Question

Specific conductance of 0.1 M \(HA\) is \(3.75 \times 10^{-4} \, \Omega^{-1} \text{cm}^{-1}\). If \(\Lambda_m^\infty (HA) = 250 \, \Omega^{-1} \text{cm}^2 \text{mol}^{-1}\), the dissociation constant \(K_a\) of \(HA\) is

• A. \(1.0\times10^{-5}\)
• B. \(2.25\times10^{-4}\)
• C. \(2.25\times10^{-5}\)
• D. 2.25×10⁻13

Hint:

For weak acids, Kₐ = (Cα²)/(1-α).

Answer 1

The Correct Option is C

\(\Lambda_m = \frac{\kappa \times 1000}{C} = \frac{3.75 \times 10^{-4} \times 1000}{0.1} = 3.75\)

\(\alpha = \frac{3.75}{250} = 0.015\)

\[ K_a = \frac{C \alpha^2}{1 - \alpha} \approx 2.25 \times 10^{-5} \]