$\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq})+\mathrm{A}^{-}(\mathrm{aq})$ The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20^{\circ} \mathrm{C}$. The dissociation constant for the acid is Given : $\mathrm{K}_{\mathrm{f}}\left(\mathrm{H}_{2} \mathrm{O}\right)=1.8 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$, molality $\equiv$ molarity

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Solutions

HA (aq) ⇌ H^{+} (aq) + A^{-} (aq)

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is

0.2 0^{\circ} C

.
The dissociation constant for the acid is
Given :

K_{f} (H_{2} O) = 1.8 K kg mol^{- 1}

, molality

\equiv

molarity

Answer: (A)

The freezing point depression fixes the van't Hoff factor through

Δ T_{f} = i K_{f} m

0.20 = i (1.8) (0.1)

, giving

i = \frac{10}{9}

. For

HA ⇌ H^{+} + A^{-}

the factor is

i = 1 + α

, so

α = \frac{1}{9}

. The dissociation constant is

K_{a} = \frac{C α ^{2}}{1 - α} = \frac{0.1 ( 1/9 ) ^{2}}{1 - 1/9} = \frac{0.1}{81} \times \frac{9}{8} = \frac{1}{720}

. The dissociation constant of the acid is

1.38 \times 1 0^{- 3}

JEE Main · 2025 · 08 Apr 2025 · Shift 2 · Single correct option