Ionic EquilibriumHard
Question
pH for the solution of salt undergoing anionic hydrolysis (say CH3COONa) is given by:
Options
A.pH = 1/2 [pKw + pKa + logC]
B.pH = 1/2 [pKw + pKa - logC]
C.pH = 1/2 [pKw + pKb - logC]
D.None of these
Solution
CH3COO- + H2O ⇋ CH3COOH + OH-
∴ [OH-] = C. h = C
or - log OH = -
[log Kw + logC - logKa] or pOH = 
[pKw - log C - pKa]
Now pH + pOH = pKw ∴ pH =
[pKw + log C + pKa]
∴ [OH-] = C. h = C
or - log OH = -
[log Kw + logC - logKa] or pOH = [pKw - log C - pKa]Now pH + pOH = pKw ∴ pH =
[pKw + log C + pKa]Create a free account to view solution
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