Discussion
The molar conductivity of a weak mono basic acid, HA at 298 K is 70 Scm2mol-1. What is the percentage ionisation of HA at 298 K?
[At infinite dilution λH+ = 340 Scm2mol-1 and λA- = 80 Scm2mol-1]
Answer: B
The percentage ionisation is 16.7 %.
By Kohlrausch's law, the limiting molar conductivity is the sum of the ionic contributions, and the degree of dissociation of a weak electrolyte is \(\alpha = \frac{\Lambda_{m}}{\Lambda^\circ_{m}}\).
\[ \begin{aligned} \Lambda^\circ_{m} &= 340 + 80 = 420 \text{ S cm}^{2}\text{ mol}^{-1} \\ \alpha &= \frac{70}{420} = 0.167 \end{aligned} \]
That is 16.7 %. 8.35 % is half of it, and 20 % would need a limiting molar conductivity of 350, which ignores the contribution of one of the ions.
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