Chapter 15Chapter 15Acid-Base EquilibriaAcid-Base Equilibria

CHEMISTRY

Acids and BasesAcids and Bases

• Arrhenius’ Definition:• Acids - are substances that produce hydrogen ions

(protons or H+) in solution.

Bases - are substances that produce hydroxide ions in solution.

Strong Acids and Strong Bases – totally ionize in solution

Weak Acids and Weak Bases – partially ionize in solution

Acid Dissociation in WaterAcid Dissociation in Water

• General Rxn. when Acid dissolves in H2O

• HCl + H2O H3O+ + Cl- acid base conj. Acid conj. base

Properties of HProperties of H22O @ 25 O @ 25 ooCC

• H2O (l) H+ (aq) + OH- (aq)

• Neutral, can act as an acid and a base

• Kw = [H+][OH-] = 1.0 x 10-14 only @ 25 oC

• Kw = water dissociation constant

Acidity vs. BasicityAcidity vs. Basicity

• If [H+] >[OH-] , solution is acidic• If [H+] <[OH-] , solution is basic

• The term pX = -log [concentration of X]

• So: pH = -log [concentration of H+]• pOH = -log [concentration of OH-]• pH = power of hydrogen; the power of H to which 10 is

raised

pHpH

• Kw = [H+][OH-] = 1.0 x 10-14 @ 25 oC

• pKw = [-log H+] + [-log OH-]= - [log 1.0 x 10-14] = 14

• pH = -log [H+]

• pOH = -log [OH-]

Properties of HProperties of H22O @ 25 O @ 25 ooCC

• pKw = - [log 1.0 x 10-14] = 14

• pH + pOH = 14

• pH = 7 pOH = 7

• pH = pOH

Things to RememberThings to Remember

• pKw = - [log 1.0 x 10-14] = 14 @ 25 oC

• pH + pOH = 14

• pH <7 ; acidic• pH > 7; basic• pH is between 0 - 14

Broensted-Lowry’s Broensted-Lowry’s DefinitionDefinition

• • Acid – is a proton (H+) donor.

• Base – is a proton (H+) acceptor.

• * Broensted-Lowry Definition is more general– It even applies to bases that have no –OH such as NH3.

TerminologiesTerminologies

• H+ = proton• OH- = hydroxide ion• H3O+ = hydronium ion

• Conjugate base –acid minus proton

• Conjugate acid – base plus proton

More TerminologiesMore Terminologies

• Conjugate acid-base pair– Consists of 2 substances related to each other by the

donation and acceptance of a single proton (H+).

• Acid Dissociation Constant (Ka)

EquationsEquations

• pH = - log [H+]• pOH = - log [OH-]• [H+] = 10 – pH

• [OH-] = 10 - pOH

• Kw = 10 - pKw

• pKw = pH + pOH• pKw = - log [Kw]• Pw = [H+][OH-]

Sample ProblemSample Problem

• At 40 oC, a solution has Kw = 2.916 x 10-14; pH = 7.51

• Calculate the following:• A. pOH of the solution• B. hydrogen ion concentration [H+]• C. hydroxide ion concentration [OH-]• D. pKw• E. Is the solution acidic basic or neutral?

EquilibriumEquilibrium

• K = [H3O+ ][Cl- ] = [H+][Cl-] [HCl][H2O] [HCl]

– H2O removed from top and bottom since H3O+ is simply H+ dissolved in water.

• Remember:Keq = [products] [reactants]

Acid StrengthAcid Strength

• Strength of acid is given by the equilibrium position of the dissociation reaction:

• HA (aq) + H2O (l) H3O+ + A-

• Strong acid – totally ionized and equilibrium lies far to the right

• Weak acid – only partially ionized and equilibrium lies far to the left

Strong Acid vs. Weak AcidStrong Acid vs. Weak Acid

• Strong Acid – yields a weak conjugate base (one that has weak affinity for proton; weaker than H2O)

• Weak Acid – yields a strong conjugate base (one that has strong affinity for proton; stronger than H2O)

ComparisonComparison

Property Strong Acid Weak Acid

Ka value Large Ka Small Ka

Equil. Position Far to the right Far to the Left

Equil. Concn [H+] = [HA]0 [H+] << [HA]0

Conj. Base Strength vs H2O

A- much weaker base than H2O

A- much stronger base than H2O

Please Note!Please Note!

• Tuesday’s experiment is Experiment 29: Choice I.

Sample ProblemsSample Problems

• Given [OH-] = 1.0 x 10-12 M, calculate pH. Is the solution basic, acidic or neutral?

• Given [H+] = 4.30 x 10-6 M, calculate pH. Is the solution basic, acidic or neutral?

• If the molarity of the acid or base is less than 10-6 M then the autoionization of water needs to be taken into account. In other words, water is the primary source of H+ and OH-, so the pH would be neutral.

Strong Acids and BasesStrong Acids and Bases

Strong Acids

• The strongest common acids are HCl, HBr, HI, HNO3, HClO3, HClO4, and H2SO4.

• are strong electrolytes.• All strong acids ionize completely in solution:

Strong Acids and BasesStrong Acids and Bases

• If the molarity of the acid or base is less than 10-6 M then the autoionization of water needs to be taken into account. In other words, water is the primary source of H+ and OH-, so the pH would be neutral.

Strong Acids and BasesStrong Acids and Bases

pH of Strong Acids and BasespH of Strong Acids and Bases

• The pH (and hence pOH) of a strong acid is given by the initial molarity of the acid.

• The pOH (and hence pH) of a strong base is given by the initial molarity of the base.

• Be careful of stoichiometric ratios!

Please Note!Please Note!

• Tuesday’s experiment is Experiment 29: Choice I.

Bronsted-Lowry Acids and Bronsted-Lowry Acids and BasesBases

• Bronsted-Lowry acids – compounds that donate a proton (H+)

• Bronsted-Lowry Bases – compounds that accept a proton (H+)

• Note that Bronsted-Lowry bases need not have the –OH group on the formula

• Weak acids are only partially ionized in solution.• There is a mixture of ions and unionized acid in solution.• Therefore, weak acids are in equilibrium:

Weak AcidsWeak Acids

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

HA(aq) H+(aq) + A-(aq)]HA[

]A][OH[ -3

aK

]HA[]A][H[ -

aK

NOTENOTE

• For Weak Acids and Weak Bases:

• USE ICE to determine H+, OH-, pH and pOH.!

Sample A ProblemSample A Problem

• A solution of 0.10 M formic acid (HCOOH) has a pH of 2.38 at 25 oC.

• A. Calculate Ka for formic acid at this temperature.

• B. What percent of this solution is ionized?

Sample ProblemSample Problem

• The Ka of acetic acid is 1.8 x 10-5.

• A. Calculate the pH of a 0.30 M solution of CH3COOH.

• B. Calculate OH- and pOH.

• C. Calculate Kb.

• Calculate % ionization.

A Simple TrickA Simple Trick

• Use of approximation: eliminates the difficulty of quadratic equations.

• Approximation is Valid if:

X_______ x 100 < 5 %

[Initial Concn.]

Relationship between KRelationship between Kaa and and

KKbb

• Ka x Kb = 1.0 x 10 -14 only at 25 oC.

pH of polyprotic acidspH of polyprotic acids

• Treat polyprotic acids as separate steps!

• #1. H2A (aq) H+ (aq) + HA- (aq) Ka1

• # 2. HA- (aq) H+ (aq) + A-2 (aq) Ka2

• Initial [H+] in Step 2 is Equil. [H+] from Step 1.• Total [H+] = SUM from Steps 1 & 2

HOMEWORKHOMEWORK

• What is the pH of a 1.00 M solution of tartaric acid, H2C4H4O6 (aq.) at 25.0 oC?

• Answer: pH = 1.49

Sample ProblemSample Problem

• The Ka of acetic acid is 1.8 x 10-5. Calculate the Kb of of CH3COOH.

Sample ProblemSample Problem

• The Ka of ammonia is 1.8 x 10-5. Calculate the pH of a 0.15 M solution of NH3.

Sample ProblemSample Problem

• Calculate the concentration of an aqueous solution of NaOH that has a pH of 11.50.

HOMEWORKHOMEWORK

• What is the pH of a 1.00 M solution of tartaric acid, H2C4H4O6 (aq.) at 25.0 oC?

• Answer: pH = 1.49

Calculating Ka from pH

• Weak acids are simply equilibrium calculations.• The pH gives the equilibrium concentration of H+.

• Using Ka, the concentration of H+ (and hence the pH) can be calculated.– Write the balanced chemical equation clearly showing the

equilibrium.

– Write the equilibrium expression. Find the value for Ka.

– Write down the initial and equilibrium concentrations for everything except pure water. We usually assume that the change in concentration of H+ is x.

Weak AcidsWeak Acids

Calculating Ka from pH

• Substitute into the equilibrium constant expression and solve. Remember to turn x into pH if necessary.

Using Ka to Calculate pH

• Percent ionization is another method to assess acid strength.

• For the reaction

Weak AcidsWeak Acids

HA(aq) + H2O(l) H3O+(aq) + A-(aq)

100]HA[

]OH[ionization %

0

3 eqm

Sample ProblemSample Problem

• A solution of NH3 in water has a pH of 10.50. What is the initial molarity of the solution?

Other Weak BasesOther Weak Bases

• Amines ex. Methylamine (CH3NH2)

• carbonate ion (CO32-)

• hypochlorite ion (ClO-1)

Weak BasesWeak Bases

• Also use ICE!

• Calculation is the same as for weak acids!

• Main difference is that you get [OH-] and pOH first.

Effects of Salts on pHEffects of Salts on pH

• Conjugate bases of strong acids have no effect on pH.• Conjugate acids of strong bases have no effect on pH.• Conjugate bases of weak acids increase pH (more basic).

• Ex. F- (aq) + H2O(l) HF (aq) + OH- (aq)

• Conjugate acids of weak bases decrease pH (more acidic).

• NH4+(aq) + H2O (l) NH3 (aq) + H3O+ (aq)

Relationship Between KRelationship Between Kaa

and Kand Kbb

Combined Effect of Cation and Anion in Solution• A cation that is the conjugate acid of a weak base will

cause a decrease in the pH of the solution.• Metal ions will cause a decrease in pH except for the

alkali metals (Grp. I) and alkaline earth metals.(Grp.II)

• When a solution contains both cations and anions from weak acids and bases, use Ka and Kb to determine the final pH of the solution.

Acid-Base Properties of Acid-Base Properties of Salt SolutionsSalt Solutions

Sample ProblemSample Problem

• Determine whether the resulting solution in water will be acidic, basic or neutral.

• A. K+ClO3-

• B. Na+CH3COO-

• C. Na2HPO4 Ka for HPO4- = 4.2 x 10-13

• D. NH4+Cl-

Sample ProblemSample Problem

• Predict whether the potassium salt of citric acid (K2

+HC6H5O7-) will form an acidic, basic or neutral

solution in water.

Polyprotic Acids• Polyprotic acids have more than one ionizable proton.• The protons are removed in steps not all at once:

• It is always easier to remove the first proton in a polyprotic acid than the second.

• Therefore, Ka1 > Ka2 > Ka3 etc.

Weak AcidsWeak Acids

H2SO3(aq) H+(aq) + HSO3-(aq) Ka1 = 1.7 x 10-2

HSO3-(aq) H+(aq) + SO3

2-(aq) Ka2 = 6.4 x 10-8

Polyprotic Acids

Weak AcidsWeak Acids

Sample ProblemSample Problem

• The solubility of CO2 in pure water at 25 oC and 0.1 atm is 0.0037 M. The common practice is to assume that all of the dissolved CO2 is in the form of carbonic acid (H2CO3), which is produced by the reaction between the CO2 and H2O.

• What is the pH of a 0.0037 M solution of H2CO3?

• Ka1 = 4.3 x 10-7

• Ka2 = 5.6 x 10 -11

AnswerAnswer

• pH = 4.4

• x1 = 4.0 x 10-5 M

• [CO3-] = 5.6 x 10-11 M

Sample ProblemSample Problem

• Calculate the pH and concentration of oxalate ion (C2O4

2-), in a 0.020 M solution of oxalic acid (H2C2O4)

Binary Acids

Acid-Base Behavior and Acid-Base Behavior and Chemical StructureChemical Structure

• Brønsted-Lowry acid is a proton donor.• Focusing on electrons: a Brønsted-Lowry acid can be

considered as an electron pair acceptor.• Lewis acid: electron pair acceptor.• Lewis base: electron pair donor.• Note: Lewis acids and bases do not need to contain

protons.• Therefore, the Lewis definition is the most general

definition of acids and bases.

Lewis Acids and BasesLewis Acids and Bases

• Lewis acids generally have an incomplete octet (e.g. BF3).

• Transition metal ions are generally Lewis acids.• Lewis acids must have a vacant orbital (into which the

electron pairs can be donated).• Compounds with -bonds can act as Lewis acids:

H2O(l) + CO2(g) H2CO3(aq)

Lewis Acids and BasesLewis Acids and Bases

End of Chapter 16End of Chapter 16Acid-Base EquilibriaAcid-Base Equilibria

Problem 1Problem 1

• Give the conjugate base of the following Bronsted-Lowry acids:

• H2SO3

• H2AsO4-

• NH4+

Problem 2Problem 2

• By what factor does [H+] change for a pH change of:

• A. 2.00 units• B. 0.50 units

Problem 3Problem 3

• Calculate [OH-] and pH for:

• A.) 1.5 x 10-3 M Sr(OH)2. Sr(OH)2 is a strong base.

• B.) a solution formed by adding 10 mL of 0.100 M HBr to 20.0 mL of 0.200 M HCl

ProblemProblem

•

• Calculate the pH of a solution made by adding 15.00 grams of NaH in enough water to make 2.5 L of solution

ProblemProblem

• Write the ionization and equilibrium expressions for HBrO2.

ProblemProblem

• A particular sample of vinegar has a pH of 2.9. Assuming acetic acid is the only acid in the vinegar, find the initial concentration of acetic acid in the vinegar.

ProblemProblem

• The acid dissociation constant for benzoic acid (HC7H5O2) is 6.3 x 10-5. Calculate the equilibrium concentrations of H3O+, C7H5O2

- and HC7H5O2 if the initial concentration of HC7H5O2 is 0.050 M.

ProblemProblem

•Calculate the pH of 0.120 M pyridine (C5H5N). Kb for pyridine is 1.7 x 10-9.

ProblemProblem

• A 0.200 M solution of a weak acid, HA is 9.4% ionized. Using this information, calculate [H+], [A-], [HA] and Ka for HA.

ProblemProblem

• An unknown salt is either NaF, NaCl, or NaOCl. When 0.05 mole of the salt is dissolved in water to form 0.500 L of solution, the pH of the solution is 8.08. What is the identity of the salt?

ProblemProblem

• Write the chemical equation and the Kb expression for the ionization of the following bases in aqueous solution:

• A. Dimethylamine (CH3)2NH

• B. Formate ion (HCOO-)

• C. Carbonate ion (CO32-)

ProblemProblem

• Calculate the molar concentration of OH- ions in a 0.075M solution of ethylamine.

• Kb of C2H5NH2 = 6.4 x 10-4.

• Calculate the pH of this solution.

ProblemProblem

• Ka for acetic acid (CH3COOH) is 1.8 x 10-5 while Ka for hypochlorous (HClO) ion is 3.0 x 10-8.

• A. Which is the stronger acid?• B. Which is the stronger conjugate base? Acetate ion

(CH3COO-) or chlorous (ClO-) ion?

• C. Calculate kb values for CH3COO- and ClO-.

Solubility vs. KSolubility vs. Kspsp

• Solubility – refers to the quantity that dissolves to form a saturated solution. Unit is gm/liter or moles/liter for molar solubility.

• - solubility if affected by temperature

• Solubility product constant – is the equilibrium constant for the equilibrium that exists between the ionic solute and its saturated aqueous solution

KKspsp

• Solubility product constant – the equilibrium constant indicating how soluble the product is in water.

• Example: CaF2 (s) Ca2+ (aq) + 2F- (aq)

• Ksp = [Ca2+][F-]2

Problem 1Problem 1

• Give the ionization equation and Ksp expression for the reaction:

• Ag2CrO4 (s) ? + ?

Problem 2Problem 2

• The Ksp for CaF2 is 3.9 x 10-11 at 25 oC. Assuming that CaF2 dissociates completely upon dissolving and that there are no other important equilibria affecting its solubility:

• a. calculate the solubility of CaF2 in moles per liter.

• b. calculate the solubility of CaF2 in grams per liter.

Problem 3Problem 3

• The Ksp for LaF3 is 2.0 x in 10-19. What is the solubility of LaF3 in water in moles per liter?

• What is the solubility of LaF3 in water in grams per liter?

AnswerAnswer

• A. 9.28 x 10-6 M

Factors Affecting SolubilityFactors Affecting Solubility

• Common-Ion Effect• Concentration

Problem 4Problem 4

• Calculate the molar solubility of CaF2 at 25 oC in a solution that is:

• A. 0.010 M in Ca(NO3)2

• B. 0.025 M in NaF

Precipitation of IonsPrecipitation of Ions

• Remember Q, the reaction quotient?

• If: Q > Ksp, prepitations occurs until Q = Ksp

Q = Ksp, equilibrium exists (saturated solution) Q < Ksp, solid dissolves until Q = Ksp.

Problem 1Problem 1

• A solution contains 1.0 x 10-12 M Ag+ and 2.0 x 10-2 M Pb2+. When Cl- is added, both AgCl and Ksp precipitate from the solution.

• What concentration of Cl- is necessary to begin the precipitation of each salt?

• Which salt precipitates first?

Insoluble ChloridesInsoluble Chlorides

• Of the common metals ions, only Ag+, Hg2 2+, Pb 2+ form insoluble chlorides.

Qualitative AnalysisQualitative Analysis

• Order of separation of ions:Cl- S2- (OH)- (PO4) 3-

NH4+

1st step: add 6M HCl

2nd step: add H2S and 0.20 M HCl

3rd step: add (NH4)2S

4th step: add (NH4)2HPO4 and NH3