QuestionElectrochemistry / degree of dissociation
The specific conductance of 0.01 M solution of acetic acid was found to be 0.015 S m⁻¹ at 25˚C. Molar conductance of acetic acid at infinite dilution is 400 × 10⁻⁴ S m² mol⁻¹. The degree of dissociation of acetic acid will be:
- (a) 0.375
- (b) 0.0375
- (c) 0.00375
- (d) 0.357
Answer: (b) 0.0375
Λm = κ/c = 0.015 S m⁻¹ / 10 mol m⁻³ = 1.5 × 10⁻³ S m² mol⁻¹. α = Λm/Λ°m = 1.5 × 10⁻³ / 400 × 10⁻⁴ = 0.0375 → (b). Trap: 0.01 M = 10 mol m⁻³.