Chemistry - Electrochemistry Question with Solution | TestHub

By Kohlrausch's law of independent migration of ions:

${\Lambda }_m^{\circ }(H_{2}S{O}_4)=2{\lambda }^{\circ }(H^{+})+{\lambda }^{\circ }(S{O}_4^{2-})=x$

${\Lambda }_m^{\circ }(K_{2}S{O}_4)=2{\lambda }^{\circ }(K^{+})+{\lambda }^{\circ }(S{O}_4^{2-})=y$

${\Lambda }_m^{\circ }(C{H}_{3}COOK)={\lambda }^{\circ }(C{H}_{3}CO{O}^{-})+{\lambda }^{\circ }(K^{+})=z$

The molar conductivity of $C{H}_{3}COOH$ at zero concentration is:

${\Lambda }_m^{\circ }(C{H}_{3}COOH)={\lambda }^{\circ }(C{H}_{3}CO{O}^{-})+{\lambda }^{\circ }(H^{+})$

Subtracting the second equation from the first:

$x-y=2{\lambda }^{\circ }(H^{+})-2{\lambda }^{\circ }(K^{+})$

$\frac{x-y}{2}={\lambda }^{\circ }(H^{+})-{\lambda }^{\circ }(K^{+})$

Adding this to the third equation:

$\frac{x-y}{2}+z=[{\lambda }^{\circ }(H^{+})-{\lambda }^{\circ }(K^{+})]+[{\lambda }^{\circ }(C{H}_{3}CO{O}^{-})+{\lambda }^{\circ }(K^{+})]$

$\frac{x-y}{2}+z={\lambda }^{\circ }(C{H}_{3}CO{O}^{-})+{\lambda }^{\circ }(H^{+})$

Therefore, ${\Lambda }_m^{\circ }(C{H}_{3}COOH)=\frac{x-y}{2}+z$