If the freezing point of 0-1M HAaq solution is -0-2046- C- then the pH of the solution is -Assume molarity - molality Kf of H2O-1-86 K kg mol-1

Δ T_f=F.pt of water F.pt of solution=0 ( 0.2046)=0.2046^∘ C Freezing point depression of a solution is given by, Δ T_f =i× K_f × m where, Δ T_f is the depress ...

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Standard XII

Chemistry

Depression in Freezing Point

If the freezi...

Question

If the freezing point of $0.1 M H A (a q)$ solution is $- {0.2046}^{\circ} C$ , then the $p H$ of the solution is :
(Assume molarity = molality)
$K_{f} o f H_{2} O = 1.86 K k g m o l^{- 1}$

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Solution

The correct option is D 2
$Δ T_{f} = F.pt of water - F.pt of solution = 0 - (- 0.2046) = {0.2046}^{\circ} C$
Freezing point depression of a solution is given by,
$Δ T_{f} = i \times K_{f} \times m$
where,
$Δ T_{f}$ is the depression in freezing point.
m is molality of the solution.
i is the van't hoff factor
$0.2046 = i \times 1.86 \times 0.1 i = 1.1$
For $H A$ :
$i = 1 + (n - 1) α 1.1 = 1 + (2 - 1) α α = 0.1$
At equilibrium:

$H A (a q .) ⇌ H^{+} (a q .) + A^{-} (a q .) C (1 - α) C α C α$
$[H^{+}]_{e q b .} = C α$
$[H^{+}]_{e q b .} = 0.1 \times 0.1 = 10^{- 2} p H = - l o g_{10} [H^{+}]$

$p H = - l o g_{10} [10^{- 2}] = 2$

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