Which out of the following is a correct equation to show change in molar conductivity with respect to concentration for a weak electrolyte, if the symbols carry their usual meaning :

\Lambda_m-\Lambda_m^\circ+AC^\frac12=0

\Lambda_m^2 C+K_a \Lambda_m^o^2-K_a \Lambda_m \Lambda_m^\circ=0

\Lambda_m-\Lambda_m^\circ-AC^\frac12=0

Electrochemistry PYQ — JEE Main 2024 · 9 Apr · Shift 2 · Q12 with Solution

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Correct answer: D

For a weak electrolyte, the degree of dissociation is related to molar conductivity by

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

Also, from Ostwald’s dilution law,

K_a = \frac{C\alpha^2}{1-\alpha}

where $C$ is the concentration.

K_a = \frac{C\left(\dfrac{\Lambda_m}{\Lambda_m^\circ}\right)^2}{1-\dfrac{\Lambda_m}{\Lambda_m^\circ}}

Simplify the denominator:

1-\frac{\Lambda_m}{\Lambda_m^\circ} = \frac{\Lambda_m^\circ-\Lambda_m}{\Lambda_m^\circ}

So,

K_a = \frac{C\Lambda_m^2}{\Lambda_m^{\circ 2}} \cdot \frac{\Lambda_m^\circ}{\Lambda_m^\circ-\Lambda_m}

K_a = \frac{C\Lambda_m^2}{\Lambda_m^\circ(\Lambda_m^\circ-\Lambda_m)}

Multiply both sides:

K_a\Lambda_m^\circ(\Lambda_m^\circ-\Lambda_m) = C\Lambda_m^2

Expand the left side:

K_a\Lambda_m^{\circ 2} - K_a\Lambda_m\Lambda_m^\circ = C\Lambda_m^2

Bring all terms to one side:

\Lambda_m^2 C - K_a\Lambda_m^{\circ 2} + K_a\Lambda_m\Lambda_m^\circ = 0

This matches Option D:

\Lambda_{\mathrm{m}}^2 \mathrm{C}-\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}}^{\circ 2}+\mathrm{K}_{\mathrm{a}} \Lambda_{\mathrm{m}} \Lambda_{\mathrm{m}}^{\circ}=0

A and C are forms of the Debye–Hückel–Onsager relation used for strong electrolytes, not weak electrolytes.
B has incorrect signs compared to the derived weak electrolyte relation.

Hence, the correct answer is D.