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The resistance of $0.1 \mathrm{M}$ weak acid $\mathrm{H} A$ in a conductivity cell is $2 \times 10^3 \mathrm{Ohm}$. The cell constant of the cell is $0.78 \mathrm{C} \mathrm{m}^{-1}$ and $\lambda_{\mathrm{m}}^{\circ}$ of acid $\mathrm{H} A$ is $390 \mathrm{~S} \mathrm{~cm}^2 \mathrm{~mol}^{-1}$. The $\mathrm{pH}$ of the solution is
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$3$
Given, $C=0.1 \mathrm{M}$
$\begin{aligned} \Lambda_{\mathrm{m}}^{\circ} & =390 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ R & =2 \times 10^3 \mathrm{ohm} \\ G^* & =0.78 \mathrm{~cm}^{-1}\end{aligned}$
Now, $K=\frac{R}{G^{\star}}=\frac{0.78}{2 \times 10^3}=3.9 \times 10^{-4}$
$\begin{aligned} \Lambda_{\mathrm{m}}^{\circ} & =\frac{K \times 1000}{C}=\frac{3.9 \times 10^{-4} \times 1000}{0.1}=3.9 \\ \alpha & =\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{3.9}{390}=10^{-2} \\ {\left[\mathrm{H}^{+}\right] } & =C \cdot \alpha=0.1 \times 10^{-2}=10^{-3}\end{aligned}$
So, $\quad \mathrm{pH}=-\log 10^{-3}=3$
$\begin{aligned} \Lambda_{\mathrm{m}}^{\circ} & =390 \mathrm{Scm}^2 \mathrm{~mol}^{-1} \\ R & =2 \times 10^3 \mathrm{ohm} \\ G^* & =0.78 \mathrm{~cm}^{-1}\end{aligned}$
Now, $K=\frac{R}{G^{\star}}=\frac{0.78}{2 \times 10^3}=3.9 \times 10^{-4}$
$\begin{aligned} \Lambda_{\mathrm{m}}^{\circ} & =\frac{K \times 1000}{C}=\frac{3.9 \times 10^{-4} \times 1000}{0.1}=3.9 \\ \alpha & =\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}=\frac{3.9}{390}=10^{-2} \\ {\left[\mathrm{H}^{+}\right] } & =C \cdot \alpha=0.1 \times 10^{-2}=10^{-3}\end{aligned}$
So, $\quad \mathrm{pH}=-\log 10^{-3}=3$
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