Answer to Question #103355 in Physical Chemistry for Ajay

Question

For the hydrolysis of a salt of a weak acid and a weak base, show that
Kh=Kw/(Ka)(Kb)

Answer 1

Let we have $MA$ a salt of weak acid and weak base. When $MA$ is dissolved in water, the interaction occurs between$M^+, A^-$  ion and water molecules.

$M^+ + A^- + H_2O ↔ HA + MOH$

At equilibrium,$K_{eq} = \frac{[HA][MOH]}{[M^+][A^-][H_2O]}$

$\implies K_{eq} [H_2O] = K_h =\frac{ [HA][MOH]}{ [M^+][A^-] }$ (where $K_h$ is hydrolysis constant.)....(1)

Ionic product of water $K_w = [H^+] [OH^-]$

And ionization constant of weak acid $K_a =\frac{[H^+] [A^-]}{[HA]}$

Ionization constant of weak base $K_b =\frac{ [M^+] [OH^-]}{[MOH]}$

$\frac{K_w}{K_a \times K_b } = \frac{ [H^+] [OH^-] [HA] [MOH]}{ [H^+] [A^-] [M^+] [OH^-] }$ $= \frac{[HA] [MOH]}{[A^-] [M^+]}$ $= K_h$ ..(from (1))

$K_h = \frac{K_w}{ K_a \times K_b}$