Question

CH₃COOH + OH^– $\rightleftharpoons$ CH₃COO^– + H₂O

a mole x mole 0

Final (a – x) mole 0 x mole

$P^{H} = P^{K_{a}} + \log\frac{x}{a - x}$

For 1/4^th neutralization,

$P_{1}^{H} = P^{K_{a}} + \log\frac{a/4}{a - a/4} = P^{K_{a}} + \log\frac{1}{3}$

For 3/4^th neutralization,

$P_{2}^{H} = P^{K_{a}} + \log\frac{3a/4}{a - 3a/4} = P^{K_{a}} + \log 3 $$$\therefore\Delta P^{H} = P_{2}^{H} - P_{1}^{H} = 2\log 3$$