Question
If the molar conductivity ( $\Lambda_{m}$ ) of a $0.050\text{ }mol{\text{ }L}^{- 1}$ solution of a monobasic weak acid is $90\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}$, its extent (degree) of dissociation will be
[Assume $\Lambda_{+}^{\circ} = 349.6\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}$ and $\Lambda_{-}^{\circ} = 50.4\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1}$.]
Options
Solution
Sol. [Key Concept: Degree of dissociation of a weak acid (α = Λₘ/Λ°ₘ)]
Degree of dissociation ( $\alpha$ ) is given as
$$\begin{matrix} & \alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} \\ & \Lambda_{m}^{\circ} = \Lambda_{+}^{\circ} + \Lambda_{-}^{\circ} \\ & \ = 349.6 + 50.4 \\ & \ = 400\text{ }S{\text{ }cm}^{2}{\text{ }mol}^{- 1} \\ & \alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}} = \frac{90}{400} = 0.225 \end{matrix}$$
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