Question #44082

An extremely dilute aqueous solution of \ce{HCl} is prepared at $298\text{ K}$ with a concentration of $1.0 \times 10^{-7}\text{ M}$. The limiting molar conductivities at $298\text{ K}$ are $\lambda^\circ_{\ce{H^+}} = 350\text{ S cm}^2\text{ mol}^{-1}$, $\lambda^\circ_{\ce{OH^-}} = 200\text{ S cm}^2\text{ mol}^{-1}$, and $\lambda^\circ_{\ce{Cl^-}} = 75\text{ S cm}^2\text{ mol}^{-1}$. The ionic product of water is $K_w = 1.0 \times 10^{-14}$. Calculate the true specific conductance ($\kappa_{sol}$) of this solution, and the apparent molar conductivity of \ce{HCl} ($\Lambda_{m, app}$) if it is calculated using the standard formula $\Lambda_{m, app} = \frac{1000 \times (\kappa_{sol} - \kappa_{\ce{H_2O}})}{C_{\ce{HCl}}}$, where $C_{\ce{HCl}}$ is the molarity of \ce{HCl} and $\kappa_{\ce{H_2O}}$ is the specific conductance of pure water. (Use $\sqrt{5} \approx 2.236$)