The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is The dissociation constant for the acid is [2025]

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Solution

Q.
$HA(aq) ⇌ H^{+} (a q) + A^{-} (a q)$

The freezing point depression of a 0.1 m aqueous solution of a monobasic weak acid HA is $0.20 ° C .$

The dissociation constant for the acid is

Given: $K_{f} (H_{2} O) = 1.8 {K kg mol}^{- 1}, molality \equiv molarity$ [2025]

1 $1.38 \times 10^{- 3}$

2 $1.1 \times 10^{- 2}$

3 $1.90 \times 10^{- 3}$

4 $1.89 \times 10^{- 1}$

Ans.
(1)

$Δ T_{f} = k_{f} m$

$0.2 = i \times 1.8 \times 0.1$

$i = \frac{0.2}{1.8 \times 0.1} = \frac{10}{9}$

Van't Hoff factor $(i)$ is related to degree of dissociation $(α)$ as:

$i = 1 + (n - 1) α$

Where $n$ is number of moles in which 1 mol of HA completely dissociates.

$\frac{10}{9} = 1 + (2 - 1) α$

$α = \frac{1}{9}$

$K_{a} = \frac{C α^{2}}{1 - α} = \frac{0.1 {(\frac{1}{9})}^{2}}{1 - \frac{1}{9}} = 1.38 \times 10^{- 3}$

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