Chapter04 Chemical Quantities and Acqueuos Reactions

Chapter 4Chemical

Quantities and Aqueous Reactions

2008, Prentice Hall

Chemistry: A Molecular Approach, 1st Ed.Nivaldo Tro

Roy KennedyMassachusetts Bay Community

CollegeWellesley Hills, MA

Tro, Chemistry: A Molecular Approach 2

Reaction Stoichiometry• the numerical relationships between chemical amounts

in a reaction is called stoichiometry• the coefficients in a balanced chemical equation

specify the relative amounts in moles of each of the substances involved in the reaction

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

2 molecules of C8H18 react with 25 molecules of O2

to form 16 molecules of CO2 and 18 molecules of H2O

2 moles of C8H18 react with 25 moles of O2

to form 16 moles of CO2 and 18 moles of H2O2 mol C8H18 : 25 mol O2 : 16 mol CO2 : 18 mol H2O


Predicting Amounts from Stoichiometry• the amounts of any other substance in a chemical

reaction can be determined from the amount of just one substance

• How much CO2 can be made from 22.0 moles of C8H18 in the combustion of C8H18?

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)2 moles C8H18 : 16 moles CO2

2188

2188 CO moles 176

HC mol 2CO mol 16HC moles 22.0


Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline

• assuming that gasoline is octane, C8H18, the equation for the reaction is:

2 C8H18(l) + 25 O2(g) 16 CO2(g) + 18 H2O(g)

• the equation for the reaction gives the mole relationship between amount of C8H18 and CO2, but we need to know the mass relationship, so the Concept Plan will be:

g C8H18 mol CO2 g CO2mol C8H18

Example – Estimate the mass of CO2 produced in 2004 by the combustion of 3.4 x 1015 g gasoline

since 8x moles of CO2 as C8H18, but the molar mass of C8H18 is 3x CO2, the number makes sense

1 mol C8H18 = 114.22g, 1 mol CO2 = 44.01g, 2 mol C8H18 = 16 mol CO2

3.4 x 1015 g C8H18

g CO2

Check:

Solution:

Concept Plan:

Relationships:

Given:Find:

g 114.22mol 1

216

2

2

188

2

188

188188

15

CO g 101.0

CO mol 1CO g 44.01

HC mol 2CO mol 16

HC g 114.22HC mol 1HC g 10.43

188

2HC mol 2

CO mol 61

g C8H18 mol CO2 g CO2mol C8H18

mol 1g 44.01


Practice• According to the following equation, how

many milliliters of water are made in the combustion of 9.0 g of glucose?

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)1. convert 9.0 g of glucose into moles (MM 180)2. convert moles of glucose into moles of water3. convert moles of water into grams (MM 18.02)4. convert grams of water into mL

a) How? what is the relationship between mass and volume?

density of water = 1.00 g/mL


Practice

OH mL 5.4

OH g 1.00OH mL 1 x

OH mole 1OH g 18.0 x

OHC mole 1 OH mole 6 x

g 10 x 80.1OHC mole 1 x OHC g 0.9

2

2

2

2

2

6126

22

61266126

According to the following equation, how many milliliters of water are made in the combustion of 9.0 g of glucose?

C6H12O6(s) + 6 O2(g) 6 CO2(g) + 6 H2O(l)


Limiting Reactant• for reactions with multiple reactants, it is likely that

one of the reactants will be completely used before the others

• when this reactant is used up, the reaction stops and no more product is made

• the reactant that limits the amount of product is called the limiting reactantsometimes called the limiting reagent the limiting reactant gets completely consumed

• reactants not completely consumed are called excess reactants

• the amount of product that can be made from the limiting reactant is called the theoretical yield


Things Don’t Always Go as Planned!

• many things can happen during the course of an experiment that cause the loss of product

• the amount of product that is made in a reaction is called the actual yieldgenerally less than the theoretical yield, never more!

• the efficiency of product recovery is generally given as the percent yield

%100yield ltheoretica

yield actualYieldPercent


Limiting and Excess Reactants in the Combustion of Methane

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)• Our balanced equation for the combustion of methane

implies that every 1 molecule of CH4 reacts with 2 molecules of O2

H

HC

H

H+

O

O

C +OO

OO+

OH H

OH H

+

11


• If we have 5 molecules of CH4 and 8 molecules of O2, which is the limiting reactant?

H

HC

H

H+

OO

OO

OO

OO

OO

OO

OO

OO

?H

HC

H

H

H

HC

H

H

H

HC

H

H H

HC

H

H

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

12


H

HC

H

H

H

HC

H

H+

OO

OO

OO

OO

OO

OO

OO

OO

H

HC

H

H

H

HC

H

H

H

HC

H

H

24

24 CO molecules 16

CH molecules 1CO molecules 2CH molecules 8

CH4(g) + 2 O2(g) CO2(g) + 2 H2O(g)

22

22 CO molecules 10

O molecules 2CO molecules 2O molecules 10

since less CO2 can be made from the O2 than the CH4, the O2 is the limiting reactant

Example 4.4Finding Limiting Reactant,

Theoretical Yield, and Percent Yield


Example:• When 28.6 kg of C are allowed to react with 88.2 kg of

TiO2 in the reaction below, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield.

(g)(s)(s)(s) CO 2 Ti C 2 TiO2


Example:When 28.6 kg of C reacts with 88.2 kg of TiO2, 42.8 kg of Ti are obtained. Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)

• Write down the given quantity and its units.

Given: 28.6 kg C88.2 kg TiO2

42.8 kg Ti produced


• Write down the quantity to find and/or its units.

Find: limiting reactanttheoretical yieldpercent yield

Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield.

TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti

17

• Write a Concept Plan:

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg

TiFind: Lim. Rct., Theor. Yld., % Yld.

Example:Find the Limiting Reactant, Theoretical Yield, and Percent Yield. TiO2(s) + 2 C(s) Ti(s) + 2 CO(g)

kgTiO2

kgC

2

2TiO g 9.877

TiO mol 1

C g .0112C mol 1

molC

molTiO2

molTi

molTi

2TiO mol 1Ti mol 1

C mol 2Ti mol 1 }smallest

amount isfrom

limitingreactant

gTiO2

gC

kg 1g 1000

kg 1g 1000

smallestmol Ti g Ti

Ti mol 1g 87.47 % Yield

yield theor.yield act. yield %

kg TiT.Y.g 000 1

kg 1

18

• Collect Needed Relationships:

1000 g = 1 kgMolar Mass TiO2 = 79.87 g/mol

Molar Mass Ti = 47.87 g/molMolar Mass C = 12.01 g/mol

1 mole TiO2 : 1 mol Ti (from the chem. equation)

2 mole C 1 mol Ti (from the chem. equation)

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg

TiFind: Lim. Rct., Theor. Yld., % Yld.CP: kg rct g rct mol rct mol Tipick smallest mol Ti TY kg Ti %Y Ti



Ti mol 104301.1TiO mol 1

Ti mol 1TiO g 79.87TiO mole 1

kg 1g 0001TiO kg 8.28 3

22

22

• Apply the Concept Plan:

InformationGiven: 28.6 kg C, 88.2 kg TiO2, 42.8 kg Ti Find: Lim. Rct., Theor. Yld., % Yld.CP: kg rct g rct mol rct mol Tipick smallest mol Ti TY kg Ti %Y Ti Rel: 1 mol C=12.01g; 1 mol Ti =47.87g; 1 mol TiO2 = 79.87g; 1000g = 1 kg; 1 mol TiO2 : 1 mol Ti; 2 mol C : 1 mol Ti


Ti mol 100791.1C mol 2Ti mol 1

C g 12.01C mole 1

kg 1g 0001C kg 8.62 3

smallest moles of TiLimiting Reactant



Ti kg 52.9 g 1000

kg 1mol 1

Ti g 47.87Ti mol 104301.1 3

Theoretical Yield





YieldPercent 100Yield lTheoretica

Yield Actual %

%9.08%100Ti kg 52.9Ti kg 42.8




• Check the Solutions:

Limiting Reactant = TiO2

Theoretical Yield = 52.9 kgPercent Yield = 80.9%

Since Ti has lower molar mass than TiO2, the T.Y. makes senseThe Percent Yield makes sense as it is less than 100%.




Practice – How many grams of N2(g) can be made from 9.05 g of NH3 reacting with 45.2 g of CuO?

2 NH3(g) + 3 CuO(s) → N2(g) + 3 Cu(s) + 3 H2O(l)



1 mol NH3 = 17.03g, 1 mol CuO = 79.55g, 1 mol N2 = 28.02 g2 mol NH3 = 1 mol N2, 3 mol CuO = 1 mol N2

9.05 g NH3, 45.2 g CuOg N2

Concept Plan:

Relationships:

Given:Find:

g 17.03mol 1

3

2

NH mol 2N mol 1

g NH3 mol N2mol NH3

g 28.02mol 1

g CuO mol N2mol CuO

CuO mol 3N mol 1 2

g 79.55mol 1

g N2smallest moles N2



units are correct, and since there are fewer moles of N2 than CuO in the reaction and N2 has a

smaller mass, the number makes sense

Check:

Solution:

23

2

3

33 N mol .2660

NH mol 2N mol 1

NH g 17.03NH mol 1NH g .059

22 N mol .1890

CuO mol 3N mol 1

CuO g 79.55CuO mol 1CuO g 5.24

22

22 N g .305

N mol 1N g 28.02N mol .1890


Solutions• when table salt is mixed with water, it seems to disappear,

or become a liquid – the mixture is homogeneous the salt is still there, as you can tell from the taste, or simply

boiling away the water

• homogeneous mixtures are called solutions• the component of the solution that changes state is called

the solute• the component that keeps its state is called the solvent

if both components start in the same state, the major component is the solvent


Describing Solutions• since solutions are mixtures, the composition can

vary from one sample to anotherpure substances have constant compositionsalt water samples from different seas or lakes have

different amounts of salt • so to describe solutions accurately, we must

describe how much of each component is presentwe saw that with pure substances, we can describe

them with a single name because all samples identical


Solution Concentration• qualitatively, solutions are often

described as dilute or concentrated

• dilute solutions have a small amount of solute compared to solvent

• concentrated solutions have a large amount of solute compared to solvent

• quantitatively, the relative amount of solute in the solution is called the concentration


Solution ConcentrationMolarity

• moles of solute per 1 liter of solution• used because it describes how many molecules

of solute in each liter of solution

L)(in solution ofamount moles) (in solute ofamount M molarity,


Preparing 1 L of a 1.00 M NaCl Solution

Example 4.5 – Find the molarity of a solution that has 25.5 g KBr dissolved in 1.75 L of solution

since most solutions are between 0 and 18 M, the answer makes sense

1 mol KBr = 119.00 g, M = moles/L

25.5 g KBr, 1.75 L solutionMolarity, M

Check:• Check

Solution:• Follow the Concept Plan to Solve the problem

Concept Plan:

Relationships:

• Strategize

Given:Find:

• Sort Information

g 119.00mol 1

M 0.122L 1.75

KBr mol 29421.0solution L

KBr moles M molarity,

KBr mol 2940.21 KBr g 119.00

KBr mol 1KBr g 5.52

g KBr mol KBr

L sol’nML

mol M


Using Molarity in Calculations• molarity shows the relationship between the

moles of solute and liters of solution• If a sugar solution concentration is 2.0 M, then

1 liter of solution contains 2.0 moles of sugar 2 liters = 4.0 moles sugar 0.5 liters = 1.0 mole sugar

• 1 L solution : 2 moles sugar

solution L 1sugar mol 2

sugar mol 2solution L 1

Example 4.6 – How many liters of 0.125 M NaOH contains 0.255 mol NaOH?

since each L has only 0.125 mol NaOH, it makes sense that 0.255 mol should

require a little more than 2 L

0.125 mol NaOH = 1 L solution

0.125 M NaOH, 0.255 mol NaOHliters, L

Check:• Check


Concept Plan:

Relationships:

• Strategize

Given:Find:


NaOH mol 0.125solution L 1

solution L 2.04 NaOH mol 0.125

solution L 1NaOH mol 552.0

mol NaOH L sol’n


Dilution• often, solutions are stored as concentrated stock

solutions• to make solutions of lower concentrations from these

stock solutions, more solvent is added the amount of solute doesn’t change, just the volume of

solutionmoles solute in solution 1 = moles solute in solution 2

• the concentrations and volumes of the stock and new solutions are inversely proportional

M1∙V1 = M2∙V2

Example 4.7 – To what volume should you dilute 0.200 L of 15.0 M NaOH to make 3.00 M NaOH?

since the solution is diluted by a factor of 5, the volume should increase by a

factor of 5, and it does

M1V1 = M2V2

V1 = 0.200L, M1 = 15.0 M, M2 = 3.00 MV2, L

Check:• Check


Concept Plan:

Relationships:

• Strategize

Given:Find:


22

11 VM

VM

L 1.00

Lmol3.00

L 200.0L

mol15.0

V1, M1, M2 V2


Solution Stoichiometry

• since molarity relates the moles of solute to the liters of solution, it can be used to convert between amount of reactants and/or products in a chemical reaction

Example 4.8 – What volume of 0.150 M KCl is required to completely react with 0.150 L of 0.175 M Pb(NO3)2 in the

reaction 2 KCl(aq) + Pb(NO3)2(aq) PbCl2(s) + 2 KNO3(aq)

since need 2x moles of KCl as Pb(NO3)2, and the molarity of Pb(NO3)2 > KCl, the volume of KCl should be more than 2x volume Pb(NO3)2

1 L Pb(NO3)2 = 0.175 mol, 1 L KCl = 0.150 mol, 1 mol Pb(NO3)2 = 2 mol KCl

0.150 M KCl, 0.150 L of 0.175 M Pb(NO3)2

L KCl

Check:• Check


Concept Plan:

Relationships:

• Strategize

Given:

Find:


23)Pb(NO L 1mol 0.175

KCl L .3500 mol .1500

KCl L 1)Pb(NO mol 1

KCl mol 2)Pb(NO L 1

mol 0.175)Pb(NO L .15002323

23

23)Pb(NO mol 1KCl mol 2

L Pb(NO3)2 mol KCl L KClmol Pb(NO3)2

mol 0.150KCl L 1

38

What Happens When a Solute Dissolves?• there are attractive forces between the solute particles holding

them together• there are also attractive forces between the solvent molecules• when we mix the solute with the solvent, there are attractive

forces between the solute particles and the solvent molecules• if the attractions between solute and solvent are strong enough,

the solute will dissolve


Table Salt Dissolving in WaterEach ion is attracted to the surrounding water molecules and pulled off and away from the crystalWhen it enters the solution, the ion is surrounded by water molecules, insulating it from other ions The result is a solution with free moving charged particles able to conduct electricity


Electrolytes and Nonelectrolytes• materials that dissolve

in water to form a solution that will conduct electricity are called electrolytes

• materials that dissolve in water to form a solution that will not conduct electricity are called nonelectrolytes


Molecular View of Electrolytes and Nonelectrolytes

• in order to conduct electricity, a material must have charged particles that are able to flow

• electrolyte solutions all contain ions dissolved in the water ionic compounds are electrolytes because they all dissociate

into their ions when they dissolve• nonelectrolyte solutions contain whole molecules

dissolved in the watergenerally, molecular compounds do not ionize when they

dissolve in water the notable exception being molecular acids


Salt vs. Sugar Dissolved in Water

ionic compounds dissociate into ions when they dissolve

molecular compounds do not dissociate when they dissolve


Acids• acids are molecular compounds that ionize when they

dissolve in water the molecules are pulled apart by their attraction for the waterwhen acids ionize, they form H+ cations and anions

• the percentage of molecules that ionize varies from one acid to another

• acids that ionize virtually 100% are called strong acidsHCl(aq) H+(aq) + Cl-(aq)

• acids that only ionize a small percentage are called weak acids

HF(aq) H+(aq) + F-(aq)


Strong and Weak Electrolytes• strong electrolytes are materials that dissolve

completely as ions ionic compounds and strong acids their solutions conduct electricity well

• weak electrolytes are materials that dissolve mostly as molecules, but partially as ionsweak acids their solutions conduct electricity, but not well

• when compounds containing a polyatomic ion dissolve, the polyatomic ion stays together

Na2SO4(aq) 2 Na+(aq) + SO42-(aq)

HC2H3O2(aq) H+(aq) + C2H3O2-(aq)


Classes of Dissolved Materials


Solubility of Ionic Compounds• some ionic compounds, like NaCl, dissolve very well in

water at room temperature• other ionic compounds, like AgCl, dissolve hardly at all

in water at room temperature• compounds that dissolve in a solvent are said to be

soluble, while those that do not are said to be insolubleNaCl is soluble in water, AgCl is insoluble in water the degree of solubility depends on the temperatureeven insoluble compounds dissolve, just not enough to be

meaningful


When Will a Salt Dissolve?

• Predicting whether a compound will dissolve in water is not easy

• The best way to do it is to do some experiments to test whether a compound will dissolve in water, then develop some rules based on those experimental resultswe call this method the empirical method


Compounds Containing the Following Ions are Generally Soluble

Exceptions(when combined with ions on the left the compound is insoluble)

Li+, Na+, K+, NH4+ none

NO3–, C2H3O2

– none

Cl–, Br–, I– Ag+, Hg22+, Pb2+

SO42– Ag+, Ca2+, Sr2+, Ba2+, Pb2+

Solubility RulesCompounds that Are Generally Soluble in Water


Compounds Containing the Following Ions are Generally Insoluble

Exceptions(when combined with ions on the left the compound is soluble or slightly soluble)

OH– Li+, Na+, K+, NH4+,

Ca2+, Sr2+, Ba2+

S2– Li+, Na+, K+, NH4+,

Ca2+, Sr2+, Ba2+

CO32–, PO4

3– Li+, Na+, K+, NH4+

Solubility RulesCompounds that Are Generally Insoluble


Precipitation Reactions• reactions between aqueous solutions of ionic

compounds that produce an ionic compound that is insoluble in water are called precipitation reactions and the insoluble product is called a precipitate

51

2 KI(aq) + Pb(NO3)2(aq) PbI2(s) + 2 KNO3(aq)


No Precipitate Formation = No Reaction

KI(aq) + NaCl(aq) KCl(aq) + NaI(aq)all ions still present, no reaction


Process for Predicting the Products ofa Precipitation Reaction

1. Determine what ions each aqueous reactant has2. Determine formulas of possible products

Exchange ions (+) ion from one reactant with (-) ion from other

Balance charges of combined ions to get formula of each product

3. Determine Solubility of Each Product in Water Use the solubility rules If product is insoluble or slightly soluble, it will precipitate

4. If neither product will precipitate, write no reaction after the arrow


Process for Predicting the Products ofa Precipitation Reaction

5. If either product is insoluble, write the formulas for the products after the arrow – writing (s) after the product that is insoluble and will precipitate, and (aq) after products that are soluble and will not precipitate

6. Balance the equation

Example 4.10 – Write the equation for the precipitation reaction between an aqueous solution of potassium carbonate and an aqueous solution of

nickel(II) chloride1. Write the formulas of the reactants

K2CO3(aq) + NiCl2(aq)

2. Determine the possible productsa) Determine the ions present

(K+ + CO32-) + (Ni2+ + Cl-)

b) Exchange the Ions(K+ + CO3

2-) + (Ni2+ + Cl-) (K+ + Cl-) + (Ni2+ + CO32-)

c) Write the formulas of the products cross charges and reduce

K2CO3(aq) + NiCl2(aq) KCl + NiCO3


3. Determine the solubility of each productKCl is soluble

NiCO3 is insoluble

4. If both products soluble, write no reactiondoes not apply since NiCO3 is insoluble


nickel(II) chloride


5. Write (aq) next to soluble products and (s) next to insoluble products

K2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)

6. Balance the EquationK2CO3(aq) + NiCl2(aq) KCl(aq) + NiCO3(s)


nickel(II) chloride


Ionic Equations• equations which describe the chemicals put into the water

and the product molecules are called molecular equations2 KOH(aq) + Mg(NO3)2(aq) 2 KNO3(aq) + Mg(OH)2(s)

• equations which describe the actual dissolved species are called complete ionic equations aqueous strong electrolytes are written as ions

soluble salts, strong acids, strong bases insoluble substances, weak electrolytes, and nonelectrolytes

written in molecule formsolids, liquids, and gases are not dissolved, therefore molecule form

2K+1(aq) + 2OH-1

(aq) + Mg+2(aq) + 2NO3

-1(aq) K+1

(aq) + 2NO3-1

(aq) + Mg(OH)2(s)


Ionic Equations• ions that are both reactants and products are called

spectator ions

2K+1(aq) + 2OH-1

(aq) + Mg+2(aq) + 2NO3

-1(aq) K+1

(aq) + 2NO3-1

(aq) + Mg(OH)2(s)

• an ionic equation in which the spectator ions are removed is called a net ionic equation

2OH-1(aq) + Mg+2

(aq) Mg(OH)2(s)


Acid-Base Reactions• also called neutralization reactions because the

acid and base neutralize each other’s properties2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)

• the net ionic equation for an acid-base reaction isH+(aq) + OH(aq) H2O(l)

as long as the salt that forms is soluble in water


Acids and Bases in Solution• acids ionize in water to form H+ ions

more precisely, the H from the acid molecule is donated to a water molecule to form hydronium ion, H3O+

most chemists use H+ and H3O+ interchangeably

• bases dissociate in water to form OH ionsbases, like NH3, that do not contain OH ions, produce OH by

pulling H off water molecules• in the reaction of an acid with a base, the H+ from the

acid combines with the OH from the base to make water• the cation from the base combines with the anion from

the acid to make the salt

acid + base salt + water

Common AcidsChemical Name Formula Uses Strength Perchloric Acid HClO4 explosives, catalyst Strong

Nitric Acid HNO3 explosives, fertilizer, dye, glue Strong

Sulfuric Acid H2SO4 explosives, fertilizer, dye, glue,

batteries Strong

Hydrochloric Acid HCl metal cleaning, food prep, ore refining, stomach acid Strong

Phosphoric Acid H3PO4 fertilizer, plastics & rubber,

food preservation Moderate

Chloric Acid HClO3 explosives Moderate

Acetic Acid HC2H3O2 plastics & rubber, food preservation, vinegar Weak

Hydrofluoric Acid HF metal cleaning, glass etching Weak Carbonic Acid H2CO3 soda water Weak

Hypochlorous Acid HClO sanitizer Weak Boric Acid H3BO3 eye wash Weak


Common BasesChemical

Name Formula Common Name Uses Strength

sodium hydroxide

NaOH lye,

caustic soda soap, plastic,

petrol refining Strong

potassium hydroxide

KOH caustic potash soap, cotton, electroplating

Strong

calcium hydroxide

Ca(OH)2 slaked lime cement Strong

sodium bicarbonate

NaHCO3 baking soda cooking, antacid Weak

magnesium hydroxide

Mg(OH)2 milk of

magnesia antacid Weak

ammonium hydroxide

NH4OH, {NH3(aq)}

ammonia water

detergent, fertilizer,

explosives, fibers Weak


HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)


Example - Write the molecular, ionic, and net-ionic equation for the reaction of aqueous nitric

acid with aqueous calcium hydroxide1. Write the formulas of the reactants

HNO3(aq) + Ca(OH)2(aq) 2. Determine the possible products

a) Determine the ions present when each reactant dissociates(H+ + NO3

-) + (Ca+2 + OH-) b) Exchange the ions, H+1 combines with OH-1 to make H2O(l)

(H+ + NO3-) + (Ca+2 + OH-) (Ca+2 + NO3

-) + H2O(l)c) Write the formula of the salt

cross the charges (H+ + NO3

-) + (Ca+2 + OH-) Ca(NO3)2 + H2O(l)


3. Determine the solubility of the saltCa(NO3)2 is soluble

4. Write an (s) after the insoluble products and a (aq) after the soluble products

HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + H2O(l)

5. Balance the equation2 HNO3(aq) + Ca(OH)2(aq) Ca(NO3)2(aq) + 2 H2O(l)


acid with aqueous calcium hydroxide



acid with aqueous calcium hydroxide6. Dissociate all aqueous strong electrolytes to get

complete ionic equation not H2O

2 H+(aq) + 2 NO3-(aq) + Ca+2(aq) + 2 OH-(aq) Ca+2(aq) +

2 NO3-(aq) + H2O(l)

7. Eliminate spectator ions to get net-ionic equation2 H+1(aq) + 2 OH-1(aq) H2O(l)

H+1(aq) + OH-1(aq) H2O(l)


Titration• often in the lab, a solution’s concentration is

determined by reacting it with another material and using stoichiometry – this process is called titration

• in the titration, the unknown solution is added to a known amount of another reactant until the reaction is just completed, at this point, called the endpoint, the reactants are in their stoichiometric ratio the unknown solution is added slowly from an

instrument called a burettea long glass tube with precise volume markings that

allows small additions of solution


Acid-Base Titrations• the difficulty is determining when there has been just

enough titrant added to complete the reaction the titrant is the solution in the burette

• in acid-base titrations, because both the reactant and product solutions are colorless, a chemical is added that changes color when the solution undergoes large changes in acidity/alkalinity the chemical is called an indicator

• at the endpoint of an acid-base titration, the number of moles of H+ equals the number of moles of OH aka the equivalence point


Titration

http://wps.prenhall.com/wps/media/objects/165/169763/Acid-BaseTitrationMovie.html


TitrationThe base solution is thetitrant in the burette.

As the base is added tothe acid, the H+ reacts withthe OH– to form water. But there is still excess acid present so the colordoes not change.

At the titration’s endpoint,just enough base has been added to neutralize all theacid. At this point the indicator changes color.


Example 4.14:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?

• Write down the given quantity and its units.

Given: 10.00 mL HCl

12.54 mL of 0.100 M NaOH


• Write down the quantity to find, and/or its units.

Find: concentration HCl, M

InformationGiven: 10.00 mL HCl

12.54 mL of 0.100 M NaOH



• Collect Needed Equations and Conversion Factors:

HCl(aq) + NaOH(aq) → NaCl(aq) + H2O(l)

1 mole HCl = 1 mole NaOH0.100 M NaOH 0.100 mol NaOH 1 L sol’n


12.54 mL of 0.100 M NaOH

Find: M HCl


solution literssolute molesMolarity


• Write a Concept Plan:

mLNaOH

LNaOH

molNaOH

NaOH L 1NaOH mol 1000.

mL 1L 0010.

molHCl

NaOH mol 1HCl mol 1


12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L

M = mol/L


mLHCl

LHCl

mL 1L 0010. HCl liters

HCl molesMolarity


NaOH mole 1HCl mol 1

L 1NaOH mol 1000

mL 1L 0.001NaOH mL 2.541

.

• Apply the Solution Map:

= 1.25 x 10-3 mol HCl


12.54 mL of 0.100 M NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L

M = mol/LCP: mL NaOH → L NaOH →

mol NaOH → mol HCl; mL HCl → L HCl & mol M

Example:The titration of 10.00 mL of HCl solution of unknown concentration requires 12.54 mL of 0.100 M NaOH solution to reach the end point. What is the concentration of the unknown HCl solution?


HCl L 010000mL 1

L 0.001NaOH mL 0.001 .



12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L




M 1250HCl L 0.01000

HCl moles 10 x 1.25Molarity-3

.


• Check the Solution:HCl solution = 0.125 M

The units of the answer, M, are correct.The magnitude of the answer makes sense since

the neutralization takes less HCl solution thanNaOH solution, so the HCl should be more concentrated.


12.54 mL NaOHFind: M HClCF: 1 mol HCl = 1 mol NaOH 0.100 mol NaOH = 1 L





Gas Evolving Reactions• Some reactions form a gas directly from the ion

exchangeK2S(aq) + H2SO4(aq) K2SO4(aq) + H2S(g)

• Other reactions form a gas by the decomposition of one of the ion exchange products into a gas and water

K2SO3(aq) + H2SO4(aq) K2SO4(aq) + H2SO3(aq)

H2SO3 H2O(l) + SO2(g)


NaHCO3(aq) + HCl(aq) NaCl(aq) + CO2(g) + H2O(l)


Compounds that UndergoGas Evolving Reactions

ReactantType

ReactingWith

Ion ExchangeProduct

Decom-pose?

GasFormed

Example

metalnS,

metal HS

acid H2S no H2S K2S(aq) + 2HCl(aq)

2KCl(aq) + H2S(g)

metalnCO3,

metal HCO3

acid H2CO3 yes CO2 K2CO3(aq) + 2HCl(aq)

2KCl(aq) + CO2(g) + H2O(l)

metalnSO3

metal HSO3

acid H2SO3 yes SO2 K2SO3(aq) + 2HCl(aq)

2KCl(aq) + SO2(g) + H2O(l)

(NH4)nanion base NH4OH yes NH3 KOH(aq) + NH4Cl(aq)

KCl(aq) + NH3(g) + H2O(l)


Example 4.15 - When an aqueous solution of sodium carbonate is added to an aqueous solution

of nitric acid, a gas evolves1. Write the formulas of the reactants

Na2CO3(aq) + HNO3(aq) 2. Determine the possible products

a) Determine the ions present when each reactant dissociates(Na+1 + CO3

-2) + (H+1 + NO3-1)

b) Exchange the anions(Na+1 + CO3

-2) + (H+1 + NO3-1) (Na+1 + NO3

-1) + (H+1 + CO3-2)

c) Write the formula of compounds cross the charges

Na2CO3(aq) + HNO3(aq) NaNO3 + H2CO3


3. Check to see either product H2S - No

4. Check to see if either product decomposes – Yes

H2CO3 decomposes into CO2(g) + H2O(l)

Na2CO3(aq) + HNO3(aq) NaNO3 + CO2(g) + H2O(l)


of nitric acid, a gas evolves


5. Determine the solubility of other productNaNO3 is soluble

6. Write an (s) after the insoluble products and a (aq) after the soluble products

Na2CO3(aq) + 2 HNO3(aq) NaNO3(aq) + CO2(g) + H2O(l)

7. Balance the equationNa2CO3(aq) + 2 HNO3(aq) NaNO3 + CO2(g) + H2O(l)


of nitric acid, a gas evolves


Other Patterns in Reactions• the precipitation, acid-base, and gas evolving

reactions all involved exchanging the ions in the solution

• other kinds of reactions involve transferring electrons from one atom to another – these are called oxidation-reduction reactionsalso known as redox reactionsmany involve the reaction of a substance with O2(g)

4 Fe(s) + 3 O2(g) 2 Fe2O3(s)


Combustion as Redox2 H2(g) + O2(g) 2 H2O(g)


Redox without Combustion2 Na(s) + Cl2(g) 2 NaCl(s)

2 Na 2 Na+ + 2 e

Cl2 + 2 e 2 Cl


Reactions of Metals with Nonmetals

• consider the following reactions:4 Na(s) + O2(g) → 2 Na2O(s)

2 Na(s) + Cl2(g) → 2 NaCl(s)

• the reaction involves a metal reacting with a nonmetal• in addition, both reactions involve the conversion of free

elements into ions 4 Na(s) + O2(g) → 2 Na+

2O– (s)

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)


Oxidation and Reduction• in order to convert a free element into an ion, the atoms

must gain or lose electronsof course, if one atom loses electrons, another must accept

them• reactions where electrons are transferred from one atom

to another are redox reactions• atoms that lose electrons are being oxidized, atoms that

gain electrons are being reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na → Na+ + 1 e– oxidationCl2 + 2 e– → 2 Cl– reduction

Leo

Ger


Electron Bookkeeping• for reactions that are not metal + nonmetal, or do

not involve O2, we need a method for determining how the electrons are transferred

• chemists assign a number to each element in a reaction called an oxidation state that allows them to determine the electron flow in the reactioneven though they look like them, oxidation states are

not ion charges!oxidation states are imaginary charges assigned based on a

set of rules ion charges are real, measurable charges


Rules for Assigning Oxidation States• rules are in order of priority1. free elements have an oxidation state = 0

Na = 0 and Cl2 = 0 in 2 Na(s) + Cl2(g)2. monatomic ions have an oxidation state equal

to their charge Na = +1 and Cl = -1 in NaCl

3. (a) the sum of the oxidation states of all the atoms in a compound is 0

Na = +1 and Cl = -1 in NaCl, (+1) + (-1) = 0


Rules for Assigning Oxidation States3. (b) the sum of the oxidation states of all the atoms in

a polyatomic ion equals the charge on the ion N = +5 and O = -2 in NO3

–, (+5) + 3(-2) = -1

4. (a) Group I metals have an oxidation state of +1 in all their compounds

Na = +1 in NaCl

4. (b) Group II metals have an oxidation state of +2 in all their compounds

Mg = +2 in MgCl2


Rules for Assigning Oxidation States5. in their compounds, nonmetals have oxidation

states according to the table below nonmetals higher on the table take priority

Nonmetal Oxidation State ExampleF -1 CF4

H +1 CH4

O -2 CO2

Group 7A -1 CCl4

Group 6A -2 CS2

Group 5A -3 NH3


Practice – Assign an Oxidation State to Each Element in the following

• Br2

• K+

• LiF

• CO2

• SO42-

• Na2O2


Practice – Assign an Oxidation State to Each Element in the following

• Br2 Br = 0, (Rule 1)

• K+ K = +1, (Rule 2)

• LiF Li = +1, (Rule 4a) & F = -1, (Rule 5)

• CO2 O = -2, (Rule 5) & C = +4, (Rule 3a)

• SO42- O = -2, (Rule 5) & S = +6, (Rule 3b)

• Na2O2 Na = +1, (Rule 4a) & O = -1, (Rule 3a)


Oxidation and ReductionAnother Definition

• oxidation occurs when an atom’s oxidation state increases during a reaction

• reduction occurs when an atom’s oxidation state decreases during a reaction

CH4 + 2 O2 → CO2 + 2 H2O-4 +1 0 +4 –2 +1 -2

oxidationreduction

http://wps.prenhall.com/wps/media/objects/165/169763/Oxidation-ReduxII.html


Oxidation–Reduction• oxidation and reduction must occur simultaneously

if an atom loses electrons another atom must take them

• the reactant that reduces an element in another reactant is called the reducing agent the reducing agent contains the element that is oxidized

• the reactant that oxidizes an element in another reactant is called the oxidizing agent the oxidizing agent contains the element that is reduced

2 Na(s) + Cl2(g) → 2 Na+Cl–(s)Na is oxidized, Cl is reduced

Na is the reducing agent, Cl2 is the oxidizing agent


Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ S + 2 NO + 4 H2O

MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O


Identify the Oxidizing and Reducing Agents in Each of the Following

3 H2S + 2 NO3– + 2 H+ S + 2 NO + 4 H2O

MnO2 + 4 HBr MnBr2 + Br2 + 2 H2O

+1 -2 +5 -2 +1 0 +2 -2 +1 -2

ox agred ag

+4 -2 +1 -1 +2 -1 0 +1 -2

oxidationreduction

oxidationreduction

red agox ag


Combustion Reactions• Reactions in which O2(g) is a

reactant are called combustion reactions

• Combustion reactions release lots of energy

• Combustion reactions are a subclass of oxidation-reduction reactions

2 C8H18(g) + 25 O2(g) 16 CO2(g) + 18 H2O(g)


Combustion Products• to predict the products of a combustion

reaction, combine each element in the other reactant with oxygen

Reactant Combustion Product

contains C CO2(g)

contains H H2O(g)

contains S SO2(g)

contains N NO(g) or NO2(g)

contains metal M2On(s)


Practice – Complete the Reactions

• combustion of C3H7OH(l)

• combustion of CH3NH2(g)


Practice – Complete the Reactions

C3H7OH(l) + 5 O2(g) 3 CO2(g) + 4 H2O(g)

CH3NH2(g) + 3 O2(g) CO2(g) + 2 H2O(g) + NO2(g)

Documents

Chapter04 Chemical Quantities and Acqueuos Reactions