Question

The equivalence conductances of sodium chloride, hydrochloric acid and sodium acetate at infinite dilution are $126.45,426.16$ and $91.0{\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}$, respectively at $25{}^o\text{C}$.
Calculate the equivalent conductance $\left({\text{in ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}\right)$ of acetic acid at infinite dilution.

Answer 1

According to Kohlrausch's law,
${\Lambda }_{\infty C{H}_{3}COONa}={\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{N{a}^{+}}=91.0...\left(i\right){\Lambda }_{\infty HCl}={\lambda }_{H^{+}}+{\lambda }_{C{l}^{-}}=426.16...\left(ii\right){\Lambda }_{\infty NaCl}={\lambda }_{N{a}^{+}}+{\lambda }_{C{l}^{-}}=126.45...\left(iii\right)$
Adding equations $\left(i\right)$ and $\left(ii\right)$ and subtracting equation $\left(iii\right)$,
${\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{N{a}^{+}}+{\lambda }_{H^{+}}+{\lambda }_{C{l}^{-}}-{\lambda }_{N{a}^{+}}-{\lambda }_{C{l}^{-}}=91.0+426.16-126.45$
$\Rightarrow {\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{H^{+}}=390.7{\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}={\Lambda }_{\infty C{H}_{3}COOH}=390.7{\text{ohm}}^{-1}{\text{cm}}^2{\text{eq}}^{-1}$