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 ×10–7 S. The pH of the solution is 4. The value of limiting molar conductivity of this weak monobasic acid in aqueous solution is Z×102 S cm–1 mol

Engineering

Chemistry

Electrolytic Conductance

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^–7 S. The pH of the solution is 4. The value of limiting molar conductivity $(\land_{m}^{o})$ of this weak monobasic acid in aqueous solution is Z×10² S cm^–1 mol^–1. The value of Z is

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C = κ $(\frac{a}{ℓ})$

⇒ 5 × 10^–7 S = κ × $(\frac{1 {cm}^{2}}{120 cm})$

$∴$ κ = 600 × 10^–7 S/cm

$∴ \land_{m} = \frac{1000 κ}{M} = \frac{1000 \times 600 \times 10^{- 7}}{15 \times 10^{- 4}} S {cm}^{2} {mol}^{- 1}$

= 40 S cm² mol^–1

pH = 4 ⇒ [H⁺] = 10^–4

$∴$ H X (aq) $⇌$ H⁺(a) + X¯ (a)

t = t_eq : (15 × 10^–4–10^–4)M 10^–4 M 10^–4 M

$∴ α = \frac{10^{- 4}}{15 \times 10^{- 4}} = \frac{\land_{m}}{\land_{m}^{0}}$

$\Rightarrow \frac{1}{15} = \frac{40}{\land_{m}^{0}}$

$∴$ 6 × 10² S cm² mol^–1

$∴$ Answer → 6