Question

An aqueous solution of 0.1 M HA shows depression in freezing point of 0.2°C. If K_f (H₂O) = 1.86 K kg mol^–1 and assuming molarity = molality, find the dissociation constant of HA.

(1) 4.50 × 10^–5

(2) 6.25 × 10^–3

(3) 5.625 × 10^–4

(4) 2.65 × 10^–4

Answer 1

Correct option is : (3) 5.625 × 10^–4

^{\(\underset{0.1(1-\alpha)}{HA} \rightleftharpoons \underset{0.1 \alpha}{H^+} + \underset{0.1 \alpha}{A^-}\)}

^{\(i = 1+\alpha\)}

^{\(\Delta T_f = iK_fm\)}

0.2 = i × 1.86 × 0.1

\(i = \frac{0.2}{0.186} = 1.075\)

\(\alpha = 0.075\)

\(K_a = \frac{0.1(\alpha)^2}{1-\alpha} \simeq 0.1(0.075)^2\)

\(= 5.625 \times 10^{-4}\)