Chemistry - Electrochemistry Question with Solution | TestHub

The relation between Molar conductance ${\mathrm{\lambda }}_{\mathrm{m}}$ and specific conductance $\mathrm{\kappa }$ is 

${\lambda }_{\mathrm{m}}=\frac{\mathrm{k}\times 1000}{\mathrm{M}}$

M = molarity

${\mathrm{\lambda }}_{\mathrm{m}}=\frac{\left(5\times {10}^{-5}\right)\times \left({10}^3\right)}{\left(2.5\times {10}^{-3}\right)}=20 \mathrm{S} {\mathrm{cm}}^2 {\mathrm{mol}}^{-1}$

Now, the degree of dissociation, $\mathrm{\alpha }=\frac{{\mathrm{\lambda }}_{\mathrm{m}}}{{\mathrm{\lambda }}^{\infty }}$

$\mathrm{\alpha }=\frac{20}{400}=\frac{1}{20}$

The acid dissociation constant, ${\mathrm{K}}_{\mathrm{a}}=\frac{{\mathrm{C\alpha }}^2}{(1-\mathrm{\alpha })}=\frac{\left(2.5\times {10}^{-3}\right)\left(\frac{1}{20}\times \frac{1}{20}\right)}{\left(\frac{19}{20}\right)}$

$= 65.789 \times {10}^{–7}$

$\approx 66 \times {10}^{–7}$