Plotting 1 / Λ _ m against cΛ _ m for aqueous solutions of a...

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

$\alpha = {Λ}_{m}{Λ}_{m}^{o}$

The equilibrium reaction is given below,

$HX\overset{}{⇌}{H}^{+} + {X}^{-}initialcAtequilibriumc - c \alphac \alphac \alpha$

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

$⇒ {K}_{a} = \dfrac{c{\left({Λ}_{m} / {Λ}_{m}^{o}\right)}^{2}}{1 - \left({Λ}_{m} / {Λ}_{m}^{o}\right)}$

$⇒ {K}_{a} = {cΛ}_{m}^{2}{Λ}_{m}^{o}\left({Λ}_{m}^{o} - {Λ}_{m}\right)$

${K}_{a}{Λ}_{m}^{o}2 - {K}_{a}{Λ}_{m}^{o}{Λ}_{m} = {cΛ}_{m}^{2}$

$\dfrac{{K}_{a}{Λ}_{m}^{o}2}{{Λ}_{m}} - {K}_{a}{Λ}_{m}^{o} = {CΛ}_{m}$

$\dfrac{{K}_{a}{Λ}_{m}^{o}2}{{Λ}_{m}} = {cΛ}_{m} + {K}_{a}{Λ}_{m}^{o}$

$\dfrac{1}{{Λ}_{m}} = \left(\dfrac{{cΛ}_{m}}{{K}_{a}{Λ}_{m}^{o}2}\right) + 1{Λ}_{m}^{o}$

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

$P = 1{Λ}_{m}^{o}$

$S = \dfrac{1}{{K}_{a}{Λ}_{m}^{o}2}$

$\dfrac{P}{S} = \left(\dfrac{1{Λ}_{m}^{o}}{\dfrac{1}{{K}_{a}{Λ}_{m}^{o}2}}\right) = {K}_{a}{Λ}_{m}^{o}$