Q.24 For a weak acid at 298 K, the molar conductivities (in ohm⁻¹ m² mol⁻¹), at infinite dilution and 0.04 mol dm⁻³ are 4.3 × 10⁻³ and 1.0 × 10⁻³, respectively.
The degree of dissociation of the acid (0.04 mol dm⁻³) at 298 K is ______.

Correct Answer: The degree of dissociation of the weak acid at 0.04 mol dm⁻³ and 298 K is
0.2326 ≈ 0.23 (or 23.3%).

Question Statement

For a weak acid at 298 K, the molar conductivities
(in ohm⁻¹ m² mol⁻¹) at infinite dilution and
at concentration 0.04 mol dm⁻³ are
4.3 × 10⁻³ and 1.0 × 10⁻³, respectively.
The degree of dissociation of the acid at 0.04 mol dm⁻³ and 298 K is ______.

This is a standard electrochemistry application of the relation between molar conductivity
and degree of dissociation for weak electrolytes.

Key Concept: Relation Between Molar Conductivity and Degree of Dissociation

For a weak electrolyte (such as a weak acid), the degree of dissociation
α is given by:

α = Λ_m / Λ_m^∘

where:

• Λ_m = molar conductivity at a given concentration
• Λ_m^∘ = molar conductivity at infinite dilution

Step-by-Step Solution

Given:

• Λ_m^∘ = 4.3 × 10⁻³
  ohm⁻¹ m² mol⁻¹
• Λ_m = 1.0 × 10⁻³
  ohm⁻¹ m² mol⁻¹

Using the formula:

α = (1.0 × 10⁻³) / (4.3 × 10⁻³) = 1.0 / 4.3 ≈ 0.2326

Degree of dissociation (fraction): α ≈ 0.23

Degree of dissociation (percentage): α ≈ 23.3%

Explanation of Possible Options

Option A: 0.023

This value is one order of magnitude smaller than the correct answer and results from
an incorrect handling of powers of ten. Hence, it is incorrect.

Option B: 0.23 (Correct)

This value correctly follows the relation
α = Λ_m / Λ_m^∘
and matches the numerical ratio of the given conductivities.

Option C: 0.43

This value overestimates the degree of dissociation and implies that nearly half of
the acid is dissociated, which is inconsistent with the data.

Option D: 0.77 or 0.78

Such high values indicate a strong electrolyte, contradicting the given information
that the substance is a weak acid.

How to Present This in an Exam Answer

For a weak electrolyte:
α = Λ_m / Λ_m^∘

Substituting values:

α = (1.0 × 10⁻³) / (4.3 × 10⁻³) ≈ 0.2326

Final Answer: Degree of dissociation, α ≈ 0.23 (≈ 23.3%).