Question

The conductivity of 0.0025 mol L\(^{-1}\) acetic acid is \(5.25 \times 10^{-5} \, \text{S cm}^{-1}\). Calculate its degree of dissociation if \(\Lambda_m^\circ\) for acetic acid is 390 S cm\(^2\) mol\(^{-1}\).

Hint:

- The degree of dissociation is the ratio of the observed molar conductivity to the theoretical molar conductivity at complete dissociation. - Molar conductivity depends on both concentration and the dissociation degree of the compound.

Answer 1

The molar conductivity (\(\Lambda_m\)) is calculated as follows: \[ \Lambda_m = \frac{k}{M} \times 1000 \] Where: - \(k = 5.25 \times 10^{-5} \, \text{S cm}^{-1}\) (given conductivity), - \(M = 0.0025 \, \text{mol L}^{-1}\) (molarity). Thus, the molar conductivity is: \[ \Lambda_m = \frac{5.25 \times 10^{-5}}{0.0025} \times 1000 = 21 \, \text{S cm}^2 \, \text{mol}^{-1}. \] The degree of dissociation (\(\alpha\)) is calculated by the formula: \[ \alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{21}{390} = 0.053. \] Thus, the degree of dissociation is 0.053.