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 ($\Lambda_0$). In this case, the limiting molar conductivity of the weak acid ($\Lambda_0$) can be calculated by summing the limiting molar conductivities of the ions it dissociates into, i.e., H⁺ and A^-:

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

Given that $\Lambda^0_{\text{H}^+} = 349.6 \text{ S cm}^2 \text{ mol}^{-1}$ and $\Lambda^0_{\text{acid}^-} = 50.4 \text{ S cm}^2 \text{ mol}^{-1}$, we can calculate $\Lambda_0$:

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

Now, we can calculate the degree of dissociation ($\alpha$) using the following formula:

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

Given that $\Lambda_m = 90 \text{ S cm}^2 \text{ mol}^{-1}$, we have:

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

Thus, the degree of dissociation is approximately 0.225.