Question

What is degree of dissociation of $CH_3COOH$ if, $\Lambda^\circ (CH_3COO^-) = 50\text{ S cm}^2\text{ mol}^{-1}$, $\Lambda^\circ (H^+) = 350\text{ S cm}^2\text{ mol}^{-1}$ and molar conductivity of $5 \times 10^{-2}\text{ M }CH_3COOH$ is $20\text{ S cm}^2\text{ mol}^{-1}$ ?

• A. $1.25 \times 10^{-4}$
• B. $5 \times 10^{-2}$
• C. $1.25 \times 10^{-2}$
• D. $5 \times 10^{-4}$

Hint:

$\alpha$ is the ratio of conductivity at a given concentration to that at infinite dilution.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the degree of dissociation ($\alpha$) of acetic acid. This requires calculating its limiting molar conductivity ($\Lambda_m^\circ$) first.
Step 2: Key Formula or Approach:
1. Kohlrausch's Law: $\Lambda_m^\circ = \lambda^\circ (cation) + \lambda^\circ (anion)$
2. Degree of dissociation: $\alpha = \frac{\Lambda_m^c}{\Lambda_m^\circ}$
Step 3: Detailed Explanation:
Given:
$\lambda^\circ (CH_3COO^-) = 50\text{ S cm}^2\text{ mol}^{-1}$
$\lambda^\circ (H^+) = 350\text{ S cm}^2\text{ mol}^{-1}$
$\Lambda_m^c = 20\text{ S cm}^2\text{ mol}^{-1}$
Calculating limiting molar conductivity ($\Lambda_m^\circ$):
\[ \Lambda_m^\circ (CH_3COOH) = \lambda^\circ (H^+) + \lambda^\circ (CH_3COO^-) \]
\[ \Lambda_m^\circ = 350 + 50 = 400\text{ S cm}^2\text{ mol}^{-1} \]
Calculating degree of dissociation ($\alpha$):
\[ \alpha = \frac{20}{400} = \frac{2}{40} = \frac{1}{20} \]
\[ \alpha = 0.05 = 5 \times 10^{-2} \]
Step 4: Final Answer:
The degree of dissociation is $5 \times 10^{-2}$.