Salt of a Weak Acid and a Strong Base:

The solution of such a salt is basic in nature. The anion of the salt is reactive. It reacts with water to form a weak acid and OH^- ions.

A^- + H₂O;   ↔   HA + OH^-

                        Weak acid

Consider, for example, the salt CH₃COONa. It ionises in water completely to give CH₃COO^- and Na⁺ ions. CH₃COO^- ions react with water to form a weak acid, CH₃COOH and OH^- ions.

CH₃COO^- + H₂O  ↔   CH₃COOH + OH^-

C(1-x)                         Cx           Cx

Thus, OH^- ion concentration increases, the solution be­comes alkaline.

Applying law of mass action,

K_h = [CH₃COOH][OH^-]/[CH₃CO^-] = (Cx×Cx)/C(1-x) = (Cx²)/(1-x) )  ...... (i)

Other equations present in the solution are:

CH₃COOH ↔ CH₃COO^- + H⁺,       K_a = [CH₃COO^-][H⁺]/[CH₃COOH]      ...... (ii)

H₂O ↔  H⁺ + OH^-,                 K_w = [H⁺][OH^-]    ....... (iii)

From eqs. (ii) and (iii),

log [OH^-] = log K_w - log K_a + log[salt]/[acid]

-pOH = -pK_w + pK_a + log[salt]/[acid]

pK_w - pOH = pK_a + log[salt]/[acid]

pH = pK_a + log[salt]/[acid]

Considering eq. (i) again,

       K_h = cx²/(1-x)      or    K_h = Ch²/(1-h)

When h is very small, (1-h) → 1

or     h² = K_h/C

or     h =  √K_h/C

[OH^-] = h × C = √(CK_h) = √(C*K_w/K_a)

[H⁺] = K_w/[OH^-]

= K_w/√(C*K_w/K_a) = √(K_a*K_w)/K_c

-log [H⁺] = -1/2log K_w - 1/2log K_a + 1/2log C

pH =  1/2pK_w +  1/2pK_a +  1/2log C

= 7 + 1/2pK_a +  1/2log C.