Chemistry - Electrochemistry Question with Solution | TestHub

Using Kohlrausch’s law,

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})={\lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-})+{\lambda }^{\circ }({\mathrm{H}}^{+})$$

We are given:

$${\Lambda }^{\circ }(\mathrm{HCl})={\lambda }^{\circ }({\mathrm{H}}^{+})+{\lambda }^{\circ }(\mathrm{C}{\mathrm{l}}^{-})=425.9\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$$

$${\Lambda }^{\circ }(\mathrm{NaCl})={\lambda }^{\circ }(\mathrm{N}{\mathrm{a}}^{+})+{\lambda }^{\circ }(\mathrm{C}{\mathrm{l}}^{-})=126.4\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$$

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COONa})={\lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-})+{\lambda }^{\circ }(\mathrm{N}{\mathrm{a}}^{+})=91.0\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$$

To find ${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})$, we can use the following relationship:

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})={\Lambda }^{\circ }(\mathrm{HCl})+{\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COONa})-{\Lambda }^{\circ }(\mathrm{NaCl})$$

Substituting the given values:

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})=425.9+91.0-126.4$$

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})=516.9-126.4$$

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})=390.5\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$$

Here, ${\Lambda }^{\circ }(\mathrm{HCl})$ and ${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COONa})$ are added, and ${\Lambda }^{\circ }(\mathrm{NaCl})$ is subtracted to eliminate $\mathrm{N}{\mathrm{a}}^{+}$ and $\mathrm{C}{\mathrm{l}}^{-}$ ions, leaving $\mathrm{C}{\mathrm{H}}_3\mathrm{CO}{\mathrm{O}}^{-}$ and ${\mathrm{H}}^{+}$ contributions.

Thus, the limiting molar conductivity of acetic acid is obtained directly from ionic conductivities.

Therefore,

$${\Lambda }^{\circ }(\mathrm{C}{\mathrm{H}}_3\mathrm{COOH})=390.5\mathrm{S}\mathrm{c}{\mathrm{m}}^2\mathrm{mo}{\mathrm{l}}^{-1}$$

Answer: (A) 390.7