Molar conductivities ($\Lambda^{\circ}_m$) at infinite dilution of $NaCl$, $HCl$, and $CH_3COONa$ are $126.4$,

Question

Molar conductivities ($\Lambda^{\circ}_m$) at infinite dilution of $NaCl$, $HCl$, and $CH_3COONa$ are $126.4$, $425.9$, and $91.0	ext{ S cm}^2	ext{ mol}^{-1}$ respectively. $\Lambda^{\circ}_m$ for $CH_3COOH$ will be:

Accepted Answer

According to Kohlrausch's law of independent migration of ions, the limiting molar conductivity of an electrolyte can be represented as the sum of the individual contributions of the anion and cation of the electrolyte.

$\Lambda^{\circ}_m(CH_3COOH) = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{CH_3COO^-}$

This can be obtained by algebraically combining the given molar conductivities of strong electrolytes:
$\Lambda^{\circ}_m(CH_3COOH) = \Lambda^{\circ}_m(CH_3COONa) + \Lambda^{\circ}_m(HCl) - \Lambda^{\circ}_m(NaCl)$
$\Lambda^{\circ}_m(CH_3COOH) = (91.0 + 425.9 - 126.4)	ext{ S cm}^2	ext{ mol}^{-1}$
$\Lambda^{\circ}_m(CH_3COOH) = 516.9 - 126.4 = 390.5	ext{ S cm}^2	ext{ mol}^{-1}$

Step-by-Step Solution

Practice All CHEMISTRY Questions