Question

The conductivity of \(0.001\,M\) solution of acetic acid is \(3.905\times10^{-5}\,S\,cm^{-1}\). Calculate its molar conductivity and degree of dissociation (\(\alpha\)). Given : \[ \lambda^\circ_{H^+}=349.6\,S\,cm^2\,mol^{-1} \] \[ \lambda^\circ_{CH_3COO^-}=40.9\,S\,cm^2\,mol^{-1} \]

Hint:

For weak electrolytes: \[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \] Always calculate \(\Lambda_m^\circ\) using Kohlrausch's law before finding the degree of dissociation.

Answer 1

Step 1: Calculate molar conductivity ($\Lambda_m$).
Using $\Lambda_m = \dfrac{\kappa \times 1000}{C}$: \[ \Lambda_m = \frac{3.905 \times 10^{-5} \times 1000}{0.001} = 39.05\,S\,cm^2\,mol^{-1} \]
Step 2: Find limiting molar conductivity ($\Lambda_m^\circ$) by Kohlrausch's law.
\[ \Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} = 349.6 + 40.9 = 390.5\,S\,cm^2\,mol^{-1} \]
Step 3: Calculate degree of dissociation ($\alpha$).
For a weak electrolyte: $\alpha = \dfrac{\Lambda_m}{\Lambda_m^\circ} = \dfrac{39.05}{390.5} = 0.10$
\[ \boxed{\Lambda_m = 39.05\,S\,cm^2\,mol^{-1},\quad \alpha = 0.10\;(10\%)} \]