The Conjugate Seesaw: Calculating the Acid Dissociation Constant of Ammonium ($NH_4^+$) from Ammonia's Base Constant

The Conjugate Seesaw: Calculating the $K_a$ of $NH_4^+$

In chemical nomenclature and acid-base theory, the relationship between a weak base and its conjugate acid is governed by a reciprocal dance mirrored in mathematical constants. When asked for the acid dissociation constant ($K_a$) of the ammonium ion ($NH_4^+$) given the base dissociation constant ($K_b$) of ammonia ($NH_3$) as $1.8 \times 10^{-5}$, we rely on the self-ionization constant of water ($K_w$).

The Mathematical Mechanism

At 25°C, the ion-product constant for water ($K_w$) is universally defined as:

$$K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}$$

When ammonia dissolves in water, it acts as a Brønsted-Lowry base:

$$NH_3 (aq) + H_2O (l) \rightleftharpoons NH_4^+ (aq) + OH^- (aq)$$

Its base dissociation constant is expressed as:

$$K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}$$

Conversely, the ammonium ion acts as a weak acid in aqueous solution:

$$NH_4^+ (aq) + H_2O (l) \rightleftharpoons NH_3 (aq) + H_3O^+ (aq)$$

Multiplying $K_a$ by $K_b$ yields $K_w$ ($K_a \times K_b = K_w$). Therefore, solving for $K_a$ is a matter of division:

$$K_a = \frac{K_w}{K_b} = \frac{1.0 \times 10^{-14}}{1.8 \times 10^{-5}} \approx 5.56 \times 10^{-10}$$

Rounded to two significant figures, the $K_a$ of $NH_4^+$ is $5.6 \times 10^{-10}$.