Chapter 5 Reactions in Aqueous Solutions. Chapter goals Understand the nature of ionic substances dissolved in water.Understand the nature of ionic substances.

Chapter 5Chapter 5

Reactions in Aqueous Reactions in Aqueous SolutionsSolutions

Chapter goalsChapter goals• Understand the nature of ionic substances Understand the nature of ionic substances

dissolved in water.dissolved in water.• Recognize common acids and bases and Recognize common acids and bases and

understand their behavior in aqueous understand their behavior in aqueous solution.solution.

• Recognize and write equations for the Recognize and write equations for the common types of reactions in aqueous common types of reactions in aqueous solution.solution.

• Recognize common oxidizing and reducing Recognize common oxidizing and reducing agents and identify common oxidation-agents and identify common oxidation-reduction reactions.reduction reactions.

• Define and use the molarity in solution Define and use the molarity in solution stoichiometry. stoichiometry.

SolutionSolution• homogeneous mixturehomogeneous mixture

• can be gas, liquid, or solidcan be gas, liquid, or solid

• solvent: component present in highest solvent: component present in highest proportionproportion

exception - waterexception - water

• solute: component(s) in solution other solute: component(s) in solution other than solventthan solvent

We will mostly study aqueous solutions: We will mostly study aqueous solutions: human body is 2/3 water. human body is 2/3 water.

ExamplesExamples

• mixture of 35% naphthalene mixture of 35% naphthalene

and 65% benzeneand 65% benzene solvent - benzenesolvent - benzene solute – naphthalenesolute – naphthalene

• mixture of 10% ethanol, 40% methanol, mixture of 10% ethanol, 40% methanol, and 50% propanoland 50% propanol

solvent - propanolsolvent - propanol solute - ethanol and methanolsolute - ethanol and methanol

ExamplesExamples

• mixture of 40% ethanol, 40% methanol, mixture of 40% ethanol, 40% methanol, and 20% butanoland 20% butanol

solvent - ethanol/methanolsolvent - ethanol/methanol mixed solventmixed solvent solute – butanolsolute – butanol

• mixture of 40% ethanol, 50% propanol, and mixture of 40% ethanol, 50% propanol, and 10% water10% water

solvent - watersolvent - water solute - ethanol and propanolsolute - ethanol and propanol

Next…Next…We will focus on compounds thatWe will focus on compounds that

produce ions in aqueous solutions. produce ions in aqueous solutions.

They are named electrolytes and may be They are named electrolytes and may be

salts, acids, or bases.salts, acids, or bases.

SaltsSaltsSalts:Salts: ionic compounds made of cations other than ionic compounds made of cations other than

HH++ and anions other than OH and anions other than OH−− or O or O22−−, , OO2222−−

NaCl: NaNaCl: Na+ + & Cl & Cl−−

KK22SOSO44: K: K++ & SO & SO4422−−

FeBrFeBr33: Fe: Fe3+3+ & Br & Br−−

ZnZn33(PO(PO44))22: Zn: Zn2+2+ & PO & PO4433−−

Ca(HCOCa(HCO33))22: Ca: Ca2+2+ & HCO & HCO33−−

ElectrolyteElectrolyte• substance that dissolves to produce an substance that dissolves to produce an

electrically conducting mediumelectrically conducting medium• forms ions in solution (dissociates/ionizes)forms ions in solution (dissociates/ionizes)• examplesexamples soluble ionic compoundssoluble ionic compounds HH22OO KBr(s) KBr(s) K K++(aq)(aq) + Br+ Br––(aq)(aq)

HH22OO Acids, HCl(g) Acids, HCl(g) H H++(aq) + Cl(aq) + Cl––(aq)(aq)

bases, NHbases, NH33 + H + H22O NHO NH44++ + OH + OH––

double arrow means equilibriumdouble arrow means equilibrium

NonelectrolytesNonelectrolytes• do not form ions in solutiondo not form ions in solution

• do not form electrically conducting media do not form electrically conducting media upon dissolutionupon dissolution

• Examples: molecular compounds Examples: molecular compounds (alcohols, sugars & acetone)(alcohols, sugars & acetone)

HH22OO CHCH33OH(l)OH(l) CH CH33OH(aq) OH(aq) N.D.N.D.

Glucose CGlucose C66HH1212OO66(s) (s) CC66HH1212OO66(aq) (aq) N.D.N.D.

Sucrose CSucrose C1212HH2222OO1111(s) (s) CC1212HH2222OO1111(aq) (aq) N.D.N.D.

N.D. = no dissociation/ionizationN.D. = no dissociation/ionization

Types of ElectrolytesTypes of Electrolytes• Strong: dissociate ~100%Strong: dissociate ~100% most ionic compounds (soluble salts), strong most ionic compounds (soluble salts), strong

acids, and strong basesacids, and strong bases

HH22OOKBr(s) KBr(s) K K++(aq)(aq) + Br+ Br––(aq)(aq)HCl(g) HCl(g) HH++(aq)(aq) + Cl+ Cl––(aq)(aq)NaOH(s) NaOH(s) Na Na++(aq) + OH(aq) + OH−−(aq)(aq)

WeakWeak: insoluble salts, weak acids and bases, : insoluble salts, weak acids and bases, water water (H2O), and certain gases (e.g. CO, and certain gases (e.g. CO22))

HH22O O dissociate only slightly in waterdissociate only slightly in waterHF(g) HHF(g) H++(aq)(aq) + F+ F––(aq)(aq)Also acetic acid, CHAlso acetic acid, CH33COOH COOH

NHNH33 + H + H22O NHO NH44++(aq) + OH(aq) + OH––(aq)(aq)

Solubility of Ionic compounds in Water:Solubility of Ionic compounds in Water: Solubility RulesSolubility Rules

Soluble CompoundsSoluble Compounds 1.1. alkali metal salts (Lialkali metal salts (Li++, Na, Na++, K, K++, Rb, Rb++…, ) …, )

except potassium perchlorateexcept potassium perchlorate 2.2. ammonium (NHammonium (NH44

++) salts) salts 3.3. all nitrates(NOall nitrates(NO33

−−), chlorates (ClO), chlorates (ClO33−−), ),

perchlorates (ClOperchlorates (ClO44−−), and acetates (C), and acetates (C22HH33OO22

−−), ), except silver acetate and potassium except silver acetate and potassium perchlorateperchlorate

4. all Cl4. all Cl−−, , BrBr−−, and , and ll−− are soluble except for are soluble except for AgAg++, Pb, Pb2+2+, and Hg, and Hg22

2+2+ salts salts 5. all SO5. all SO44

2−2− are soluble except for Pb are soluble except for Pb2+2+, Sr, Sr2+2+, , and Baand Ba2+ 2+ saltssalts

Solubility of Ionic compounds in Water: RulesSolubility of Ionic compounds in Water: Rules

Insoluble or slightly soluble CompoundsInsoluble or slightly soluble Compounds

6.6. metal oxides (Ometal oxides (O22−−) except those of the ) except those of the alkali metals, Caalkali metals, Ca2+2+, Sr, Sr2+2+, and Ba, and Ba2+2+

7.7. hydroxides (OHhydroxides (OH−−) except those of the ) except those of the alkali metals, Baalkali metals, Ba2+2+, Sr, Sr2+2+, and NH, and NH44

++. Calcium . Calcium

hydroxide is slightly solublehydroxide is slightly soluble

8.8. carbonates, phosphates, sulfides, and carbonates, phosphates, sulfides, and sulfites except those of the alkali metals and sulfites except those of the alkali metals and the ammonium ion (NHthe ammonium ion (NH44

++))

9. for salts of Cr9. for salts of Cr22OO7722−−, P, P3−3−, , CrOCrO44

22−−, , CC22OO4422−−,,

assume they are insoluble except for IA assume they are insoluble except for IA metals and NHmetals and NH44

++ salts salts

Precipitation Reactions:Precipitation Reactions:A Driving Force in Chemical ReactionsA Driving Force in Chemical Reactions

• formation of insoluble solid (formation of insoluble solid (precipitateprecipitate, , ppt) is a common reaction in aqueous ppt) is a common reaction in aqueous solutions:solutions:

• reactants are generally water-soluble reactants are generally water-soluble ionic compoundsionic compounds

• once substances dissolve in water they once substances dissolve in water they dissociate to give the appropriate cations dissociate to give the appropriate cations and anionsand anions

• if the cation of one compound forms an if the cation of one compound forms an insoluble compound with the anion of insoluble compound with the anion of another, precipitation will occuranother, precipitation will occur

Precipitation Reaction:Precipitation Reaction:A Double Replacement (A Double Replacement (MetathesisMetathesis) Reaction) Reaction• Both ionic compounds trade partner ionsBoth ionic compounds trade partner ions ____________________ | || | AB(aq) + CD(aq) AB(aq) + CD(aq) AD(s) + CB(aq) AD(s) + CB(aq) |_______||_______| AD is an insoluble or slightly soluble salt AD is an insoluble or slightly soluble salt

AA++, B, B−−, C, C++, and D, and D−− are ions are ions

AgNOAgNO33(aq) + NaCl(aq) (aq) + NaCl(aq) AgCl(s)AgCl(s) + NaNO + NaNO33(aq)(aq)

weak electrolyteweak electrolyte (unionized (unionized precipitateprecipitate))

Precipitation Reaction:Precipitation Reaction:A Double Replacement (A Double Replacement (MetathesisMetathesis) Reaction) Reaction

A (solid) A (solid) precipitateprecipitate is formed. is formed.

Example: complete and balance theExample: complete and balance the equation equation

(NH(NH44))33POPO44(aq) + MgSO(aq) + MgSO44(aq) (aq) MgPOMgPO44(s) + NH(s) + NH44SOSO44(aq)(aq)

we will write the right subscripts laterwe will write the right subscripts later

Using the solubility rules, predict if at least oneUsing the solubility rules, predict if at least one

product is going to be insoluble in water.product is going to be insoluble in water.

According to rule 8, MgPOAccording to rule 8, MgPO44 (subscripts not (subscripts not

right) is not soluble in water. right) is not soluble in water.

Ions are MgIons are Mg2+2+, PO, PO4433−−, NH, NH44

++, and S, and SOO4422−−; ;

subscriptssubscripts(NH(NH44))33POPO44(aq) + MgSO(aq) + MgSO44(aq) (aq) Mg Mg33(PO(PO44))22(s) + (NH(s) + (NH44))22SOSO44(aq)(aq)

balancingbalancing2(NH2(NH44))33POPO44(aq) + 3MgSO(aq) + 3MgSO44(aq) (aq) Mg Mg33(PO(PO44))22(s) + 3(NH(s) + 3(NH44))22SOSO44(aq)(aq)

Example: complete and balance theExample: complete and balance the equationequation

• NaNa22SOSO44(aq) + BaBr(aq) + BaBr22(aq) (aq)

NaNa22SOSO44(aq) + BaBr(aq) + BaBr22(aq) (aq) BaSOBaSO44(s) + NaBr(s) + NaBr

NaNa22SOSO44 + BaBr + BaBr22 BaSOBaSO44(s)(s) + 2NaBr(aq) + 2NaBr(aq)

driving forcedriving force = formation of insoluble = formation of insoluble barium sulfate (barium sulfate (precipitateprecipitate))

• Os(NOOs(NO33))55(aq) + Rb(aq) + Rb22S(aq) S(aq)

Os(NOOs(NO33))55 + Rb + Rb22S S Os Os22SS55(s)(s) + RbNO + RbNO33(aq)(aq)

2 Os(NO2 Os(NO33))55 + 5 Rb + 5 Rb22S S Os Os22SS55(s)(s) + 10 RbNO + 10 RbNO33

driving forcedriving force = form. of insoluble Os = form. of insoluble Os5+5+ sulfide sulfide

((precipitateprecipitate))

Net Ionic Equations: Spectator Ions Net Ionic Equations: Spectator Ions The equationThe equation

AgNOAgNO33(aq) + NaCl(aq) (aq) + NaCl(aq) AgCl(s)AgCl(s) + NaNO + NaNO33(aq)(aq)

is not quite correct, because three salts are is not quite correct, because three salts are dissociated in ions while AgCl is a precipitate.dissociated in ions while AgCl is a precipitate.

AgAg++(aq) + (aq) + NONO33−−(aq) + Na(aq) + Na++(aq) (aq) + Cl+ Cl−−(aq) (aq) AgCl(s)AgCl(s) + + NaNa++(aq) + NO(aq) + NO33

−−(aq)(aq)

before reaction after reactionbefore reaction after reaction

NaNa++ and NO and NO33−− are present on both sides of equation, i.e., are present on both sides of equation, i.e.,

before and after reaction. They are called before and after reaction. They are called spectator ionsspectator ions; ; they do not participate in net reaction; they they do not participate in net reaction; they can be removedcan be removedfrom the equation, but they remain in the solution.from the equation, but they remain in the solution.

AgAg++(aq) + Cl(aq) + Cl−−(aq) (aq) AgCl(s) AgCl(s) is the net ionic equation is the net ionic equation

Net Ionic Equations: Spectator Ions Net Ionic Equations: Spectator Ions For two previous examples:For two previous examples:

2(NH2(NH44))33POPO44(aq) + 3MgSO(aq) + 3MgSO44(aq) (aq) Mg Mg33(PO(PO44))22(s) + 3(NH(s) + 3(NH44))22SOSO44(aq)(aq)

6NH6NH44++(aq) + 2PO(aq) + 2PO44

3−3−(aq) + 3Mg(aq) + 3Mg2+2+(aq) + 3SO(aq) + 3SO442−2−(aq) (aq) MgMg33(PO(PO44))22(s)(s) + +

beforebefore reaction reaction 6NH6NH44++(aq) + 3SO(aq) + 3SO44

2−2−(aq) (aq)

after reactionafter reaction

3Mg3Mg2+2+(aq) + 2PO(aq) + 2PO443−3−(aq) (aq) MgMg33(PO(PO44))22(s)(s) is the net equation is the net equation

spectator ions are eliminated from the equationspectator ions are eliminated from the equation ==============================================

NaNa22SOSO44(aq) + BaBr(aq) + BaBr22(aq) (aq) BaSOBaSO44(s)(s) + 2NaBr(aq) + 2NaBr(aq)

2Na2Na++(aq) + SO(aq) + SO442−2−(aq) + Ba(aq) + Ba2+2+(aq) + 2Br(aq) + 2Br−−(aq) (aq) BaSOBaSO44(s)(s) + 2Na + 2Na++(aq) + 2Br(aq) + 2Br−−

BaBa2+2+(aq) + SO(aq) + SO442−2−(aq) (aq) BaSO BaSO44(s)(s) net ionic equation net ionic equation

Net Ionic Equations: Spectator Ions Net Ionic Equations: Spectator Ions

For the metathesis reactionFor the metathesis reaction

2 KF(aq) + Pb(NO2 KF(aq) + Pb(NO33))22(aq) (aq) PbF PbF22((ss) + 2 KNO) + 2 KNO33(aq)(aq)

formula unit equationformula unit equation

spectator ions are eliminatedspectator ions are eliminated

2K2K++ + 2F + 2F–– + Pb + Pb2+2+ + 2 NO + 2 NO33–– PbF PbF22 + 2K + 2K++ + 2NO + 2NO33

––

ionic equationionic equation

PbFPbF22 is the precipitate is the precipitate

2F2F––(aq) + Pb(aq) + Pb2+2+(aq) (aq) PbF PbF22(s)(s)

net ionic equationnet ionic equation

Net Ionic Equations: Spectator Ions Net Ionic Equations: Spectator Ions

NHNH44Cl(aq) + KNOCl(aq) + KNO33(aq) (aq) NH NH44NONO33(aq) + KCl(aq)(aq) + KCl(aq)

NHNH44++ + Cl + Cl–– + K + K++ + NO + NO33

–– NH NH44++ + NO + NO33

–– + +

KK++ + Cl + Cl––

all ions are spectatorsall ions are spectators; all can be cancelled; all can be cancelled

no net ionic equationno net ionic equation

no driving force for reactionno driving force for reaction

N.R. (there is no reaction)N.R. (there is no reaction)

The two salts are just dissolved in water.The two salts are just dissolved in water.

Acids and Bases Acids and Bases AcidAcid

• Arrhenius definitionArrhenius definition substance that ionizes in water to produce substance that ionizes in water to produce

HH++, hydrogen ion, and hence increases the , hydrogen ion, and hence increases the concentration of this ionconcentration of this ion

HCl(aq) HCl(aq) H H++(aq) + Cl(aq) + Cl––(aq)(aq)

• Brønsted-Lowry definitionBrønsted-Lowry definition substance capable of donating Hsubstance capable of donating H++

HCl + HHCl + H22O O H H33OO++ + Cl + Cl––(aq)(aq)

Acids and Bases Acids and Bases BaseBase• Arrhenius definitionArrhenius definition substance that increases the concentration of substance that increases the concentration of

OHOH–– in aqueous solution in aqueous solution KOH(aq) KOH(aq) K K++(aq) + OH(aq) + OH––(aq) (aq)

NHNH33 + H + H22O NHO NH44++ + OH + OH––

• Brønsted/Lowry definitionBrønsted/Lowry definition substance capable of accepting Hsubstance capable of accepting H++

KOH(aq) KOH(aq) K K++(aq) + OH(aq) + OH––(aq) (aq)

OHOH–– + H + H++ H H22O O (OH(OH–– from NaOH accepts H from NaOH accepts H++))

NHNH33 + H + H22O NHO NH44++ + OH + OH–– (NH(NH3 3 accepts Haccepts H++))

Water can act as both an acid and a base: it Water can act as both an acid and a base: it is an is an amphotericamphoteric substance substance

HClOHClO44 + H + H22O O H H33OO++ + ClO + ClO44––

acid acid basebase

(accepts H(accepts H++ from HClO from HClO44))

NHNH33 + H + H22O NHO NH44++ + OH + OH––

base base acidacid

(donates H(donates H++ to NH to NH33))

Strong AcidsStrong Acidsdissociate ~100%dissociate ~100%• HCl, HBr, HI (no HF) hydro…ic acidHCl, HBr, HI (no HF) hydro…ic acid

• HNOHNO33 nitric acid nitric acid

• HClOHClO33 chloric acid (moderate) chloric acid (moderate)

• HClOHClO44 perchloric acid perchloric acid

• HH22SOSO44 (first proton) sulfuric acid (first proton) sulfuric acid

HH22SOSO44(aq) (aq) H H++(aq) + HSO(aq) + HSO44−−(aq)(aq)

22ndnd weak:weak: HSO HSO44−−(aq) (aq) HH++(aq) + SO(aq) + SO44

2−2−(aq)(aq)

Weak AcidsWeak Acids• dissociate <100%dissociate <100%

• most other acidsmost other acids

HFHF hydrofluoric acidhydrofluoric acid

HCNHCN hydrocyanic acidhydrocyanic acid

HNOHNO22 nitrous acidnitrous acid

CHCH33COCO22HH acetic acid acetic acid

HH22COCO33 carbonic acid (both protons)carbonic acid (both protons)

HH33POPO44 phosphoric acid (all protons)phosphoric acid (all protons)

HH22SOSO33 sulfurous acid (both protons)sulfurous acid (both protons)

oxalic acid Hoxalic acid H22CC22OO44(aq) H(aq) H++(aq) + HC(aq) + HC22OO44−−(aq)(aq)

Strong BasesStrong Basesdissociate ~100%dissociate ~100%

• alkali metal hydroxidesalkali metal hydroxides

LiOHLiOH, NaOH, KOH, RbOH, NaOH, KOH, RbOH

name: lithium hydroxidename: lithium hydroxide

• hydroxide ofhydroxide of

Ca Ca(OH)Ca Ca(OH)22 calcium hydroxide calcium hydroxide

Ba Ba(OH)Ba Ba(OH)22

Sr Sr(OH)Sr Sr(OH)22

Ammonia, NHAmmonia, NH33, is a weak base, is a weak base

Neutralization ReactionsNeutralization Reactions• acid + OH-ctg. base acid + OH-ctg. base salt + water salt + water

(a double replacement reaction)(a double replacement reaction)

HF(aq) + KOH(aq) HF(aq) + KOH(aq) KF(aq) + H KF(aq) + H22OO

HF(aq) + KHF(aq) + K++(aq) + OH(aq) + OH––(aq) (aq) K K++(aq) + F(aq) + F––(aq) + H(aq) + H22OO

HF(aq) + OHHF(aq) + OH––(aq) (aq) F F––(aq) + H(aq) + H22O O net ionic net ionic

spectator ions are eliminated from equationspectator ions are eliminated from equation

HF is a weak acid and HCl is a strong acidHF is a weak acid and HCl is a strong acid

• acid + non-OH-ctg base acid + non-OH-ctg base salt salt

HCl(aq) + NHHCl(aq) + NH33(aq) (aq) NH NH44Cl(aq)Cl(aq)

HH++(aq) + Cl(aq) + Cl––(aq) + NH(aq) + NH33(aq) (aq) NH NH44++(aq) + Cl(aq) + Cl––(aq)(aq)

HH++(aq) + NH(aq) + NH33(aq) (aq) NH NH44++(aq) (aq) net ionic equationnet ionic equation

Neutralization ReactionsNeutralization Reactions• Strong acid + strong base Strong acid + strong base salt + water salt + water

(a double replacement reaction)(a double replacement reaction)

HClOHClO33(aq) + NaOH(aq) (aq) + NaOH(aq) NaClO NaClO33(aq) + H(aq) + H22OO

chloric acidchloric acid

HH++(aq) + ClO(aq) + ClO33−(aq)−(aq) + + Na Na++(aq) + OH(aq) + OH––(aq) (aq)

NaNa++(aq) + ClO(aq) + ClO33––(aq) + H(aq) + H22OO

HH++(aq) + OH(aq) + OH––(aq) (aq) H H22O O net ionic equationnet ionic equation

spectator ions, spectator ions, ClOClO33−− + + Na Na++, , are eliminated from are eliminated from

equationequation

Formation of a Weak Acid or Base as a Formation of a Weak Acid or Base as a Driving ForceDriving Force(another double replacement reaction)(another double replacement reaction)

• HNOHNO33(aq) + KCN(aq) (aq) + KCN(aq) HCN(aq) + KNO HCN(aq) + KNO33(aq)(aq)

HH++(aq) + NO(aq) + NO33––(aq) + K(aq) + K++(aq) + CN(aq) + CN––(aq) (aq) HCN (aq) + HCN (aq) +

KK++(aq) + NO(aq) + NO33––(aq)(aq)

HH++(aq) + CN(aq) + CN––(aq) (aq) HCN(aq) HCN(aq) (a weak acid)(a weak acid)

• NHNH44Cl + NaOH(aq) Cl + NaOH(aq) NH NH44OH + NaCl(aq)OH + NaCl(aq)

NHNH44++(aq) + Cl(aq) + Cl––(aq)(aq) + Na + Na++(aq) + OH(aq) + OH––(aq) (aq) NH NH44OH OH

NaNa++(aq) + Cl(aq) + Cl––(aq)(aq)

NHNH44++(aq) + OH(aq) + OH––(aq) (aq) NH NH44OH OH (a weak base)(a weak base)

NHNH44OH is NHOH is NH33 in water, i.e., NH in water, i.e., NH33 + H + H22OO

When no Weak Electrolytes are FormedWhen no Weak Electrolytes are Formed

• HNOHNO33(aq) + KCl(aq) (aq) + KCl(aq) HCl(aq) + KNO HCl(aq) + KNO33(aq)(aq)

HH++(aq) + NO(aq) + NO33––(aq) + K(aq) + K++(aq) + Cl(aq) + Cl––(aq) (aq)

HH++(aq) + Cl(aq) + Cl––(aq) + K(aq) + K++(aq) + NO(aq) + NO33––(aq) (aq)

There is no net reaction: There is no net reaction: N.R. No driving forceN.R. No driving force

All ions are spectators.All ions are spectators.

• BaClBaCl22(aq) + 2NaOH(aq) (aq) + 2NaOH(aq) Ba(OH) Ba(OH)22(aq) + 2NaCl(aq)(aq) + 2NaCl(aq)

BaBa2+2+(aq) + 2Cl(aq) + 2Cl––(aq)(aq) + 2Na + 2Na++(aq) + 2OH(aq) + 2OH––(aq) (aq)

BaBa2+2+(aq) + 2OH(aq) + 2OH––(aq) + 2Na(aq) + 2Na++(aq) + 2Cl(aq) + 2Cl––(aq)(aq)

There is no net reaction: There is no net reaction: N.R. No driving forceN.R. No driving force

Gas Forming Reactions (a Driving Force)Gas Forming Reactions (a Driving Force)Some of the weak acids and bases that are formed at Some of the weak acids and bases that are formed at

double replacementdouble replacement reactions decompose to form a reactions decompose to form a

gas and watergas and water

COCO22

NaNa22COCO33(aq) + 2HCl(aq) (aq) + 2HCl(aq) HH22COCO33(aq) + 2NaCl(aq)(aq) + 2NaCl(aq)

HH22COCO33(aq) (aq) H H22O + COO + CO22(g)(g)

NaNa22COCO33(aq) + 2HCl(aq) (aq) + 2HCl(aq) HH22O + COO + CO22(g)(g) + 2NaCl(aq) + 2NaCl(aq)

SOSO22

NaNa22SOSO33(aq) + 2HCl(aq) (aq) + 2HCl(aq) HH22SOSO33(aq) + 2NaCl(aq)(aq) + 2NaCl(aq)

HH22SOSO33(aq) (aq) H H22O + SOO + SO22(g)(g)

NaNa22SOSO33(aq) + 2HCl(aq) (aq) + 2HCl(aq) HH22O + SOO + SO22(g)(g) + 2NaCl(aq) + 2NaCl(aq)

Redox (Oxidation-Reduction) ReactionsRedox (Oxidation-Reduction) Reactions• involve transfer of electron(s)involve transfer of electron(s)

• oxidation: loss of electron(s)oxidation: loss of electron(s)

• reduction: gain of electron(s)reduction: gain of electron(s)

• some can be identified when an some can be identified when an

uncombined element is a reactant or a productuncombined element is a reactant or a product

• eg. eg. Zn(s)Zn(s) + Cu + Cu2+2+(aq) (aq) Zn Zn2+2+(aq) + (aq) + Cu(s)Cu(s)

• Zn Zn Zn Zn2+2+

• Zn(s) Zn(s) Zn Zn2+2+(aq) + 2 e(aq) + 2 e––, oxidation, oxidation

• CuCu2+2+ Cu Cu

• CuCu2+2+(aq) + 2e(aq) + 2e–– Cu(s), reduction Cu(s), reduction

Single Displacement ReactionsSingle Displacement Reactions

• Zn(s) + CuClZn(s) + CuCl22(aq) (aq) Cu(s) + ZnCl Cu(s) + ZnCl22(aq)(aq)

• Zn oxidized to ZnZn oxidized to Zn2+2+

• CuCu2+2+ reduced to Cu reduced to Cu

• occurs because zinc is more active than occurs because zinc is more active than coppercopper

• ClCl22(g) + CuBr(g) + CuBr22(aq) (aq) Br Br22(l)(l) + CuCl+ CuCl22(aq)(aq)

• Br oxidized from BrBr oxidized from Br–– to Br to Br22

• Cl reduced from ClCl reduced from Cl22 to Cl to Cl––

• Cl is more active than BrCl is more active than Br

Oxidation NumbersOxidation Numbers• also an accounting toolalso an accounting tool• very usefulvery useful• oxidation numbers of all atoms in oxidation numbers of all atoms in

substance add up to charge on substancesubstance add up to charge on substancee.g. (e.g. (charge of speciescharge of species))

zerozero for Al for Al22(SO(SO44))33 and H and H33POPO44

+1+1 for NH for NH44++

––22 for Cr for Cr22OO772–2–

Assigning Oxidation Numbers, ONAssigning Oxidation Numbers, ON• ON = 0 for all atoms in any substance in ON = 0 for all atoms in any substance in

most elemental form, Na(s), Zn(s), Hg(l) most elemental form, Na(s), Zn(s), Hg(l) HH22(g), Cl(g), Cl22(g), I(g), I22(s), O(s), O22(g), C(s), P(g), C(s), P44(s), S(s), S88(s) (s)

• ON = charge for monatomic ions (ON = charge for monatomic ions (NaNa++, S, S22−−))

• ON = –1 for F in all compoundsON = –1 for F in all compounds

• ON = ON = –2 for O–2 for O in compounds, usually in compounds, usually

– exceptions: peroxide, Oexceptions: peroxide, O222–2–, ON = –1, ON = –1

– superoxide, Osuperoxide, O22––, ON = –1/2, ON = –1/2

• ON = ON = +1 for H+1 for H in compounds, usually in compounds, usually

– exception: ON = –1 in metallic hydridesexception: ON = –1 in metallic hydrides

Assigning Oxidation Numbers, ONAssigning Oxidation Numbers, ON

• ON = ON = +1 for alkali metals+1 for alkali metals in compounds in compounds

• ON = ON = +2 for alkaline+2 for alkaline earth metals in earth metals in compoundscompounds

• ON = ON = +3 for Al+3 for Al in compounds in compounds

• ON = ON = −1 for −1 for Cl, Br, and ICl, Br, and I (iodine) in binary (iodine) in binary compounds except for those with oxygen compounds except for those with oxygen (in these cases they variable positive)(in these cases they variable positive)

• F is always −1F is always −1

Assign ON to Each Atom in the Assign ON to Each Atom in the Following SubstancesFollowing Substances

WClWCl66

x + 6(–1) = 0x + 6(–1) = 0

x –6 = 0x –6 = 0

x = +6x = +6

––11??

??

NaNa22SS22OO33

+2 + 2x – 6 = 0+2 + 2x – 6 = 0

2x = 6 – 22x = 6 – 2

x = +2 ON of Sx = +2 ON of S

––22+1+1

??

NaNa22SS44OO88

+2 + 4x –16 = 0+2 + 4x –16 = 0

4x = 16 – 24x = 16 – 2

+14 +7+14 +7 x = x = —— = —— ON of S 4 24 2

––22+1+1

CrCr22OO772–2–

2x –14 = 2x –14 = –2–2

2x = 14 – 22x = 14 – 2

+12+12 x = x = —— = +6 ON of Cr 22

––22??

HH22CC22OO44

+2 + 2x – 8 = 0+2 + 2x – 8 = 0

2x = 8 – 22x = 8 – 2

+6+6 x = x = —— = +3 ON of C 22

––22+1+1

MoBrMoBr55++

x – 5 = +1x – 5 = +1

x = 5 + 1 x = 5 + 1

x = 6 ON of Mo = +6x = 6 ON of Mo = +6

––11??

Oxidizing and Reducing AgentsOxidizing and Reducing Agents

In every redox reaction there is (at least) a In every redox reaction there is (at least) a

Reducing agentReducing agent (the one that is oxidized) and (the one that is oxidized) and

an an oxidizing agentoxidizing agent (the one that is reduced) (the one that is reduced)

ON increases ON decreasesON increases ON decreases

The species is The species isThe species is The species is

oxidized reduced oxidized reduced

+7+7+6+6+5+5+4+4+3+3+2+2+1 +1 00

−−11−−22−− 33−− 44−− 55−− 66−− 77

• Activity (Electromotive) Series Activity (Electromotive) Series for metalsfor metals

LiLiKK

BaBaCaCaNaNaMgMgAlAlMnMnZnZnCrCrFeFeCoCoNiNiSnSnPbPb(H(H22))CuCuAgAgHgHgPtPtAuAu

Activity Activity increasesincreases

Activity Activity decreasesdecreases

ExamplesExamples

• Complete and balance each of the Complete and balance each of the following chemical equationsfollowing chemical equations

A free and chemically active metal displacing A free and chemically active metal displacing a less active metal from a compounda less active metal from a compound• Mg + FeClMg + FeCl33 Mg + FeClMg + FeCl33 Fe + MgCl Fe + MgCl22

3Mg(s) + 2FeCl3Mg(s) + 2FeCl33(aq) (aq) 2Fe(s) + 3MgCl 2Fe(s) + 3MgCl22(aq)(aq)Mg Mg Mg Mg2+2+, oxidized; Mg reducing agent, oxidized; Mg reducing agentFeFe3+3+ Fe, reduced; Fe Fe, reduced; Fe3+3+ oxidizing agent oxidizing agent

• Sn + CrFSn + CrF33 Sn is less reactive than Cr Sn is less reactive than Cr Sn + CrFSn + CrF33 No ReactionNo Reaction

• Pb(s) + Au(ClOPb(s) + Au(ClO33))33(aq) (aq) Pb(s) + Au(ClOPb(s) + Au(ClO33))33(aq) (aq) Au(s) + Pb(ClO Au(s) + Pb(ClO33))22(aq)(aq) 33Pb(s) + 2Au(ClOPb(s) + 2Au(ClO33))33(aq) (aq) 2Au(s) + 3Pb(ClO 2Au(s) + 3Pb(ClO33))22(aq)(aq) Pb is oxidized AuPb is oxidized Au+3+3 is reduced is reduced

LiLiKK

BaBaCaCaNaNaMgMgAlAlMnMnZnZnCrCrFeFeCoCoNiNiSnSnPbPbHH22

CuCuHgHgAgAgPtPtAuAu

A free and chemically active metal displacing A free and chemically active metal displacing

a less active metal from a compounda less active metal from a compound

• Zn + CrBrZn + CrBr33 Zn + CrBrZn + CrBr33 Cr + ZnBr Cr + ZnBr22

3Zn(s) + 2CrBr3Zn(s) + 2CrBr33(aq) (aq) 2Cr(s) + 3 ZnBr 2Cr(s) + 3 ZnBr22(aq)(aq)

• Zn oxidized to ZnZn oxidized to Zn2+2+; Zn reducing agent; Zn reducing agent

• CrCr3+3+ reduced to Cr; Cr reduced to Cr; Cr3+3+ oxidizing agent oxidizing agent

• Ag(s) + Hg(NOAg(s) + Hg(NO33))22 No ReactionNo Reaction Ag is less reactive than HgAg is less reactive than Hg

LiLiKK

BaBaCaCaNaNaMgMgAlAlMnMnZnZnCrCrFeFeCoCoNiNiSnSnPbPbHH22

CuCuHgHgAgAgPtPtAuAu

A free and chemically active metal displacingA free and chemically active metal displacing

Hydrogen from acids or waterHydrogen from acids or water• Fe + HBr Fe + HBr Fe + HBr Fe + HBr H H22 + FeBr + FeBr33

0 +1 0 +30 +1 0 +3

2Fe + 6HBr 2Fe + 6HBr 3H 3H22 + 2 FeBr + 2 FeBr33

Fe oxidized to FeFe oxidized to Fe3+3+; Fe reducing agent; Fe reducing agentHH++ reduced to H reduced to H22; H; H++ oxidizing agent oxidizing agent

• Cu + HBr Cu + HBr Cu less active than H Cu less active than H22

Cu + HBr Cu + HBr No Reaction No Reaction

LiLiKK

BaBaCaCaNaNaMgMgAlAlMnMnZnZnCrCrFeFeCoCoNiNiSnSnPbPbHH22

CuCuHgHgAgAgPtPtAuAu

A free and chemically active metal displacingA free and chemically active metal displacingHydrogen from acids or waterHydrogen from acids or water• K(s) + HK(s) + H22O(l) O(l) 0 +1 +1 00 +1 +1 0 2K(s) + 2H2K(s) + 2H22O(l) O(l) 2 KOH(aq) + H 2 KOH(aq) + H22(g)(g)K oxidized to KK oxidized to K++; K reducing agent; K reducing agentHH++ reduced to H reduced to H22; H; H++ oxidizing agent oxidizing agent

• Ag(s) + HAg(s) + H22O(l) O(l) Ag less active than H Ag less active than H22

Ag(s) + HAg(s) + H22O(l) O(l) No ReactionNo Reaction • Ni(s) + HNi(s) + H22SOSO44(aq) (aq) 0 +1 +2 00 +1 +2 0 Ni(s) + HNi(s) + H22SOSO44(aq) (aq) NiSO NiSO44(aq) + H(aq) + H22(g)(g)Ni oxidized to NiNi oxidized to Ni2+2+; Ni reducing agent; Ni reducing agentHH++ reduced to H reduced to H22; H; H++ oxidizing agent oxidizing agent

LiLiKK

BaBaCaCaNaNaMgMgAlAlMnMnZnZnCrCrFeFeCoCoNiNiSnSnPbPbHH22

CuCuHgHgAgAgPtPtAuAu

• Which one of the following metals could Which one of the following metals could

be used safely for lining a tank intended be used safely for lining a tank intended

for storage of sulfuric acid?for storage of sulfuric acid?

HH22SOSO44

• aluminumaluminum

• ironiron

• chromiumchromium

• mercurymercury

• coppercopper

• tintin

LiLiKK

BaBaCaCaNaNaMgMgAlAlMnMnZnZnCrCrFeFeCoCoNiNiSnSnPbPbHH22

CuCuHgHgAgAgPtPtAuAu

Nonmetal Activity SeriesNonmetal Activity Series

FF

ClCl

BrBr

II

same order as in periodic tablesame order as in periodic table

F > Cl > Br > IF > Cl > Br > I

Complete and BalanceComplete and Balance• ClCl22 + FeBr + FeBr33 ClCl22(g) + FeBr(g) + FeBr33(aq) (aq) Br Br22(l) + FeCl(l) + FeCl33(aq)(aq)

00 −1 0 −1−1 0 −1

3Cl3Cl22(g) + 2FeBr(g) + 2FeBr33(aq) (aq) 3 Br 3 Br22(l) + 2 FeCl(l) + 2 FeCl33(aq)(aq)

ClCl22 oxidizing agent; oxidizing agent; BrBr–– reducing agent reducing agent

• II22(s) + NaF(aq) (s) + NaF(aq) I I22 is less active than F is less active than F22

II22(s) + NaF(aq) (s) + NaF(aq) No ReactionNo Reaction

• FF22(g) + NaCl(aq) (g) + NaCl(aq) 0 0 −1 o −1−1 o −1

FF22(g) + 2 NaCl(aq) (g) + 2 NaCl(aq) Cl Cl22(g) + 2 NaF(aq)(g) + 2 NaF(aq) FF22 oxidizing agent; Cl oxidizing agent; Cl–– reducing agent reducing agent

• BrBr22 + FeCl + FeCl33 Br less active than Cl Br less active than Cl BrBr22 + FeCl + FeCl33 No ReactionNo Reaction

Identifying Oxidizing and Reducing AgentsIdentifying Oxidizing and Reducing Agents

0 +3 0 +20 +3 0 +2

3Zn(s) + 2CrBr3Zn(s) + 2CrBr33(aq) (aq) 2Cr(s) + 3 ZnBr 2Cr(s) + 3 ZnBr22(aq)(aq)• Zn oxidized to ZnZn oxidized to Zn2+2+; Zn reducing agent; Zn reducing agent• CrCr3+3+ reduced to Cr; Cr reduced to Cr; Cr3+3+ oxidizing agent oxidizing agent

0 +1 +2 00 +1 +2 0

Ni(s) + HNi(s) + H22SOSO44(aq) (aq) NiSO NiSO44(aq)) + H(aq)) + H22(g)(g)Ni oxidized to NiNi oxidized to Ni2+2+; Ni reducing agent; Ni reducing agentHH++ reduced to H reduced to H22; H; H++ oxidizing agent oxidizing agent

0 -1 0 -10 -1 0 -1

FF22(g) + 2 NaCl(aq) (g) + 2 NaCl(aq) Cl Cl22(g) + 2 NaF(aq)(g) + 2 NaF(aq)

FF22 reduced; Cl reduced; Cl–– oxidized oxidized

FF22 oxidizing agent; Cl oxidizing agent; Cl–– reducing agent reducing agent

Identifying Oxidizing and Reducing AgentsIdentifying Oxidizing and Reducing Agents

• The device for testing breath for the presence The device for testing breath for the presence of alcohol is based on the following reaction. of alcohol is based on the following reaction. Identify the oxidizing and reducing agentsIdentify the oxidizing and reducing agents

ON: –1 +6ON: –1 +6

3CH3CH33CHCH22OH(aq) + 2OH(aq) + 2CrCr22OO772–2–((aq) + 16Haq) + 16H++

ethanol ethanol orange-redorange-red

+3+3

3CH3CH33COCO22H(aq) + 4 H(aq) + 4 CrCr3+3+(aq) + 11H(aq) + 11H22OO

acetic acid acetic acid greengreen

ethanol is oxidized (reducing ethanol is oxidized (reducing agentagent))

CrCr22OO772– 2– (dichromate ion) is reduced (oxidizing (dichromate ion) is reduced (oxidizing agt.agt.))

Balancing redox equationsBalancing redox equations

ON:ON: +6+6 in acidic media (presence of H in acidic media (presence of H++))

CrCr22OO772–2–(aq) + Fe(aq) + Fe2+2+(aq) (aq) CrCr3+3+(aq) + Fe(aq) + Fe3+3+(aq) (aq)

CrCr22OO772– 2– + 6e + 6e−− 2Cr2Cr3+3+ (this the reduction) (this the reduction)

FeFe2+2+ FeFe3+3+ + e + e−− (this is the oxidation) (this is the oxidation)

CrCr22OO772– 2– + 6e + 6e−− + 14H + 14H++ 2Cr2Cr3+3+ (to have +6 = +6, charge) (to have +6 = +6, charge)

CrCr22OO772– 2– + 6e + 6e−− + 14H + 14H++ 2Cr2Cr3+3+ + 7H + 7H22O (to balance H & O)O (to balance H & O)

CrCr22OO772– 2– + 6e + 6e−− + 14H + 14H++ 2Cr2Cr3+3+ + 7H + 7H22O O to equal # ofto equal # of

66(Fe(Fe2+2+ FeFe3+3+ + e + e−−) ) electrons, 6electrons, 6

CrCr22OO772– 2– + 14H + 14H++ + 6Fe + 6Fe2+2+ 2Cr2Cr3+3+ + 7H + 7H22O + O + FeFe3+3+

Balancing redox equationsBalancing redox equations

ON:ON: +7 +4+7 +4 in basic media (OH in basic media (OH−−) ) +4 +6+4 +6

MnOMnO44––(aq) + SO(aq) + SO33

22−−(aq) (aq) MnOMnO22(s) + SO(s) + SO4422−−(aq) (aq)

MnOMnO44– – + 3e + 3e−− MnOMnO22 (this the reduction) (this the reduction)

SOSO3322−− SOSO44

22−− + 2e + 2e−− (this is the oxidation) (this is the oxidation)

MnOMnO44– – + 3e + 3e−− MnOMnO22 + 4OH + 4OH−− (to equal charges, (to equal charges, −4−4))

SOSO3322−− + 2OH+ 2OH−− SOSO44

22−− + 2e + 2e−− (to equal charges, (to equal charges, −4−4))

22(MnO(MnO44– – + 3e + 3e−− +2H +2H22O O MnOMnO22 + 4OH + 4OH−−) ) to equal H, O,to equal H, O,

33(SO(SO3322−− + 2OH+ 2OH−− SOSO44

22−− + 2e + 2e−− + H + H22O)O) and electronsand electrons

2MnO2MnO44– – + H + H22O + 3SOO + 3SO33

22−− 2MnO2MnO22 + 2OH + 2OH−− + 3SO + 3SO4422−−

Measuring Concentrations of Compounds in Measuring Concentrations of Compounds in SolutionsSolutions

Concentration TermsConcentration Terms

Parts PerParts Per• Hundred (percent, %)Hundred (percent, %) weight/weight, %(w/w) (weight/weight, %(w/w) (most commonmost common))

mass solute (g)mass solute (g)———————— x 100mass solution (g) mass solution (g)

volume/volume, %(v/v)volume/volume, %(v/v)

V solute (mL) V solute (mL) liquid solute in liquid solute in———————— x 100 liquid solvent V solution (mL)V solution (mL)

Parts PerParts Per

• Hundred (percent, %)Hundred (percent, %) weight/volume, %(w/v)weight/volume, %(w/v)

mass solute (g)mass solute (g)———————— x 100V solution (mL)V solution (mL)

Learning CheckLearning Check A solution is prepared by mixing 15.0 g of NaA solution is prepared by mixing 15.0 g of Na22COCO3 3

and 235 g of Hand 235 g of H22O. The final V of solution is 242 O. The final V of solution is 242 mL. Calculate the %w/w and %w/v concentration mL. Calculate the %w/w and %w/v concentration of the solution. of the solution.

g solution = 15.0 g Nag solution = 15.0 g Na22COCO33 + 235 g H + 235 g H22O = 250. gO = 250. g

15.0 g solute15.0 g solute

%w/w%w/w = = ——————— x 100 = 6.00% NaNa22COCO33

250. g solution

15.0 g solute15.0 g solute

%w/v%w/v = = ——————— x 100 = 6.20% NaNa22COCO33

242 mL solution

Molarity, Molarity, MMThe Molarity, The Molarity, MM, usually known as the molar , usually known as the molar concentration of a solute in a solution, is theconcentration of a solute in a solution, is thenumber of number of moles of solute per liter (1000 mL) moles of solute per liter (1000 mL) of solution of solution / or mmoles per mL of solution./ or mmoles per mL of solution.

To calculate it we need moles of soluteTo calculate it we need moles of soluteand V(in liters) of solution (or mmol and mL)and V(in liters) of solution (or mmol and mL)and to divideand to divide

mol solute mmol solutemol solute mmol soluteMM = = —————— = ——————— V(L) solution V(mL) solution

Calculation of Molarity, Calculation of Molarity, MMWhat is the molarity of 500. mL NaOH solution ifWhat is the molarity of 500. mL NaOH solution if

it contains 6.00 g NaOH? it contains 6.00 g NaOH?

500. mL 500. mL 1000 = 0.500 L (volume in liters) 1000 = 0.500 L (volume in liters)

FW (NaOH) = 40.0 g/mol (FW (NaOH) = 40.0 g/mol (from periodic tablefrom periodic table))

How many moles of NaOH?How many moles of NaOH? 1 mol NaOH1 mol NaOH6.00 g6.00 g x x —————— = 0.150 mol NaOH 40.0 g NaOH

mol solute 0.150 molmol solute 0.150 molMM = = —————— = ————— = 0.300 M NaOH V(L) solution 0.500 L

(0.300 mol in 1 L or 0.300 mmol in 1 mL)

Formality, Formality, FFis the same as molarity, is the same as molarity, but referred to ionicbut referred to ionic

compoundscompounds in aqueous solution in aqueous solution

FW (formula weights) of solute per L of solutionFW (formula weights) of solute per L of solution

1 FW1 FW# FW = g solute x # FW = g solute x ————— = # of FW = # of moles g solute

# FW (# of formula weights) # FW (# of formula weights)

FF = = —————— V(L) solution (volume in liters)

Formality = Molarity

Molality, Molality, mmis the amount is the amount (moles) of solute per kg of (moles) of solute per kg of

solventsolvent (usually but not necessarily water). (usually but not necessarily water).

What is the What is the mm of a solution prepared by of a solution prepared by

dissolving 25.3 g Nadissolving 25.3 g Na22COCO33 in 458 g water? in 458 g water?

458 g = 0.458 kg (458 g = 0.458 kg (after dividing by 1000after dividing by 1000)) 1 mol Na1 mol Na22COCO33

25.3 g25.3 g Na Na22COCO33 x x —————— = 0.239 mol Na2CO3

106.0 g NaNa22COCO33

mol solute 0.239 molmol solute 0.239 molmm = = —————— = ————— = 0.522 m Na2CO3 kg H2O 0.458 kg

(0.522 mol in 1 kg H2O)

Mole Fraction, Mole Fraction, XX

is the amount is the amount (mol) of a given component of a(mol) of a given component of a

solution per mol of solution. solution per mol of solution.

Here we need # moles of every componentHere we need # moles of every component

(solute(s) and solvent) and the total(solute(s) and solvent) and the total

e.g. for a solution with ne.g. for a solution with n11, n, n22, n, n33, … mol, … mol

nnii n nii mol fraction mol fraction XXii = = —————— = —— (no units) nn11+ n+ n22+ n+ n33 + … n + … ntt

ni: moles of component i (1, 2, 3, …)nt: total number of moles

How many g of NaCl are contained in How many g of NaCl are contained in 250.0 mL of 0.2193 250.0 mL of 0.2193 MM NaCl solution? NaCl solution?

250.0 mL 250.0 mL 1000 = 0.2500 L 1000 = 0.2500 L

Now, Now, MM as a conversion factor as a conversion factor 0.2193 mol0.2193 mol0.2500 L x 0.2500 L x ————— = 0.05480.0548 mol NaClNaCl 1 L sol’n

grams out of moles and formula weight (58.44) 58.44 g NaCl58.44 g NaCl0.0548 mol x 0.0548 mol x —————— = 3.204 g NaCl3.204 g NaCl 1 mol NaCl

Making SolutionsMaking Solutions• Consider the making of 1.00 L of 1.00 Consider the making of 1.00 L of 1.00 MM NaCl NaCl

solution.solution.• need 1.00 mol NaCl or 58.5 g NaClneed 1.00 mol NaCl or 58.5 g NaCl• dissolve 58.5 g NaCl in 1.00 L water?dissolve 58.5 g NaCl in 1.00 L water?

NO!!NO!!• dissolve 58.5 g NaCl in less than 1.000 L water dissolve 58.5 g NaCl in less than 1.000 L water

and dilute to a total volume of 1.000 Land dilute to a total volume of 1.000 L

Example: Describe the preparation of 300.0 Example: Describe the preparation of 300.0 mL of 0.4281 mL of 0.4281 MM silver nitrate solution. silver nitrate solution.300.0 mL 300.0 mL 1000 = 0.3000 L AgNO 1000 = 0.3000 L AgNO33 FW = 169.97 g/mol FW = 169.97 g/mol

0.4281 mol0.4281 mol0.3000 L x 0.3000 L x ————— = 0.12840.1284 mol AgNOAgNO33

1 L sol’n 169.97 g169.97 g 0.1284 mol x 0.1284 mol x ————— = 21.82 g21.82 g AgNOAgNO33

1 mol AgNO3

• Dissolve 21.82 g AgNODissolve 21.82 g AgNO33 in 300.0 mL water?in 300.0 mL water?

NO!!NO!!• Dissolve 21.82 g AgNODissolve 21.82 g AgNO33 in water and dilute to in water and dilute to

300.0 mL300.0 mL..

How many milliliters of 2.00 M HNOHow many milliliters of 2.00 M HNO33 contain contain

24.0 g HNO24.0 g HNO33??

HNOHNO33 FW = 63.0 g/mol FW = 63.0 g/mol How many moles?How many moles?

1 mol HNO1 mol HNO33

24.0 g HNO24.0 g HNO33 x x —————— = 0.381 mol HNO0.381 mol HNO33

63.0 g63.0 g HNO3

Now the Now the MM with the volume on top to get mL with the volume on top to get mL 1 L sol’n 1000 mL0.381 mol x 0.381 mol x —————— x ———— = 191 mL 2.00 mol HNO3 1 L sl’n

How many grams of AlClHow many grams of AlCl33 are needed to are needed to

prepare 25 mL of a 0.150 M solution?prepare 25 mL of a 0.150 M solution?

25 mL 25 mL 1000 = 0.025 L FW (AlCl 1000 = 0.025 L FW (AlCl33) = 133.5 g/mol) = 133.5 g/mol

All at once:All at once:

V(L) of sol’n and mol and FW V(L) of sol’n and mol and FW MM to calculate to calculate to calculate to calculate

mol of AlClmol of AlCl33 g of AlCl g of AlCl33

0.025 L x 0.025 L x 0.150 mole0.150 mole x x 133.5 g 133.5 g = 0.500 g AlCl = 0.500 g AlCl33

1 L1 L 1 mole 1 mole

DilutionDilutionthe process of decreasing the concentrationthe process of decreasing the concentration

of solutes in a solution by addition of solvent of solutes in a solution by addition of solvent

or another solution that does not contain the or another solution that does not contain the

same solutes same solutes

• volume increases and concentration volume increases and concentration decreases.decreases.

concentrated diluted solutionconcentrated diluted solution

volumetric flask: calibrated to contain, tcvolumetric flask: calibrated to contain, tc

pipet: calibrated to deliver, tdpipet: calibrated to deliver, td

Example: Calculate the concentration of a Example: Calculate the concentration of a solution made by diluting 25.0 mL of 0.200 M solution made by diluting 25.0 mL of 0.200 M methanol, CHmethanol, CH33OH, solution to 100.0 mL.OH, solution to 100.0 mL.

Key for calculations: moles of solute takenKey for calculations: moles of solute taken

from the concentrated solution are the same from the concentrated solution are the same

in the dilutedin the diluted solution (only solvent is added.)solution (only solvent is added.)

moles = concentration x V = moles = concentration x V = MM x V x V

CCcc x V x Vcc = C = Cdd x x VVdd MMcc x V x Vcc = = MMdd x x VVdd

one of the four is unknownone of the four is unknown

c: concentrated d: dilutedc: concentrated d: diluted

L or mL can be used for volume L or mL can be used for volume

Example: Calculate the concentration of Example: Calculate the concentration of a solution made by diluting 25.0 mL of a solution made by diluting 25.0 mL of 0.200 M methanol, CH0.200 M methanol, CH33OH, to 100.0 mL.OH, to 100.0 mL.

MMcc x V x Vcc = = MMdd x x VVdd

MMdd is the unknown is the unknown

25.0 mL x 0.200 M = 100.0 mL x 25.0 mL x 0.200 M = 100.0 mL x MMdd

25.0 mL x 0.200 M25.0 mL x 0.200 MMMdd = = ————————— = 0.0500 M ————————— = 0.0500 M 100.0 mL100.0 mL

0.0500 mole/L0.0500 mole/L

How many mL of a 0.515 M NaBr solution How many mL of a 0.515 M NaBr solution must be diluted to produce 500.0 mL of a must be diluted to produce 500.0 mL of a 0.103 M NaBr solution?0.103 M NaBr solution?

MMcc x V x Vcc = = MMdd x x VVdd VVc c is the unknownis the unknown

VVcc x 0.515 M = 500.0 mL x 0.103 M x 0.515 M = 500.0 mL x 0.103 M

500.0 mL x 0.103 M500.0 mL x 0.103 MVVcc = = ————————— = 100.0 mL————————— = 100.0 mL 0.515 M0.515 M

Serial DilutionsSerial DilutionsExample: Calculate the M of NaOH in a solution Example: Calculate the M of NaOH in a solution made by diluting 25.0 mL of 0.928 M NaOH to made by diluting 25.0 mL of 0.928 M NaOH to 200.0 mL and then diluting 50.0 mL of the 200.0 mL and then diluting 50.0 mL of the

second solution to 100.0 mL. second solution to 100.0 mL. MMcc x V x Vcc = = MMdd x x VVdd

First dilution:First dilution: 25.0 mL x 0.928 M25.0 mL x 0.928 MMMdd = = ————————— = 0.116 M ————————— = 0.116 M 200.0 mL200.0 mL

Second dilution:Second dilution: 50.0 mL x 0.116 M50.0 mL x 0.116 MMMdd = = ————————— = 0.0580 M ————————— = 0.0580 M 100.0 mL100.0 mL

Solution StoichiometrySolution StoichiometryUse of Use of MM, V, and coefficients in equations to , V, and coefficients in equations to

calculate any amount of reagent or product.calculate any amount of reagent or product.

• Example: What volume of 0.273 Example: What volume of 0.273 MM potassium chloride solution is required to potassium chloride solution is required to react with exactly 0.836 mmol of silver react with exactly 0.836 mmol of silver nitrate?nitrate?

• KCl(aq) + AgNOKCl(aq) + AgNO33(aq) (aq)

• KCl(aq) + AgNOKCl(aq) + AgNO33(aq) (aq) KNO KNO33(aq) + AgCl(s)(aq) + AgCl(s)

• driving force, formation of precipitate AgCldriving force, formation of precipitate AgCl

• Key: work with mmol of KCl and AgNOKey: work with mmol of KCl and AgNO33

Solution StoichiometrySolution Stoichiometry

What volume of 0.273 What volume of 0.273 MM KCl solution is required KCl solution is required

to react with exactly 0.836 mmol of AgNOto react with exactly 0.836 mmol of AgNO33 ? ?

KCl(aq) + AgNOKCl(aq) + AgNO33(aq) (aq) KNO KNO33(aq) + AgCl(s)(aq) + AgCl(s)

First, calculate mmol of KCl:First, calculate mmol of KCl:

1 mmol KCl1 mmol KCl0.836 mmol AgNO0.836 mmol AgNO33 x x—————— = 0.836 mmol —————— = 0.836 mmol 1 mmol AgNO1 mmol AgNO33 KCl KCl

Second, calculate volume of KCl solution:Second, calculate volume of KCl solution: 1 mL1 mL0.836 mmol KCl x 0.836 mmol KCl x ———— = 3.06 mL of solution ———— = 3.06 mL of solution 0.273 0.273 mmol (KCl)mmol (KCl)

TitrationTitration

BuretBuret

TitrantTitrant

Erlenmeyer (conical) FlaskErlenmeyer (conical) Flask

TitrateTitrate

TitrationTitration

• buret: calibrated td, fine tip to deliver small buret: calibrated td, fine tip to deliver small volumes accurately, stopcock for flow volumes accurately, stopcock for flow controlcontrol

• Erlenmeyer flask: sloped walls allow Erlenmeyer flask: sloped walls allow swirling of solutions without spilling or swirling of solutions without spilling or splashingsplashing

• Titrant: solution containing reagent that Titrant: solution containing reagent that will react with sample in well known will react with sample in well known mannermanner

• Titrate: solution containing the sampleTitrate: solution containing the sample

TitrationTitration• equivalence pointequivalence point: point in a titration at : point in a titration at

which the exact stoichiometric amount of which the exact stoichiometric amount of titrant has been added to react with the titrant has been added to react with the titratetitrate

• end pointend point: the point in a titration at which : the point in a titration at which the the indicatorindicator changes, titration stopped changes, titration stopped here and volume of titrant read.here and volume of titrant read.

• ideally, end point = equivalence pointideally, end point = equivalence point• reality, not so, error introduced, hopefully reality, not so, error introduced, hopefully

small errorsmall error• Four parameters: V, Four parameters: V, MM of titrant and V, of titrant and V, MM of of

titrate. Usually one is unknown.titrate. Usually one is unknown.

TitrationTitration

Example: Calculate the concentration of Example: Calculate the concentration of

hydrochloric acid in a solution if 35.0 mL of hydrochloric acid in a solution if 35.0 mL of

it required 28.9 mL of 0.178 it required 28.9 mL of 0.178 MM potassium potassium

hydroxide solution for titration.hydroxide solution for titration.

HCl: titrated, HCl: titrated, V known, V known, MM unknown unknown

(in this problem)(in this problem)

KOH: titrant, KOH: titrant, V and V and MM known known

TitrationTitration

BuretBuret

0.178 0.178 MM KOH KOH

Erlenmeyer (conical) FlaskErlenmeyer (conical) Flask

35.0 mL sample of HCl Sol’n, 35.0 mL sample of HCl Sol’n, unknown unknown M + drops indicatorM + drops indicator

HCl + KOH HCl + KOH KCl + H KCl + H22OO

TitrationTitrationStrategyStrategy: mmol of KOH are calculated first.: mmol of KOH are calculated first.Second, by using stoichiometry coefficientsSecond, by using stoichiometry coefficientsmol of HCl are calculated. mol of HCl are calculated. Finally, Finally, MM of HCl is calculated with mmol and V(mL) of HCl is calculated with mmol and V(mL)

0.178 mmol KOH 1 mmol HCl0.178 mmol KOH 1 mmol HCl28.9 mLx28.9 mLx———————— x————— = 5.14 mmol ———————— x————— = 5.14 mmol

1 mL 1 mmol KOH HCl1 mL 1 mmol KOH HCl

Based on the Based on the end point concept (mol ratio)end point concept (mol ratio)

5.14 mmol5.14 mmolMMHClHCl = = ————— = 0.147 ————— = 0.147 MM (mol/L) (mol/L) 35.0 mL (mmol/mL)35.0 mL (mmol/mL)

Example: Calculate the concentration of Example: Calculate the concentration of arsenic acid in a solution given that a 25.0 arsenic acid in a solution given that a 25.0 mL sample of that solution required 42.2 mL mL sample of that solution required 42.2 mL of 0.274 of 0.274 MM potassium hydroxide for titration. potassium hydroxide for titration.

• HH33AsOAsO44 + KOH + KOH K K33AsOAsO44 + H + H22OO• HH33AsOAsO44 + 3 KOH + 3 KOH K K33AsOAsO44 + 3 H + 3 H22OO

MM of H of H33AsOAsO44 is the unknown. is the unknown.

1 mol H1 mol H33AsOAsO44 reacts with 3 mol KOH. That is reacts with 3 mol KOH. That is

the mole ratio.the mole ratio.

We can use mL and mmol instead of L and molWe can use mL and mmol instead of L and mol

Example: Calculate the concentration of Example: Calculate the concentration of arsenic acid in a solution given that a 25.0 arsenic acid in a solution given that a 25.0 mL sample of that solution required 42.2 mL mL sample of that solution required 42.2 mL of 0.274 of 0.274 MM potassium hydroxide for titration. potassium hydroxide for titration.

• HH33AsOAsO44 + 3 KOH + 3 KOH K K33AsOAsO44 + 3 H + 3 H22OO

0.274 mmol KOH 1 mmol H0.274 mmol KOH 1 mmol H33AsOAsO44

42.2 mLx42.2 mLx———————— x————— = 3.85 mmol ———————— x————— = 3.85 mmol 1 mL 3 mmol KOH 1 mL 3 mmol KOH HH33AsOAsO44

3.85 mmol3.85 mmolM M HH33AsOAsO44 = = ————— = 0.154 ————— = 0.154 MM (mol/L) (mol/L) 25.0 mL (mmol/mL)25.0 mL (mmol/mL)

Purity Analysis: A 1.034 g-sample of impure Purity Analysis: A 1.034 g-sample of impure oxalic acid is dissolved in water and an acid-oxalic acid is dissolved in water and an acid-base indicator added. The sample requires 34.47 base indicator added. The sample requires 34.47 mL of 0.485 mL of 0.485 MM NaOH solution to reach the NaOH solution to reach the equivalence point. What is the mass of Hequivalence point. What is the mass of H22CC22OO44

and what is its mass percent in the sample?and what is its mass percent in the sample?

• HH22CC22OO44 + 2 NaOH + 2 NaOH Na Na22CC22OO44 + 2 H + 2 H22OO

StrategyStrategy: mol of H: mol of H22CC22OO44 out of mol of NaOH (by out of mol of NaOH (by

using V and using V and MM). Then, mol of H). Then, mol of H22CC22OO44 will be will be

used to calculate g of Hused to calculate g of H22CC22OO44 and, hence, % of it and, hence, % of it

in the sample. in the sample.

Coefficients are 1 for HCoefficients are 1 for H22CC22OO44 and 2 for NaOH. and 2 for NaOH.

Oxalic acid: Oxac M W = 90.04 g/molOxalic acid: Oxac M W = 90.04 g/mol

HH22CC22OO44 + 2 NaOH + 2 NaOH Na Na22CC22OO44 + 2 H + 2 H22OO

0.485 mol 1 mol Oxac0.485 mol 1 mol Oxac34.47 mLx34.47 mLx————— x—————— = 0.00836 mol ————— x—————— = 0.00836 mol 1000 mL 2 mol NaOH Oxac1000 mL 2 mol NaOH Oxac

Based on the Based on the end point concept (mol ratio)end point concept (mol ratio)

90.04 g90.04 g0.008360.00836 mol mol x x ———— = 0.753 g oxalic acid ———— = 0.753 g oxalic acid 1 mol 1 mol

0.753 g oxac0.753 g oxac%Oxac = ——————— x 100 = 72.8%%Oxac = ——————— x 100 = 72.8% 1.034 g of sample1.034 g of sample

Molar Mass of an Acid: A 1.056 g of a pure acid, Molar Mass of an Acid: A 1.056 g of a pure acid, HA, is dissolved in water and an acid-base HA, is dissolved in water and an acid-base indicator added. The solution requires 33.78 mL indicator added. The solution requires 33.78 mL of 0.256 of 0.256 MM NaOH solution to reach the NaOH solution to reach the equivalence point. What is the molar mass of equivalence point. What is the molar mass of the acid? the acid? We don’t knowWe don’t know what A iswhat A is

• HA + NaOH HA + NaOH NaA + H NaA + H22OO

StrategyStrategy: mol of HA out of mol of NaOH (by : mol of HA out of mol of NaOH (by

using V and using V and MM). Then, mol of HA and g of HA ). Then, mol of HA and g of HA

will be used to calculate the molar mass.will be used to calculate the molar mass.

Coefficients are 1 for HA and 1 for NaOH.Coefficients are 1 for HA and 1 for NaOH.

Molar Mass of HAMolar Mass of HA

33.78 mL NaOH solution = 0.03378 L 33.78 mL NaOH solution = 0.03378 L

HA + NaOH HA + NaOH NaA + H NaA + H22OO

0.256 mol NaOH 1 mol HA0.256 mol NaOH 1 mol HA0.03378 Lx0.03378 Lx———————— x————— = 8.65x10———————— x————— = 8.65x10-3-3

1 L sol’n 1 mol NaOH mol HA1 L sol’n 1 mol NaOH mol HA

1.056 g HA1.056 g HAMolar mass, M W = Molar mass, M W = ——————— = 122 g/mol——————— = 122 g/mol 8.65x108.65x10-3-3 mol HA mol HA

122 grams per mol 122 grams per mol

Vinegar: A 25.0-mL sample of vinegar (which Vinegar: A 25.0-mL sample of vinegar (which contains the weak acetic acid, CHcontains the weak acetic acid, CH33COCO22H) H)

requires 28.33 mL of a 0.953 requires 28.33 mL of a 0.953 MM solution of NaOH solution of NaOH for titration to the equivalence point. What mass for titration to the equivalence point. What mass of the acid, in grams, is in the vinegar sample of the acid, in grams, is in the vinegar sample and what is the and what is the MM of acetic acid in the vinegar? of acetic acid in the vinegar?

• CHCH33COCO22H + NaOH H + NaOH NaCH NaCH33COCO22 + H + H22OO

StrategyStrategy: mol of CH: mol of CH33COCO22H (HOAc) out of mol of H (HOAc) out of mol of

NaOH (by using V and NaOH (by using V and MM). Then, g of HOAc will ). Then, g of HOAc will

be calculated with the molar mass. The be calculated with the molar mass. The MM will will

be calculated by using the volume of vinegar.be calculated by using the volume of vinegar.

Coefficients are 1 for HOAc and 1 for NaOH.Coefficients are 1 for HOAc and 1 for NaOH.

CHCH33COCO22H: HOAc M W = 60.05 g/molH: HOAc M W = 60.05 g/mol

0.953 mol 1 mol HOAc0.953 mol 1 mol HOAc28.33 mLx28.33 mLx————— x—————— = 0.0270 mol ————— x—————— = 0.0270 mol 1000 mL 1 mol NaOH HOAc1000 mL 1 mol NaOH HOAc

60.05 g60.05 g0.02700.0270 mol mol x x ——— = 1.62 g acetic acid in vinegar——— = 1.62 g acetic acid in vinegar 1 mol 1 mol

For the molarity, VFor the molarity, Vvinegarvinegar = 25.0 mL = 0.0250 L = 25.0 mL = 0.0250 L (sol’n)(sol’n)

0.0270 mol HOAc0.0270 mol HOAcMM HOAc = ————————— = 1.08 HOAc = ————————— = 1.08 MM 0.0250 L sol’n0.0250 L sol’n

Problem:Problem: To analyze an iron-containing To analyze an iron-containing compound, you convert all the iron into Fecompound, you convert all the iron into Fe2+2+ in in aqueous solution and then titrate the solution aqueous solution and then titrate the solution with standardized KMnOwith standardized KMnO44. The balanced-net ionic . The balanced-net ionic

equation isequation isMnOMnO44

−− + 5Fe + 5Fe2+2+ + 8H + 8H++ Mn Mn2+2+ + 5Fe + 5Fe3+3+ + 4H + 4H22OO

A 0.598-g sample of the iron compound requires A 0.598-g sample of the iron compound requires 22.25 mL of 0.0123 22.25 mL of 0.0123 MM KMnO KMnO44 solution for titration solution for titration

to the equivalence point. What is the mass % of to the equivalence point. What is the mass % of iron in the sample? iron in the sample? Strategy:Strategy:

mL and M of MnOmL and M of MnO44−− to to mol of mol of MnOMnO44

−− and Feand Fe2+2+

Coefficients are 1 and 5 for Coefficients are 1 and 5 for MnOMnO44−− and Feand Fe2+2+ respectively respectively

FeFe2+2+ and MnO and MnO44−− (mole ratio is 5 to 1):(mole ratio is 5 to 1):

0.0123 mol MnO0.0123 mol MnO44−− 5 mol Fe 5 mol Fe2+2+

22.25 mLx 22.25 mLx ———————— x————— = 1.37x10———————— x————— = 1.37x10−3−3

1000 mL 1 mol 1000 mL 1 mol MnOMnO44−− mol Fe mol Fe2+2+

55.85 g Fe55.85 g Feg of iron: 1.37 x 10g of iron: 1.37 x 10−3−3 mol Fe mol Fe2+2+ x ————— = 0.0765 x ————— = 0.0765 1 mol Fe1 mol Fe2+2+ g Fe g Fe

Now, for the mass % of iron:Now, for the mass % of iron:

0.0765 g Fe0.0765 g Fe ———————— ———————— x 100 = 12.8% Fex 100 = 12.8% Fe 0.598 g sample0.598 g sample

NormalityNormality• equivalents (eq) solute per liter solutionequivalents (eq) solute per liter solution• milliequivalents (meq) solute per mL solutionmilliequivalents (meq) solute per mL solution

eq solute meq soluteeq solute meq soluteN = N = —————— = —————— —————— = —————— (calculated by dividing)(calculated by dividing) V(L) sol’n V(mL) sol’nV(L) sol’n V(mL) sol’n

• equivalent = 1 equivalent weightequivalent = 1 equivalent weight

• Equivalent Weight (EW); given in g/eqEquivalent Weight (EW); given in g/eq

acid/baseacid/base: the mass of a substance required to furnish : the mass of a substance required to furnish or react with exactly 1 mol Hor react with exactly 1 mol H++

redox reactionsredox reactions: the mass of substance able to gain or : the mass of substance able to gain or lose 1 mol elose 1 mol e−−

Equivalent weightEquivalent weight

HClHCl• HCl HCl H H++ + Cl + Cl––

• 1 mol HCl 1 mol HCl 1 mol H 1 mol H++, , EW = MW EW = MWHH22SOSO44

• HH22SOSO44 2H 2H++ + SO + SO442–2–

• 1 mol H1 mol H22SOSO44 2 mol H 2 mol H++, , EW = MW/2EW = MW/2HHnnA (in general)A (in general) • EW = MW/n, n = # HEW = MW/n, n = # H++ in molecule of acid in molecule of acidKOHKOH• KOH KOH K K++ + OH + OH––

• OHOH–– + H + H++ H H22OO• , 1 mol KOH reacts with 1 mol H, 1 mol KOH reacts with 1 mol H++, , EW = FW EW = FW • continued with polyhydroxides…continued with polyhydroxides…

Equivalent weightEquivalent weight

Fe(OH)Fe(OH)33 • 1 mol Fe(OH)1 mol Fe(OH)33 3OH 3OH––

• 1 mol reacts with 3 mol H1 mol reacts with 3 mol H++, , EW = FW/3 EW = FW/3M(OH)M(OH)nn (in general) (in general)• EW = FW/n, n = # OHEW = FW/n, n = # OH–– in formula unit of base in formula unit of base

Redox:Redox: +7 0+7 0

MnOMnO44–– + 5e + 5e–– Mn Mn2+2+ Zn Zn Zn Zn2+2+ + 2 e + 2 e––

FW FW MnOMnO44–– AW Zn AW Zn

EW = EW = ————— EW = ——— ————— EW = ——— 5 2 5 2

Example: Calculate the normality of barium Example: Calculate the normality of barium hydroxide in a solution made by dissolving 0.991 hydroxide in a solution made by dissolving 0.991 g of barium hydroxide in water and diluting to g of barium hydroxide in water and diluting to 100.0 mL.100.0 mL.Ba(OH)Ba(OH)22 FW 171.32 g mgFW 171.32 g mg EW = EW = —— = ———— = 85.66 —— = 85.66 ———— = ———— = 85.66 —— = 85.66 —— 2 2 eq meq2 2 eq meq

1 eq Ba(OH)1 eq Ba(OH)22 0.991 g Ba(OH)0.991 g Ba(OH)2 2 x x —————— = 0.0116 eq Ba(OH)—————— = 0.0116 eq Ba(OH)22

85.66 g 85.66 g Ba(OH)Ba(OH)22

0.0116 eq Ba(OH)0.0116 eq Ba(OH)22

N Ba(OH)N Ba(OH)22 = = ————————— = 0.116 N Ba(OH)————————— = 0.116 N Ba(OH)22 0.1000 L sol’n0.1000 L sol’n

Example: Describe the preparation of 250.0 mL Example: Describe the preparation of 250.0 mL of 0.100 of 0.100 NN oxalic acid solution from the solid. oxalic acid solution from the solid. Oxalic acid is HOxalic acid is H22CC22OO44..• 1 mol oxalic acid 1 mol oxalic acid 2 mol H 2 mol H++

• EW = 1/2 MW = 1/2(90.04) = 45.03g/eqEW = 1/2 MW = 1/2(90.04) = 45.03g/eq• 0.100 0.100 NN indicates 0.100 eq OA/L sol’n indicates 0.100 eq OA/L sol’n

0.100 eq OA0.100 eq OA0.2500 L0.2500 L x x —————— = 0.0250 —————— = 0.0250 eq OAeq OA 1 L sol’n1 L sol’n

45.03 g OA 45.03 g OA 0.0250 eq OA x 0.0250 eq OA x —————— = 1.13 g OA—————— = 1.13 g OA 1 eq OA1 eq OA • Dissolve 1.13 g of oxalic acid in water and dilute Dissolve 1.13 g of oxalic acid in water and dilute

to a total volume of 250.0 mLto a total volume of 250.0 mL

Example: What is the normality of a 0.300 Example: What is the normality of a 0.300 MM arsenic acid solution?arsenic acid solution?• HH33AsOAsO44

MWMW• 1 mol H1 mol H33AsOAsO44 3 mol H 3 mol H++ EW = EW = ———— 33 g Hg H33AsOAsO44 g H g H33AsOAsO44 g H g H33AsOAsO44 eq = eq = ————— = ———— = 3 x ————— = 3 mol ————— = ———— = 3 x ————— = 3 mol EW EW HH33AsOAsO44 MW MW MW H MW H33AsOAsO44 H H33AsOAsO44

33

eq moleq mol N = N = —— —— MM = ——— Then, = ——— Then, V(L) V(L)V(L) V(L)

N HN H33AsOAsO44 = 3 x= 3 x M H M H33AsOAsO44 = 3 x 0.300 = 0.900 = 3 x 0.300 = 0.900 NN

CONCLUSION: For an Acid HCONCLUSION: For an Acid HnnA or a A or a

Base M(OH)Base M(OH)nn

NN = n x = n x MM

n = # Hn = # H++ in molecule of acid or # OH in molecule of acid or # OH–– in in formula unit of baseformula unit of base

By definition, By definition, 1 eq1 eq of anything will react of anything will react with exactly with exactly 1 eq1 eq of anything else of anything else

Equivalence PointEquivalence PointThe point in a titration at whichThe point in a titration at which

eq titrant = eq titrateeq titrant = eq titratemeq titrant = meq titratemeq titrant = meq titrate

Example: Calculate the concentration of Example: Calculate the concentration of phosphoric acid in a solution given that a 25.0 phosphoric acid in a solution given that a 25.0 mL sample of that solution required 42.2 mL mL sample of that solution required 42.2 mL of 0.274 of 0.274 NN potassium hydroxide for titration. potassium hydroxide for titration.• HH33POPO44 + 3 KOH + 3 KOH K K33POPO44 + 3 H + 3 H22OO• at Eq. Pt., meq Hat Eq. Pt., meq H33POPO44 = meq KOH = meq KOH

meq Hmeq H33POPO44 = meq KOH = meq KOH

(mL H(mL H33POPO44)()(N N HH33POPO44) = ( mL KOH)() = ( mL KOH)(NN KOH) KOH)

(25.0 mL)(X ) = (42.2 mL)(0.274 (25.0 mL)(X ) = (42.2 mL)(0.274 NN ) )X = 0.463 X = 0.463 NN

Now, Now, NN = 3x = 3x MM, then , then MM = = NN/3/3X = 0.463/3 = 0.154 X = 0.463/3 = 0.154 MM

Example: Calculate the % oxalic acid (HExample: Calculate the % oxalic acid (H22CC22OO44) )

in a solid given that a 1.709 g sample of the in a solid given that a 1.709 g sample of the solid required 24.9 mL of 0.0998 solid required 24.9 mL of 0.0998 NN potassium potassium hydroxide for titration.hydroxide for titration.• HH22CC22OO44 + 2KOH + 2KOH K K22CC22OO44 + 2H + 2H22OO Due to the two protons of HDue to the two protons of H22CC22OO44 (OA), (OA),• EW = MW/2 = 45.03g/eq = 45.03mg/meqEW = MW/2 = 45.03g/eq = 45.03mg/meq

g OAg OA % OA = % OA = ————— x 100————— x 100 g sampleg sample

Strategy: calculate eq of NaOH that are the sameStrategy: calculate eq of NaOH that are the samefor OA; then, calculate g of OA with eq and EW for OA; then, calculate g of OA with eq and EW of OAof OA

At equivalence point,At equivalence point, eq OA = eq KOH eq OA = eq KOH data of KOH sol’ndata of KOH sol’n 0.0998 eq KOH 0.0998 eq KOH eq OA = 0.0249 L x eq OA = 0.0249 L x ——————— = ——————— = 0.00249 eq0.00249 eq 1 L sol’n1 L sol’n

45.03 g OA45.03 g OA 0.00249 eq OA x 0.00249 eq OA x ——————— = ——————— = 0.112 g OA0.112 g OA 1 eq OA1 eq OA

0.112 g OA 0.112 g OA % OA = % OA = ——————— x 100 = ——————— x 100 = 6.55%6.55% 1.709 g sample1.709 g sample