Chemistry - Ionic Equilibrium Question with Solution | TestHub

$\underset{\underset{\text{Second}\text{case}}{\underset{\text{First}\text{case}}{}}}{}\underset{\underset{\text{20\%}\text{yellow}}{\underset{\text{80\%}\text{yellow}}{\text{Acid}\text{form}}}}{\text{HIn}}\rightleftharpoons {\text{H}}^{\oplus }\text{+}\underset{\underset{\text{80\%}\text{red}}{\underset{\text{30\%}\text{red}}{\text{Base}\text{form}}}}{{\text{In}}^{-}}$
First case
$\mathrm{p}{\mathrm{H}}_1=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{l}\mathrm{o}\mathrm{g}\frac{\left[\mathrm{I}{\mathrm{n}}^{-}\right]}{\left[\mathrm{H}\mathrm{I}\mathrm{n}\right]}$
$=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{20}{80}\right)$
$=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{l}\mathrm{o}\mathrm{g}1/4$…(i)
Second case
$\mathrm{p}{\mathrm{H}}_2=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{l}\mathrm{o}\mathrm{g}\left(\frac{80}{20}\right)$
$=\mathrm{p}{\mathrm{K}}_{\mathrm{a}}+\mathrm{l}\mathrm{o}\mathrm{g}4$…(ii)
From equations (i) and (ii), we get
$\mathrm{p}{\mathrm{H}}_2-\mathrm{p}{\mathrm{H}}_1=2\mathrm{l}\mathrm{o}\mathrm{g}4=4\mathrm{l}\mathrm{o}\mathrm{g}2$
$=4\times 0.3=1.2$