Chemistry - Electrochemistry Question with Solution | TestHub

${\mathrm{\lambda }}_{{\mathrm{X}}^{-}}^0\approx {\mathrm{\lambda }}_{{\mathrm{Y}}^{-}}^0$
$\Rightarrow {\mathrm{\lambda }}_{{\mathrm{H}}^{+}}^0+{\mathrm{\lambda }}_{{\mathrm{X}}^{-}}^0\mathrm{ }\approx {\mathrm{\lambda }}_{{\mathrm{H}}^{+}}^0+{\mathrm{\lambda }}_{{\mathrm{Y}}^{-}}^0$
$\Rightarrow {\mathrm{\lambda }}_{\mathrm{H}\mathrm{X}}^0\approx {\mathrm{\lambda }}_{\mathrm{H}\mathrm{Y}}^0$...... (i)
Also,$\frac{{\mathrm{\lambda }}_{\mathrm{m}}}{{\mathrm{\lambda }}_{\mathrm{m}}^0}=\mathrm{\alpha }$,
So${\mathrm{\lambda }}_{\mathrm{m}}\left(\mathrm{H}\mathrm{X}\right)=\mathrm{ }{\mathrm{\lambda }}_{\mathrm{m}}^0{\mathrm{\alpha }}_1$and${\mathrm{\lambda }}_{\mathrm{m}}\left(\mathrm{H}\mathrm{Y}\right)=\mathrm{ }{\mathrm{\lambda }}_{\mathrm{m}}^0{\mathrm{\alpha }}_2$
(Where${\mathrm{\alpha }}_1$and${\mathrm{\alpha }}_2$are degree of dissociation of HX and HY respectively)
Now, given that
${\mathrm{\lambda }}_{\mathrm{m}}\left(\mathrm{H}\mathrm{Y}\right)=10{\mathrm{\lambda }}_{\mathrm{m}}\left(\mathrm{H}\mathrm{X}\right)$
$\Rightarrow {\mathrm{\lambda }}_{\mathrm{m}}^0{\mathrm{\alpha }}_2=10\times {\mathrm{\lambda }}_{\mathrm{m}}^0{\mathrm{\alpha }}_1$
${\mathrm{\alpha }}_2=10{\mathrm{\alpha }}_1$...... (ii)
${\mathrm{K}}_{\mathrm{a}}=\frac{\mathrm{C}{\mathrm{\alpha }}^2}{1-\mathrm{\alpha }}$, But$\mathrm{\alpha }\ll 1$, therefore${\mathrm{K}}_{\mathrm{a}}=\mathrm{C}{\mathrm{\alpha }}^2$
$\Rightarrow \frac{{\mathrm{K}}_{\mathrm{a}}\left(\mathrm{H}\mathrm{X}\right)}{{\mathrm{K}}_{\mathrm{a}}\left(\mathrm{H}\mathrm{Y}\right)}=\frac{0.01{\mathrm{\alpha }}_1^2}{0.1{\mathrm{\alpha }}_2^2}=\frac{0.01}{0.1}\times {\left(\frac{1}{10}\right)}^2=\frac{1}{1000}$
$\Rightarrow [\log ({\mathrm{K}}_{\mathrm{a}}\left(\mathrm{H}\mathrm{X}\right)-\log \left({\mathrm{K}}_{\mathrm{a}}\left(\mathrm{H}\mathrm{Y}\right)\right]=-3$
$\Rightarrow \mathrm{p}{\mathrm{K}}_{\mathrm{a}}\left(\mathrm{H}\mathrm{X}\right)-\mathrm{p}{\mathrm{K}}_{\mathrm{a}}\left(\mathrm{H}\mathrm{Y}\right)=3$