Which of the following is the correct equatio | vedclass

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(D) For a weak electrolyte $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_{a}$ is given by $K_{a} = \frac{\alpha^2 C}{1 - \alpha}$

Vedclass · Accepted Answer

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

Vedclass · Answer

(D) For a weak electrolyte $HA \rightleftharpoons H^{+} + A^{-}$,the dissociation constant $K_{a}$ is given by $K_{a} = \frac{\alpha^2 C}{1 - \alpha}$. Since $\alpha = \frac{\Lambda_{m}}{\Lambda_{m}^{\circ}}$,we substitute this into the expression: $K_{a} = \frac{(\Lambda_{m} / \Lambda_{m}^{\circ})^2 C}{1 - (\Lambda_{m} / \Lambda_{m}^{\circ})} = \frac{\Lambda_{m}^2 C}{\Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m})}$. Rearranging this gives: $K_{a} \Lambda_{m}^{\circ}(\Lambda_{m}^{\circ} - \Lambda_{m}) = \Lambda_{m}^2 C$. $K_{a} (\Lambda_{m}^{\circ})^2 - K_{a} \Lambda_{m} \Lambda_{m}^{\circ} = \Lambda_{m}^2 C$. Therefore,the correct equation is $\Lambda_{m}^2 C + K_{a} \Lambda_{m} \Lambda_{m}^{\circ} - K_{a} (\Lambda_{m}^{\circ})^2 = 0$.

$c^{1/2} / (mol \, L^{-1})^{1/2}$	$\Lambda_m / S \, cm^2 \, mol^{-1}$
$0.000198$	$148.61$
$0.000309$	$148.29$
$0.000521$	$147.81$
$0.000989$	$147.09$

Which of the following is the correct equation representing the change in molar conductivity with respect to concentration for a weak electrolyte,where the symbols carry their usual meaning?

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