Electrochemistry

The dissociation of a weak electrolyte ($\mathrm{HA}$) in solution can be represented as: $\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{A}^{-}(\mathrm{aq})$ The degree of dissociation ($\alpha$) for a weak electrolyte is given by: $\alpha = \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$ Where $\Lambda_{\mathrm{m}}$ is the molar conductivity at a given concentration $C$, and $\Lambda_{\mathrm{m}}^{\circ}$ is the molar conductivity at infinite dilution.

The dissociation constant ($K_{\mathrm{a}}$) is described by: $K_{\mathrm{a}} = \dfrac{\alpha^2 C}{1 - \alpha}$ Rewriting this equation in terms of $\Lambda_{\mathrm{m}}$ and $\Lambda_{\mathrm{m}}^{\circ}$: $K_{\mathrm{a}} = \dfrac{\left(\dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C}{1 - \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}}$ By multiplying out and simplifying the above expression: $\left( \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right)^2 C + K_{\mathrm{a}} \left( \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right) = K_{\mathrm{a}}$ Substitute $\alpha = \dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$: $\left(\dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C + K_{\mathrm{a}} \left(\dfrac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right) - K_{\mathrm{a}} = 0$ Rewriting by multiplying through by $\left( \Lambda_{\mathrm{m}}^{\circ} \right)^2$: $\Lambda_{\mathrm{m}}^2 C + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} - K_{\mathrm{a}} \left( \Lambda_{\mathrm{m}}^{\circ} \right)^2 = 0$ Thus, the correct equation that describes the change in molar conductivity with respect to concentration for a weak electrolyte is given by: Option D: $\Lambda_{\mathrm{m}}^2 C - K_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2} + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} = 0$