Chapter 17: Acid-base equilibria

Chemistry 1062: Principles of Chemistry II

Andy Aspaas, Instructor

Acid-ionization equilibria

• Weak acids such as acetic acid involve the following equilibrium when dissolved in water:

HC2H3O2(aq) + H2O(aq) = H3O+(aq) + C2H3O2-(aq)

• Since acetic acid is a weak acid, only a small amount of its molecules are ionized, and this equilibrium lies far to the left

• General weak acid equilibrium: HA + H2O = H3O+ + A-

• Ka: acid-dissociationconstant

– Related to equilibrium constant (Kc)

[HA]

]][AO[HO][H

O][HA][H

]][AO[H

32

2

3

ca

c

KK

K

Calculating acid-dissociation constant

• Most Ka values were obtained by recording the pH of a weak acid solution at equilibrium

• To solve the problem, the molarity of the weak acid solution must be known (this refers to how the solution was prepared, and can be considered the initial value on an ICE table)

• x = [H3O+] = [A-] = antilog(-pH)

• If x is sufficiently small, an approximation can be used in the Ka equation, (Ca – x) ca

• Degree of ionization: percentage of solute that’s in ionized form: ([A-]/Ca) x 100%

Calculating Ka

• A 0.025 M solution of lactic acid has a pH of 2.75.

– Calculate Ka

– Calculate the degree of ionization.• Construct ICE table

• Construct Ka equation

• Calculate [H3O+] from pH

• Calculate Ka

• Calculate degree of ionization

Using Ka in calculations

• What is the pH of a 0.10 M acetic acid solution? Ka = 1.7 x 10-5

• Follow same procedure as last slide, in constructing ICE table and Ka expression

• If Ca / Ka > 100, you can simplify Ka expression to just Ka = (x2 / Ca)

Solving for pH when the approximation is not valid

• What is the pH of an aqueous 0.0030 M solution of pyruvic acid, Ka = 1.4 x 10-4

• If Ca / Ka < 100, the approximation is not valid, and Ka = x2 / (Ca – x) must be solved with the quadratic equation

ax 2 bx c 0

x b b2 4ac2a

Polyprotic acids

• Polyprotic acids have more than one acidic proton• Separate Ka values for each acidic proton• Ka for the most acidic proton, Ka1 is generally much

larger than Ka2

– Proton concentration as a result of Ka2 equilibrium is negligible, can be ignored in pH calculations

• For triprotic acids like H3PO4, Ka3 is even smaller than Ka2

Polyprotic acid calculations

• What is the pH of a 0.25 M solution of sulfurous acid (H2SO3)? What is the concentration of sulfite ion, SO3

2- at equilibrium? Ka1 = 1.3 x 10-2, Ka2 = 6.3 x 10-8

• Solve for pH by just using Ka1 reaction

• Solve for [SO32-] by using equilibrium values from Ka1

reaction as initial values in Ka2 equilibrium

Base-ionization equilibria

• Kb (the base-ionization constant) can be found by a method similar to that of Ka

• B + H2O = HB+ + OH-

• Weak bases are nearlyall amines

– All contain nitrogenwith lone pair that acceptsproton from water

[B]

]][OH[HBO][H

O][B][H

]][OH[HB

2

2

cb

c

KK

K

Weak base pH calculation

• What is the pH of a 0.20 M solution of ammonia in water? Kb = 1.8 x 10-5

• Use an ICE table and construct a Kb equation to solve for [OH-]

• Convert to [H3O+] using Kw, and then to pH

– Or, convert to pOH and subtract from 14 to get pH

Acid-base properties of salt solutions

• A salt is an ionic compound resulting from an acid-base reaction

• NaCl (aq) is obtained by neutralizing NaOH and HCl

• Consider the individual ions produced in a salt to determine whether a salt solution is acidic or basic

• Hydrolysis of an ion: acid-base reaction of an ion with water

Predicting whether a salt is an acid or base

• Is an aqueous solution of NH4Cl acidic or basic?• Consider reaction of individual ions with water

NH4+ + H2O = NH3 + H3O+

Cl- + H2O = no reaction• Ammonium can donate a proton to water, so it will

make the solution acidic• Chloride does not accept a proton from water, since

HCl is a strong acid• Therefore, the solution will be acidic

Generalizations about salt pH

• A salt of a strong base and strong acid is neutral since no ions will react with water

• A salt of a strong base and a weak acid is basic (the anion is strongly basic and will produce hydroxide)

• A salt of a weak base and a strong acid is acidic (the cation is strongly acidic and will produce hydronium)

• A salt of a weak base and a weak acid can be either acidic or basic. If Ka of the cation is larger than Kb for the anion, the solution is acidic, and vice-versa.

pH of salt solutions

• The Ka and Kb of a conjugate acid-base pair are related

• Ex. for the HCN / CN- conjugate acid-base pair, both hydrolysis equilibria can be drawn

HCN + H2O = H3O+ + CN- Ka

CN- + H2O = HCN + OH- Kb

• The sum of these two equilibria is water’s autoionization equilibrium, 2H2O = H3O+ + OH- Kw

• When the two reactions are added, the equilibrium constants are multiplied, so KaKb = Kw

pH of salt solutions

• What is the pH of a 0.015 M solution of sodium benzoate? Ka of benzoic acid is 6.3 x 10-5

• Consider the hydrolysis reactions of each of the ions

• Calculate Kb of sodium benzoate from the Ka of its conjugate acid

• Solve for [OH-], convert to pOH and then pH

Common-ion effect

• Addition of another solute to a weak acid or base equilibrium can disrupt the equilibrium

• Ex. if HCl is added to a solution of acetic acid:

HC2H3O2 + H2O = C2H3O2- + H3O+

• Since HCl is a strong acid, it will add H3O+ to the system, and the equilibrium above will shift to the left (Le Chatelier’s principle)

• The ionization of weak acids is repressed by the addition of strong acid

• The same applies to the addition of extra conjugate base to the equilibrium: if acetate were added, the ionization of acetic acid would be repressed

Common-ion calculation

• What is the concentration of formate ion, CHO2-, in a

solution of 0.10 M HCHO2 and 0.20 M HCl? Ka = 1.7 x 10-4

• Set up the equilibrium with the weak acid on the left

HCHO2 + H2O = CHO2- + H3O+

• Initial concentrations of HCHO2 and H3O+ will be as given in the problem

• An approximation can be used if x is sufficiently smaller than either of the concentrations

Buffers

• Buffer: a solution of a weak acid and its conjugate base, or a weak base and its conjugate acid

• Ability to resist changes in pH• In a solution of equimolar amounts of a weak acid

and its conjugate base…– If acid is added, the conjugate base will accept

the protons to neutralize it– If base is added, the weak acid will supply

protons to neutralize it

pH of a buffer

• What is the pH of a buffer prepared by adding 30.0 mL of 0.15 M acetic acid to 70.0 mL of 0.20 M sodium acetate? Ka = 1.7 x 10-5

• Calculate initial molarities

• Use those as initial values in ICE and solve like a common-ion problem

• Use the approximation if x is sufficiently small

Adding a strong acid or base to a buffer

• First use stoichiometry to determine immediate effects of adding strong acid/base

• Ex. adding HCl to an acetic acid/sodium acetate buffer solution

HC2H3O2 + H2O = C2H3O2- + H3O+

• Assume that acetate will be stoichiometrically reduced in moles by the moles of H3O+ added

• Then allow the solution to equilibrate using new calculated concentrations

Adding a strong acid or base to a buffer

• Using the buffer in the last example, calculate the effect of adding 9.5 mL of 0.10 M HCl

• Calculate moles of H3O+ added to solution, subtract this from the base, and add this to the acid

• Calculate new molarities and solve the equilibrium equation as before

Henderson-Hasselbalch equation

• In buffer solutions made with weak acids and their salt, the equilibrium concentrations of HA and A- differ very little from their initial values

• Using the same assumption that x is very small, the equation can be rearranged

• pH = pKa + log ( [A-] / [HA] )

– for weak acid / salt buffers only! (the most common kind)

Acid-base titration curves

• Titration curve: plot of pH vs volume of acid or base added to a solution

• Equivalence point: point in a titration where a stoichiometric amount of reactant has been added

Strong acid, strong base

• Strong acid / strong base titration: equivalence point is at pH 7.0 because the salt of a strong acid and strong base is neutral

• Calculating points on a titration curve for a strong acid strong base is a matter of stoichiometry: just subtract molar amounts of acid and base to find the concentration of unreacted H3O+

• What is the pH of a solution in which 15 mL of 0.10 M NaOH has been added to 25 mL of 0.10 M HCl?

Titration of a weak acid by a strong base

• What is the pH at the equivalence point when 25 mL of 0.10 M HF is titrated by 0.15 M NaOH? Ka = 6.8 x 10-4

• In this case the pH is basic at the equivalence point• To find the pH of the equivalence point, first find the

number of moles of the salt formed if the acid were completely deprotonated, and use total volume to get its concentration

• Then, find the pH just as before.• Similar calculations can be used for titration of a

weak base by a strong acid.