JEE Main 2025 Apr 8 Shift 2, Chemistry Q9: Colligative Properties

JEE Main2025Apr 8, Shift 2Chemistry

$H A (a q) ⇌ H^{+} (a q) + A^{-} (a q)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is 0.20 $^{\circ}$ C. The dissociation constant for the acid is

Given: $K_{f} (H_{2} O) = 1.8$ K kg mol $^{- 1}$ , molality $\equiv$ molarity

A
$1.38 \times 1 0^{- 3}$
B
$1.1 \times 1 0^{- 2}$
C
$1.90 \times 1 0^{- 3}$
D
$1.89 \times 1 0^{- 1}$

Solution

Using $Δ T_{f} = i K_{f} m$ : $0.20 = i \times 1.8 \times 0.1$ , which gives $i = \frac{0.20}{0.18} = \frac{10}{9}$ .

For a weak monobasic acid dissociating as $H A ⇌ H^{+} + A^{-}$ , the van't Hoff factor is $i = 1 + α$ , so the degree of dissociation is $α = i - 1 = \frac{10}{9} - 1 = \frac{1}{9}$ .

The dissociation constant is $K_{a} = \frac{C α ^{2}}{1 - α}$ , where $C = 0.1$ m.

Substituting: $K_{a} = \frac{0.1 \times ( 1/9 ) ^{2}}{1 - 1/9} = \frac{0.1/81}{8/9} = \frac{0.1 \times 9}{81 \times 8} = \frac{1}{720}$ .

$K_{a} \approx 1.38 \times 1 0^{- 3}$ , matching option (1).

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JEE Main 2025 Apr 8 Shift 2, Chemistry Q9: Colligative Properties

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