Plotting 1 / Λ m against cΛ m for aqueous solutions of a monobasic weak acid ( HX ) resulted in a…

The degree of dissociation is the ratio of molar conductivity to the limiting molar conductivity.

$\mathrm{\alpha }=\frac{{\mathrm{\Lambda }}_{\mathrm{m}}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}$

The equilibrium reaction is given below,

$$\begin{gathered} \mathrm{HX}\stackrel{}{\rightleftharpoons }{\mathrm{H}}^{+}+{\mathrm{X}}^{-} \\ \mathrm{initial} \mathrm{c} \\ \mathrm{At} \mathrm{equilibrium} \mathrm{c}-\mathrm{c\alpha } \mathrm{c\alpha } \mathrm{c\alpha } \end{gathered}$$

The acid dissociation constant, ${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{c\alpha }}^2}{1-\mathrm{\alpha }}$

$\Rightarrow {\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{c}{\left({\mathrm{\Lambda }}_{\mathrm{m}}/{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}\right)}^2}{1-\left({\mathrm{\Lambda }}_{\mathrm{m}}/{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}\right)}$

$\Rightarrow {\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{c\Lambda }}_{\mathrm{m}}^2}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}\left({\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}-{\mathrm{\Lambda }}_{\mathrm{m}}\right)}$

${\mathrm{K}}_{\mathrm{a}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}^2-{\mathrm{K}}_{\mathrm{a}}{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}{\mathrm{\Lambda }}_{\mathrm{m}}={\mathrm{c\Lambda }}_{\mathrm{m}}^2$

$\frac{{\mathrm{K}}_{\mathrm{a}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}^2}{{\mathrm{\Lambda }}_{\mathrm{m}}}-{\mathrm{K}}_{\mathrm{a}}{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}={\mathrm{C\Lambda }}_{\mathrm{m}}$

$\frac{{\mathrm{K}}_{\mathrm{a}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}^2}{{\mathrm{\Lambda }}_{\mathrm{m}}}={\mathrm{c\Lambda }}_{\mathrm{m}}+{\mathrm{K}}_{\mathrm{a}}{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$

$\frac{1}{{\mathrm{\Lambda }}_{\mathrm{m}}}=\left(\frac{{\mathrm{c\Lambda }}_{\mathrm{m}}}{{\mathrm{K}}_{\mathrm{a}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}^2}\right)+\frac{1}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}$

From the above line equation, slope and y-intercept can be written as follows,

$\mathrm{P}=\frac{1}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}$

$\mathrm{S}=\frac{1}{{\mathrm{K}}_{\mathrm{a}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}^2}$

$\frac{\mathrm{P}}{\mathrm{S}}=\left(\frac{\frac{1}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}}{\frac{1}{{\mathrm{K}}_{\mathrm{a}}{{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}}^2}}\right)={\mathrm{K}}_{\mathrm{a}}{\mathrm{\Lambda }}_{\mathrm{m}}^{\mathrm{o}}$