Question

If the molar conductivity ($\Lambda_m$) of a 0.050 mol $L^{–1}$ solution of a monobasic weak acid is 90 S $cm^{2} mol^{–1}$, its extent (degree) of dissociation will be:
[Assume: $\Lambda^0$ = 349.6 S $cm^{2} mol^{–1}$ and $\Lambda^0_{\text{acid}}$ = 50.4 S$ cm^{2} mol^{–1}$]

• A. 0.125
• B. 0.225
• C. 0.215
• D. 0.115

Hint:

For weak electrolytes, use \(\alpha = \frac{\Lambda_m}{\Lambda^0_m}\) to find the degree of dissociation from molar conductivities.

Answer 1

The Correct Option is B

The degree of dissociation ($\alpha$) is defined as the ratio of the molar conductivity at a given concentration ($\Lambda_m$) to the limiting molar conductivity at infinite dilution ($\Lambda_0$). For the weak acid, the limiting molar conductivity ($\Lambda_0$) is the sum of the limiting molar conductivities of its constituent ions, H⁺ and A^-:

$$\Lambda_0 = \Lambda^0_{\text{H}^+} + \Lambda^0_{\text{acid}^-}$$

Using the given values $\Lambda^0_{\text{H}^+} = 349.6 \text{ S cm}^2 \text{ mol}^{-1}$ and $\Lambda^0_{\text{acid}^-} = 50.4 \text{ S cm}^2 \text{ mol}^{-1}$, $\Lambda_0$ is calculated as:

$$\Lambda_0 = 349.6 + 50.4 = 400.0 \text{ S cm}^2 \text{ mol}^{-1}$$

The degree of dissociation ($\alpha$) is then calculated using:

$$\alpha = \frac{\Lambda_m}{\Lambda_0}$$

With $\Lambda_m = 90 \text{ S cm}^2 \text{ mol}^{-1}$, the calculation is:

$$\alpha = \frac{90}{400} = 0.225$$

The degree of dissociation is approximately 0.225.