CommIon

CHM 1046
General Chemistry II
Dr. Michael Blaber

Additional Aspects of Aqueous Equilibria

The Common-Ion Effect

A weak acid or weak base will partially ionize in aqueous solution. For example,

Weak acids often have K_a values < 1.0 x 10^-3
Only a small percentage of the acid dissociates in aqueous solution

Ionic compounds (i.e. salts) dissociate completely in aqueous solution

Some salts may contain ions derived from weak or strong acids, or bases

Ions derived from strong acids or bases will not alter the pH (the conjugate base or conjugate acid of a strong acid or strong base have no tendency to either accept or donate a proton)

Ions derived from weak acids or bases will have a certain tendency to either donate or accept a proton (i.e. the reverse process to the ionization reaction that produced them), and will affect the pH of the sample

What happens if a salt is added to a solution of a weak acid, and that salt contains a conjugate base to the weak acid?

Example: Acetic acid (HC₂H₃O₂)

The ionization of acetic acid, a weak acid, is as follows:

HC₂H₃O₂(aq) ó C₂H₃O₂^-(aq) + H⁺(aq)

Now let's add some Sodium Acetate salt.

Sodium acetate will dissociate completely in solution:

NaC₂H₃O₂(aq) à Na⁺(aq) + C₂H₃O₂^-(aq)

We have increased the acetate conjugate base concentration (without increasing the concentration of H⁺)
Le Chatelier's principle would predict that the equilibrium will shift to the left in order to minimize the perturbing effects of the added concentration of conjugate base:

HC₂H₃O₂(aq) ß C₂H₃O₂^-(aq) + H⁺(aq)

This shift to the left will also result in a decrease in the H⁺ concentration

A reduction in the [H⁺] will raise the pH of the solution (i.e. be more basic)
A shift to the left means that less acid will dissociate

The "Common Ion Effect":
The dissociation of a weak electrolyte is decreased by adding to the solution a strong electrolyte (i.e. a salt) that has an ion in common with the weak electrolyte

A similar type of result would be observed for the ionization of a weak base and the addition of a salt that represents the conjugate acid

For example, ammonia (NH₃) is a weak base that ionizes in H₂O to produce ammonium ion (NH₄⁺) and hydroxide ion (OH^-)

NH₃(aq) + H₂O(l) ó NH₄⁺(aq) + OH^-(aq)

The addition of ammonium sulfate salt (NH₄)₂SO₄ would add ammonium ions (the conjugate acid).

This would shift the equilibrium concentration to the left

This would reduce the [OH^-] concentration and lower the pH
Note that the sulfate ion, SO₄^2-, is the conjugate base of a strong acid (H₂SO₄, sulfuric acid) and therefore has no basic properties (and is a spectator ion in the above ionization reaction of ammonia)

Acetic acid (CH₃COOH) is a weak acid with the following ionization reaction:

CH₃COOH + H₂O Û H₃O⁺ + CH₃COO^- K_a = 1.8 x 10^-5

What is the pH of a solution that is 0.5M in acetic acid and 2.5M in sodium acetate, CH₃COONa?

1.8 x 10^-5 = [CH₃COO^-][H₃O⁺]/[CH₃COOH]

If XM = amount of acetic acid that dissociates, then X M of H₃O⁺ and CH₃COO^- ions are formed. The added sodium acetate is a salt containing the cation of a strong base (Na⁺) and an anion of a weak acid (CH₃COO^-). This will dissociate completely in water to produce 2.5M each Na⁺ ion and CH₃COO^-. Thus, at equilibrium, the concentration of CH3COO^- will be the sum of the amount produced by dissociation of the acetic acid, plus the amount from the dissociation of the sodium salt, i.e. [CH₃COO^-] = (X + 2.5)M

1.8 x 10^-5 = (X + 2.5)(X)/(0.5 - X)

X² + 2.50X - 0.9 x 10^-5 = 0

This is a quadratic, and

X = 3.60 x 10^-6M = [H₃O⁺] = [H⁺]

pH = -log[H⁺] = -log(3.60 x 10^-6) = 5.44

Can we use the shortcut to the quadratic?

0.5M acetic acid / 1.8 x 10^-5 = 27,777 (which is greater than 100)

Therefore, the amount of acid that dissociates is small compared to the starting concentration. Thus, (0.5M - X) » (0.5M). If this is the case, then (X + 2.5) would also » (2.5M).

1.8 x 10^-5 = (X + 2.5)(X)/(0.5 - X) » (2.5)(X)/(0.5)

9.0 x 10^-6 = 2.5X

X = 3.60 x 10^-6

This is the same answer as above when using the quadratic.

What is the pH of an aqueous solution that contains 0.15M NH₃ and 0.05M (NH₄)₂SO₄? The K_b for NH₃ is 1.8 x 10^-5.

The 0.05M of (NH₄)₂SO₄ salt will dissociate completely to produce 0.1M of NH₄⁺

(NH₄)₂SO₄(aq) à 2NH₄⁺(aq) + SO₄^2-(aq)

Ammonia will ionize to produce NH₄⁺ and OH^- according to the value for K_b

NH₃(aq) + H₂O(l) ó NH₄⁺(aq) + OH^-(aq)

and at equilibrium:

Set X = amount of NH₃ that will ionize as the reaction goes to equilibrium

The added salt is not directly altering the concentration of either NH₃ or OH^-
Therefore if X M of NH₃ ionize, then X M of OH^- are produced at equilibrium
The concentration of NH₄⁺ at equilibrium is equal to the amount of NH₃ that ionizes, plus the added amount of NH₄⁺ from the salt (X + 0.1M)
The concentration of NH₃ at equilibrium is equal to the starting concentration minus the amount that ionized (0.15M - X)

We can now set up the equilibrium expression:

K_b = 1.8 x 10^-5 = (X + 0.1)(X) / (0.15 - X)

1.8 x 10^-5 (0.15 - X) = X² + 0.1X

2.70 x 10^-6 - 1.8 x 10^-5 X = X² + 0.1X

0 = X² + 0.1X + 1.8 x 10^-5 X - 2.70 x 10^-6

0 = X² + 0.100X - 2.70 x 10^-6

This is a quadratic; a = 1, b = 0.1, c = -2.70 x 10^-6. The solution(s) for X are:

X = 2.7 x 10^-5
X = -0.100
X cannot be negative (otherwise the concentration of OH^- at equilibrium would be a negative number), thus, X = 2.7 x 10^-5

Therefore, at equilibrium:

[OH^-] = 2.7 x 10^-5 M
[NH₄⁺] = (X + 0.1M) ~ 0.100M
[NH₃] = (0.15 - X) ~ 0.150M
pOH = -log[OH^-] = 4.57
pH = (14 - pOH) = 9.43

How does this pH compare to that expected if no (NH₄)₂SO₄ salt were added?

NH₃(aq) + H₂O(l) ó NH₄⁺(aq) + OH^-(aq)

and at equilibrium:

The initial concentration of NH₃ is 0.15M

X = amount of NH₃ that ionizes as the reaction reaches equilibrium
Therefore, at equilibrium, [OH^-] = [NH₄⁺] = X M, and the concentration of NH₃ at equilibrium = (0.15 - X)

K_b = 1.8 x 10^-5 = X² / (0.15 - X)

1.8 x 10^-5 (0.15 - X) = X²

2.70 x 10^-6 - 1.8 x 10^-5X = X²

X² + 1.8 x 10^-5X - 2.70 x 10^-6 = 0

This is a quadratic. The solutions are:

X = 1.63 x 10^-3
X = -1.65 x 10^-3
X cannot be negative (otherwise we would be producing NH₃ from nothing)

Therefore at equilibrium:

[NH₃] = 0.15 - 1.63 x 10^-3 = 0.148M
[OH^-] = 1.63 x 10^-3
[NH₄⁺] = 1.63 x 10^-3
pOH = -log[OH^-] = 2.79
pH = 14 - pOH = 11.2