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Chemistry · Ch 6 — Equilibrium

Relation between Ka and Kb

6.11.5

Relation between Ka and Kb

The Relationship Between KaK_a and KbK_b

The strength of an acid is measured by its acid dissociation constant KaK_a, and the strength of a base by its base dissociation constant KbK_b. For any conjugate acid-base pair, these two constants are not independent — they are linked by a simple and powerful relation that involves the ionisation constant of water, KwK_w.

Consider the conjugate pair NH4+NH_4^+ (acid) and NH3NH_3 (base). Each species undergoes its own equilibrium with water.

Acid dissociation of NH4+NH_4^+:

NH4+(aq)+H2O(l)⇌H3O+(aq)+NH3(aq)NH_4^+(aq) + H_2O(l) \rightleftharpoons H_3O^+(aq) + NH_3(aq)

Ka=[H3O+][NH3][NH4+]=5.6×10−10K_a = \frac{[H_3O^+][NH_3]}{[NH_4^+]} = 5.6 \times 10^{-10}

Base dissociation of NH3NH_3:

NH3(aq)+H2O(l)⇌NH4+(aq)+OH−(aq)NH_3(aq) + H_2O(l) \rightleftharpoons NH_4^+(aq) + OH^-(aq)

Kb=[NH4+][OH−][NH3]=1.8×10−5K_b = \frac{[NH_4^+][OH^-]}{[NH_3]} = 1.8 \times 10^{-5}

Now add these two reactions. The NH4+NH_4^+ and NH3NH_3 cancel on opposite sides, leaving:

2H2O(l)⇌H3O+(aq)+OH−(aq)2H_2O(l) \rightleftharpoons H_3O^+(aq) + OH^-(aq)

This is the autoionisation of water, whose equilibrium constant is KwK_w:

Kw=[H3O+][OH−]=1.0×10−14K_w = [H_3O^+][OH^-] = 1.0 \times 10^{-14}

When two reactions are added to give a net reaction, the equilibrium constant of the net reaction is the product of the equilibrium constants of the individual reactions. Therefore:

Ka×Kb=([H3O+][NH3][NH4+])×([NH4+][OH−][NH3])=[H3O+][OH−]=KwK_a \times K_b = \left( \frac{[H_3O^+][NH_3]}{[NH_4^+]} \right) \times \left( \frac{[NH_4^+][OH^-]}{[NH_3]} \right) = [H_3O^+][OH^-] = K_w

Numerically:

(5.6×10−10)×(1.8×10−5)=1.0×10−14(5.6 \times 10^{-10}) \times (1.8 \times 10^{-5}) = 1.0 \times 10^{-14}

This is not a coincidence. It is a general result.

Ka×Kb=KwK_a \times K_b = K_w

Important

This relation holds for any conjugate acid-base pair. If you know KaK_a for the acid, you immediately know KbK_b for its conjugate base, and vice versa. A strong acid (large KaK_a) will have a weak conjugate base (small KbK_b), and a strong base (large KbK_b) will have a weak conjugate acid (small KaK_a).

An Alternative Derivation

The same result can be obtained by starting from the general base-dissociation equilibrium for a base BB:

B(aq)+H2O(l)⇌BH+(aq)+OH−(aq)B(aq) + H_2O(l) \rightleftharpoons BH^+(aq) + OH^-(aq)

Kb=[BH+][OH−][B]K_b = \frac{[BH^+][OH^-]}{[B]}

(The concentration of water is constant and is absorbed into KbK_b.)

Multiply numerator and denominator by [H+][H^+]:

Kb=[BH+][OH−][H+][B][H+]K_b = \frac{[BH^+][OH^-][H^+]}{[B][H^+]}

Group the terms:

Kb=[OH−][H+]×[BH+][B][H+]K_b = \frac{[OH^-][H^+] \times [BH^+]}{[B][H^+]}

The first product [OH−][H+][OH^-][H^+] is KwK_w. The second fraction [BH+][B][H+]\frac{[BH^+]}{[B][H^+]} is the reciprocal of the acid dissociation constant for the conjugate acid BH+BH^+:

Ka=[H+][B][BH+]so1Ka=[BH+][B][H+]K_a = \frac{[H^+][B]}{[BH^+]} \quad \text{so} \quad \frac{1}{K_a} = \frac{[BH^+]}{[B][H^+]}

Therefore:

Kb=Kw×1KaK_b = K_w \times \frac{1}{K_a}

which rearranges to:

Ka×Kb=KwK_a \times K_b = K_w

The pK Relationship

Taking the negative logarithm (base 10) of both sides of Ka×Kb=KwK_a \times K_b = K_w gives:

−log⁡(Ka)+(−log⁡(Kb))=−log⁡(Kw)-\log(K_a) + (-\log(K_b)) = -\log(K_w)

pKa+pKb=pKwpK_a + pK_b = pK_w …

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