Applications of Aqueous Equilibria

AP Chemistry

Chapter 15

This chapter covers

• More acid-base equilibria

• Solubility of salts

• Formation of complex ions

Common ion

• Some solutions contain not only a weak acid…

• …but also the salt of the weak acid:

• Acetic acid

• Sodium acetate

• These form a solution with powerful results

Suppose a solution contains HF (Ka = 7.2x10-4) and NaF

NaF (s) + H2O Na+ (aq) + F-(aq)

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

F- is called the “common ion”Thinking in terms of LeChatelier’s principle, does the presence of a common ion have any effect?

The result will be fewer H+ ions and a higher pH.

With more F-, the equilibrium will not shift as dramatically in the direction of the products.

This is the Common Ion Effect.

15.1 The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2 M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0M HF (Ka=7.2x10-4) and 1.0 M NaF.

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

1.0 0 1.0-x x x

1.0-x x 1.0+x

Ka=7.2x10-4 = x(1.0+x) 1.0-x

x = [H+] = 7.2x10-4

7.2x10-4 100 = 0.072% 1.0

Buffered Solutions

• A buffered solution is one that resists a change in its pH when either H+ or OH- is added

• Our blood contains a buffering system of HCO3

- and H2CO3

• If it weren’t so, we would die!

• Our blood must stay in a very narrow pH range.

2 common buffers

• A solution of a weak acid plus the soluble salt of the weak acid

• Sodium acetate• Acetic acid

• A solution of a weak base plus the soluble salt of the weak base

• Ammonia• Ammonium chloride

Using LeChatalier’s Principle, we can determine what happens to an acid when added to a buffer.

HC2H3O2 H+ + C2H3O2-

Adding a strong acid, reacts with the anion and shifts the reaction to the formation of the weak acid.

HC2H3O2 H+ + C2H3O2-

Adding a strong base, reacts with the H+ forming water and shifts the reaction to the formation of more H+.

Using LeChatalier’s Principle, we can determine what happens to an acid when added to a buffer.

15.2 a buffered solution contains 0.50 M acetic acid (Ka=1.8x10-5) and 0.50 M sodium acetate. Calculate the pH of this solution.

HC2H3O2 H+ + C2H3O2-

.50 0 .50-x +x +x

.50-x x .50+x

Ka = 1.8x10-5 = x (.50+x) .50-x

x = [H+] = 1.8x10-5 pH = 4.74

15.3 Calculate the change in pH that occurs when 0.010 mol solid NaOH is added to 1.0 L of the buffered solution in #15.2. Compare the pH change with that which occurs when 0.010 mol solid NaOH is added to 1.0 L of water.

-x +x +x.49-x x .51+x

Ka = 1.8x10-5 = x (.51+x) .49-x

x = [H+] = 1.7x10-5

pH = 4.76

HC2H3O2 + OH- H2O + C2H3O2-

.50 .010 .50-.010 -.010 +.010 +.010 .49 0 .010 .51

pH = .02

Notice, we first had to take care of the reaction between the strong base and the weak acid, then we worked the equilibrium problem!

HC2H3O2 H+ + C2H3O2-

.49 .51

Just How Does Buffering Work?

• As we add OH-, the weak acid is the best source of H+ ions….

• OH- + HA H2O + A-

• The OH- ions are not allowed to accumulate and are thus replaced by A- ion

The equilibrium expression is…

HA H+ + A-

Ka = [H+][A-] [HA]

…and can be written as: [H+] = Ka [HA] [A-]

So the pH is determined by this ratio. When OH- is added, HA is converted to A-

A change in [HA]/[A-] is usually very small and [H+] and pH remain relatively constant.If [H+] is added then H+ + A- HA and free [H+] do not accumulate.If [HA] and [A-] are large compared with [H+] which is added, little pH change occurs

Calculate [H+] in 0.10 M HF (Ka = 7.2x10-4) and 0.30 M NaF.

[H+] = Ka [HA] [A-]

[H+] = 7.2x10-4 [.10] [.30] [H+] = 2.0x10-4 M

Another way to handle this is to take the –log of both sides of the 1st equation. Trust me, you’ll like this.

-log[H+] = -log Ka [HA] [A-]

pH = pKa + log [A-] [HA]

pH = pKa + log [base] [acid]

If this base/acid ratio = 1, the better the buffer and, as a bonus, pH = pKa!

15.4 Calculate the pH of a solution containing 0.75 M lactic acid (HC3H5O3, Ka=1.4x10-4) and 0.25 M sodium lactate.

pH = pKa + log [A-] [HA]

pH = -log(1.4x10-4) + log [.25] [.75]

pH = 3.38

15.5 A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution.

pH = pKa + log [A-] [HA]

pH = -log(5.6x10-10) + log [.25] [.40]

pH = 9.05

NH3 NH4+ + OH-

Ka = Kw = 10-14 =5.6x10-10

Kb 1.8x10-5

15.6 Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution form example 15.5

NH3 NH4+ + OH-

.25 .40 0-.10 +.10 +.10

.15 .50 .10

Ka = Kw = 10-14 =5.6x10-10

Kb 1.8x10-5

pH = pKa + log [A-] [HA]

pH = -log(5.6x10-10) + log [.15] [.50]

pH = 8.73

Buffer Capacity

• The amount of protons or hydroxide ions the buffer can absorb without significant change in pH.

• It is determined by the base/acid ratio.

15.7 Calculate the change in pH that occurs when 0.010 mol gaseous HCl is added to 1.0 L of each of the following solutions:SolutionA: 5.00 M HC2H3O2 and 5.00 M C2H3O2

-

pH = pKa + log [5.00] [5.00]

pH = -log 1.8x10-5 + 0 = 4.74 initially

C2H3O2- + H+ HC2H3O2

5.00 .010 5.00-.010 -.010 +.0104.99 5.01pH = pKa + log [4.99] [5.01]

pH = 4.74 + -.00174 = 4.74

Solution B: .050 M HC2H3O2 and .050 M C2H3O2

pH = pKa + log [.050] [.050]

pH = -log 1.8x10-5 + 0 = 4.74 initially

C2H3O2- + H+ HC2H3O2

.050 .010 .050-.010 -.010 +.010 .04 .060pH = pKa + log [.04] [.06]

pH = 4.74 + -.176 = 4.56

When the concentrations of [HA] and [A-] are large and the addition of acid or base is small…

…there is very little effect on the pH As the concentrations of [HA] and [A-] are smaller, the addition of acid or base has a greater effect on pH.

Best Buffer

• To produce the most effective buffer…

• [HA] = [A-]

• pKa of weak acid should be as close as possible to the desired pH.

pH = pKa + log [A-] [HA]

4.00 = pKa + 0

A pH of 4.00 is wanted

Ka = 1.0x10-4

15.8 A chemist needs a solution buffered at pH 4.30 and can choose from the following acids (and their sodium salts):

a.Chloroacetic acid (Ka=1.35x10-3)

b.Propanoic acid (Ka=1.3x10-5)

c. Benzoic acid (Ka=6.4x10-5)

d.Hypochlorous acid (Ka=3.5x10-8)

Calculate the ratio [HA]/[A-] required for each system to yield a pH of 4.30. Which system will work best?[H+] = Ka [HA] [A-]

4.30=-log[H+]so

[H+]=5.0x10-5

5.0x10-5 = [HA] Ka [A-]

5.0x10-5 = [HA]1.35x10-3 [A-].00375.0x10-5 = [HA]1.35x10-5 [A-]3.85.0x10-5 = [HA]6.4x10-5 [A-].785.0x10-5 = [HA]3.5x10-8 [A-]1400