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

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, \(\quad \mathrm{pH}=-\log 10^{-3}=3\)