Question

If the molar conductivity ( Λ m ) of a 0.050 m o l L − 1 solution of a monobasic weak acid is 90 S c m 2 m o l − 1 , its extent (degree) of dissociation will be [Assume Λ ∘ + = 349.6 S c m 2 m o l − 1 and Λ ∘ − = 50.4 S c m 2 m o l − 1 . ]

Answer 1

0.225

Answer 2

To determine the extent (degree) of dissociation ( $\alpha$ ) of a weak acid, we use the relationship between its molar conductivity at a given concentration ( $\Lambda_m$ ) and its molar conductivity at infinite dilution ( $\Lambda_m^\circ$ ).

1. Calculate Molar Conductivity at Infinite Dilution ( $\Lambda_m^\circ$ ):

According to Kohlrausch's Law of Independent Migration of Ions, the molar conductivity of an electrolyte at infinite dilution is the sum of the molar conductivities of its constituent ions at infinite dilution. For a monobasic weak acid (HA), it dissociates into H⁺ and A⁻ ions.

Given:

$\Lambda^\circ_+(H^+) = 349.6 \text{ S cm}^2 \text{ mol}^{-1}$

$\Lambda^\circ_-(A^-) = 50.4 \text{ S cm}^2 \text{ mol}^{-1}$

Therefore, the molar conductivity of the weak acid at infinite dilution is:

$\Lambda_m^\circ = \Lambda^\circ_+(H^+) + \Lambda^\circ_-(A^-)$

$\Lambda_m^\circ = 349.6 \text{ S cm}^2 \text{ mol}^{-1} + 50.4 \text{ S cm}^2 \text{ mol}^{-1}$

$\Lambda_m^\circ = 400.0 \text{ S cm}^2 \text{ mol}^{-1}$

2. Calculate the Extent (Degree) of Dissociation ( $\alpha$ ):

The degree of dissociation is given by the formula:

$\alpha = \frac{\Lambda_m}{\Lambda_m^\circ}$

Given:

Molar conductivity of the solution ( $\Lambda_m$ ) = $90 \text{ S cm}^2 \text{ mol}^{-1}$

Molar conductivity at infinite dilution ( $\Lambda_m^\circ$ ) = $400.0 \text{ S cm}^2 \text{ mol}^{-1}$

Substitute the values into the formula:

$\alpha = \frac{90 \text{ S cm}^2 \text{ mol}^{-1}}{400.0 \text{ S cm}^2 \text{ mol}^{-1}}$

$\alpha = 0.225$

The concentration of the solution (0.050 mol L⁻¹) is not required for this calculation as the molar conductivity ( $\Lambda_m$ ) at that concentration is already provided.

Explanation of the solution:

1. Calculate the molar conductivity at infinite dilution ( $\Lambda_m^\circ$ ) using Kohlrausch's Law: $\Lambda_m^\circ = \Lambda^\circ_+(H^+) + \Lambda^\circ_-(A^-) = 349.6 + 50.4 = 400.0 \text{ S cm}^2 \text{ mol}^{-1}$.

2. Calculate the degree of dissociation ( $\alpha$ ) using the formula: $\alpha = \frac{\Lambda_m}{\Lambda_m^\circ} = \frac{90}{400.0} = 0.225$.