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 \( \alpha \):

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

Answer 2

The degree of dissociation of acetic acid is:

• Option 1: 0.021
• Option 2: 0.21
• Option 3: 0.012
• Option 4: 0.12

Solution Explanation:

Given:

k = 1.65 × 10^-4 S cm^-1

C = 0.02 M

λ_H+^∘ = 349.1 S cm² mol^-1

λ_CH3COO-^∘ = 40.9 S cm² mol^-1

Calculate Λ_m:

Λ_m = k / C = (1.65 × 10^-4) / 0.02 = 0.00825 S cm² mol^-1

Calculate Λ^∘:

Λ^∘ = λ_H+^∘ + λ_CH3COO-^∘ = 349.1 + 40.9 = 390 S cm² mol^-1

Calculate α (degree of dissociation):

α = Λ_m / Λ^∘ = 0.00825 / 390 ≈ 0.021

Correct Answer:

Option 1: 0.021