Question

The degree of dissociation of acetic acid is:

• A. 0.021
• B. 0.21
• C. 0.012
• D. 0.12

Hint:

The degree of dissociation ($\alpha$) is the ratio of the molar conductivity of the solution ($\Lambda_m$) to the limiting molar conductivity ($\Lambda^\circ$).

Answer 1

The Correct Option is A

Given:

• $k = 1.65 \times 10^{-4} \text{ S cm}^{-1}$,
• $C = 0.02 \text{ M}$,
• $\lambda^\circ_{\text{H}^+} = 349.1 \text{ S cm}^2 \text{mol}^{-1}$,
• $\lambda^\circ_{\text{CH}_3\text{COO}^-} = 40.9 \text{ S cm}^2 \text{mol}^{-1}$.

1. Calculate $\Lambda_m$:

   $\Lambda_m = \frac{k}{C} = \frac{1.65 \times 10^{-4}}{0.02} = 0.00825 \text{ S cm}^2 \text{mol}^{-1}$

2. Calculate $\Lambda^\circ$:

   $\Lambda^\circ = \lambda^\circ_{\text{H}^+} + \lambda^\circ_{\text{CH}_3\text{COO}^-} = 349.1 + 40.9 = 390 \text{ S cm}^2 \text{mol}^{-1}$

3. Calculate $\alpha$:

   $\alpha = \frac{\Lambda_m}{\Lambda^\circ} = \frac{0.00825}{390} \approx 0.021$