If ${\mathrm{\Lambda }}_{\mathrm{c}}$ and ${\mathrm{\Lambda }}^{\mathrm{\infty }}$ are the molar conductivities ofa weak acid at concentration c and its equilibrium constant may be computed by using the expression

The degree of dissociation of the weak acid is $\alpha =\frac{{\mathrm{\Lambda }}_c}{{\mathrm{\Lambda }}^{\mathrm{\infty }}}$

Also since $\begin{array}{cccc}\mathrm{HA}& \rightleftharpoons & {\mathrm{H}}^{+}+& {\mathrm{A}}^{-}\\ c(1-\alpha )& & \mathrm{c}\alpha & \mathrm{c}\alpha \end{array}$ we have

$K_{\mathrm{a}}=\frac{\left[{\mathrm{H}}^{+}\right]\left[{\mathrm{A}}^{-}\right]}{\left[\mathrm{HA}\right]}=\frac{\left(c\alpha \right)\left(c\alpha \right)}{c(1-\alpha )}=\frac{c{\alpha }^2}{1-\alpha }=\frac{c{\left({\mathrm{\Lambda }}_c/{\mathrm{\Lambda }}^{\mathrm{\infty }}\right)}^2}{1-\left({\mathrm{\Lambda }}_c/{\mathrm{\Lambda }}^{\mathrm{\infty }}\right)}=\frac{c{\mathrm{\Lambda }}_{\mathrm{e}}^2}{{\mathrm{\Lambda }}^{\mathrm{\infty }}\left({\mathrm{\Lambda }}^{\mathrm{\infty }}-{\mathrm{\Lambda }}_c\right)}$