Question

Conductivity of a 0.00241 M acetic acid solution is \(7.896 \times 10^{-5}\) S cm\(^{-1}\). If \(\Lambda^{\circ}\) for acetic acid is \(390.5\) S cm\(^2\) mol\(^{-1}\), calculate its degree of dissociation (\(\alpha\)).

Hint:

For weak electrolytes: \[ \alpha = \frac{\Lambda_m}{\Lambda^{\circ}} \] Molar conductivity increases with dilution due to increased dissociation.

Answer 1

Step 1: Molar conductivity (\(\Lambda_m\)) is given by: \[ \Lambda_m = \frac{\kappa \times 1000}{C} \] where \(\kappa = 7.896 \times 10^{-5}\) S cm\(^{-1}\) \(C = 0.00241\) M
Step 2: Substitute the values: \[ \Lambda_m = \frac{7.896 \times 10^{-5} \times 1000}{0.00241} \] \[ \Lambda_m \approx 32.76 \ \text{S cm}^2 \text{ mol}^{-1} \]
Step 3: Degree of dissociation (\(\alpha\)) is given by: \[ \alpha = \frac{\Lambda_m}{\Lambda^{\circ}} \]
Step 4: Substitute the values: \[ \alpha = \frac{32.76}{390.5} \] \[ \alpha \approx 0.084 \]
Step 5: Therefore, the degree of dissociation is approximately: \[ \alpha = 8.4% \]