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  4. Lambda°m for NaCl, HCl and NaAc are 126.4, 425.9 and 91.0 S cm2 mol-1 respectively. Lambda° for HAc is

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Q. $\Lambda^{\circ}_m$ for $NaCl, HCl$ and $NaAc$ are $126.4, 425.9$ and $91.0 \,S \,cm^2 \,mol^{-1}$ respectively. $\Lambda^{\circ}$ for $HAc$ is

Electrochemistry

Solution:

$\Lambda^{\circ}_{m(HAc)} = \lambda^{\circ}_{H^+} + \lambda^{\circ}_{Ac^-}$
$= \lambda^{\circ}_{H^+} + \lambda^{\circ}_{Cl^-} +\lambda^{\circ}_{Ac^-}+ \lambda^{\circ}_{Na^+} - \lambda^{\circ}_{Cl^-} -\lambda^{\circ}_{Na^+}$
$=\Lambda^{\circ}_{m(HCl)} + \Lambda^{\circ}_{m(NaAc)} - \Lambda^{\circ}_{m(NaCl)}$
$ = (425.9 + 91.0 - 126.4)S \,cm^2\,mol^{-1}$
$ = 390.5\,S\,cm^2\,mol^{-1}$