Question

${\Lambda }^O$ for $NaCl,HCl$ and ${CH}_{3}COONa$ are $126.0,426.0$ and $91.0S{cm}^2{mol}^{-1}$ respectively. Calculate ${\Lambda }^O$ for ${CH}_{3}COOH$.

Answer 1

According to Kohlraush's law,

${{\Lambda }^0}_{NaCl}={\lambda }_{{Na}^{+}}+{\lambda }_{{Cl}^{-}}=126.0⟶\left(i\right)$

${{\Lambda }^0}_{HCl}={\lambda }_{H^{+}}+{\lambda }_{{Cl}^{-}}=426.0⟶\left(ii\right)$

${{\Lambda }^0}_{C{H}_{3}COONa}={\lambda }_{{C{H}_{3}COO}^{-}}+{\lambda }_{{Na}^{+}}=91.0⟶\left(iii\right)$

On adding ${eq}^n\left(ii\right)\&\left(iii\right)$ and subtracting ${eq}^n\left(i\right)$, we have

${\lambda }_{H^{+}}+{\lambda }_{{Cl}^{-}}+{\lambda }_{{C{H}_{3}COO}^{-}}+{\lambda }_{{Na}^{+}}-({\lambda }_{{Na}^{+}}+{\lambda }_{{Cl}^{-}})=426.0+91.0-126.0$

${\lambda }_{H^{+}}+{\lambda }_{{Cl}^{-}}+{\lambda }_{{C{H}_{3}COO}^{-}}+{\lambda }_{{Na}^{+}}-{\lambda }_{{Na}^{+}}-{\lambda }_{{Cl}^{-}}=390.0$

${\lambda }_{{C{H}_{3}COO}^{-}}+{\lambda }_{H^{+}}=390.0$

$\Rightarrow {{\Lambda }^0}_{C{H}_{3}COOH}=390.0S{cm}^2{mol}^{-1}$