Question

The molar conductivity of 0.007 M acetic acid is 20 S cm² mol⁻¹. What is the dissociation constant of acetic acid? Choose the correct option. 

• A. 2.50×10⁻⁵ mol L⁻¹
• B. 1.75×10⁻⁴ mol L⁻¹
• C. 2.50×10⁻⁴ mol L⁻¹
• D. 1.75×10⁻⁵ mol L⁻¹

Answer 1

The Correct Option is D

To find the dissociation constant (\(K_a\)) of acetic acid using its molar conductivity, we can follow these steps:

1. Determine the molar conductivity at infinite dilution (\(\Lambda_m^{\infty}\)). For acetic acid (CH₃COOH), we approximate this by adding the molar conductivities of the ions of acetic acid: \(\Lambda_m^{\infty} = \lambda_{\text{H}^+}^{\infty} + \lambda_{\text{CH}_3\text{COO}^-}^{\infty}\).
2. Calculate the degree of dissociation (\(\alpha\)) using the formula: \(\alpha = \frac{\Lambda_m}{\Lambda_m^{\infty}}\) where \(\Lambda_m\) is the given molar conductivity (20 S cm² mol⁻¹).
3. Use the relationship between the dissociation constant and the degree of dissociation: \(K_a = C \alpha^2\) where \(C\) is the concentration of acetic acid (0.007 M).
4. Assuming typical values for \(\Lambda_m^{\infty}\) for acetic acid (e.g., \(391 \, \text{S cm}^2 \text{mol}^{-1}\)), calculate \(\alpha\):

\[ \alpha = \frac{20}{391} = 0.0512 \] 

1. Now substitute the values into the equilibrium expression for \(K_a\):

\[ K_a = 0.007 \times (0.0512)^2 = 1.75 \times 10^{-5} \, \text{mol L}^{-1} \]

Thus, the dissociation constant of acetic acid is 1.75 × 10⁻⁵ mol L⁻¹, which matches option d.

Option	Value	Explanation
(a)	2.50×10−5 mol L−1	Does not match calculation
(b)	1.75×10−4 mol L−1	Dissociation constant too large
(c)	2.50×10−4 mol L−1	Dissociation constant too large
(d)	1.75×10−5 mol L−1	Matches the correct calculation