If the depression in freezing point of $0.1$ molal $H A (a q .)$ is $- {0.2046}^{o} C$ , then pH of the solution is: $[K_{f} (H_{2} O) = {1.86}^{o} C {m o l}^{- 1} k g]$ :

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Solution

The correct option is D $2$
Reaction involve-

$H A_{(a q)} ⟶ H_{(a q)}^{+} + A_{(a q)}^{-}$
$C - C α$ $C α$ $C α$

we have,

$Δ T_{f} = i K_{f} m$

$0.2046 = i \times 1.86 \times 0.1$

$i = 1.1$
we also have,
$α = \frac{i - 1}{n - 1}$

$α = \frac{1.1 - 1}{2 - 1}$

$α = 0.1$
we have,
$p H = - l o g (α C) = - l o g (0.1 * 0.1) = - l o g (0.01)$
$p H = 2$

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Similar questions

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If the depression in freezing point of 0.1 molal HA(aq.) is −0.2046oC, then pH of the solution is: [Kf(H2O)=1.86oCmol−1kg] :

The correct option is D 2Reaction involve-HA(aq)⟶H+(aq)+A−(aq)C−Cα Cα Cαwe have,ΔTf=iKfm0.2046=i×1.86×0.1i=1.1we also have,α=i−1n−1α=1.1−12−1α=0.1we have, pH=−log(αC)=−log(0.1∗0.1)=−log(0.01)pH=2

If the depression in freezing point of $0.1$ molal $H A (a q .)$ is $- {0.2046}^{o} C$ , then pH of the solution is: $[K_{f} (H_{2} O) = {1.86}^{o} C {m o l}^{- 1} k g]$ :