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The measured freezing point depression for a $0.1 \mathrm{m}$ aqueous $\mathrm{CH}_{3} \mathrm{COOH}$ solution is $0.19^{\circ} \mathrm{C}$. The acid dissociation constant $K_{a}$ at this concentration will be (Given, $K_{f}$ the molal cryoscopic constant $=1.86 \mathrm{K}$ $\mathrm{kg} \mathrm{mol}^{-1}$ )
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Verified Answer
The correct answer is:
$4 \times 10^{-5}$
$\begin{aligned} \Delta T_{f} &=i \times K_{t} \times m \\ \therefore & i=\frac{\Delta T_{f}}{K_{t} \times m}=\frac{0.19}{1.86 \times 0.1}=1.02 \end{aligned}$
Again from, $\alpha=\frac{i-1}{n-1}=\frac{1.02-1}{2-1}$ $=0.02=2.0 \times 10^{-2}$
$$
\begin{aligned}
\text { for } \mathrm{CH}_{3} \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+} \\
K \mathrm{a} &=\mathrm{C} \alpha^{2} \\
&=0.1 \times\left(2 \times 10^{-2}\right)^{2} \\
&=4 \times 10^{-5}
\end{aligned}
$$
Again from, $\alpha=\frac{i-1}{n-1}=\frac{1.02-1}{2-1}$ $=0.02=2.0 \times 10^{-2}$
$$
\begin{aligned}
\text { for } \mathrm{CH}_{3} \mathrm{COOH} & \rightleftharpoons \mathrm{CH}_{3} \mathrm{COO}^{-}+\mathrm{H}^{+} \\
K \mathrm{a} &=\mathrm{C} \alpha^{2} \\
&=0.1 \times\left(2 \times 10^{-2}\right)^{2} \\
&=4 \times 10^{-5}
\end{aligned}
$$
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