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Depression in Freezing Point

The freezing ...

Question

The freezing point of a $0.08$ molal solution of $N a H S O$ is $- {0.372}^{0}$ C. Calculate the dissociation constant for the reaction.
$H S O_{4} ⇌ H^{\oplus} + S O_{4}^{2 -}$
$K_{f}$ for water = $1.86 K m^{- 1}$ .

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Solution

$N a H S O_{4} \to N a^{+} + {H S O_{4}}^{-}$
$0.08$ $0.08$
$\Rightarrow △ T_{f} = k_{f m}$

$\Rightarrow m = \frac{△ T_{f}}{k_{f}} = \frac{0.372}{1.86} = 0.2$

${H S O_{4}}^{-} \to H^{+} + {S O_{4}}^{2 -}$
At $+ = 0$ $0.08$ $0$ $0$
At eqb $(0.08 - x)$ $x$ $x$

$\Rightarrow (0.08 - x) + 2 x = 0.2$
$\Rightarrow x = 0.04$

$\Rightarrow k = \frac{[H^{+}] [{S O_{4}}^{2 -}]}{[{H S O_{4}}^{-}]} = \frac{(0.04) (0.04)}{(0.04)} = 0.04$

Hence, the answer is $0.04.$

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