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, we start by using the relationship between molar conductivity at a given concentration (\(Λ_m\)) and the limiting molar conductivity (\(Λ_m^0\)). Acetic acid is a weak electrolyte, so the expression to calculate the molar conductivity at infinite dilution (\(Λ_m^0\)) is given by:

• For acetic acid, \(Λ_m^0 = Λ_{m}^{H^+} + Λ_{m}^{CH_3COO^-}\)

However, for simplicity, we'll use the known value \(Λ_m^0\) for acetic acid, which is approximately \(390 \text{ S cm}^2 \text{ mol}^{-1}\).

Given data:

• Molar conductivity of acetic acid, \(Λ_m = 20 \text{ S cm}^2 \text{ mol}^{-1}\).
• Concentration, \(C = 0.007 \text{ M}\).

Using the formula for the degree of dissociation (\(α\)):

\[ α = \frac{Λ_m}{Λ_m^0} \]

Substitute the values:

\[ α = \frac{20}{390} ≈ 0.05128 \]

The dissociation constant (\(K_a\)) is calculated using the formula:

\[ K_a = Cα^2 \]

Substitute the values of \(C\) and \(α\):

\[ K_a = 0.007 × (0.05128)^2 ≈ 1.75 × 10^{-5} \text{ mol L}^{-1} \]

Therefore, the dissociation constant of acetic acid is \(1.75 × 10^{-5} \text{ mol L}^{-1}\). This matches the correct option.