Following limiting molar conductivities are given as ${\Lambda }_m^{\circ }(\mathrm{H}{}_2\mathrm{SO}{}_4)=x{\text{S cm}}^2{\text{mol}}^{-1}$ ${\Lambda }_m^{\circ }(\mathrm{K}{}_2\mathrm{SO}{}_4)=y{\text{S cm}}^2{\text{mol}}^{-1}$ ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOK})=z{\text{S cm}}^2{\text{mol}}^{-1}$ ${\Lambda }_m^{\circ }$ (in ${\text{S cm}}^2{\text{mol}}^{-1}$) for $\mathrm{CH}{}_3\mathrm{COOH}$ will be :

Solution & Explanation

\textbf{Answer:} (D) $\frac{x-y}{2}+z$ \textbf{Solution:} By Kohlrausch's law, ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOH})={\lambda }^{\circ }(\mathrm{CH}{}_3\mathrm{COO}{}^{-})+{\lambda }^{\circ }(\mathrm{H}{}^{+})$. Express each ionic contribution from the given electrolytes (per mole of each ion): ${\lambda }^{\circ }(\mathrm{H}{}^{+})=\frac{1}{2}{\Lambda }_m^{\circ }(\mathrm{H}{}_2\mathrm{SO}{}_4)-\frac{1}{2}{\lambda }^{\circ }(\mathrm{SO}{}_4{}^{2-})=\frac{1}{2}x-\frac{1}{2}{\lambda }^{\circ }(\mathrm{SO}{}_4{}^{2-})$ ${\lambda }^{\circ }(\mathrm{SO}{}_4{}^{2-})={\Lambda }_m^{\circ }(\mathrm{K}{}_2\mathrm{SO}{}_4)-2{\lambda }^{\circ }(\mathrm{K}{}^{+})=y-2{\lambda }^{\circ }(\mathrm{K}{}^{+})$ ${\lambda }^{\circ }(\mathrm{CH}{}_3\mathrm{COO}{}^{-})={\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOK})-{\lambda }^{\circ }(\mathrm{K}{}^{+})=z-{\lambda }^{\circ }(\mathrm{K}{}^{+})$ Combining so that the $\mathrm{K}{}^{+}$ terms cancel: ${\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOH})={\Lambda }_m^{\circ }(\mathrm{CH}{}_3\mathrm{COOK})+\frac{1}{2}{\Lambda }_m^{\circ }(\mathrm{H}{}_2\mathrm{SO}{}_4)-\frac{1}{2}{\Lambda }_m^{\circ }(\mathrm{K}{}_2\mathrm{SO}{}_4)=z+\frac{x}{2}-\frac{y}{2}=\frac{x-y}{2}+z$