The equivalent conductances of two strong electrolytes at infinite dilution in $\mathrm{H}_2 \mathrm{O}$…
The equivalent conductances of two strong electrolytes at infinite dilution in $\mathrm{H}_2 \mathrm{O}$ (where ions move freely through a solution) at $25^{\circ} \mathrm{C}$ are given below:
$\wedge^{\circ} \mathrm{CH}_3 \mathrm{COONa}=91.0 \mathrm{~S} \mathrm{~cm}^2$ /equiv
$\wedge_{\mathrm{HCl}}^{\circ}=426.2 \mathrm{~S} \mathrm{~cm^2}$ / equiv
What additional information/quantity one needs to calculate $\wedge^{\circ}$ of an aqueous solution of acetic acid?
$\wedge^{\circ}$ of $\mathrm{NaCl}$
$\wedge^{\circ}$ of $\mathrm{CH}_3 \mathrm{COOK}$
The limiting equivalent conductance of $\mathrm{H}^{+}\left(\wedge_{\mathrm{H}^{+}}^{\circ}\right)$
$\wedge^{\circ}$ of chloroacetic acid $\left(\mathrm{C} / \mathrm{CH}_2 \mathrm{COOH}\right)$
Solution
From Kohlrausch's law
$\Lambda_{\mathrm{CH}_3 \mathrm{COOH}}^{\circ}=\Lambda_{\mathrm{CH}_3 \mathrm{COONa}}^{\circ}+\Lambda_{\mathrm{HCl}}^{\circ}-\Lambda_{\mathrm{NaCl}}^{\circ}$
Hence, (A) is the correct answer.