Question

Calculate \( \Lambda_m^0 \) for acetic acid and its degree of dissociation (\( \alpha \)) if its molar conductivity is 48.1 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).
Given that
\( \Lambda_m^0 (\text{HC}) = 426 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{NaCl}) = 126 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \),
\( \Lambda_m^0 (\text{CH}_3\text{COONa}) = 91 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).

Hint:

The degree of dissociation \( \alpha \) indicates the fraction of the total molecules that dissociate into ions in a solution.

Answer 1

First, we need to calculate the molar conductivity at infinite dilution, \( \Lambda_m^0 \), for acetic acid, which is the sum of the conductivity contributions from the ions of acetic acid and sodium acetate: \[ \Lambda_m^0 (\text{Acetic acid}) = \Lambda_m^0 (\text{HC}) - \Lambda_m^0 (\text{NaCl}) = 426 - 126 = 300 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}. \] Next, we can calculate the degree of dissociation \( \alpha \) by comparing the observed molar conductivity of acetic acid to the expected conductivity at infinite dilution: \[ \alpha = \frac{\Lambda_m}{\Lambda_m^0}. \] Substitute the given values: \[ \alpha = \frac{48.1}{300} \approx 0.16. \] Thus, the degree of dissociation \( \alpha \) is 0.16, and the molar conductivity at infinite dilution for acetic acid is 300 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).