Question

The molar conductivity of 0.027 M methanoic acid is 40.42 S cm2 mol–1. The value of dissociation constant of this acid is (Given: \(\lambda^0_{H^+} = 349.6~\text{S cm}^2 \text{mol}^{-1},~\lambda^0_{HCOO^-} = 54.6~\text{S cm}^2 \text{mol}^{-1}\))

• A. \(1.5 \times 10^{-5}\)
• B. \(6.0 \times 10^{-5}\)
• C. \(4.5 \times 10^{-4}\)
• D. \(3.0 \times 10^{-4}\)

Hint:

Use \(\Lambda_m = \alpha \Lambda^0_m\) and \(K_a = \alpha^2 C\) for weak electrolyte dissociation constant problems.

Answer 1

The Correct Option is D

Molar conductivity at infinite dilution: \[ \Lambda^0_m = \lambda^0_{H^+} + \lambda^0_{HCOO^-} = 349.6 + 54.6 = 404.2~\text{S cm}^2 \text{mol}^{-1} \] Degree of dissociation (\(\alpha\)) = \(\frac{\Lambda_m}{\Lambda^0_m} = \frac{40.42}{404.2} = 0.1\) \[ K_a = \alpha^2 \cdot C = (0.1)^2 \cdot 0.027 = 0.00027 = 2.7 \times 10^{-4} \approx 3.0 \times 10^{-4} \]