POa(aq) has which of thefollowing values for normality'/

A. 0.10 N H3POa(aq)B. 0.30 N H3PO4(aq)C. 0.60 N H3PO4(aq)D. 0.90 N H3POa(aq)

53. When 0.1 moles NaH2PO4 and 0.2 moles Na2HPO4are mixed in 100 mL, what is the pH of the solution?

A. pH . PKal + PKa22

B. pKaZ > pH > PKul + PKu22

g. pKa2 + pKa3 > pH > pKa2

2

D. pH t PKa2 + PKa32

5 4. How many mL of 0.40 M H2SO4(aq) would require thesame amount of base to reach full neutralization aswould 25 mL 0.60 M H3POa(aq)?

A. 25.00 mL 0.40 M H2SOa(aq)B. 37.50 mL 0.40 M H2SOa(aq)C. 50.00 mL 0.40 M H2SO4(aq)D . 56.25 mL 0.40 M H2SOa(aq)

5 5. Which titration curve represents the complete titrationofphosphoric acid?

A.

mL strong base added

5 6. What is the phosphate concentration in Flask 2 after 4imL of 0.30 N NaOH has been added?

A. 0.050 M PO+3-(aq)

B. 0.100MPO+3-(aq)C. 0.450 M PO43-(aq)

D. 0.900 M PO+3-(aq)

pH

B.

pH

C.

pH

D.

pH

mL strong base added

mL strong base added

mL strong base added

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Passage lX (Questions 57 - 63)

Polyprotic acids are acids that contain more than oneacidic hydrogen. A typical example is sulfuric acid (H2SOa).

The first proton is easily removed, while the second proton isharder to remove. Sulfuric acid is a strong acid with respectto its first proton, but weak with respect to its second proton.The titration curve for a polyprotic acid looks like twoseparate titration curves that have been connected. Theendpoint of the curve for the first proton is the starting pointof the curve for the second proton. A student titrates anunknown diprotic acid with strong base. The titration curvefor the experiment is shown in Figure 1.

mL Tinant added

Figure 1

The exact pH and exact volume of titrant used are notknown, but the graph is proportional throughout the durationof the titration. The student repeats the experiment threetimes, and the graph in Figure I represents the best results ofthe three trials.

5 7. The unknown acid can BEST be categorized as which ofthe following?

A. A strong monoprotic acidB. A diprotic acid with both protons strongC. A diprotic acid with one strong proton and one

weak protonD . A diprotic acid with both protons weak

5 8. Which of the following relationships is NOT true aboutthe unknown acid?

A. pKat <pKa2B. pH at first equivalence point > pKalC . pH at second equivalence point > pKn2D. pH at first equivalence point > pKu2

59. In Figure 1, the titrant can best be described as a:

A. strong acid.B. strong base.

C . weak acid.

D. weak base.

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6 0. Which change requires that the MOST titrant be addedto the solution?

A . Going from a pH < pKat to a pH > pKa2B. Going from the first equivalence point to the

second equivalence pointC . Going from a pH > pKat to a pH <pKa2D. Going from the initial point to the first equivalence

point

61 . Il after the addition of 5 mL of titrant, the pH of thesolution is less than the pKul of a diprotic acid, whatvolume of titrant is required to reach the secondequivalence point from the fully protonated state?

A . Between 5 and 10 mLB. Between 10 and 15 mLC . Between 15 and 20 mLD. More than 20 mL

6 2. What is true the predominant species and any otherspecies present between points d and e on the titrationctrrve?

A. A2- (the fully deprotonated form) is all that ispresent.

B. A2- is present along with some HA- (the partiallyprotonated form).

C. HA- (the partially protonated form) is all that ispresent.

D. HA- is present along with some H2A (the fullyprotonated form).

6 3. Which of these points in Figure 1 is NOT describedcorrectly below?

A. Point b is where [H2A] = [HA-].B. Point c is the first equivalence point.C. Point d is where pH = pK62.

D. Point f is where tHA-l = ;62-1.

Passage X (Questions 64 - 70)

Polyequivalent bases are bases that can neutralize morethan one acidic hydrogen per base molecule. A typicalexample of a diequivalent base is carbonate (CO32-;. thefirst proton added corresponds to the second proton removed.Carbonate is a stronger base than its conjugate acidbicarbonate (HCO3-), although both are considered weakbases, because the pK6 values are 3.61 and 7.63 respectively.The titration curve in Figure I is obtained when calciumcarbonate (CaCO3) is titrated with a strong acid, such as

0.5 1.0 1.5 2.O

Equivalents strong acid added

Figure 1

The exact pH and exact volume of titrant used in thistitration are not provided, but the graph is proportionalthroughout its duration. The nine points marked along thetitration curve note some of the key transitional points.Despite the fact that the pH is recorded for the titration of thebase (rather than the pOH), the titration curve is stillpredictable.

Among the key points recorded are the two equivalencepoints, and the two points at which the pH = pKu (both thefirst and second pKu points are marked). It is interesting tonote that when pH = pKat, pOH = pK62. The other pointsof the curve are interesting in that they represent differentpoints at which the pH is predictable from the trend in the pHand the amount of acid added.

64. At which point is the pH equal

A. Point cB. Point dC. Point fD. Point h

to the value of pK61?

6 5. Which of the following relationships is NOT true?

A. pKat + pK62 = l{B. pKaZ + pK62 -- 14

C. pHinitiuly > pKa2D . PHut 2nd equivalence < PKa l

hydrochloric acid.

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6 6. How can the pH of solution be determined at point d?

A. pH = 7.0

B. pH - PKul + PKu22

c. pH = ^4pK.rP

* (pKrfD. pH =2(pKat-pKa2)

6 7. What is NOT true at point e on the titration curve?

A . Fewer than 1.5 equivalents of HCI have been added.

B. The pH is less than 6.37.

C . The pH is greater than pKul .

D. The pH is less than pK22.

6 8. In what pH range is the concentration of bicarbonate theGREATEST?

A. pH less than 3.67B. pH between 3.61 and7.00C. pH between pH 7.00 and 10.33D. pH greater than 10.33

6 9 . Between which two points does the pH of the solutionchange by the GREATEST amount?

A . Between point c and point dB. Between point d and point e

C . Between point e and point fD . Between point f and point g

7 0. Which of the following graphs shows changing pOH a.a function of equivalents strong acid when calciui:.carbonate is titrated by hydrochloric acid?

Equivalents strong acid Equivalents strong acii

Equivalents strong acid Equivalents strong a.-

Passage Xl (Questions 71 - 76)

When titrating an acid with a strong base, an indicator is

added to the solution to indicate when the equivalence point

has been reached. An indicator works by changing color at or

near the equivalence point ofthe titration. Because indicators

can lose or gain a proton, they are susceptible to the effects

ofvarying solution pH during titration. A conjugate acid and

its conjugate base have different colors (although in some

cases, one of the two is colorless.) With most indicators,

both the conjugate acid and conjugate base are colored

species. When both species are colored, it is often harder to

see the color change, unless there is a pronounced difference

between the two colors. An example with the hypothetical

indicator HQ, and its dissociation to conjugate base Q-, is

shown in Reaction 1. Assume HQ is colorless and Q- is

orange.

Reaction 1

When the conjugate base (Q-) is the predominant species

in solution, it appears orange. When the conjugate acid (HQ)

is the predominant species in solution, the solution appears

clear. When the pH of the solution is equal to the pKs of the

indicator, there are equal parts of HQ and Q- in solution, so

the solution has an orange hue' Indicator are used in low

concentration, so they don't become visibly detectable until

at least one-tenth of the indicator is in its deprotonated

(orange conjugate base) form. This occurs when the solutionpH iJ approximately one pH unit below the pKa of the

indicator. Equation 1 lists the active range of an indicator.

PH(at equivalence) = PKa(indicator) + 1'

Equation 1

Because it is desirable to have the color change near the

equivalence point of the species being titrated, Equation 1

helps when selecting an indicator. If the equivalence pH is

not known, then the pH at equivalence can be approximated

as pKu1u.i6; + 3 (where the pKu is for the acid being titrated)'

This means that often the pKu of the indicator is about three

pH units higher than the pKu of the acid being titrated. This

holds true only if the pK' for the acid is between 4 and 10' It

is not affected by concentration factors. Figure 1 shows a

titration curve for the titration of a weak acid by strong base

and the active range of an indicator:

HQ(aq) -.-5- H:O+(aq)

Clear

+ Q (aq)

Orange

Equiva

pH

t\

c:ao

rE *l8;J€.2'd

mL titrant added

Figure 1

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The change in shading represents the increase in th9\-intensity ofthe orange color ofthe solution. The equivalence

occurs in the middle of the color change band'

7 L. The BEST indicator to use for the titration of an acid

with a pKu of 5.0 would be one having a pKa of:

A . 1.5.B. 5.0.c. 8.5.D. 11.5.

72, An indicator in a concentration that is too high could

have what effect on a solution?

A. It could interfere with the acid and/or base

properties of the solution being titrated.

B. It could change the viscosity of the solution being

titrated, resulting in a non-homogeneous solution'

C . The color change could be too extreme to be useful'

D. The color change could be too subtle to detect.

7 3. What is the ratio of the conjugate base to conjugate acid

at two pH units above the PKu?

A. 100: 1

B. 2:1C. l:2D. 1:100

7 4. Agood indicator has which of the following properties?

A. It should be transparent throughout the titration'

B. It should be unreactive with respect to acid-base

reactions.

C . It should be a strong acid or a strong base.

D . It should be a weak acid or a weak base.

7 5. For which of these titrations is the pH at equivalence

the same, regardless of concentration?

A . The titration of a weak acid by a strong base

B. The titration of a strong acid by a weak base

C . The titration of the first proton of a weak diprotic

acid by a strong base

D. The titration of the second proton of a weak

diprotic acid bY a strong base

7 6. Which BEST explains why the indicator pKu may be

one unit off from the equivalence pH?

A. A difference of one pH unit has an insignificant

effect on color at the pH of the indicator's pKa'

B. The pH changes rapidly at equivalence'

C . At the equivalence point for the acid, pH = PKa'

D. An indicator's color change occurs at only one very

specific PH value.

point

333 GO ON TO THE NEXT PAGE

Passage Xll (Questions 77 - 83)

Indicators have two main purposes in chemistry. Thefirst use is as an aid in determining the equivalence point in atitration experiment. If the pH at equivalence is known, thenan indicator that changes color at or near that pH value can beused to detect the equivalence point. The ideal indicator hasits pKu value within + I of the pH at equivalence.

The second use of an indicator is to approximate the pHof an unknown aqueous solution by examining the color ofthe indicator in a sample of the solution. The indicatorchanges color at a pH value approximately one unit from itspKu value, so if the pH of solution is within one pH unit ofthe indicator's pKs value, then the pH can be estimated.Table 1 shows the relationship between pH and the colorassociated with three indicators.

pH Indicator I Indicator II Indicator IIII Red Yellow Purole2 Red Yellow Purple3 Red Yellow Violet4 Red Yellow Fuchsia5 Red Yellow Clear6 Red-oranse Yellow Clearl Orange Yellow Clear8 Manso Yellow Clear9 Yellow Chartreuse Clear

10 Yellow Green Clearll Yellow Aouamarine Cleart2 Yellow Blue Clear13 Yellow Blue Clear

Table IFor the titration of a strong acid by strong base, the pH

at the equivalence point is 7.0, while for the titration of aweak acid by a strong base, the pH at equivalence is greaterthan 7.0. Because the pH at equivalence is different, adifferent indicator is required for the two titrations, althoughfor a weak acid with a pKu of 3 or less, the indicator usedlnthe titration of a strong acid may work.

The equilibrium distribution of the deprotonated andprotonated forms of an indicator obeys Equation 1, where Ind-is the deprotonated form and H-Ind is the protonated form.

PHsolution = PKa(indicalorl + log llLH-tnd

Equation 1

7 7. Which indicator could be used to determine theequivalence point in the titration of 1.0 M benzoic acidby 1.0 M KOH(aq)?

The pKu of H5C6COOH is 4.21.

A. Indicator II onlyB. Indicator III onlyC. Both Indicatorl and IndicatorllD . Both Indicator II and Indicator trI

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78. Which combination of colors is NOT possible for asolution?

A. I: red II: yellow III: purpleB. I: mango II: yellow III: violetC . I: red II: yellow IfI: fuchsiaD. I: yellow II: aquamarine III: clear

7 9 . If the ratio of blue species to clear species within anindicator equilibrium is 1000:l at pH = 6, what is thepK2 of the indicator, given that the deprotonated formabsorbs visible lighr?

A. 3

B. 4c. 9

D. 10

8 0. What is the pH of a solution that is clear when IndicatorIII is added, faint chartreuse when Indicator II is added.and faint mango when Indicator I is added?

A. 7.0B.'7.5c. 8.0D. 8.5

81 . Which of the following accurately describes the pK,values for the three indicators?

A. I: 6.86B. I: 8.94C . I:1.21D. I: 8.34

82. The active rangefollowing?

A.pH2topH6B.pH3topH4C.pH3topH5D.pH4topH6

II: 9.87II:7.21II: 8.28II:9.66

III:4.22III: 5.98III:4.11III: 3.85

for Indicator III is which of the

83. The best description of the absorbance of lightassociated with Indicator III is:

A. The protonated form absorbs light with a L6u* of426 nm, while the deprotonated form absorbs lightwith a )"*u,. of 339 nm.

B. The protonated form absorbs light with a l"pa* of339 nm, while the deprotonated form absorbs lightwith a 1,.6" of 426 nm.

C. The protonated form absorbs light with a )"p.x of598 nm, while the deprotonated form absorbs lightwith a l,-a* of 339 nm.

D. The protonated form absorbs light with a L.11ax of339 nm, while the deprotonated form absorbs lightwith a l"*u" of 598 nm.

Passage Xlll (Questions 84 - 89)

In titration, it is common to use indicators to signify a

certain pH for the solution. Indicators are made, mostcommonly, from organic dyes that gradually change colorwithin a given pH range. A color change is observed,because the indicator in its protonated state is one color and

in its deprotonated state is a different color. Reaction 1

shows the equilibrium for a generic indicator.

Hlnd(aq) + H2O(l) s- H3O+(aq) + Ind-(aq)

Reaction 1

Table 1 below lists a series of indicators, the pKu of the

indicator, the active range for visual detection, and the

respective colors of the protonated form (acid) and

deprotonated form (base). The Henderson-Hasselbalchequation describes the relationship between the members of aconjugate pair. When the concentrations of the acid and baseare equal, the pH equals the pKu. When this occurs, the colorof the solution is an average of the colors listed in Table l.

Table 1

Depending on the concentrations and species beingtitrated, there is one ideal indicator. To detect the equivalencepoint for a titration, a small portion of indicator is added to

solution. Ideally, the equivalence point of the titrationshould be equal to the pK6 of the indicator. At the very least,

the equivalence pH must fall within the indicator's activerange. When a strong acid or strong base is titrated with a

strong titrant, the pH at the equivalence point is always equalto 7.0. For a weak acid titration, the pH at equivalence can

be estimated by taking an average of the pKa of the acidbeing titrated and the pH of the titrant base.

84. Which of the following does NOT form a bluesolution?

A. Bromphenol blue in a pH = 7.0 bufferB. Bromcresol green in an aqueous ammonia solutionC . Thymol blue in an acetic acid solutionD. Bromthymol blue in a hydroxide solution

fndicator pKa pH range Acid Base

Alizarin Yellow 10.8 10.0 - 1r.6 Yellow Red

Phenolphthalein 8.8 8.0 - 9.6 Clear Pink

Thymol Blue 8.4 7.6 - 9.2 Yellow Blue

Cresol Red 8.0 1.2 - 8.8 Yellow RdBromthvmol Blue 6.8 6.0 - 1.6 Yellow Blue

Chlorophenol Blue 5.6 4.8 - 6.4 Yellow Red

Bromcresol Green 4.4 3.1 - 5.1 Yellow Blue

Methvl Oranqe 3.9 3.2 - 4.6 Orange Yellow

Bromphenol Blue -1- I 3.0 - 4.4 Yellow Blue

Ervthrosin B 2.8 2.2 - 3.6 Orange Yellow

Thymol Blue 2.0 t.2 - 2.8 Red Yellow

Cresol Red t.6 r.0 - 2.2 Red Yellow

Methvl Violet 0.8 0.0 - 1.6 Yellow Violet

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8 5 . Which of the following indicators CANNOT be used forthe titration of ammonia by hydrochloric acid?

A. PhenolphthaleinB. Bromcresol green

C. Methyl orangeD. Bromphenol blue

86. If KCN has pKb = 4.68 and KF has PKu = 10.83,

which indicator is NOT the ideal choice for thefollowing proposed titrations?

A. 1.00 M KF(aq) titrated by 1.00 M HCl(aq) with an

indicator of cresol red

B. 0.01 M KF(aq) by 0.01 M HCI(aq) with an

indicator of methyl violetC. 1.00 M KCN(aq) by 1.00 M HCI(aq) with an

indicator of bromcresol green

D. 0.01 M KCN(aq) by 0.01 M HCI(aq) with an

indicator of chlorophenol blue

8 7. Which of the following indicator : color correlations is

NOT correct for biological pH (pH =1.4)?

A. Methyl violet : VioletB. Methyl orange : YellowC. Thymol blue : BlueD. Bromcresol green : Blue

8 8. Which of the following titrations requires thymol blueindicator?

A. A low concentration of strong base titrated by astrong acid

B. A low concentration of weak base titrated by a

sfrong acidC. A highly concentrated strong

strong acidD. A highly concentrated weak

strong acid

8 9. Which of the following indicators should be chosen toidentify the equivalence point of a strong acid titrated bya strong base?

A. Methyl violetB. Methyl orangeC. Bromthymol blueD. Alizarin yellow

base titrated by a

base titrated by a

Passage XIV (Questions 90 - 96)

A pH stick is a device that can approximate the pH of anaqueous solution by reference to a color blend band. It is aplastic stick with indicators attached to specific segments ofthe stick. Each indicator has a range of color change thatextends to either side of a central pKu value. When the pH ofthe solution is less than the pKu of the indicator, theindicator exists predominantly in its protonated state. Whenthe pH of the solution is greater than the pK" of theindicator, the indicator exists predominantly in itsdeprotonated state. An indicator is ideal when both theprotonated and deprotonated species are colored. When thespecies are primary colors, the change is easier to observe.

Typical pH sticks come with three to four indicators,such as those listed in Table 1. Each solution must be testedwith a separate pH stick. A pH stick cannot be used for morethan one test, because the acidity of the first test solutionaffects other solutions that are added to the pH stick.

Indicator PKaProtonated

ColorDeprotonated

ColorBromcresol Green 4.31 Yellow Blue

Methyl Red 5.21 Red YellowBromthymol Blue 6.18 Yellow Blue

Phenolphthalein 8.',79 Clear Magenta

Table IA student uses a pH stick with four indicator markers to

approximate the pH of five separate solutions. Table 2 liststhe results for the five separate solutions. Each sequence ofcolors represents what is observed for the four indicators inorder of increasing pKu.

Table 2

An indicator is used to detect the endpoint in a titration.An ideal indicator for a titration has its pKu equal to theequivalence pH of the titration. This is the ideal condition,but in practice, the exact pH at equivalence often cannot bedetermined. The general rule for titration is that the pH atequivalence should be within one of the pKo of the indicator.

9 0. What is the pH range of the pH stick?

A. 3.4 ro 8.8B. 3.4 to 9.8C. 4.4 to 8.8D. 4.4 ro 9.8

91 . It is NOT possible to estimate pH for:

A. Solution 1 only.

B. Solution I and Solution 2 only.C . Solution 2 and Solution 3 only.D. Solution 1 and Solution 4 only.

92. An aqueous solution that has a hydroxide concentrationof 1.0 x 10-6 M would show what colors?

A. Yellow with bromcresol green

B. Red with methyl red

C. Blue with bromthymol blueD. Clear with phenolphthalein

9 3 . What is the approximate pH of Solution 5?

A. 4

B. 5

c. 6

D.1

9 4. When HCI is added to Solution 2, the colors on the pHstick do not change. How can this be explained?

A. The hydronium concentration is too high for thepH to be affected by the addition of HCl.

B. The hydronium concentration is too low for the pHto be affected by the addition of HCl.

C. The solution is a buffer made from a carboxylicacid and its carboxylate conjugate base.

D . The solution is a buffer made from an amine and itsammonium conjugate acid.

9 5. Which indicator could be used in the titration of a weakacid by a strong base?

A. Bromcresol green

B. Methyl red

C. Bromthymol blueD. Phenolphthalein

9 6. Which of the following statements is INVALID?

A. A solution that turns bromthymol blue to bluewould turn Phenolphthalein to magenta.

B. A solution cannot show two green marks on thepH stick.

C. A solution that turns methyl red to yellow wouldturn bromcresol green to blue.

D. The pH stick can estimate pH best when that valuefalls between 4.21 and 5.31.

Solution BromcresolGreen

MethylRed

BromthymolBlue

Phenol-phthalein

I Yellow Red Yellow Clear

2 BIue Orange Yellow Clear

3 Blue Yellow Greenish-blue

Clear

4 Blue Yellow Blue Magenta

5Greenish-

blueReddish-orange

Yellow Clear

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Passage XV (Questions 97 - 100)

A common class of acids in organic chemistry is the

carboxylic acids. Their acidity is attributable to the electron-

withdrawing nature of the carbonyl group through resonance'

Short-chain carboxylic acids are water-soluble. As the alkyl

chain length increases, the hydrophilicity decreases, making

carboxylic acids of five carbons or more rather insoluble'

Long-chain fatty acids are often used as surfactants because of

their insolubility.

Phenols constitute another common class of acids in

organic chemistry. Their acidity is attributable to the

electron-withdrawing nature of the benzene ring through

resonance. Because aromatic rings are less electron-

withdrawing than carbonyl groups, phenols are weaker acids

than their carboxylic acid counterparts. Most phenols are

insoluble in water in their protonated state. Table 1 lists a

few examples of carboxylic acids and phenols, along withtheir pKu values.

Table 1

Because of solubility constraints, carboxylic acids and

amines are more often involved in aqueous buffering than

phenols. However, it is possible to titrate phenols, because

their conjugate base is water-soluble, allowing for the

reaction to be monitored.

9 7. Which of these mixtures produces the MOST effective

buffer for pH 4.0?

A. H3CCO2H + H3CCO2NaB. CI3CCO2H + CI3CCO2NaC. O2NC6HaCO2H + O2NC6H4CO2NaD. C6H5CO2H + C6H5CO2Na

-Pe

0.64

Organic Acid

o

.r..Aot3.40

4.21.

7.18

10.01

Copyright O by The Berkeley Review@ cct THAT'S ENOUGH CHEM FOR NOW.

9 8. For the titration of p-nitrophenol, which of the

following reagents would be best?

A. KOHB. Potassiump-nitrophenoxideC. PhenolD. HCl

9 9 . What is the pH of a solution made by mixing 20 mL

0.10 M phenol with 10 mL 0.10 M KOH(aq)?

A. 10.0B. 1.0c. 5.8D. 5.5

1 0 0. Which of the following graphs represents the titration

of 50 mL 0.1 M p-nitrophenol by 0.20 M KOH(aq)?

25mL 0.20 M KOt{aq)

25mL 0.20 M KOHaq)

50mL 0.20 M KOF{aq)

50mL 0.2014 KOHaq)

1.B 2.A 3.86.D ',1.8 8.D

11. C 12. B 13. A16. B r1. C 18. C21. C 22, C 23, C26. C 27. D 28. A31. C 32. C 33. C36. D 31. A. 38. A41. c 42. D 43. D46. D 4-1. B 48. A5r. B 52. D 53. C56. A 51. D 58. D61. D 62. B 63. D66. B 61. B 68. C'n. c 12. A 73. A76. B 77. A 78. B81. A 82. C 83. C86. B 87. C 88. D

91. D 92. C 93. B96. A 97. D 98. A

4.8 5. 89.D 10.A

14. B 15. C19. A 20. D24. D 25. D29. B 30. c34. D 35. D39. C 40. c44. B 45. B49. A 50. c54. D 55. B59. B 60. A64. C 65. B69. A 70. A14. D 75. C'79. A 80. D84. C 85. A89. C 90. B94. C 95. D99. A 100. A

Buffers and Titration Passage Answers

1- Choice B is correct. The mixture is composed of NaHCO3 and H2CO3. Without considering how much of eachcomponent is present in solution, recognize first that they are a conjugate pair, meaning tiat the solution is abuffer. This means that the pH is close to the pKu. Carbonic acid is dlp.-toti", so be sure you understand that thefirst proton is involved in this conjugate paia and that the pH shouli be close to pKu1. This makes choice Bthe best candidate. If you wish to solve for the exact value, the first step is to convert 0.839 grams NaHCO3(s)into moles: (0.839 grams)(-l rnale-) = 0.01 moles NaHCo3. The number of moles of H2Co3 = (0.10 L)(0.10 M) =84 grams0.01 moles H2CO3. The pH can be found using the Henderson-Hasselbalch equation:

pH = pKu*logMol"tconjugate pH=pKal *to* u91g!I{!5 =6.4+logll =6.4+logr=6.4M.t"r .-rj"g"t" *l.t I r --or '-o Moles H2CO3 - .01

Since the log of 1 is 0, the pH of the solution is equal to pKu1, 6.4. The best answer is choice B.

Choice A is correct. Even though the solution is a buffer, the addition of HCI decreases the pH slightly. Abuffer resists extreme pH changes, but a small change is often observed. The addition of an acid to the bufferedsolution lowers the pH. Since the pH is initially 7.27, the final pH value must be lower than 7.27. The onlyanswer choice less than 7.21 is choice A,7.1,4.

Choice B is correct. To make a buffer, a weak acid and its conjugate base must be mixed. Benzoic acid is a weakacid, so it must be mixed with benzoate, its conjugate base. The desired pH is equal to the pKu of benzoic acid,so according to the Henderson-Hasselbalch equation, equal parts of benioate an-d benzoic acid must be mixed.This can be accomplished either by adding one-half equivaient (in terms of moles) of strong base (NaOH) toconvert half of the benzoic acid into benzoate, or by adding an equivalent amount (in terms of motes) of benzoateto the benzoic acid solution. Choice A has the conjugate pair uaaed together in an equal gram ratio, not equalmole ratio. This does generate equal mole portions, so the pH is not eqrial to the pKu. This etiminates choice A.The remaining three dhoices involve the mixture of benzoic acid with i strong base, so they must be mixed in amanner that half-titrates the benzoic acid. The only answer with half as much strong base as weak acid(benzoic acid) is choice B. This mixture yields equal molar portions of the two componentJof the conjugate pair(benzoic acid and benzoate), so the best answer is choice B.

Choice B is correct. The concentration of the titrant strong base (0.2 M NaOH) is twice that of the weak acidHF (0.1 M), so to reach equivalence, only half the volume of strong base is required. There are initially 50 mLof 0.1 M HF present, so only 25 mL of 0.2 M NaOH are required to reach the equivalence point. The target pHfor the solution is 3.3, which happens to be the pKu of uf. fnis means that the pH of ,ol.rtion is equal io pkuof HF, which is true when [HF] = [F ]. This occurs when the HF(aq) is half-titrat"i. rf 25 mL 0.2 M lriaOHlaqy isrequired for full titration, then 12.5 mL is required for half titration. Choice B is the best answer.

Choice B is correct. This question takes more than the usual amount of effort to answer. There are two factors toconsider: dilution and reactivity. Addition of sodium hydroxide solution converts some of the acetic acid toacetate, and it dilutes the solution. After the completion of the reaction, 3.0 mmole of H3CCO2- are present in30 mL of aqueous solution. From here on, it's strictly a matter of calculation, beginning with 3 mmqle = 0.1 M

30mLChoice B is the best answer. The reaction chart below shows how the moles were detir-mined.

,

J.

4.

5.

HOAc6mmols-3 mmols3mmols

+ NaOH3 mrnols-3 mmols

0

HzoXXX

XXX

XXX

+ H3CCO2-0

+3mmols3mmols

6. Choice D is correct. A buffer is prepared by mixing (in this case) a weak acid with a half-molar equivalent o{strong base. This is also referred to as half-titrating a weak acid. Choice A is eliminated, because twice asmuch acid as weak base has been added. Choice B is eliminated, because both species are acids. Choice C iseliminated, because both species are bases. In choice D, half of an equivalent of ,iror-,g base is added to a weakacid. This results in a buffer, so the best answer is therefore answer choice D.

Copyright @ by The Berkeley Review@ 334 Section V Detailed Explanations

8.

7. Choice B is correct. The solution is acidic, so the pH is less than 7.0. A11 of the answer choices are less than7,so nothing is eliminated. For a weak acid with pKu between 2 and 12 in an aqueous solution where [HA];,i11u1 isgreater than Ku, use the shortcut equation to determine the pH. The pKu is 5.0, and [HA] is 0.1 M.

pH =|nKa -llog tHAl =l1s.o; -t./."r(0.10) = 2.5 -L..r) =2.5 + 0.5 = 3.0

The pH is 3.0, so the best answer is choice B.

Choice D is correct. A buffer results when a solution contains roughly equal concentrations of weak acid and itsconjugate base. This can be achieved by mixing the components of the conjugate pair in a roughly one-to-oneratio, as is observed in choices A and B. In choice C, the acid is diprotic, so the 1.5 equivalents of strong basecompletely remove the first proton (to form HCO3-) and then pull off a second proton from half of thebicarbonate ions. The result is a solution with equal parts HCO3- and COr2-. Because these are a conjugatepair, the solution forms a buffer. This eliminates choice C. In choice D, methyl amine is capable of gainingonly one proton, so the 1.5 equivalents of HCI completely converts the weak base (H3CNH2) into its conjugateacid (H3CNH3+), with a leftover of 0.5 equivalents of HCl. The solution is a mixture of weak acid and strongacid, which does not result in a buffer. The best answer is thus choice D.

Choice D is correct. Because the solution contains more weak acid than conjugate base, the pH is less than thepKu of the acid. This eliminates choices A and B. According to Equation 1, it is a log relationship, so the bestanswer is choice D. Plugging into Equation 1 would yield pH = pKa + log 7 / 2 = pKu - log 2.

Choice A is correct. The pH can be regulated (maintained at a relatively constant value) by a buffer. A bufferis composed of a weak acid and its conjugate base. Because the pH of the buffer must be close to (within one unitof) the pKu of the acid, the pH becomes harder to regulate as the weak acid becomes stronger. To regulate thepH at 1.0 requires an acid with a pKu no greater than 2.0. An acid with a pKu of 2.0 or less would be one of thestrongest weak acids. Increasing acid strength diminishes its ability to buffer. You should note that Table 1

does not contain any acids that could be used to buffer at 1.0. The best answer is thus choice A.

Choice C is correct. A buffer serves to maintain relatively constant pH. Choice A is eliminated, because anaqueous solution where the hydronium and hydroxide concentrations are equal has a pH of T.0,notnecessarily a

constant pH. A buffer solution is one in which the ratio of the acid and base in the conjugate pair never exceeds10 : 1, but that does not address the hydronium-to-hydroxide ratio. Consider a pH = 4.0buffer, for instance. AtpH = 4.0, the hydronium concentration is 10-4 M, and the hydroxide concentration is 10-10 M. This results in aratio of 1,000,000 : 1. This is far greater than 10:1, so choice B is eliminated. Choice C is the best, because arelatively constant pH implies that the hydronium concentration is relatively constant. If hydroniumconcentration is relatively constant, so is hydroxide concentration. This makes choice C the best answer.Choice D is eliminated, because the pH should be within one unit of the pKu value for the weak acid, not thehydronium-to-hydroxide ratio. The best answer is thus choice C.

Choice B is correct. The solution before adding the hydroxide has a 15 : 10 mole ratio of weak acid to conjugatebase. Addition of 1.00 mL of KOH converts some of the weak acid into conjugate base so that the ratio of 15 : 10

becomes 14 ; 77, still in favor of the weak acid. Because the weak acid is slightly more concentrated than itsconjugate base, the pH of the solution is slightly lower than the pKu. The pKu is 4.74, so the best answer ischoice B.

Choice A is correct. The largest ratio of conjugate base to weak acid is found in the solution with a pH greaterthan the pKu of the weak acid by the largest amount. Formic acid (HCO2H) has a pKa of 3.64. In choice A,when the solution pH is 4.00, the pH is 0.36 greater than the pKu. Hypochlorous acid (HCIO) has a pKu of7.49. In choice B, when the solution pH is 7.00, the pH is 0.49less than the pKu. There is more yeak acid thanconjugate base, which eliminates choice B. Hypobromous acid (HBrO) has a pKu of 8.67. In choice C, when thesolution pH is 8.50, the pH is 0.77 less than the pKu. There is more weak acid than conjugate base, whicheliminates choice C. Ammonium ion (NH+*) has a pKu o19.26. In choice D, when the solution pH is 9.50, thepH is 0.24 greater than the pKu. 0.36 is greater than 0.24, so the best answer is choice A.

11.

9.

10.

t2-

13.

Copyright @ by The Berkeley Review@ 339 Section V Detailed Explanations

'1.4.

15.

Choice B is correct. Addition of water to a buffer equally dilutes the weak acid and its conjugate base. Theresult is that the ratio of base to acid has not changed, so the buffer does not shift. The equilibrium has notbeen disturbed, so the reaction exhibits no net shift. The pH remains constant. The best answer is choice B.

Choice C is correct. Ammonium (NHa+) has a pKu value of 9.26 as given in Table 1. This means that when theacid and conjugate b_u:"_ut9 present in equal molar concentration, the-pH of the solution is 9.26. To make the pHof the solution equal 9.5, the conjugate base must be in excess to suci a degree that the log of the base-to-acidratio is 0.24. Choice A lists the conjugate base (ammonia) in excess, so choice A is vahdl Hydrocyanic acid(HCN) has a pKu value of 9'32 as given in Table 1. This means that when the acid and co.,lugate base arepresent in equal molar concentration, the pH of the solution is 9.32. To make the pH of the solltion equal 9.5,the conjugate base must be in excess to such a degree that the log of the base-to-acid ratio is 0.18. Choice B liststhe conjugate base (sodium cyanide) in excess, so choice B is valid. From choice A, we know that an ammoniumand ammonia solution must have excess ammonia for the pH to equal 9.5. If one-half equivalent of the strongacid HCI were added to one equivalent of ammonia, then the pH would equal the pru p.zoy. The excess HCIu.dd"9- lowers the pH, so the value is less than 9.5. This makes choice C invatid and thus the answer youshould choose. From choice D, we know that an hydrocyanic acid and cyanide solution must have excesscyanide for the pH to equal 9.5. If one-half equivalent of the strong base NaOH were added to one equivalentof hydrocyanic acid, then the pH would equal the pKu Q.32). Ti-,"

"""us NaOH added raises the pH to avalue greater than 9.32 and approximately equal to 9.5, making choice D valid. Choose answer C.

Choice B is correct. One half-equivalent of base when added to one equivalent of carbonic acid removes thefirst proton from half of the carbonic acid molecules present in soluti,on. This results in a solution of halfcarbonic acid and half bicarbonate ([H2Cog] = tH-o3-l). When the concentration of acid equals theconcentration of conjugate base, pH equals pKul for carbonic acid, which is 6.37, as given in the passag". Wt

"r,one and one-half equivalents of base are added to one equivalent of carbonic acid, tie first proton is c[mpletelyremoved from all of the carbonic acid molecules present in solution. The remaining half of an equivalentcontinues to remove the second proton from half of the bicarbonate molecules present in iolution. This iesults ina solution where [HCO3-] = [CO32-]. When the concentration of bicarbonate equals the concentration ofcarbonate, pH equals pKu2 for carbonic acid, which is 10.83, as given in the purrig". This makes choice Bcorrect. When one and one-half equivalents of base are added to one equivalent of lhosphoric acid, the firstproton is completely removed from all of the phosphoric acid molecules piesent in solution. The remaining halfof an equivalent continues to remove the second proton from half of the dihydrophosphate present in solution.This results in a solution where [H2POa-] = [HPoa2-1. When this equality irotds tr,re, tn" pH equals pKu2 forphosphoric acid. From Table 2, we know that Ku2 for phosphoric acid is e .Z x 10-8, which

"qrrit", to a pKu2value of 8 - log 6.2, which is a little over 7. This does r-,ot *ute a pH of 10.83, so choice C is eliminated. \iVhentwo and one-half equivalents of base are added to one equivalent of phosphoric acid, the first an4 secondPtglolt are completely removed from all of the phosphoric acid molecules piesent in solution. The remaininghalf of an equivalent removes the third proton from half of the hydrophosphate in solution. This results in asolution where [HfOnz-1 = [POa3-1. \Alhen this equality holds true, the pH equals pKu3 for phosphoric acid.From Table 2, we know that Ku3 for phosphoric acid is 4.8 x 10-13, which equates to a pKu3 value of tg - log 4.g,which is a little over 12. This does not make a solution with a pH of 10.83,1o choice D is eliminated.

\7. Choice C is correct. One and one-half equivalents of base when added to one equivalent of oxalic acidcompletely removes the first proton from all of the oxalic acid molecules present in soluiion and continues on toremove the second proton from half of the bioxalate present in solution. This results in a solution where[HC2oa-] = [Czoq2-]. When this equali_ty holds true, ihe pH equals pKu2 for oxalic acid. From Table 2, weknow that Ku2 for oxalic acid is 5.1 x 10-5, which equates to a pKl2 ,rul,r"-of 5 - log 5.1, which is less than Z, sochoice A is eliminated- One-half equivalent of base when added to one equivalent bf carbonic acid removes thefirst proton from half of the carbonic acid present in solution. This produces a solution where [H2CO3] =[HCo3-]. When this equality holds true, the pH equals pK21 for carbonic acid, which is 6.27, as given in thepassage. Choice B is eliminated. One-half equivalent of base when added to one equivalent of hipochlorousacid removes the proton from half of the hypochlorous acid present in solution. This results in a solution where[HCIO] = [ClO-]' \A/hen the concentration oJ acid equals the concentration of conjugate base, the pH equals pKufor HCIO, which is 7.49 as given in the table, Choiie C is therefore correct. Onl-taff equivaleni of acid whenadded to one equivalent of pyridine adds a proton to half of the pyridine present in solution. This results in asolution where [C5H5N] = [C5H5NH+]. \zVhen this is true, the pH

"qnuir pKu for C5H5N, which is 5.16, as

given in Table 1. Choice D is eliminated.

't_6.

Copyright @ by The Berkeley Review@ 340 Section V Detailed Explanations

18. Choice C is correct. The pKu value for formic acid (HCO2H) is 3.78. In order to have a pH = 4.00, the solutionmust be mixed with excess conjugate base (HCO2-). This makes choice A valid. The pKu value for aniliniumcation (C6H5NH3+) is 4.64. Lr order to have a pH = 4.00, the solution must be mixed with excess conjugate acid(C6H5NH3+). This makes choice B valid. The pK6 value for benzoic acid (C6H5CO2H) is 4.19. In order tohave a pH = 4.00, the solution must be mixed with excess conjugate acid (C6H5COZH). This makes choice Cinvalid. If excess C6H5CO2- were mixed with C6H5CO2H, then the pH of the solution would be greater than4.19, which is greater than 4.00. This makes choice C invalid. You should select choice C. The pKu value forpyruvic acid (H3CCOCO2H) is 3.89. In order to have a pH = 4.00, the solution must be mixed with excess

conjugate base (H3CCOCO2-). This makes choice D valid.

Choice A is correct. Blood that is rich in carbon dioxide is also rich in H2CO3. Since carbonic acid is acidic, thepH decreases as carbonic acid is added to solution. The final solution thus has a pH lower than7.4. This makes

choice A the correct choice. Because the amount of acid is increasing, the phosphate buffer system is affected.The additional hydronium ion present is absorbed by the phosphate buffer, thus increasing the amount ofH2POa- present and decreasing the amount of HPO42- and PO43- present in solution. This eliminates choice B.

If llzCbi is added to solution, it equilibrates by making hydronium ion (H3O+) and bicarbonate anion (COrz-;.This results in increases in the concentrations of both bicarbonate and hydronium ions. This eliminates choices

C and D.

Choice D is correct. The concentration of the conjugate base is greater than the concentration of the acid whenthe pH is greater than the pKu. When the pH is greater than the pKu, the log of the ratio of base to acid is apositive value, indicating that the ratio of conjugate base to acid is greater than 1.0. The pKu for HF is 3.17, so

at pH = 3.00, the pH is less than the pKu. This eliminates choice A. The value for pKul for H2CO3 is 6.37, so

at pH - 6.00, the pH is less than the pKu1. This eliminates choice B. The value for pKu2 for H2C2Oa is 5 - log5.1 (which is roughly 4.3), so at pH = 4.00, the pH is less than the pKu2. This eliminates choice C. The valuefor pKu2 for H3PO4 is 8 - log 6.2 (which is roughly 7.2), so at pH = 8.00, the pH is greater than the pKu2. Thisresults in the base being more concentrated than the acid, making choice D the correct choice.

Choice C is correct. The value of pKul is 6.37 and pKu2 is 10.33 for carbonic acid (H2CO3). The pH is greaterthan pKu1, so the conjugate base (bicarbonate) is in greater concentration than the acid (carbonic acid). Thismakes choice A valid. The pH is less than pKu2, so the conjugate acid (bicarbonate) is in greater concentrationthan the base (carbonate). This makes choice B valid. The value for pKu2 is7.20 and pKu3 is 12.32 forphosphoric acid. The pH is greater than pKu2, so the conjugate base (HfOnZ-; is in greater concentration thanthe acid (H2POa-). This means that choice C is invalid, making it the correct choice for the question. The pHis less than pKuj, so the conjugate acid (UPO42-) is in greater concentration than the base (POna-;. This makeschoice D valid.

Choice C is correct. To reach equivalence, it takes (0.03 L)(0.20 M) = 0.000 moles of NaOH. Formula weight is

grams per mole equivalent, so the unknown acid has a formula weight of 1.0 g/O.OO6 moles = 767 grams per mole.

From the choices given, the only acid that is close to 1.67 grams per mole is p-nitro benzoic acid(O2NC6HaCOZH). To verify your answer, add up the atomic weights of the molecule, O = 12, N = 14, C = 1,2

and H = 1 (all in grams per mole). The sum is 46 (for NO2) + 76 (for CoH+) + 45 (for CO2H) = 167 grams Petmole, so the answer is definitely C.

Choice C is correct. The formula weight of a compound is measured in grams per mole. The moles are

determined from 0.040 liters x 0.25 M = 0.010 moles. The formula weight is 1'0 8/0.010 moles = 100 grams per

mole. Pick choice C for best results.

Choice D is correct. Since the graph begins at a high pH, the initial solution must be basic. This rules outchoices A and B. The equivalence point is at pH = 4.6, which is < 7. This indicates that the neutra.lizedproduct is a weak acid, so the titrant must be a strong acid. (Because the equivalence point is not 7.0, thecompound must be a weak base). The graph represents D, the titration of a weak base by strong acid.

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20.

21.

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25.

24.

Copyright @ by The Berkeley Review@ 341 Section V Detailed Explanations

25.

26.

27.

Choice D is correct. To predict the pH near the equivalence point, which is what you're doing when you choosean indicator, you should use the Henderson-Hasselbalch equation. At 2 pH units beyond the pKu of the acid,the ratio of conjugate base to acid is 100:1, which means that the reaction is almost at equivalence. This is thepoint at which the indicator should start to show some color change. In this question, the pKu for the weakacid is 7.20. The best indicator is active around pH = 9.20. This is choice D, phenolphthalein, which is activebetween 8.0 and 9.6. You may remember using phenolphthalein in titration labs from general chemistry.

Choice C is correct. A buffer is formed when a conjugate pair (acid and base) are present in solution in roughlyequal molar amounts. To achieve this, you can: (1) mix the conjugate acid and base evenly, (2) add the acid andhalf-titrate it (add a half molar equivalent of strong base), or (3) add the base and titrate it halfway (addhalf an equivalent of strong base). Choice A is a case of full titration, which leads to the complete conversionof the weak acid to its conjugate base, so that is out. Choice B is a case of over-titrating by adding double theamount of base needed. In choice C, the weak acid (H3CCO2H) is added to a half-equivalent of strong base,which results in a buffered solution. Pick C. Choice D is an acid with an acid, which doesn't react.

Choice D is correct. The best choice for a buffer of pH = 8.5 is a weak acid with a pKa close to 8.5 mixed insolution with its conjugate base. The acid 4-ethanoylphenol (H3CCOC6HaOH) has a pKu of 8.4, which isclosest to the 8.5 value, so choose D. To make the buffer equal to 8.5 (slightly to the basic side of 8.4), thereshould be a slight excess of the conjugate base relative to 4-ethanoylphenol. The log of the ratio of conjugatebase to weak acid should be 0.1, a positive number, so the ratio of conjugate base to weak acid must be greaterthan 1.

Choice A is correct. The formula weight, also known as the empirical weight, is derived from the mass ofcompound per mole of equivalent. Adding up the molecular mass of the compound yields 90 grams,/mole.Because there are two equivalents of protons per oxalic acid, the formula weight is 45 grams/mole. Pick A. Besure not to pick B by mistake.

Choice B is correct. If the titration required 30.0 mL of 0.10 M KOH(aq) to neutralize 25.0 mL of 0.10 M HA(aq),then according the relationship MacidVacid = MbaseVbar", the molarity of the acid must be 0.12 M. However,seven drops (although the exact number is irrelevant) of HCI must be added to back titrate to the equivalencepoint. This means that the molarity of the unknown weak acid is actually a little less than 0.12 M, makingchoice B the best answer.

Choice C is correct. As a general rule, the pH at the equivalence point is either three units above or below thepKu for the acid depending on the type of titration. This would mean that the pH at the two equivalencepoints differs by approximately 6 pH units. This makes choice C the best choice.

To calculate the pH exactly for the two titrations, one can plug into the shortcut equation for both the conjugatebase (formed at point b) and conjugate acid (formed at point d). At the equivalence point, the volume is doubleits original value, so the concentration is half of its initial value. At point b, there is 0.05 M OAc-, and at pointd, there is 0.05 M HOAc. From Figure 1, the pKu for HOAc is roughly 5.0. This means that pK6 for OAc- isroughly 9.0 The two calculations are shown below:

ForHOAcatpointd:pH=lpKu-ltoglHAl =1tS.Ol -1log(0.05) =2.5-1(-t.3)=2.5+0.65=3.152' 2- 2 2- 2

ForOAc-atpointb: pOH=lpKu-ltoglA-l =ftO.Ol-llog(0.05) =4.5-lf-r.:l =4.5+0.65 =5.15.'.pH=8.85' 2' 2 - 2 2 - 2The difference in pH is 5.7, so the best answer is choice C.

Choice C is correct. The pH of a solution depends on the concentration and strength of the reagents in solution.The highest pH at equivalence results from the presence of the strongest conjugate base (since all of theconjugate bases are in equal concentration). At equivalence, choices B and D are both acids, so they are botheliminated (pH at equivalence is less than 7.0). Methyl ammonium is less acidic than acetic acid based on therelative pKu general values given in the passage. The stronger base is the conjugate base of the weaker acid,making methylamine the more basic compound between acetate and the amine. The greater hydroxideconcentration results from the titration of 0.10 M H3CNH3C1, so choice A is eliminated, and thus choice C isthe best answer.

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Copyright @ by The Berkeley Review@ Section V Detailed Explanations

32. Choice C is correct. When the pH is greater than the pKu, then according to the Henderson-Hasselbalchequation, the conjugate base is in greater concentration than the acid. This eliminates choices A and B.

Because the pH and pKu are negative log values, if pH > pKu, then [H*] < Ku. Choice C is correct.

Choice C is correct. Initially in the titration of H3CCO2H by NaOH (at point a), the concentration ofH3CCO2H is 0.10 M. At the equivalence point in the titration of H3CCO2Na by HCI (point d), theconcentration of H3CCO2H is 0.05 M, due to dilution that occurs when the aqueous base solution is added. Notein the graph below that the final volume is 50 mL, not 25 mL. The concentrations of H3CCO2H between the twopoints in question differ by a factor of 2. This means that the pH of the two solutions differs by -log{2. Withthe H3CCO2H concentration being greater at the initial point of the titration of H3CCO2H by NaOH than theequivalence point of the titration of H3CCO2Na by HCl, the pH is lower at point a (the initial point of thetitration of H3CCO2H by NaOH). The best answer is therefore choice C. The graph below shows the data:

33.

pH

Start with 25 mL0.10 M II3CCO2-

Start with 25 mL0.10 M H3CCO2H

Finish with 50 mL0.05 M II3CCO2-

i Start with 50 mLi 0.05 M H3cco2H

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35.

35.

mL titrant solution added 25.0

Choice D is correct. By having a lower pKu value, H3CCO2H is more acidic than CH3NH3+. Because it ismore acidic, H3CCO2H buffers at a lower pH than CH3NH3+, so choice A is eliminated. Because it is moreacidic, H3CCO2H has a conjugate base that is weaker (and thus has a higher pKb value) than the conjugatebase of CH3NH3+. Choice B is thus eliminated. Because it is more acidic, H3CCO2H dissociates more thanCH3NH3+, so choice C is eliminated. Because it is more acidic, H3CCO2H is a better electron pair acceptorthan CH3NH3+ making choice D correct. The point of this question is to view the many different ways inwhich relative acidity can be expressed.

Choice D is correct. Because it is more acidic, H3CCO2H yields a lower pH value than CH3NH3+ of equimolarconcentration, so choice A is valid and consequently eliminated. Because it is more acidic, H3CCO2H has a

conjugate base with a higher pK6 value than the conjugate base of CH3NH3+. Choice B is valid and thuseliminated. Because it is more acidic, H3CCO2H produces more conjugate base than CH3NH3+ when added towater, so choice C is valid and consequently eliminated. Because it is more acidic, H3CCO2H is a better protondonor than CH3NH3+, so choice D is not true. The correct answer is choice D.

Choice D is correct. Given that the pKa value for HF is lower than that of H3CCO2H, HF is a stronger acid

than H3CCO2H. This means that equimolar HF and H3CCO2H solutions result in a lower pH for the HFsolution. This makes choice A invalid and thus eliminated. \Alhen equimolar HF and H3CCO2H are titratedequally, they both have conr.erted into the same amount of conjugate base in the buffer region, meaning that [F-]equals [HgCCOZ-]. This eiiminates choices B and C. When both HF and H3CCO2H are half-titrated, the pHof solution is lower for the FIF solution, because the pH equals the pKu of the acid and the pKu value for HF is

lower than that of H3CCO2H. This meeins that pH(HF solution) < pKa(acetic acid) which means that [H+]1gpsolution) > Ka(acetic acid)- This makes choice D the best choice'

97. Choice A is correcl The initiai pH is greatest in the solution with the lowest concentration of the weakest

acid. H3CCO2H is the rt-eaker acid of the two choices, and 0.010 M is the lowest of the concentrations. Thismeans that choice A is correcl

Copyright @ by The Berkelev Rerierva 343 Section V Detailed Explanations

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Choice A is correct. When the pH of the solution exceeds the pKu for the acid, then there must be an excess ofconjugate base relative- to the conjugate acid, according to the Henderson-Hasselbalch equation. Thiseliminates choices C and D. The pH is the negative log of the [H3O+] and the pKu is the negaiive log of theKa. If PH > pKu, then the [HeO+] must be less than Ku. This makes choice A iorrect. To make this clearer,when pH = PKa, then [H3O+] = Ka. As the pH increases, the [H3O+] decreases, while both pKu and Ku remainconstant. The [H3O+] is thus less than Ku. The derivation from the equation for Ku is shown below:

rcu =[H]o.llai ro Ku, =]f whenpH>pKu,#L, 1.0,so- x-a > 1 and Ka > [H:o*l" tHAl [Hso'l tHA] tHAl 1Hrorl -

Choice C is correct. In all three weak acid titrations, the volume of sodium hydroxide solution added to reachthe equivalence point was 25.0 milliliters, so choices A and B are eliminated. The higher the acidconcentration initially, the more conjugate base that forms at the equivalence point. The higher theconcentration of conjugate base, the higher pH for the solution at the equivalence point. Pick choice C.

Choice C is correct. If the base concentration is doubled while the acid concentration remains constant, then thebase is twice as concentrated as the acid. Half as much titrant (base) is required, so the equivalence point isreached in half the base volume. This eliminates choices B and D. The acid is still HCI (i strong acid), andthe titrant base is still NaOH (a strong base), so the pH is still 7.0 at equivalence, and the .rrrr" iti1 has thesame shape (lip-free and sigmoidal). The best answer is choice C.

Choice C is correct. Choice A is eliminated, because a strong acid and strong base do not make a buffer whenmixed together. Choice B is eliminated, because NaOH is not a weak base. Choice D is eliminated, because,f" pH changes throughout the titration; but it is at the equivalence point that the pH changes drastically.Choice C is the best answer, because the log scale means that as long as the concentrations are 100 timesdifferent, then the linear difference is 2.0 on the log scale. This in turn *"a.r that the slopes are equal throughmost of the titration, excePt near the equivalence point. Perhaps the answer would be betier if it m;ntioned thedifferent concentrations of the titrant bases in each trial. The best answer is not always perfect.

Choice D is correct. Ammonia is a weak base, so choices A and C are eliminated, because the curves do not showthe initial dip in pH (lip-o-weakness). The greater the concentration of ammonia initially, the greater thepH initially, so the 1.00 M NH3 has a higher starting pH than the 0.010 M NH3. The correct choice is D.

Choice D is correct. The greatest Ku value correlates with the strongest acid. Of the acids in the experiment,only HCI shows a strong acid titration curve (lip-free) and an equivalence pH of 7, so choice D is the answer.

Choice B is correct. The initial pH in the titration of HCN is greater than the initial pH in the titration ofHCIO, according to the titration curve in the passage. This means that the initial pH in the titration of HCNmust be greater than 4.23, which eliminates choice A. Because HCN is acidic, the initial pH must be below 7.0,which eliminates choice D. The pH at the equivalence point in the titration of HCIO ls less than the pH atthe equivalence point in the titration of HCN, according to the titration curve in the passage. This means thatthe pH at the equivalence point in the titration of HCIO must be less than 11.01, which eliminates choice C.The only choice that remains is choice B, so choice B is the best answer. The numbers are reasonable, because5.16 is greater than 4.23 (and 4.74 if you look at the titration curve), and 10.08 fits between 8J2 and 11.01 in theequivalence pH data.

Choice B is correct. Statement is invalid, as the curves in Figure 1 show. Strong acids fully dissociate uponaddition to water, so the pH is low in the beginning, It remains relatively constant, because pH is a log scale.It is the strong acid curve that has a plateau in the beginning. A buffer is defined ur u ro,tghly equai molarmixture of a weak acid and its conjugate base. Based on the definition, a strong acid when mixed wiih a strongbase does not form a buffer. Do not be fooled by the flat region of the strong acid titration curve, which isattributed to mathematics, not buffering. Statement II is valid. Statement IiI holds true for strong acids, butnot for weak acids. For instance, a weak acid with a pKu greater than 7,has a pH greater than seven for mosiof the buffer region, which is observed before the equivalence point. For a weak acid titration curve, the pH isgreater than 7 at equivalence, so there must be a point on the curve before the equivalence point that is greate:than 7. Only statement II is valid, so choice B is correct.

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Copyright @ by The Berkeley Review@ 344 Section V Detailed Explanations

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Choice D is correct. Because 0.10 M HIO has a greater pH than 0.10 M HCN, HIO must be a weaker acid thanHCN, and the titration curve associated with HIO would have all of the points greater than the titrationcurve of HCN. The pKu of HCN is greater than 8.72 (the half-equivalence pH is greater than the equivalencepH in the HOAc titiation curve). fhus, the pKu of HIO mustbe greater thund.7z, which makes choice Aincorrect. The pH at equivalence in the titration of HCN is 11.01, so the pH at equivalence in the titration ofHIO must be greater than 11.01, and choice B is thus eliminated. The pH of 0.10 M HIO must have a pH lessthan7.0, because although it is a weak acid, it is still acidic. The pH cannot be7.2!, so choice C is eliminated.The pKu of HCN is 9.32, so the pK" of HIO must be greater than 9.32. The pH at equivalence must be at least apH unit less than 13 (the pH of 0.10 M KOH), so the difference between the pKu and the pH at equivalence isless than three pH units. This makes choice D a true statement. The other way of confirming choice D is to lookat the trend whereby as the acid gets weaker, the difference between pK" and the pH at equivalence getssmaller. The difference for HCN is 11.01 - 9.32 = 1.69. For HIO, the difference should be less than 1.69.

Choice B is correct. When 10 mL 0.10 M HCIO are mixed with 10 mL 0.15 M KCIO, a buffer is formed. \zVhenwater is added to a buffer, the acid and conjugate base are diluted equally, so the pH does not change. The pHconsequently remains at7.64. The best answer is choice B.

Choice A is correct. Because CN- is a stronger base than CIO- (HCN is a weaker acid than HCIO), the titrationcurve for NaCN should have a higher initial pH, higher pH at the half-titrated point (because the pKu forHCN is greater than the pKu for HCIO), and higher pH at the equivalence point. This can be seen only in thetitration curve in choice A. Choices C and D should have been eliminated immediately, because theirrespective graphs show that the two acids have the same pKu, which according to Figure 1 in the passage,they don't have.

Choice A is correct. The stronger the acid, the lower its pKu value. HCI is the strongest acid, so it has thelowest (and only negative) pKu value. Choices C and D are eliminated. From the titration data, it can be seenthat the next strongest acid is HOAc, meaning that the pKu for HOAc should be just ahead of the pKa for HCIin the sequence. This makes choice A the best answer.

50. C is correct. The moles of NaOH can be calculated from the moles of the acid (H+) that were neutralized, sincethe amount of base added is to the point of neutralization, where the moles acid equals the moles base. Keep inmind that sulfuric acid is diprotic and that the units given are units of normality. Normality already takesinto account the two equivalents of hydronium ion per sulfuric acid. Given in the passage, the number of molesof acid (H+) is (.025 LX0.20 M) = 0.005 moles. To reach equivalence, this must be the moles of NaOH as well.The question asks for concentration, so the moles NaOH must be divided by the volume of solution.

Molaritv = moles = 0.005 mole = 0.25M' liter 0.02 LThus, the concentration of NaOH(aq) is 0.25 M. Answer C is the best answer for you to choose. Do what is best!

51. Choice B is correct. The first proton is always the easiest to remove (that is, pKal is smaller or more acidicthan pK62). Reversing this wording gives the correct answer B. The first proton must be easiest to remove,because it comes off first. This also a rare case where the correct choice contains the word "always."

52. Choice D is correct. Normality is based on the molarity of equivalent base required to neutralize all of theacidic protons. Since there are three (3) acidic protons on H3PO4, the normality is three times the molarity.

NormalitY = 3(0'3 M) = 0.90 N H3POa'

Pick choice D if you want to be a supemova of chemistry wisdom. Or pick it because it's the right answer.

53. Choice C is correct. The conversion from H3POa to H2PO4- inv_olves the loss of the first proton, so the acidity iscalculated using pKu1. The conversion from H2POa- to HPO42- involves the loss of tne second proton, so theacidity is calculated using pKu2. Mixing NaH2POa and Na2HPO4 yields H2POa- and HPO42-, so the pH ofthe solution is close to pKu2. Because there is more of HPOnz- in solution than H2PO4-, the pH is greater thanpKu2. This eliminates choices A and B. If the solution contained pure HPO42-, the pH would be an average ofpK62 and pKaa. The presence of H2PO4- lowers the pH, making the best answer choice C. Using theHenderson-Hasselbalch equation would give a pH = pKa2 +log2 = pKu2 + 0.3.

Copyright O by The Berkeley Review@ 345 Section V Detailed Explanations

54. Choice D is correct. Sulfuric acid has only two (2) acidic protons, whereas phosphoric acid has three (3) acidicprotons to neutralize. Thus, phosphoric acid requires only two-thirds the amount of base that is required forsulfuric acid. The number of moles of phosphoric acid is (25 mL)(0.6 M) = 15 mmoles H3POa. This means thatthere are 45 mmoles of H+ to neutralize. The question asks how many mL of 0.40 M H2SOa(aq) contains this

amount of H+. 0.40 M H2SOa(aq) = 0.8 M H+. This means that the number of mL ir, 45^mr-n9le, which equals0.8 M

56.25rnL 0.40 M H2SO4, given as answer choice D.

Choice B is correct. Phosphoric acid is a triprotic acid, so choices A and C can be eliminated. Because the firstproton of phosphoric acid is weak, the start of the curve should have a cusp (lip-o-weakness). It is only inchoice B that the cusp is present, so the best answer is choice B.

Choice A is correct. Flask 2 is initially filled with 40.0 mL of 0.30 N H3POa. The concentration is equivalent to0.10 M H3POa. Adding 40 mL of 0.30 N NaOH completely neutralizes all three protons on phosphoric acid,leaving only the conjugate base PO43-1aq) in solution. All of the moles of phosphoric acid(H3PO4) presentinitially are converted into phosphate (POnr-;. The addition of 40 mL of NaOH solution also doubles thevolume of the solution, increasing it from 40 mL to 80 mL. This cuts the concentration in half. Had the volumeremained 40 mL, and if all of the H3POa were converted into PO43-, then the concentration of PO43-(aq) wouldhave been 0.10 M. Flowever, the final concentration after the extra 40 mL of solution is accounted for is 0.5 x 0.10M POa3-(aq). That equals 0.050 M POa3-(aq), which is choice A.

Choice D is correct. Because there are two vertical inflection points on the titration curve, the acid has twoequivalence points and thus is diprotic. Because of the lip-o-weakness (initial cusp in the titration curve), thefirst proton is associated with a weak acid. The second proton is always weaker than the first proton, so bothprotons are weak. Choice D is the best answer

Choice D is correct. The first pKu is always lower than the second pKa by definition, meaning that choice A isvalid for all polyprotic acids. The first equivalence point (point c) is greater than the point where pH = pKat(point b). This makes choice B a valid statement and thus eliminates it. The second equivalence point (point e)is greater than the point where pH = pKu2 (point d). This makes choice C a valid statement and thuseliminates it. The first equivalence point (point c) is less than the point where pH = pKaZ (point d). Thismakes choice D an invalid statement and thus makes it the best-answer. To go from the first equivalence pointto the point at which the pH equals pKu2 (where [HA-] = [A2-]), base must be added to the solution. Thismeans that pKu2 is greater than the pH at the first equivalence point (addition of base increases the pH),confirming choice D.

Choice B is correct. To carry out the titration, a strong base must be added to react with the unknown weak acid.The pH increases as base is added, and according to the curve in Figure 1, the pH does increase when movingleft to right. The best answer is choice B. A weak base is not strong enough to carry out the titration.

Choice A is correct. To go from one equivalence point to another requires that one equivalent (whatever exactquantity that may be) of base titrant be added. Choices B and D can both be eliminated, because they requirean equal amount of base. It takes one equivalent of titrant to go from pH = pKal to pH = pKa2, so in choice A,more than one equivalent is required. In choice C, less than one equivalent is required. This makes the bestanswer choice A. On the graph, to go from one labeled point to the next labeled point, requires one-half of an

equivalent. So starting to before point b and finishing after point d clearly is more than one equivalent.

Choice D is correct. When pH = pKal (true at point a), exactly one-half of an equivalent of titrant has beenadded to solution. To reach the second equivalence point, two equivalents must be added. If 5 mL does not reachthe point at which pH equals pKu1, then one-half of an equivalent must be greater than 5 mL. This means thatone equivalent is greater than 10 mL and thus, two equivalents are greater than 20 mL. More than 20 mL isrequired to reach the second equivalence point, so the best answer is choice D.

Choice B is correct. Point d is the point at which pH = p_Ka2 (where [HA-] = tA2-l). Point e is the secondequivalence point (at which all of the species present is A2-). Between the two points (d and e), it is safe toassume that 42- is the predominant species with some HA- present. This is best described by choice B.

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Copyright @ by The Berkeley Review@ 346 Section V Detailed Explanations

63. Choice D is correct. Point b is the point at which pH = pKa1, so [H2A] = [HA-]. This makes choice A valid.Point c is the first vertical inflection point, so it is the first equivalence point. Choice B is valid. Point d is thepoint at which pH = pKa2, because [HA-] = [A2-]. Choice C is valid. Point f represents a point where excessbase is being added and all of the species exists in the fully deprotonated form (A2-). Pick D to feel good.

b (pH > pKu2)

The passage is easier when all of the points in Figure 1 are labeled.

a (All exists as COq2-)

g(pH = pKuz)

d (First equivalence point, all exists as HCO3-)

(pH =pK.r)

B (PH < PKur)

h (Second equivalence point,all exists as H2CO3)

i (excess acid added)

2.0 Equivalents strong acid added

Choice C is correct. The pH equals the pKu of the acid when the titration is half way to the equivalence point.The two points marked by x represent the points at which pH = pKu. The value of pKul is lower than pKu2, sothe correct answer is point f, choice C. Point c is where the pH = pKa2, point d is the first equivalence point(where the species exists as HCO3-), and point h is the second equivalence point (where the species exists as

H2CO3).

Choice B is correct. Conjugate pairs have pK values that sum to 14. In the case of a diprotic acid, the firstproton to be lost in an acid dissociation reaction corresponds to the second (last) proton to be gained in a basehydrolysis reaction. This means that the following conjugate pair relationships hold true.

1.51.00.5

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HrCO3(ar) S H*(uq) + HCO3-(ar)Pxlz

HCo3-(ae) =--EL H+(aq) + cor2-(at)PKilr

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The conjugate pair relationship yields pKal + pKbZ = 14 and pKa2 + pKbl = 14. This makes choice A valid andchoice B invalid. The question asks you to seek a statement that is not true, so choice B is best. According toFigure 1, the initial pH is the highest point, and it is greater than pK32, making choice C a valid statement.According to Figure L, the second equivalence point is lower than all pKu values, including pK21. This makeschoice D a valid statement.

Choice B is correct. In a fashion similar to the way in which we determine the isoelectric point of an aminoacid, the middle equivalence point can be determined by knowing that it lies exactly in the middle of the twopKu values. The pH of the first equivalence point is thus found by averaging pKal and pKu2. The pH cannot be7.0 for this titration, and the Kn when dealing with the acidity of bicarbonate would involve pKu2 (because itis the second proton to be lost). The best answer for this question is choice B.

Choice B is correct. At point e of the titration curve, the amount of HCI added is greater than one equivalentbut less than one and one-half equivalents. This makes choice A a valid statement. The pH should be lowerthan the pH of point c (the point at which pH = pKaZ), so choice D is a valid statement. The value of pK31 is6.37, so choices B and C are contradictory answers. The pH is greater than the pH at point f (the point atwhich pH = pKut), which makes choice B invalid, and thus the best answer. It is critical in this passage thatyou recognize where pKul and pK32 lie.

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Copyright @ by The Berkeley Review@ 347 Section V Detailed Explanations

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Choice C is correct. Bicarbonate is in its greatest concentration at the first equivalence point. This point liesmidway between pKul and pKa2. From the passage/ the values for pK61 and pK62 are 3.67 and 7.63respectively. The values for pKnl and pKu2 are thereforc 6.37 and 10.33, respectively. The bicarbonateconcentration is greatest midway between 6.37 and 10.33. The best answer is choice C.

Choice A is correct. The pH change between points c and d corresponds to the change between pK22 and the firstequivalence point. This is greater than the incremental change from point d to e (first equivalence to a pointwith pH greater than pKul), greater than the incremental change from point e to f (a point with pH barelygreater than pKul to a point where pH = pKat), and greater than the incremental change from point f to g (apoint with pH = pKal to a point where pH is just less than pKat). The best answer is choice A. It may be easierto answer this question by observing the titration curve and noting the change in pH (y-axis) between the twopoints in question.

Choice A is correct. Changing the labeling of the y-axis from pH to pOH does not change the shape of thegraph. The base is still weak, so choice B can be eliminated. The pOH starts lower than 7 (because pH startsgreater than 7), so choices C and D are eliminated. The best answer is choice A, because the graph flipsvertically about the pH or pOH = 7line when pH and pOH are interchanged.

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Choice C is correct. Because the indicator has to be active at the equivalence point, the pKu of the indicatorhas to be greater then the pKu of the acid being titrated. This is because the equivalence pH is higher thanthe pKu for the acid being titrated. Because the acid being titrated is weak, the pH at equivalence is greaterthan 7,0. It is stated in the passage that the pKu values should differ by about 3. This makes the best choice C(which differs by 3.5 pH units) the best. Choice C is roughly three units above the pKu for the acid and isgreater than 7.0.

Choice A is correct. An indicator concentration that is too high results in a second competing acid that affectsthe titration and a faint color in the solution that is permanent. If the indicator were in concentration equal tothe acid being titrated, then the curve would look like a diprotic acid. While the color is affected by theexcess indicator, the color change is certainly not going to be too minimal, so choice D gets eliminated. It is a

color change, which involves frequency of light. A higher indicator concentration affects the intensity of thelight, so the color change would just be more intense, but every bit as detectable. This eliminates choice C.While the viscosity may change with the addition of the indicator, this should not affect the reactionequilibrium, only the rate. The best answer is choice A, because the indicator can react with the titrant base.You may recall from titration experiments in general chemistry lab that you add only a few drops of indicatorto the solution.

Choice A is correct. When the pH is greater than the pKu, the solution has more conjugate base than acid, so

choices C and D are eliminated. The difference between the concentrations is 2 on the log scale, so on a linearscale, it can't be 2 : 1, (it must be 100 : 1). This eliminates choice B and confirms that choice A is the correctanswer. Using the Henderson-Hasselbalch equation, the value can be determined.

/J.

pH=pKa+togJ4l=pKa +2* loqlA-l =z= [A-] =102=1oo" tHAl tHAl

74. Choice D is correct. A good indicator changes color (thus it cannot be transparent) and it must be present in lou-

concentration in solution. This eliminates choice A. If it is not affected by acid-base reactions, it can't react in a

way that indicates a pH change. Choice B is eliminated. Because the indicator must undergo some acid-basechemical reaction, it is best when the indicator is a weak acid or weak base, so it does not affect the pH much.If it were a strong acid or strong base would alter the pH and affect the titration. The best answer is choice D.

75. Choice C is correct. The question calls for a titration in which the pH at equivalence is the same, regardless oithe concentration of reactants. Choices A and B can be eliminated, because the pH of weak acids and base'varies with the concentration. The pH at equivalence when the first proton of a weak diprotic acid has beer.removed is the average of pKul and pKu2. You most likely can recall this from determining the isoelectric pHof an amino acid. It applies to any polyprotic acid, including the amino acids. Because the pKu values do nc":

change, the pH at equivalence does not change, so the same pH at equivalence is observed. The best answer i.choice C. The second proton of a diprotic acid is affected by the concentration, so choice D is eliminated.

Copyright @ by The Berkeley Review@ 34A Section V Detailed Explanations

76. Choice B is correct. Choice C should be eliminated first, because the pH at equivalence is greater than the pKuof the acid. Choice D should also be eliminated, because the color changes over a range, not at one exact value.At the equivalence pH (when the indicator is changing color), a change of one pH unit is substantial. Thiseliminates choice A. The best answer is choice B, because the pH changes so rapidly that one drop of titrantcan change the pH by more than one pH unit at the equivalence point. The indicator can determine this valueto the accuracy of the one drop.

Choice A is correct. The titration of 1.0 M benzoic acid by 1.0 M potassium hydroxide has an equivalence pHvalue greater than 7.0, because the acid is weak and the base is strong, so the conjugate base that is formed atequivalence makes the pH higher than7. The conjugate base, benzoate, has a pK6 value of 9.79. Tl,:tis confirmsthat benzoate is a weak base. The best indicator must have a color change range greater than 7.0. Indicator Igoes from reddish orange to mango as pH increases from 6 to 8, so Indicator I has a range that is too low.Indicator I is eliminated. Indicator I may seem close, but as the benzoic acid titration reaches equivalence, theindicator would have already turned yellow, so no color change occurs at equivalence. Indicator II goes fromchartreuse to aquamarine as pH increases from 9 to 1.1., so Indicator II is a good choice. The pH at theequivalence point falls between 9 and 11, in all likelihood. Indicator III goes from violet to clear as pHincreases from 3 to 5, so Indicator III is eliminated. Only hrdicator II will work, so the best answer is choice A.

If you wish to solve for the pH at equivalence, keep in mind that the conjugate base has formed and that thesolution's volume has increased, diluting all components. The pOH at the equivalence point is found using theshortcut equation.

pOH=lpru-llogtA-l =Lp.zo1-ltog(0.5)=n.39-11-0.3)=4.89+0,15=5.04.'.pH=8.962 2- 2 2- 2

This means that the pH at equivalence is about 9.0. The 0.5 M conjugate base is determined, because when thestrong base was added to the benzoic acid, it not only converted the benzoic acid to conjugate base, it alsodiluted the solution. With an equivalence pH of 9.0, only indicator II can work. A faster way to get theapproximate pH at equivalence is to average the pKa of the weak acid and the pH of the titrant. This yieldsan approximate value. In this case, the average of 4.27 and 14 is 9.1. The approximate value is usually off byabout 0.15, which is close enough for our purposes.

Choice B is correct. The combination of colors that is nof possible for a solution is the one that does not have apH value at which the three colors may exist. In choice A, Indicator I is red at pH values of 5 or less, IndicatorII is yellow at pH values of 8 or less, and Indicator III is purple at pH values of 2 or less. At a pH value of 2 orless, the color combination in choice A is possible. In choice B, Indicator I is mango at a pH value of 8, IndicatorII is yellow at pH values of 8 or less, and Indicator III is violet at a pH value of 3. The solution cannotsimultaneously be pH = 3 and pH = 8, so the color combination in choice B is not possible. In choice C, Indicator Iis red at pH values of 5 or less, Indicator II is yellow at pH values of 8 or less, and Indicator III is fuchsia at apH value of 4. At a pH value of 4, the color combination in choice C is possible. In choice D, Indicator I isyellow at pH values of 9 or greater, Indicator II is aquamarine at a pH value of 11, and Indicator III is clear at

pH values of 5 or greater. At a pH value of 11, the color combination in choice D is possible. Choice B is thebest answer choice.

79. Choice A is correct. If the ratio of blue species to clear species within the indicator equilibrium is 1000 : 1 at pH= 6, and the deprotonated form absorbs visible light (which makes the deprotonated form blue), then at a pHof 6, the indicator is predominantly in the deprotonated form. This means that the pKu value is less than 6.0.

This eliminates choices C and D. Because the ratio is 1000:1, the log value is 3, making the pKu 3 units less

than the pH. The equation is as follows:

pH = pKa(indicator)*,.tTH## = pKa(indicator) + log 1000

6.0 = pK6l6dicator) + log 1000 = PKa(indicator) + 3

Thus, the PKa(indicator) = 3.0, so the best answer is choice A.

80. Choice D is correct. The key colors are faint mango with Indicator I (which implies that the pH is just greaterthan 8) and faint chartreuse with Indicator II (which implies that the pH is just less than 9). The best optionfor pH is thus choice D, 8.5, which falls between 8 and 9. Pick D to feel a rainbow of happy colors shine down.

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Copyright @ by The Berkeley Review@ 349 Section V Detailed Explanations

81. Choice A is correct. The color change range for Indicator I is from 6 to 8, so the pKu value for Indicator I isroughly 7. The color change range for hrdicator II is from 9 to 11, so the pKu value for Indicator II is roughly 10.The color change range for Indicator III is from 3 to 5, so the pKu value for Indicator III is roughly 4. The pKuvalue for Indicator I is incorrect in choices B and D, the pKu value for Indicator II is incorrect in choices B and C,and the pKu value for Indicator III is incorrect in choice B. The best answer is thus choice A.

Choice C is correct. In Indicator III the color change of detectability is from violet (at pH of 3) to the first signof clear (at a pH near 5). The pH range is thus best described as from 3 to 5" which makes choice C the bestanswer.

Choice C is correct. Indicator III is purple when protonated (at low pH) and clear when deprotonated (at highpH). This means that the absorbance by the deprotonated form is beyond the visible range, which eliminateschoices B and D. The visible range is 400 nm to 700 nm, so only the 339 nm absorbance can apply to thedeprotonated species. Because the protonated form is purple, the highest absorbance (I-u*) must be in therange of the complementary color of purple, which is yellow. Yellow light lies in the middle of the visiblespectrum/ so it must have a value somewhere in the middle between 400 and 700 nm. A value of.426 nm is lowenough to be associated with violet light, so the value of 598 nm must be the best choice for the absorbance ofthe protonated species. The best answer is choice C. As a point of interest, there is no known indicator whichgoes from colored to clear when deprotonated, because the negative charge usually lowers the transition energybetween the ground state and first excited state in a highly conjugated organic molecule. A highly conjugatedorganic molecule is typical for an indicator.

82.

83.

84. Choice C is correct. Bromphenol blue at pH = 7.0 has a pH value greater than the pKu, so the species is presentin its anionic (conjugate base) form. The base form of bromphenol blue according to Table 1 is blue, so choice A iseliminated. Bromcresol green in ammonia solution has a pH > 7.0, so the pH value is greater than the pK2.This means that the species is present in its anionic (conjugate base) form. The base form of bromcresol greenaccording to Table 1 is blue, so choice B is eliminated. Thymol blue in acetic acid solution has a pH < 7.0, so thepH value is less than the pKu, and the species is present in its cationic (conjugate acid) form. The acid form ofthymol blue according to Table 1 is also yellow, so choice C is correct. Bromthymol blue in hydroxide solutionhas a pH > 7.0, so the pH value is greater than the pKu. This means that the species is present in its anionic(conjugate base) form. The base form of bromthymol blue according to Table 1 is blue, so choice D is eliminated.

Choice A is correct. The titration of ammonia with hydrochloric acid forms an acid at the equivalence point.This means that the pH at the equivalence point is less than 7.0. Of the choices, only phenolphthalein has apKu greater than 7.0, so phenolphthalein cannot be used as the indicator. The pH at the equivalence pointshould equal the pKu of the indicator, so bromcresol green (pKa = 4.4), methyl orange (pKa = 4.0), andbromphenol blue (pKa = 3.7) should all be close enough together that they can all work. The better of theindicators depends on the initial concentration of ammonia. Choice A is correct.

Choice B is correct. The ideal indicator has a pKu close to the pH at equivalence. To solve this question, thepH at equivalence must be determined for each answer choice. In choices A and B, HF is formed at equivalence.The pK6 for HF is 3.17. In choice A, the pH of 1.00 M HCl is 0. The average of the pKu for HF and the titrantpH leads to a pH at equivalence of approximately 1.6. Cresol red has a pKa of 7.6, so choice A has a validindicator for the titration. In choice B, the pH of 0.01 M HCI is 2. The average of the pK" for HF and thetitrant pH leads to a pH at equivalence of approximately 2.6. Methyl violet has a pKu of 0.8, so choice B hasan invalid indicator for the titration. Choice B is the best answer.

If you are highly observant, you'll note that as the concentration of the species decreases, the equivalence pHgets closer to 7. This means that choice B requires an indicator with a higher pKu than choice A. This is alsotrue when comparing choices C and D. Chlorophenol blue has a higher pKu than bromcresol green, so choices Cand D fit the desired trend. To verify this, let's consider the pH at equivalence for the remaining choices.

In choices C and D, HCN is formed at equivalence. The pKu for HCN is 9.32. In choice C, the pH of 1.00 M HCIis 0. The average of the pKa for HCN and the titrant pH leads to a pH at equivalence of approximately 4.7.Bromcresol green has a pKa of 4.4, so choice C has a valid indicator for the titration. In choice D, the pH of0.01 M HCI is 2. The average of the pKu for HCN and the titrant pH leads to a pH at equivalence ofapproximately 5.7. Chlorophenol blue has a pKu of 5.6, so choice D has a valid indicator for the titration.

85.

86.

L

Copyright @ by The Berkeley Review@ 350 Section V Detailed Explanations

87. Choice C is correct. When the pH is greater than the pKu of the indicator, the conjugate base is thepredominant species in solution, so the solution assumes the color of the conjugate base. \rVhen the pH is lessthan the pKu of the indicator, the conjugate acid is the predominant species in solution, so the solution assumesthe color of the conjugate acid. The pKu values for the four choices are: methyl violet (pKa = 0.8), methylorange (pKa = 4.0), thymol blue (pKa = 8.4), and bromcresol green (pKu = 4.4). At a pH of.7.4, methyl violetexists in its violet conjugate base form, so choice A is valid. At a pH of 7.4, methyl orange exists in its yellowconjugate base form, so choice B is valid. At a pH of 7.4, thymol blue exists in its yellow conjugate acid form, sochoice C is invalid. At a pH of T.A,bromcresol green exists in its blue conjugate base form, so choice D is valid.The only choice that does not correctly correlate the indicator with the solution color is C.

Choice D is correct. Thymol blue has a pKu value of 2.0, so it changes color in a highly acidic medium. \Alhentitrating a strong base with a strong acid, the equivalence point is 7.0, so choices A and C are eliminated. Thetitration of a weak base by a strong acid leads to an equivalence pH value less 7.0, so choice thymol blue isappropriate for choices B and D. A pKu value of 2.0 is rather low, so the conjugate acid present at theequivalence point must dissociate readily and be in high concentration. The higher the concentration of weakbase initially, the higher the concentration of weak acid at the equivalence point. To ensure that theequivalence pH is as low as 2.0, the weak base and titrant strong acid should both be highly concentrated. Thebase must be very weak with a pKU of 10.0 or greater to be a reasonable choice for this problem. Given thelimited information, the best answer is choice D.

Choice C is correct. A strong acid when titrated by a strong base shows an equivalence pH of 7.0. The indicatorchosen should have a pKu value near 7.0. It is okay if the pKu is as high as 8.0, but not any greater than that.The pKu values for the four choices are: methyl violet (pKu = 0.8), methyl orange (pKa = 4.0), bromthymol blue(PKa = 6.8), and alizarin yellow (pKa = 10.8). Only bromthymol blue has a pKa value close to7.0, so onlybromthymol blue changes color at the equivalence point. The best answer is choice C.

88.

89.

90. Choice B is correct. The pH range of the stick is found by considering the pH range of each indicator attachedto the pH stick. Indicators have a pH range (color change band) of approximately pKullr.,dicator) + 1, dependingon the colors and intensity of the indicator. The four indicators have respective ranges of 3.37 to 5.37,4.21 to6.2I,5.78 to 7.78, and 7.79 to 9.79. This means that the overall range of the pH stick is 3.37 to 9.79, whichmakes choice B, 3.4 to 9.8, the best answer.

9L. Choice D is correct. If bromcresol green appears yellow, the pH of the solution is less than 3.37 (4.37 -l). Atthis pH, the other indicators would be protonated, so the pH could not be determined from any of theindicators. Because Solution 1 appears yellow with bromcresol green, it is not possible to know the exact pH.Solution 1 could have any pH value less than 3.37. If phenolphthalein appears magenta, the pH of thesolution is greater than9.79 (8.79 +1). At this pH, the other indicators would be deprotonated, so the pH couldnot be determined from any of the indicators. Because Solution 4 appears magenta with phenolphthalein, it isnot possible to know the exact pH. Solution 4 could have any pH value greater than9.79. The pH of Solution 1

and Solution 4 cannot be approximated from the pH stick, so the best answer is choice D.

92. Choice C is correct. An aqueous solution with a hydroxide concentration of 1.0 x 10-6 M has a pOH of 6.0 andtherefore a pH of 8.0. The pH of the solution is more than one unit greater than 4.37 (the pKu of bromcresolgreen), so the bromcresol green indicator will turn blue. Choice A can be eliminated. The pH of the solution ismore than one unit greater than 5.21 (the pKu of methyl red), so the methyl red indicator will turn yellow.Choice B can be eliminated. The pH of the solution is more than one unit greater than 6.78 (the pKu ofbromthymol blue), so the bromthymol blue indicator will tum blue. Choice C is the best answer. The pH of thesolution is roughly one unit less than 8.79 (the pKu of phenolphthalein), so the phenolphthalein indicatorappears clear. Choice D can be eliminated.

93, Choice B is correct. Solution 5 is greenish-blue with bromcresol blue, so the pH of Solution 5 is greater than 4.37and less than 5.37. Solution 5 is reddish-orange with methyl red, so the pH of Solution 5 is less than 5.21 butgreater than 4.21. The pH of Solution 5 falls between 4.37 and 5.2L, so choice B is the best answer. Only whenthe color of the indicator is a composite of the protonated and deprotonated colors can the pH of the solution beapproximated. Bromthymol blue and phenolphthalein are purely the protonated color, so they were not usefulin approximating the pH of Solution 5.

Copyright @ by The Berkeley Review@ 351 Section V Detailed Explanations

94. Choice C is correct. When HCI is added to a solution, hydronium ions are released. If the colors of the

indicators do not change, that means that the pH does not change significantly, and thus the hydronium ionconcentration does t ot chut g" significantly. There are three explanations for the pH not changingsignificantly. One reason pH does not change is that the hydronium concentration is so high that any change inhydronium is negligible on the log scale. A second reason pH does not change is that the hydroxideconcentration is so nlgn tnat any change in hydroxide concentration is negligible on the log scale. The last

reason is that the solution is a buffer. Because Solution 2 appears orange with methyl red indicator, the

approximate pH is around 5.2. This does not describe a solution rich in hydronium or hydroxide' The best

explanation is tnat Solution 2 is a buffer with a pH around 5. The pKu for a carboxylic acid is around 5, so the

best answer is choice C.

Choice D is correct. Remember in choosing an indicator for titration that the pKu of the indicator must be

within +1 of the pH at equivalence. When titrating a weak acid with a strong base, the products the,weak

conjugate base an^d water. An aqueous weak base solution forms at equivalence, which has a pH greater than 7,

ro thu lndi.ator should have a pku value greater than7. The best answer is phenolphthalein, choice D.

Choice A is correct. A solution that turns blue with bromthymol blue has a pH greater than 7 '78 (pK" = 6.68, so

the pure blue color starts at 6.68 + 1). A pH greater than7.78 does not guarantee that the pH is greater than

8.7i, t:ne pKu for phenolphthalein. This means that the solution may or may not turn magenta withphenolphtihatein. Ctroi." A ir invalid. If a solution tums green with bromcresol green, the pH is roughly 4.37'

if a solution turns bromthymol blue green, the pH is roughly 6.68. The pH cannot simultaneously be 4.37 and

6.68, so the stick cannot simultaneouily hurre two green marks. Choice B is a valid statement. A solution that

turns methyl red yellow has a pH greater than 6.21 (pKa = 5.21,so the pure yellow color starts at5.21' + 1). ApH greater than 6.2t guarantees that the pH is greater than 4.31, the pKu for bromcresol green' This means

inat tne solution *.rri t.r* blue with bromcresol green. Choice C is a valid statement. When the pH of the

solution falls between4.27 and5.37, it falls into the color blend range of two separate indicators. As a result,

the pH can be approximated with twice the accuracy. Choice D is a valid statement.

Choice D is correct. The best choice for a buffer of pH = 4.0 is a weak acid with a pKu close to 4.0 mixed with itsconjugate base. Benzoic acid has a pKu of 4.27 and nitrobenzoic acid has a pKu of 3.40. Benzoic acid is closer, so

choose D. If the buffer is to Ue a.O (the acidic side of 4.21), there should be a slight excess of benzoic acid

relative to the conjugate base.

Choice A is correct. To titrate a weak acid (such as p-nitrophenol), you must add a strong base. The only strong

base listed among the choices is KOH. Choose A, and move on.

Choice A is correct. The addition of 10 mL 0.10 M KOH(aq) to 20 mL 0.10 M phenol results in the half-titration

of the phenol. At the halfway point, pH = pK2. The pKu for phenol is 10.0, so choice A is the best answer.

Choice A is correct. This titration is of a weak acid with a strong base that is twice as concentrated (0.20 M

base with 0.10 M weak acid), so one-half the volume of the strong base is used. One-half of 50 mL amounts to

25 mL at equivalence. This eliminates choice B and D. Since the titrant is a strong base, the neutralized

product is aieak base, so the equivalence pH is greater than 7. Only the titration curve in choice A shows pH

it equivalence and pKu for the phenol as being greater than7.

95.

96.

97.

98.

99.

100.

Copyright O by The Berkeley Review@ 352 Section V Detailed DxPtanations

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