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Question:

The conductance of a 0.0015 M aqueous solution of a weak monobasic acid was determined by using a conductivity cell consisting of platinized Pt electrodes. The distance between the electrodes is 120 cm with an area of cross section of 1 cm². The conductance of this solution was found to be 5 × 10⁻⁵ S. The pH of the solution is 4. The value of limiting molar conductivity (Λ⁰ₘ) of this weak monobasic acid in aqueous solution is Z × 10² S cm² mol⁻¹. The value of Z is:

Solution:

The conductivity k = G/a
k = 5 × 10⁻⁵ S × 120 cm / 1 cm²
k = 0.00006 S/cm
The molar conductivity Λₘ = 1000k/C
Λₘ = 1000 cm³/L × 0.00006 S/cm / 0.0015 mol/L
Λₘ = 40 S cm²/mol
pH = 4
[H⁺] = 10⁻⁴ M
The degree of dissociation α = Λₘ/Λ⁰ₘ
Λ⁰ₘ = Λₘ/α
Λ⁰ₘ = 40 S cm²/mol / 0.06667
Λ⁰ₘ = 6 × 10² S cm²/mol
But Λ⁰ₘ = Z × 10² S cm²/mol
Hence, Z = 6