Equivalent conductances of BaCl_2,H_2SO_4 and | vedclass

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(A) According to Kohlrausch's law,the equivalent conductance at infinite dilution is the sum of the equivalent conductances of the constituent ions.Fo

Vedclass · Accepted Answer

$\Lambda_1^\infty + \Lambda_2^\infty - \Lambda_3^\infty$

Vedclass · Answer

(A) According to Kohlrausch's law,the equivalent conductance at infinite dilution is the sum of the equivalent conductances of the constituent ions.For $BaCl_2$: $\Lambda_1^\infty = \lambda_{eq}(Ba^{2+}) + \lambda_{eq}(Cl^-)$For $H_2SO_4$: $\Lambda_2^\infty = \lambda_{eq}(H^+) + \lambda_{eq}(SO_4^{2-})$For $HCl$: $\Lambda_3^\infty = \lambda_{eq}(H^+) + \lambda_{eq}(Cl^-)$We need to find the equivalent conductance of $BaSO_4$ at infinite dilution,which is $\Lambda_{BaSO_4}^\infty = \lambda_{eq}(Ba^{2+}) + \lambda_{eq}(SO_4^{2-})$.By adding the expressions for $BaCl_2$ and $H_2SO_4$,we get: $\Lambda_1^\infty + \Lambda_2^\infty = \lambda_{eq}(Ba^{2+}) + \lambda_{eq}(Cl^-) + \lambda_{eq}(H^+) + \lambda_{eq}(SO_4^{2-})$.To isolate $\lambda_{eq}(Ba^{2+}) + \lambda_{eq}(SO_4^{2-})$,we subtract the expression for $HCl$ $(\Lambda_3^\infty)$:$\Lambda_{BaSO_4}^\infty = (\Lambda_1^\infty + \Lambda_2^\infty) - \Lambda_3^\infty$.

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

Equivalent conductances of $BaCl_2$,$H_2SO_4$ and $HCl$ at infinite dilution are $\Lambda_1^\infty$,$\Lambda_2^\infty$ and $\Lambda_3^\infty$ respectively. Then,the equivalent conductance of $BaSO_4$ at infinite dilution will be:

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