Ionic EquilibriumMedium
Question
A 0.1 M acetic acid solution is titrated against 0.1 M-NaOH solution. What would be the difference in pH between 1/4 and 3/4 stages of neutralization of the acid?
Options
A.2 log(0.75)
B.2 log(0.25)
C.log 3
D.2 log 3
Solution
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 1/4th neutralization,
$P_{1}^{H} = P^{K_{a}} + \log\frac{a/4}{a - a/4} = P^{K_{a}} + \log\frac{1}{3}$
For 3/4th neutralization,
$P_{2}^{H} = P^{K_{a}} + \log\frac{3a/4}{a - 3a/4} = P^{K_{a}} + \log 3 $$$\therefore\Delta P^{H} = P_{2}^{H} - P_{1}^{H} = 2\log 3$$
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