A solution of monobasic acid of concentration 0.02 M has conductivity 0.2 \Omega–1m–1 at 25ºC. The molar conductivity at infinite dilution is 4 × 10–2 \Omega–1 m2 mol

Engineering

Chemistry

pH Calculation for Acids and Bases

pH Scale and Relative Strength of Acids

Electrolytic Conductance

Question

A solution of monobasic acid of concentration 0.02 M has conductivity 0.2 $Ω$ ^–1m^–1 at 25ºC. The molar conductivity at infinite dilution is 4 × 10^–2 $Ω$ ^–1 m² mol^–1. Find pH of solution.

2.7

3.7

2.3

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K = 0.2 $Ω$ ^–1 m^–1 = 2 × 10^–3 $Ω$ ^–1 cm^–1

$\land_{m}^{o}$ = 4 × 10^–2 $Ω$ ^–1 m²mol^–1 = 400 $Ω$ ^–1 cm² mol^–1

$∴ \land_{m}^{} = \frac{1000 k}{M} = \frac{1000 \times 2 \times 10^{- 3}}{0 .02} = 100 Ω^{- 1} {cm}^{2} {mol}^{- 1}$

$∴ α = \frac{\land_{m}}{\land_{m}^{o}} = \frac{100}{400} = \frac{1}{4}$

$∴$ HA (aq.) $⇌$ H⁺ (aq.) + A^¯ (aq.)

t = 0 0.02 M 0 0

$t = t_{eq} 0 .0 2 (1 - \frac{0 .02}{4} M) = 0 .0 2 \times \frac{1}{4} M \frac{0 .02}{4} M$

$∴ [H^{+}] = \frac{0 .02}{4} M = 0 .00 5 M$

$∴$ pH = – log (5 × 10^–3) = 3 – 0.7 = 2.3