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

Concept: Molar conductivity is defined as the conductance of the volume of solution containing one mole of electrolyte placed between two electrodes one centimetre apart. It is related to conductivity by: \[ \Lambda_m=\frac{\kappa \times 1000}{C} \] For weak electrolytes, \[ \alpha=\frac{\Lambda_m}{\Lambda_m^\circ} \] where \[ \Lambda_m^\circ = \lambda^\circ_{H^+} + \lambda^\circ_{CH_3COO^-} \]

Step 1: Calculate molar conductivity Given, \[ \kappa=3.905\times10^{-5}\,S\,cm^{-1} \] \[ C=0.001\,M \] Using \[ \Lambda_m=\frac{\kappa \times1000}{C} \] \[ \Lambda_m= \frac{3.905\times10^{-5}\times1000}{0.001} \] \[ \Lambda_m=39.05\,S\,cm^2\,mol^{-1} \]

Step 2: Calculate limiting molar conductivity \[ \Lambda_m^\circ = 349.6+40.9 \] \[ \Lambda_m^\circ = 390.5\,S\,cm^2\,mol^{-1} \]

Step 3: Calculate degree of dissociation \[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} \] \[ = \frac{39.05}{390.5} \] \[ =0.10 \]

Final Answer \[ \boxed{ \Lambda_m = 39.05\,S\,cm^2\,mol^{-1} } \] \[ \boxed{ \alpha=0.10 } \] or \[ \boxed{ 10\% } \]