Question

The specific conductance of 0.0025 M acetic acid is \(5 × 10^{–5} S\ cm^{–1}\) at a certain temperature. The dissociation constant of acetic acid is ___________ \(× 10^{–7}.\) (Nearest integer) Consider limiting molar conductivity of \(CH_3COOH\) as \(400\  S cm^2 mol^{–1}\).

Answer 1

Correct Answer: 66

We are given: \[ k = 5 \times 10^{-5} \, \text{S cm}^{-1}, \quad C = 0.0025 \, \text{M}, \quad \Lambda_{\text{m}} \, \text{(limiting molar conductivity)} = 400 \, \text{S cm}^2 \text{mol}^{-1} \] Step 1: Calculate molar conductivity \[ \Lambda_{\text{m}} = \frac{k \times 1000}{C} \] Substitute the given values: \[ \Lambda_{\text{m}} = \frac{5 \times 10^{-5} \times 1000}{0.0025} = \frac{5 \times 10^{-2}}{2.5 \times 10^{-3}} = 20 \, \text{S cm}^2 \text{mol}^{-1} \] Step 2: Degree of dissociation \[ \alpha = \frac{20}{400} = \frac{1}{20} \] Step 3: Calculate dissociation constant \( K_a \) \[ K_a = \frac{C \alpha^2}{1 - \alpha} \] Substitute the values: \[ K_a = \frac{0.0025 \times \left(\frac{1}{20}\right)^2}{1 - \frac{1}{20}} = \frac{0.0025 \times \frac{1}{400}}{\frac{19}{20}} = \frac{0.0025 \times 10^{-6}}{19/20} = 66 \times 10^{-7} \] Thus, the dissociation constant \( K_a \) is \( 66 \times 10^{-7} \).