Question

The conductivity of a weak acid HA of concentration 0.001 mol L\(^{-1}\) is \( 2.0 \times 10^{-5} \, \text{S cm}^{-1} \). If \( \Lambda_m^\circ ({HA}) = 190 \, \text{S cm}^2 \text{mol}^{-1} \), the ionization constant (\( K_a \)) of HA is equal to _______ \( \times 10^{-6} \).

• A. 24
• B. 48
• C. 12
• D. 45

Hint:

The ionization constant for a weak acid can be calculated using the formula \( K_a = \frac{C \Lambda_m^\circ}{\Lambda} \), where \( C \) is the concentration, \( \Lambda_m^\circ \) is the molar conductivity at infinite dilution, and \( \Lambda \) is the measured conductivity.

Answer 1

The Correct Option is C

\[ A_m = 1000 \times \frac{\kappa}{M} \] \[ = 1000 \times \frac{2 \times 10^{-5}}{0.001} = 20 \, S \, cm^2 \, mol^{-1} \] \[ \Rightarrow \alpha = \frac{A_m}{A_m^o} = \frac{20}{190} = \left( \frac{2}{19} \right) \] \[ {HA} \rightleftharpoons {H}^+ + {A}^- \] \[ \begin{array}{c} 0.001(1 - \alpha) \quad 0.001\alpha \quad 0.001\alpha \end{array} \] \[ K_a = 0.001 \left( \frac{\alpha^2}{1 - \alpha} \right) = \frac{0.001 \times \left( \frac{2}{19} \right)^2}{1 - \left( \frac{2}{19} \right)} \] \[ = 12.3 \times 10^{-6} \]