Acid/base dissociation constant

An acid/base dissociation constant is a measure of the strength of an aqueous acid/base.

A weak acid dissociates partially in an aqueous solution to give the hydroxonium ion and a conjugate base.

HA(aq)+H_2O(l)\rightleftharpoons H_3O^+(aq)+A^-(aq)

Since water is the solvent, its activity approximately equals to one, and the equilibrium constant is given by:

K_a=\frac{\left [ H_3O^+ \right ][A^-]}{[HA]}\; \; \; \; \; \; \; \; 12

Ka is called the acid dissociation constant. The higher the value of Ka of an acid, the greater the extent of dissociation of the acid (i.e., the stronger the acid). It is difficult to determine accurately the Ka of an acid that completely ionises in water (a strong acid), since [HA] → 0. Hence, Ka is a useful measure only for weak acids.

A weak base dissociates partially in an aqueous solution to give a conjugate acid and the hydroxide ion.

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

Since water is the solvent, its activity approximately equals to one, and the equilibrium constant is given by:

K_b=\frac{[BH^+][OH^-]}{[B]}\; \; \; \; \; \; \; \; 13

Kb is called the base dissociation constant. The higher the value of Kb of a base, the greater the extent of dissociation of the base (i.e., the stronger the base). Similar to acids, it is difficult to determine accurately the Kb of a base that completely ionises in water (a strong base), since [B] → 0. Hence, Kb is a useful measure only for weak bases.

Note that BH+ is an acid with the following dissociation equilibrium:

BH^+(aq)+H_2O(l)\rightleftharpoons B(aq)+H_3O^+(aq)

K_a=\frac{[B][H_3O^+]}{[BH^+]}\; \; \; \; \; \; \; \; 14

Multiplying eq14 with eq13,

K_aK_b=[H_3O^+][OH^-]\; \; \; \; \; \; \; \; 15

Substituting eq11 from the previous article in eq15

K_aK_b=K_w\; \; \; \; \; \; \; \; 16

Eq16 expresses the relationship for all conjugate acid-base pairs.

 

Question

Given that Ka for HCN is 6.2×10-10 at 25oC, calculate the equilibrium constant for

CN^-(aq)+H_2O(l)\rightleftharpoons HCN(aq)+OH^-(aq)

Answer

K_b=\frac{K_w}{K_a}=\frac{10^{-14}}{6.2\times 10^{-10}}=1.6\times 10^{-5}

 

 

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