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

The molar conductivity of acetic acid at infinite dilution, \( \Lambda_m^0 \), is determined by summing the conductivity contributions of its constituent ions and those of sodium acetate. The calculation is as follows:\[\Lambda_m^0 (\text{Acetic acid}) = \Lambda_m^0 (\text{HC}) - \Lambda_m^0 (\text{NaCl}) = 426 - 126 = 300 \, \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1}.\]The degree of dissociation, \( \alpha \), is then computed by relating the observed molar conductivity of acetic acid to its theoretical conductivity at infinite dilution:\[\alpha = \frac{\Lambda_m}{\Lambda_m^0}.\]Substituting the provided values yields:\[\alpha = \frac{48.1}{300} \approx 0.16.\]Consequently, the degree of dissociation \( \alpha \) for acetic acid is 0.16, with a molar conductivity at infinite dilution of 300 \( \Omega^{-1} \, \text{cm}^2 \, \text{mol}^{-1} \).