The conductivity of $0.001\, mol\, L^{-1}$ solution of a weak monobasic acid is $4.0 × 10^{-5}\,S\, cm^{-1}$. Calculate its dissociation constant if $Λ_m^o$, for weak acid is $400.0\, S\, cm^2\, mol^{-1}$.

$1.1 × 10^{-5}\, mol\, L^{-1}$

The correct answer is Option (4) → $1.1 × 10^{-5}\, mol\, L^{-1}$

Given:

• Conductivity, $\kappa = 4.0 \times 10^{-5}\ \text{S cm}^{-1}$
• Concentration, $c = 0.001\ \text{mol L}^{-1}$
• Limiting molar conductivity, $\Lambda_m^0 = 400.0\ \text{S cm}^2\text{ mol}^{-1}$

Step 1: Calculate molar conductivity

$\Lambda_m = \frac{\kappa \times 1000}{c} = \frac{4.0 \times 10^{-5} \times 1000}{0.001} = 40\ \text{S cm}^2\text{ mol}^{-1}$

Step 2: Degree of dissociation

$\alpha = \frac{\Lambda_m}{\Lambda_m^0} = \frac{40}{400} = 0.1$

Step 3: Dissociation constant

For a weak monobasic acid:

$K_a = \frac{c\alpha^2}{1-\alpha} = \frac{0.001 \times (0.1)^2}{0.9} = \frac{1.0 \times 10^{-5}}{0.9} \approx 1.1 \times 10^{-5}$