If $0.1 M H_{2} S O_{4} (a q)$ solution shows freezing point $- {0.3906}^{o} C$ then what is the $K_{2}$ for $H_{2} S O_{4}$ ?
(Assume $m = M$ and $K_{f (H_{2} O)} = 1.86 K k g m o l^{- 1})$

0.122

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0.0122

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1.11 \times 10^{- 3}

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None of these

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Solution

The correct option is B $0.0122$
(b) $Δ T_{f} = K_{f} . m . i \Rightarrow 0.3906 = i \times 1.86 \times 0.1,$
$i = 2.1$
first step of $H_{2} S O_{4}$ is strong so
$H_{2} S O_{4} \to H^{+} + H S O_{4}^{-}$ ;
0 C c(I - a)
$H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}$
c(I - a) Ca ca
$i = \frac{C (1 - a) + C a + C a + C}{C} \Rightarrow 2.1$ ;
$a = 0.1$
Now, $H S O_{4}^{-} ⇌ H^{+} + S O_{4}^{2 -}$
C(1 - a) (Ca + C) Ca
$K_{a 2} = \frac{C (1 + a) \times C a}{C (1 - a)}$
$\Rightarrow \frac{1.1 \times 0.1 \times 0.1}{0.9} = 0.0122$

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