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-
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,
Kh = [CH3COOH][OH-]/[CH3CO-] = (Cx×Cx)/C(1-x) = (Cx2)/(1-x) ) ...... (i)
Other equations present in the solution are:
CH3COOH ↔ CH3COO- + H+, Ka = [CH3COO-][H+]/[CH3COOH] ...... (ii)
H2O ↔ H+ + OH-, Kw = [H+][OH-] ....... (iii)
From eqs. (ii) and (iii),
log [OH-] = log Kw - log Ka + log[salt]/[acid]
-pOH = -pKw + pKa + log[salt]/[acid]
pKw - pOH = pKa + log[salt]/[acid]
pH = pKa + log[salt]/[acid]
Considering eq. (i) again,
Kh = cx2/(1-x) or Kh = Ch2/(1-h)
When h is very small, (1-h) → 1
or h2 = Kh/C
or h = √Kh/C
[OH-] = h × C = √(CKh) = √(C*Kw/Ka)
[H+] = Kw/[OH-]
= Kw/√(C*Kw/Ka) = √(Ka*Kw)/Kc
-log [H+] = -1/2log Kw - 1/2log Ka + 1/2log C
pH = 1/2pKw + 1/2pKa + 1/2log C
= 7 + 1/2pKa + 1/2log C.