Electrochemistry

Concept:Use the Debye-Hückel-Onsager equation to determine limiting molar conductivities from given data, then apply Kohlrausch’s law to find $\Lambda_m^0$ for AgCl, and finally relate conductivity to solubility in a dilute saturated solution.Explanation:For a strong electrolyte, $\Lambda_m = \Lambda_m^0 - b\sqrt{c}$. For two concentrations, $b = \frac{\Lambda_{m,1} - \Lambda_{m,2}}{\sqrt{c_2} - \sqrt{c_1}}$. Given: $\sqrt{0.01}=0.1$, $\sqrt{0.04}=0.2$, so $\sqrt{c_2}-\sqrt{c_1}=0.1$. For NaNO$_3$: $b = \frac{111-101}{0.1}=100$, $\Lambda_m^0 = 111 + 100\times0.1 = 121\ \mathrm{S\,cm^2\,mol^{-1}}$. For NaCl: $b = \frac{117-107}{0.1}=100$, $\Lambda_m^0 = 117 + 100\times0.1 = 127\ \mathrm{S\,cm^2\,mol^{-1}}$. For AgNO$_3$: $b = \frac{125-116}{0.1}=90$, $\Lambda_m^0 = 125 + 90\times0.1 = 134\ \mathrm{S\,cm^2\,mol^{-1}}$. By Kohlrausch’s law: $\Lambda_m^0(\mathrm{AgCl}) = \Lambda_m^0(\mathrm{AgNO_3}) + \Lambda_m^0(\mathrm{NaCl}) - \Lambda_m^0(\mathrm{NaNO_3})$ $= 134 + 127 - 121 = 140\ \mathrm{S\,cm^2\,mol^{-1}}$. For a very dilute saturated solution, $\Lambda_m = \Lambda_m^0 = \frac{\kappa \times 1000}{c}$. Given $\kappa = 1.40\times10^{-6}\ \mathrm{S\,cm^{-1}}$, $c = \frac{1.40\times10^{-6}\times 1000}{140} = 10^{-5}\ \mathrm{mol\,L^{-1}}$. Thus solubility $X = 10^{-5}\ \mathrm{mol\,L^{-1}}$. $\log_{10}(X^{-1}) = \log_{10}\big((10^{-5})^{-1}\big) = \log_{10}(10^{5}) = 5$.Answer:5