Molar conductance of 0.01 M acetic acid was found to be 16 × 10-4 S m2 mol-1 at 30°C. Molar conductance of H+ and CH3COO- ions at infinite dilution are 350 × 10-4 S m2 mol-1 and 50 × 10-4 S m2 mol-1, respectively at same temperature. What percentage of acetic acid is dissociated at that concentration ?
0.040
The degree of dissociation of a weak electrolyte is found by comparing its molar conductance at the working concentration with its value at infinite dilution:
\(\alpha = \frac{\Lambda_m}{\Lambda_m^{\circ}}\).
Step 1 — find \(\Lambda_m^{\circ}\) using Kohlrausch's law. At infinite dilution each ion contributes independently, so
\(\Lambda_m^{\circ} = \lambda^{\circ}_{H^{+}} + \lambda^{\circ}_{CH_3COO^{-}} = (350 + 50) \times 10^{-4} = 400 \times 10^{-4}\) S m2 mol-1.
This step is necessary because \(\Lambda_m^{\circ}\) for a weak acid cannot be obtained by extrapolating measurements — the conductance rises too steeply near infinite dilution — so it must be assembled from the individual ionic conductivities.
Step 2 — take the ratio.
\(\alpha = \frac{16 \times 10^{-4}}{400 \times 10^{-4}} = 0.040\).
So the degree of dissociation is 0.040, that is 4% — and the answer is quoted as the fraction 0.040.
The result is chemically sensible: acetic acid is a weak acid, so only a few per cent of it is ionised at 0.01 M. As a check, \(K_a = \frac{c\alpha^{2}}{1-\alpha} \approx 0.01 \times (0.04)^{2} = 1.6 \times 10^{-5}\), close to the accepted value of \(1.8 \times 10^{-5}\).
Note also that the very high value of \(\lambda^{\circ}\) for H+ reflects the Grotthuss mechanism, in which the proton hops along hydrogen bonds rather than migrating bodily through the solution.
Hence the degree of dissociation is 0.040.
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