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Grade: 12


prove that pH=1/2[pKw+pKa+log C] for weak acid strong base salt solution

3 years ago

Answers : (1)

Vikas TU
12284 Points
Derivation for Strong base and Weak acid reaction.

A- + H2O;   \rightleftharpoons   HA + OH-

     Here HA is weak acid considered.

As 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  \rightleftharpoons   CH3COOH + OH-

C(1-x)                         Cx           Cx

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

From 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 \rightleftharpoons CH3COO- + H+,       

Ka = [CH3COO-][H+]/[CH3COOH]      ...... (ii)

H2\rightleftharpoons  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=1/2[pKw+pKa+log C].


3 years ago
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