8.3: Weak Acid and Base Equilibria

Introduction

Weak acids and bases do not completely dissociate in water. As a result, their equilibria involve the undissociated acid or base, along with its conjugate base or acid. This leads to equilibrium expressions involving $ K_a $ (acid dissociation constant) for weak acids and $ K_b $ (base dissociation constant) for weak bases. Acid Dissociation Constant Base Dissociation Constant

Weak Acid Equilibria

Acid Dissociation Constant ( $ K_a $ )

For a weak acid, $ HA $ , in water: $$ HA(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + A^-(aq) $$ The acid dissociation constant, $ K_a $ , is defined as: $$ K_a = \frac{[H_3O^+][A^-]}{[HA]} $$ A smaller $ K_a $ indicates a weaker acid.

Calculating $ [H_3O^+] $ and pH

To calculate the hydronium ion concentration ( $ [H_3O^+] $ ) and pH for a weak acid solution, we often use an ICE Tables|ICE Table.

Example: Calculate the pH of a 0.10 M solution of acetic acid ( $ CH_3COOH $ ), given that $ K_a = 1.8 \times 10^{-5} $ .

$$ K_a = \frac{[H_3O^+][CH_3COO^-]}{[CH_3COOH]} = \frac{x^2}{0.10 - x} = 1.8 \times 10^{-5} $$ Approximation: Since $ K_a $ is small, assume $ x « 0.10 $ , so $ 0.10 - x \approx 0.10 $ . Approximation Techniques $$ x^2 \approx 1.8 \times 10^{-6} $$ $$ x = \sqrt{1.8 \times 10^{-6}} = 1.34 \times 10^{-3} \approx [H_3O^+] $$ $$ pH = -log[H_3O^+] = -log(1.34 \times 10^{-3}) = 2.87 $$

Percent Dissociation

Percent dissociation is the percentage of the weak acid that dissociates in solution. $$ Percent \ Dissociation = \frac{[A^-]{equilibrium}}{[HA]{initial}} \times 100% $$
In the acetic acid example above, the percent dissociation is: $$ \frac{1.34 \times 10^{-3}}{0.10} \times 100% = 1.34% $$

Weak Base Equilibria

Base Dissociation Constant ( $ K_b $ )

For a weak base, $ B $ , in water: $$ B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq) $$ The base dissociation constant, $ K_b $ , is defined as: $$ K_b = \frac{[BH^+][OH^-]}{[B]} $$ A smaller $ K_b $ indicates a weaker base.

Calculating $ [OH^-] $ and pOH

Similar to weak acids, we can use an ICE Tables|ICE Table to calculate the hydroxide ion concentration ( $ [OH^-] $ ) and pOH for a weak base solution.

Example: Calculate the pH of a 0.15 M solution of ammonia ( $ NH_3 $ ), given that $ K_b = 1.8 \times 10^{-5} $ .

$$ K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = \frac{x^2}{0.15 - x} = 1.8 \times 10^{-5} $$ Approximation: Since $ K_b $ is small, assume $ x « 0.15 $ , so $ 0.15 - x \approx 0.15 $ . $$ x^2 \approx 2.7 \times 10^{-6} $$ $$ x = \sqrt{2.7 \times 10^{-6}} = 1.64 \times 10^{-3} \approx [OH^-] $$ $$ pOH = -log[OH^-] = -log(1.64 \times 10^{-3}) = 2.79 $$ $$ pH = 14 - pOH = 14 - 2.79 = 11.21 $$

Percent Protonation

Percent protonation is the percentage of the weak base that protonates in solution. $$ Percent \ Protonation = \frac{[BH^+]{equilibrium}}{[B]{initial}} \times 100% $$
In the ammonia example above, the percent protonation is: $$ \frac{1.64 \times 10^{-3}}{0.15} \times 100% = 1.09% $$

$ K_a $ and $ K_b $ Relationship for Conjugate Acid-Base Pairs Ka and Kb Relationships

For a conjugate acid-base pair, the product of $ K_a $ and $ K_b $ is equal to the Ionic product of water Kw| $ K_w $ . $$ K_a \times K_b = K_w = 1.0 \times 10^{-14} \ at \ 25^\circ C $$ This relationship is useful for calculating $ K_b $ if $ K_a $ is known (or vice versa) for a conjugate pair. This is especially useful when you need to determine the Base Strength and Conjugate Acid Strength|base strength of a conjugate base given the Acid Strength and Conjugate Base Strength|acid strength of the conjugate acid.

Example: If $ K_a $ for $ HA $ is $ 1.0 \times 10^{-5} $ , then $ K_b $ for $ A^- $ is: $$ K_b = \frac{K_w}{K_a} = \frac{1.0 \times 10^{-14}}{1.0 \times 10^{-5}} = 1.0 \times 10^{-9} $$