Question

The equivalent conductances of sodium choride, hydrochloric acid and sodium acetate at infinite dilution are $126.45,426.16$ and $91.0oh{m}^{-1}c{m}^{2}e{q}^{-1}$, respectively, at ${25}^{\circ }C$. Calculate the equivalent conductance of acetic acid at infinite dilution.

Answer 1

According to Kohlrausch's law,
${\wedge }_{\infty C{H}_{3}COONa}={\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{N{a}^{+}}=91.0....\left(i\right)$
${\wedge }_{\infty HCl}={\lambda }_{H^{+}}+{\lambda }_{C{l}^{-}}=426.16.....\left(ii\right)$
${\wedge }_{\infty NaCl}={\lambda }_{N{a}^{+}}+{\lambda }_{C{l}^{-}}=126.45....\left(iii\right)$
Adding eqs. (i) and (ii) and subtracting eq. (iii)
${\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{N{a}^{+}}+{\lambda }_{H^{+}}+{\lambda }_{C{l}^{-}}-{\lambda }_{N{a}^{+}}-{\lambda }_{C{l}^{-}}$
$=91.0+426.16-126.45$
${\lambda }_{C{H}_{3}CO{O}^{-}}+{\lambda }_{H^{+}}={\wedge }_{\infty C{H}_{3}COOH}=390.7oh{m}^{-1}c{m}^{2}e{q}^{-1}$.