QJEEM24Ch664f: JEE Main 2024 Chemistry - Class 12 JEEM24Ch664f Solution: Which out of the following is a correct equation to show change …

The dissociation of a weak electrolyte ( $\mathrm{HA}$ ) in solution can be represented as:

$\mathrm{HA}(\mathrm{aq}) \rightleftharpoons \mathrm{H}^{+}(\mathrm{aq}) + \mathrm{A}^{-}(\mathrm{aq})$

The degree of dissociation ( $\alpha$ ) for a weak electrolyte is given by:

$\alpha = \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$

Where $\Lambda_{\mathrm{m}}$ is the molar conductivity at a given concentration $C$ , and $\Lambda_{\mathrm{m}}^{\circ}$ is the molar conductivity at infinite dilution.

The dissociation constant ( $K_{\mathrm{a}}$ ) is described by:

$K_{\mathrm{a}} = \frac{\alpha^2 C}{1 - \alpha}$

Rewriting this equation in terms of $\Lambda_{\mathrm{m}}$ and $\Lambda_{\mathrm{m}}^{\circ}$ :

$K_{\mathrm{a}} = \frac{\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C}{1 - \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}}$

By multiplying out and simplifying the above expression:

$\left( \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right)^2 C + K_{\mathrm{a}} \left( \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}} \right) = K_{\mathrm{a}}$

Substitute $\alpha = \frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}$ :

$\left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right)^2 C + K_{\mathrm{a}} \left(\frac{\Lambda_{\mathrm{m}}}{\Lambda_{\mathrm{m}}^{\circ}}\right) - K_{\mathrm{a}} = 0$

Rewriting by multiplying through by $\left( \Lambda_{\mathrm{m}}^{\circ} \right)^2$ :

$\Lambda_{\mathrm{m}}^2 C + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} - K_{\mathrm{a}} \left( \Lambda_{\mathrm{m}}^{\circ} \right)^2 = 0$

Thus, the correct equation that describes the change in molar conductivity with respect to concentration for a weak electrolyte is given by:

Option D:

$\Lambda_{\mathrm{m}}^2 C - K_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2} + K_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ} = 0$