GAS LAWS & SOLUTIONS

KINETIC MOLECULAR THEORY (KMT) (pp. 426-429)

I. The kinetic molecular theory was developed to help predict the behavior of gases. It describes thebehavior of gases at the molecular level.

2. This theory is based on assumptions about a theoretical gas known as an I1H;;)q L coAS

3. The assumptions are:

a. Ideal gas particles are so small that the volume of the individual particles if they were at rest isessentially zero when compared with the total volume of the gas.

b. Ideal gas particles are in constant, rapid, random motion, moving in straight lines in all directions untilthey collide with other particles.

c. There are no attractive or repulsive forces between particles and collisions between particles are elastic.

d. The average kinetic energy of the particles is directly proportional to the absolute temperature (measuredin Kelvins).

FOUR VARIABLES DESCRIBE A GAS

Temperature

I. The temperature of a gas determines the average kinetic energy of the particles.

2. While the average kinetic energy of a collection of gases at a given temperature will be same, thevelocity at which they travel will not. This is because the mass values of the various gases are different.

3. Temperature is usually measured in either Celsius or Kelvin. Both are related to one another:

Volume

I. A derived unit: ~ X ~S X \12. Gas volumes are expressed four different ways:

.3fY\

Pressure

I. The force that the gas particles exert over a unit area:FvR.C.G'

fftr=ssu/(<= s: UNIT ~"'A

2. Pressure is a measure of the total force exerted by the moving particles of a gas as they collide with thewalls of the container.

3. Atmospheric pressure is measured in terms of: cd-IVt.

10(,.3 J:{

Standard Temperature and Pressure (STP)

1. In order to study the effects of changing temperature and pressure on a gas, one must have a standard forcomparison.

2. STP represents a pressure and temperature that are fairly easy to reproduce:

T s: a 7.3 J( oCe A,.:JD /,00 a.t-~

f Iol-Wl 373K l!l,.:J o 10/, 3 ~p~.:=

a13~ A.v 0 l\oO~

CJ-13K -4..)0 7~O,O I11nA WJd13 II: "\"-'D 7~(}. 0 .h.l\' r

Particles

I. Gases with equal volumes under the same conditions of pressure and temperature have equal numbers ofparticles.

2. This is Avo GA~ 's LA\,J and it holds true only for gases.

3. The amount of gas is measured in Wl 0 Ie4. The volume of one mole of gas at STP is known as the MO LAl V()LlJ/VI~

I VV\~:- a.~.Lj LGAS LAWS (pp. 403-415)

I. The gas laws are a series of mathematical relationships that relate the following variables:

~I - Tbl'V\P8Z,A 'NfL l;-

V- vo t.vm E-

f- Pt~s,ul-l?

V\ - t'\10LE~

Charles' Law

1. Relates volume and temperature.

2. As the temperature decreases, the volume of a gas decreases. As temperature increases, the volume of agas increases.

3. The volume of a gas at constant pressure is directly proportional to the absolute temperature.

4. Charles' Law helped prove the existence of absolute zero.

5. Mathematically this is:

V, Yz..TJ.

***** A sample of gas occupies 24.0m3 at IOO.OK.

V, 'V-z.s: -

Tz.-Vz....

z t{Oo]

Determine its volume at 400.0K.

***** Gas in a balloon occupies 2.50L at 300.0K.

VI \[ L.=:

T, T2..Q,S-O L

At what temperature will the balloon expand to 7.50L?

( 3elO, 0) (1 . .r0 )

01·50Ia. ::

7, S-o l::. 9(Jo. K

30G.OIL I~OJ.

9,00 )((0 KBoyle's Law

I. The volume of a gas at constant temperature is inversely proportional to the pressure.

2. Like Charles' Law, we can write:

P, V. ~ P,- VL

***** The gas in a balloon has a volume of 4.00L at IOO.OkPa. The balloon is released into the atmosphereand the gas in it expands to a volume of 8.00L. Determine the pressure on the balloon at the new volume.

PIV, =~(4.00)( 100.0)

1-

.: P'2. (6-.0d)

(~•00) ( I00 .0]

(8' ·00 J

***** Ifthe pressure of a 2.50m3 sample of gas is 1.50 atm, what volume will the gas occupy if thepressure is changed to 7.50 atm?

~ V I ~ P2. \f 2-

( I •5 0 ')( Ol ' S-O) z: (1.~o 1V d-

(I. so) (0I.$1:J)_

7. )"0

Gay-Lussac's Law

I. The pressure of a given gas varies with the Kelvin temperature when the volume remains constant.

2. This is expressed as:

P, P-z..-

T, TL***** The pressure of a gas in a tank is 324.24kPa at 295.0K.temperature is raised to 333.0K.

Pr P2-:: -/2.-

Determine the gas pressure if the

T,

={J-z...

33.3

***** A gas in a sealed container has a pressure of 125.0kPa at 30.0oC. Determine the temperature in thecontainer ifthe pressure is increased to 201.0kPa.

P, P; T _(z.ul)(303)

--I .., ~ -, ILJT, do

'leQ

303.0

lol- 0:: -

I~

Dalton's Law (pp. 420 - 425)

I. The total pressure in a gas mixture is the sum of the partial pressures of the individual components.

2. The M~n '"' L P..'2k--:s.sUIZl.?' of a gas is the pressure of an individual gas in a gas mixture thatcontributes to the total pressure of the mixture.

3. This is expressed mathematically as:

PrDTAL:: PT:: ~ of p~-f Ps -I •• ,

4. Dalton's law is often used to determine the pressure of a gas collected over water.

***** Hydrogen gas is collected over water at a total pressure of 95.0kPa. The volume of gas collected is28.0mL at 25.0oe. Determine the partial pressure of the hydrogen gas if the water vapor pressure is3.17kPa.

p·rt)"; ;! P,..j'L- + P;.fz.o

q S .0 llc(. :: (J,-h + .3.' 1)c~

-Rf 2-:: q 1 ,If k Ib.

***** A gas is collected over water at 50.0De and a barometric pressure of I05.00kPa. Determine thepressure of the gas if the water vapor pressure is 12.34kPa.

PruT ~ ~aJ ·f PI-h. 0

10S. o0 .kPtIl " 10J-I I;), a if

fjaS .: q ;;J. & (p:: q ;). 7)c ~

COMBINED GAS LAW

1. The combined gas law states the relationship among pressure, volume and temperature of a fixedamount of gas:

i~ V,

2. The combined gas law allows one to work out problems involving more variables that change.

***** A gas at 11O.0kPa and 30.0oe fills a flexible container with an initial volume of2.00L. Ifthetemperature is raised to so.o-c and the pressure increased to 440.0kPa, what is the new volume?

T,!' e;l1,3.,. 3.0 :: .30 .3

T2'= Jli3+ 81J:- 3.53

(110) ( t.ooo') (Ifl(O) V l::

3.5 3

:58';) L

***** At O.OOoCand 1.00 atm pressure, a sample of gas occupies 30.0mL. If the temperature is increasedto 30.0oC and the entire gas sample is transferred to a 20.0mL container, what will be the gas pressureinside the container?

T, ~ d-/3 -10 s: ';)'73 ((

/2.. ~ ~ 7"3 -+ 30 s: :lO 3. J{.

P,V, :: f2- v'zTz-

.;173

IDEAL GAS LAW (pp. 415-420)

1. The number of moles is the fourth variable that can be used along with pressure, volume andtemperature to describe a gas.

2. The ideal gas law is: N;:: n f2... Tp ~ N'-£SSlJe..(;--

\/: VO LU /l'\ (;

T; nll1 Pt:--.eA7VR.-l::-

{l : MVLE"j

3. Ifpressure is expressed in atmospheres, then:

L'JYY\wwl /(

4. If pressure is expressed in kPa, then:

R.:: [,,3/4

***** Determine the number of moles of gas in a 3.00L container at 300.0K and a pressure of 1.50 atm.

Pv= fl Q./

(\:py (1,.$)(3)~T ~

::

C. oill) ( 300)

***** Determine the Celsius temperature of2.49 moles of gas contained in a l.OOL container at a pressureof 143kPa.

N= nQ..T

T :: PV (P/3)(1'.- ::flfL (~,l(q)( 4', 3/ LI) s: ~,q, K

il.:- 0

Ci" 313

SOLUTIONS

WHAT ARE SOLUTIONS?

Characteristics of Solutions

1. Solutions are homogeneous mixtures containing two or more substances called the solute and thesolvent.

2. The ,sOLUTF is the substance that dissolves.

3. The SoLV&.J T is the dissolving medium.

4. A solution may exist as a gas, liquid or solid depending on the state of its solvent.

5. Solutions can contain more than one solute.

6. Some combinations of substances can readily form solutions and others do not.

7. A substance that dissolves in a solvent is said to be SCJI..v&L/.;;' in that solvent.

8. A substance that does not dissolve in a solvent is said to be ,bJSuLlJ ,g, L/;; in that solvent.

9. Two liquids that are not soluble in each other are said to be rmfl1I'sC., &rL t:10. Two liquids that are soluble in each other are said to be M J sc, 6L(;;'

Solvation in Aqueous Solutions

1. To form a solution, solute particles must separate from one another and the solute and solventparticles must mix.

2. Forces of attraction exist between the pure solute particles, between the pure solvent particles, andbetween the solute and solvent particles.

3. When a solid solute particle is placed in a solvent, the solvent particles completely surround thesurface of the solid solute.

4. If the forces of attraction between the solvent and solvent particles are greater than the attractiveforces holding the solute particles together, the solvent particles pull the solute particles apart andsurround them.

5. These surrounded solute particles then move away from the solid solute out into the solution.

6. The process of surrounding solute particles with solvent particles to form a solution is called~LvA T7(nJ . When it occurs in water it is called 1-1y D{lj:\TION

7. Lit-IE t>ISSOLVES Ll\(G"in a specific solvent.

is the general rule used to determine whether salvation will occur

8. Water will dissolve ionic compounds such as sodium chloride and polar compounds such assucrose, but not nonpolar substances such as oil.

Factors That Affect Rate of Solvation

I. For salvation to occur the solute and solvent particles must come in contact with one another.

2. There are three ways to increase the collisions between solute and solvent particles which thenincreases the rate of solvation:

a. A e, ITA 'n ,,-'6 fH E m I 'f. .,1/-eJ:

b . .:t"NC~""'f\ s I ~ G n+G ScJW'"i"E"'5 SuZi==l-\q; Ml.::.~

Heat of Solution

1. Energy is needed to overcome the forces of attraction of the solute particles for each other andsolvent particles for each other.

2. This process is l:"'N DO n-t Gilm I.C

3. When the solute and solvent particles mix, they are attracted to one another and energy is released.This process is ic""!<o·nitl2..n11 C

4. The overall energy change that occurs during solution formation is called thet-ltt\T of ~LV n .,J

Solubility

1. Only a limited amount of solute can dissolve in a solvent at a given set of conditions.

2. .)OwS ILI T'1 refers to the maximum amount of solute that will dissolve in a given amount ofsolvent at a specified temperature and pressure.

3. It is expressed in terms of: J SOLVn::

/DOj SOI..Y/;1J I

4. When a solute dissolves, its particles mix in with the solvent particles. As time passes, the spacesavailable to the solute particles decrease and they collide with one another and the original crystal.Some of the solvated particles will re-attach themselves to the crystal.

5. As long as the rate of salvation remains greater than the rate of crystallization, salvation willcontinue.

6. Depending on the amount of solute present, the two rates may eventually equalize and no moresolute appears to dissolve.

7. A state of dynamic equilibrium is reached. As one particle is solvated, another crystallizes. Theoverall amount of dissolved solute remains constant.

8. Such a solution is said to be a Sl.\nJe4Tb---n SOLI) 77U,J ; it contains the maximum amount ofdissolved solute for a given amount of solvent at a specific temperature and pressure.

9. An U~eA n l) SOLUT] VAlis one that contains less dissolved solute for a giventemperature and pressure.

Factors That Affect Solubility

1. Pressure affects the solubility of gases. The greater the pressure, the more gas that can bedissolved in a liquid.

2. Solubility also depends on the nature of the solute and the solvent.

3. Temperature affects the solubility of all substances. For most substances, increasing thetemperature of the solvent increases the solubility of the solute.

4. A SllP/':/l!i.AnJeA-n-t:> can be formed using temperature effects. This type of solutioncontains more dissolved solute than a saturated solution at the same temperature.

5. Pressure affects the solubility ofa gas. The solubility ofa gas in any solvent increases as itsexternal pressure increases.

Solubility Curves

I. Solubility curves can be used to predict the mass of a solute that will dissolve in IOOgof water atvarious temperatures.

***** Using the solubility curves provided, answer the following questions:

a. Determine the solubility of potassium nitrate (KN03) at 60°C.

II S J-b. Determine which salt, ammonium chloride (NH4CI) or potassium chloride (KCI) has a greater

solubility at OOC.

c. Determine which of the following salt's solubility in water is most temperature dependent: sodiumnitrate (NaN03), potassium chloride (KCI) or sodium chloride (NaCl).

t--.I~JJ03

***** A saturated solution of NaN03 in IOOgof water at 60°C is cooled to 10°C. Determine the mass, ingrams, of solute that will precipitate out of solution.

***** A solution ofNH4CI in IOOgof water at 60°C contains 52.0g of solute. This solution isvtJ S4TVRAW D

***** A solution of HCI in lOOg of water at 40°C contains 61.0g of solute. This solution is..5,+nJt<A n:-:D

***** A solution of KCI03 in IOOgof water at 80°C contains 45.0g of solute. This solution is~uP8Z.~Tk~

SOLUTION CONCENTRATION

1. The totJCk~'TlO,J ofa solution is a measure of how much solute is dissolved in a specificamount of solvent or solution.

2. Concentration may be described qualitatively as eitherCUI\lCk'")J~ 11::J:> or D IL u 1l:::but these are not useful to a chemist.

Using Percent to Describe Concentration

I. Concentration expressed as a percent is a ratio of a measured amount of solute to a measuredamount of solution.

2. Percent by mass is the ratio of the solute's mass to the solution's mass:MASS SOLvTl:;-

MASS SoLuno,.J(100)

MAS~ o~ SoLVn ON .: W\A Sso;:: ScJLu IE" + flIIA.n CFF !oLV cxr r=***** What is the percent by mass ofNaHC03 in a solution containing 20.0g NaHC03 dissolved in 600.0mL of water? I./.1 OJ

j I~l-O .: £toO.O n1 L 't Id~)O'>1 L z: ~OO. OJ

~O. 0 :: & io. OJ

;/0.00l.t LtJ, 0 (IUU) .:

***** A student has 1500.0g of a bleach solution. The percent by mass of the solute, sodium hypochlorite(NaOCI) is 3.62%. Determine the mass ofNaOCI in the solution.

X

x t: s tf. ~J-3. Percent by volume usually describes solutions in which both solute and solvent are liquids.

"'OLuM G .sou.) 113"'(IIJD)

SOLUnIJ,J

VVL\)YYlLE- af=' S()L-vnoA!.:- \/OLUm~ SOLlJn: i "0 LV M~ SO Lv'l:?J 'T***** Determine the percent by volume of ethanol in a solution that contains 35.0 mL of ethanol dissolvedin 115.0 mL of water.

3$,0- ----

(H'S-+ 3 s-)

***** A student has 100.0 mL of30.0% aqueous solution of oil in gasoline. Determine the volumes of theoil and the gasoline present in the solution.

/00,0('0 (»)

')( .:: 30, OM L 0 I L

loa. 0 - .ie:!. () == 70,0 w. Ll 3.00 z

Molarity

1. MOLA~ \ T'-f is the number of moles of solute dissolved per liter of solution.

2. M

3. To calculate a solution's molarity, one must know the volume of the solution and the amount ofdissolved solute:

(11OLl? s 0 F -SOL.v n=

***** Determine the molarity of an aqueous solution containing 40.0g of glucose (C6H1206) in 450.0 mLof solution.

+se.o WI L:- O.l.{SO L

1V\u--Q Ci. I~IL0 (, = -to, 0J CeJI,z. O~ \ I",~ c.. I-I'L O,~• ;J:J-:J h1JI -I PO. 1"-1 Ct, !l,LO(; -

• ol..Ol ;).l\.loLI~ \ 'r-( - -- O. lfq3 11f...4S0 -

***** A bleach solution ofNaOCI has a concentration ofO.128M. Determine the mass ofNaOCI in thesolution if its volume is 1.00 L.

. rz 8r.o o t,

y- 128 MuJdL ~ 11'uJ Jr", a cJ 171/,1./ tfs tJ~oC/

In-,,.f «l« Gel

Preparing Molar Solutions

I. Many lab solutions are prepared by diluting stock solutions purchased from a supply company.

2. To determine the volume of stock solution that must be diluted use:

MJ\j, .;:.M z, \/2-~ ,YloiAlLn't .).TClC-\c:.. SuL.\JDOrJ

I/OWWlG DC S:m(¥ .5:OLV·noIJ \]a. I' VOLU"ftI,e D1Lun; ,sQLUno"'-'

***** Determine the volume, in mL, of a 3.00M KI stock solution needed to make 0.300 L ofa 1.25M KIsolution. M ' I M V

, '\I r::: z, t:

{~,oo)\J, : (l'dS)(,30o)

(I. J.S)(. 3(0)

(Yj a.: IV(OLAe. t rf DiLVTE

./;).S L iss ; l

***** Determine the molarity ofa dilute solution made by starting with 20.0 mL ofa 3.50M stock solutionand diluting it to 100.0 mL.

M/ VI :- M2- v;(3 • .5)( so. 0): 1\1d ( 100.0")

(~.s) (1.0, 0")

(IOO,O)

COLLIGATIVE PROPERTIES OF SOLUTIONS

z: 0, "70 fY/

1. Solutes affect some of the physical properties of their solvents.

2. Most of these effects are caused by the number of particles in the solution rather than the type ofparticle in solution.

3. Physical properties of solutions that are affected by the number of particles, but not the identity ofdissolved solute particles are called WLLI~A'n v(; fMPI.:<-a:n(;;,5

4. Colligative properties include:

Electrolytes and Colligative Properties

I. Ionic compounds are called electrolytes because they dissociate in water to form a solution thatconducts an electric current.

2. Some molecular compounds also ionize in water.

3. Electrolytes that produce many ions are called strong electrolytes while those that produce only afew ions in solution are called weak electrolytes.

4. Compare the dissolution of 1 mole ofNaCI and I mole ofCaCh in water:

.IONS

Vapor Pressure Lowering

I. Vapor pressure is he pressure exerted in a closed container by liquid particles that have escapedthe liquid's surface and entered the gaseous state.

2. Adding a nonvolatile solute (one that has little tendency to become a gas) to a solvent lowers thesolvent's vapor pressure.

3. This is because the solvent's surface is now a combination of solvent particles and solute particles.There are fewer solvent particles available to escape from the surface so fewer particles enter thegaseous state and the vapor pressure is lowered.

4. Thus, IfAJb2. P.et.~ utI,; LOW!£/N <:>isdue to the number of solute particles in solution.

Boiling Point Elevation

I. When the temperature of a solution containing a nonvolatile solute is raised to the boiling point ofthe pure solvent, the resulting vapor pressure is still less than atmospheric pressure and thesolution will not boil.

2. The solution must be heated to a higher temperature to supply the additional kinetic energy neededto raise the vapor pressure to atmospheric pressure.

3. The temperature difference between a solution's boiling point and a pure solvent's boiling point iscalled f.IolLIN 6 Po"..1 T I:::l.l.::-V An 0 Ai

maLA L BOILI,J II Pol p. T"

k/b :: l-LL:>fAnv;J CO,\! S'/'f\N T (07'1'\11)It can be expressed as:4.

at t:: blSSoCltlno,J r:A C-TVR..

(~ OF PMnc.u;:-s)

***** Determine the new boiling point ofa 2.75m solution ofNaOH in water.

II16 z: ~,df t<b N8-J brs: (.SI2.)(,;l)(~7~) z: d.3/C,"C

Freezing Point Depression

I. In a solution, the solute particles interfere with the attractive forces among the solvent particles.

2. This prevents the solvent from entering the solid state at its normal freezing point.

3. The freezing pint of a solution is always lower than that of a pure solvent.

4. A solution's FttEEtl,J6 Po/,...\T DEfJ{l£Sf6.er/, is the difference between its freezing point andthe freezing point of its pure solvent.

5. It can be expressed as: ~ If :: .l,\dt' 1<'+<:At ': DISSOCI4 no~ i=At..YOe.. (.f:t OF P..4.~Tl(LbS")

W\ z: MOLAUN M= .sOLVTlu,'-f

It!,.. (0 ~~1)" = h-\OLAL ~~/JoJ(' i'ollJT 1>l?PflksSlo,J evNSmAi'T /r.

***** Determine the freezing point of a O.029m aqueous solution of sodium chloride (NaC\).

NaCI(s) - Na" (aq) + CI- (aq)

( •0 2-q) ( 01) ( ~I n,) =Mr- :: M dt/<t =

~f = 0.00 - .101 8'J o= - 0,1/ C

***** A lab technician determines the boiling point elevation of an aqueous solution ofa nonvolatile,nonelectrolyte to be 1.12oC. Determine the solution's molality.

ciTb :: MeA ~ kb1.1a :: ( I) (. S 12..) !vi

I, I 2.- ::•s I 2..M -

Osmosis and Osmotic Pressure

1. 05 t\10S l S is the diffusion of solvent particles across a semi permeable membrane from anarea of higher solvent concentration to an area oflower solvent concentration.

HETEROGENEOUS MIXTURES

1. Not all mixtures are solutions.

2. Heterogeneous mixtures contain substances that exist in distinct phases.

Suspensions

1. A SuSPeJ.SIO,J is a mixture containing particles that settle out if left undisturbed.

2. Suspensions can also be separated using a filter.

1. Particles in a suspension are larger than atoms, while those in solution are atomic-scale in size.

2. A CO L.LO ID is a heterogeneous mixture of intermediate size particles. These are betweenthe size of solution particles and suspension particles.

3. Colloids are classified according to the phases of their dispersed particles and dispersing mediums.

Brownian Motion

1. Colloid particles make jerky, random movements. This erratic movement of colloid particles iscalled B.'2..0WN14tJ MOll 0 ""

2. Brownian motion results from collisions of particles of the dispersion medium with the dispersedparticles. These collisions prevent the colloid particles from settling out of the mixture.

The Tyndall Effect

I. Whereas concentrated colloids are often cloudy or opaque, dilute colloids sometimes appear asclear as solutions.

2. Dilute colloids appear to be homogeneous because their dispersed particles are so small that theycannot be seen by the unaided eye.

3. Dispersed colloidal particles are large enough to scatter light, a phenomenon known as the TyndallEffect.

4. A solution will never scatter light.

ACIDS & BASES

AN INTRODUCTION

Properties of Acids and Bases

1. Acidic solutions taste sour. A base tastes bitter and feels slippery.

2. Acids react with some metals to form hydrogen.

in (s) + J.. H CJ (1Mb) -~ In CJfJ (~) of lI."J.fJJ

3. Metal carbonates and hydrogen carbonates react with aqueous solutions of acids to produce carbondioxide.

4. Aqueous solutions of acids turn blue litmus paper pink. Aqueous solutions of bases turn redlitmus paper blue.

5. Both acid and base solutions will conduct electricity.

Ions in Solution

1. All aqueous solutions contain hydrogen ions (W) and hydroxide ions (OH-). The relative amountsof the two ions determine whether an aqueous solution is acidic, basic or neutral.

2. An acidic solution contains more hydrogen ions than hydroxide ions. A basic solution containsmore hydroxide ions than hydrogen ions.

MiD 64Mi3. Water self-ionizes as follows:

This simplifies to:

l/tJ 0 (1)

4. Water is neutral since equal numbers of H" and OH- ions are present.

The Arrhenius Model of Acids and Bases

I. The Arrhenius model states that an acid is a substance that contains hydrogen and ionizes toproduce hydrogen ions in an aqueous solution:

STRENGTHS OF ACIDS AND BASES

Strengths of Acids

I. The strength of an acid depends upon how completely it ionizes. -t --2. A .sTJlOAi6 J)CI D will ionize completely. ,~t!kt;)f 1-/;.0(1) -7 1-130 rfl}J -f CI (o.r

Other strong acids are:,f(~10l/) HrJ03) I+J., HZ.S04, H£rj I-1C10...2

3. An acid that ionizes only artially in dilute aqueous solutions is called a tJl.::-r:JiC 4ClJ;>H(LI-h ()z. Cc<i) of-- i-fz.O () £,? !h 0-1 ({L~) j- Cz..Ha ().e. - (P-j)

Other weak acids gre:

Strengths of Bases

1. S17lJJiJ6 B/.ISl:;:....s dissociate entirely into metal ions and hydroxide ions in aqueous

NOL014s~~tio~ AI"" t (~)1 OJ.l-raj)Other strong bases are:

I~OI-J, IQbO~, CsOl-l

2. W8t ~ e4.sES

C l~J ",U+2- .,.. Hz.0 (~)dissociate partially in dilute aqueous solutions.

~ (2L-/-3 tJf.b T (~) f 0/1 -rNj))

WHAT IS pH?

pH and pOH

1. The acidity of a solution is based on the concentration of H+ ions in the solution. Since thisnumber often small and expressed in scientific notation, chemists adopted an easier way based oncommon logarithms.

2. This scale, the f ~ .seAL~

f \-\~ - lcc. [\4+Jj ~

, is based on the following:

.nJ M 0 LI-}..e IT y

***** Determine the pH of solutions having the following H+ ion concentrations:

a. [W] = 1.00 x 10·2M

F4:7 -1o<J (1.0 'l/ 0 - ~) := ;). 00

b. [W] = 3.00 x 10·6M

c. [W] = 8.20 x 1O-6M

3. Sometimes chemists use a pOH scale to express the alkalinity (basicity) of a solution>

4. The relationship between pH and pOH is:

***** Determine the pH and pOH of aqueous solutions having the following ion concentrations:

a. [OH-] = 1.00 x 1O-6M

pOl4 <: -loj (,.C() ''''O-~)~PII;: lef· 00- c. 00 ~

b. [OH-] = 6.50 x 10-4M

f (J /-1 t: - J~ (~,So ¥. ( 0 - C{) z:

fll-:: IV. 00 ~ ,3. /9 .:c. [W] = 3.60 x 1O-9M

pl/::-AJ (3,(;'OyrO-1)::. ~~<f

;; O}/.: I If·00 - If. r.tt{ -= s..5 f..:,

d. [W] = 0.025M

3.19

/0, f-j

~.i'!J (.OOl,s): I. c 0

/l/,OO - I.•(PO = /;), V 0

5. One can calculate the [H+] and [OH-] concentrations ofa solution if the pH is known.

***** Determine the [W] and [OH-] concentrations of a solution that has a pH = 7.40.

po~ = ~·~O- 7. C/O

LH-t J s: 10

10- (".(;0

::

z-7

~, S ( Y to fY1

NEUTRALIZATION

The Reaction Between Acids and Bases

1. A neutralization reaction is a reaction in which an acid and a base react in aqueous solution toproduce a salt and water.

2. A salt is an ionic compound made up of a cation from a base and an anion from an acid.

Acid - Base Titration

I. The stoichiometry of an acid-base neutralization reaction is the same as that of any other reactionthat occurs in solution.

2. Stoichiometry provides the basis for a procedure called TITIZA no;J ,which is used todetermine the concentrations of acidic and basic solutions.

3. Titration is a method for determining the concentration of a solution by reacting a known volumeof the solution with a solution of known concentration.

4. The titration process continues until the reaction reaches the stoichiometric point. Thestoichiometric point is the point at which the moles of H+ ions from the acid equal the moles ofOH- ions from the base.

5. The stoichiometric point is known as the GQIJI ALaJCE 'FtJ/N T of the titration.

4

12

10

pH g

+- ---.Equivelence PointAll ecld converted to conjugate bsee

6 Helf-equivelence PointpH = pKa = 4.75

2

OTO--~--'~-~-~~--T--~1~2--~-~1r'--~-~20Tihn\ "'.'"rno (rrL)

6. Chemists often use a chemical dye to detect the equivalence point rather than a pH meter.

7. Chemical dyes whose colors are affected by acidic and basic solutions are calledf\-CtD- I3A$G J:NDICI!fTOR....J .

8. The point at which the indicator used in a titration changes color is called the~7iD ?olAfr of the titration. The color change of the indicator selected for an acid-

base titration should coincide closely with the equivalence point of the titration.

9. Various indicators are needed because the equivalence point is not always when pH = 7.00. Theequivalence point can be higher or lower depending on the strengths of the acids and basesinvolved.

PHASE DIAGRAMS - HEATING CURVES - ENERGY DIAGRAMS:::Q..\~6 ~

I. A phase diagram looks like: 'Pe~O'(jsr 0~~o 'eP(J\.),I\~ \...-\ t-l~

/ 1'\-1'''-

'~

c.R.mCi.\LFOI,.Jr

\

..1lY1.&mSI8LE TO LIQUEFY)I1tf €>AS /.).'3.ovi: TJ-IIJ

{>OIJJr

LI 90 1.D

SbLIO

2. A heating/cooling curve looks like:

Tern p eratur e

gas

liquid

1 aporiz ati onB oilin g temperature not change,

due to he at of vaporization

Melting, t emper etu e not change,s id due to he at. of fusi en

Temperature risestime

3. An energy diagram looks like:

EXOTHERMIC activatedcomplex--------------

1 activationPotential...,:._e_n_e_rg_y__ .energy reactants(kJ)

----------------------~--~~reaction pathway

Fof2. AN /;:f,U::oTti E bl'\lC

PfUJU3S. )11-1 ~ f!Wove..JeNetl6 'f Flt-.llS' HES

1-4-1&H ~ Il-{ArJ n+ 1:REAc..If.lMI &J...i02..G'f

30 Solubility Curves for Selected Solutes

150

Q)-::l-oU)

••••oU)E~G

/IV /III /

/v f v

/ V;/V 1/./J i

NaN 03/ / / \'..1-Y / nn"'3

/

/V /~

~<, HCI / UH"C "-/ V

\~

/ -:

\ N V"'" NH3/ ~ ></ I< CI"

~,/ ----- V--

> V V-- =<;~~ '-- L----- Na ---

Y' x •....• /'

/----V...--1' c>: <. -:

:::=--V---....></ ...... xcu )3

---- ---- ---'"!' <,»: v <, I'-....-......r-, .>:<;-----

~ r<-.........----- /S( )2 ----- r--.

140130

120

110

70

60

50

40

3020

10

o 10 20 30 40 50 60 70 80 90 100

Temperature (DC)

Chemistry Transparency 30© 1989 Prentice Hall, Inc.