Chemistry - Ionic Equilibrium Question with Solution | TestHub

By$\mathrm{\alpha }=\sqrt{\frac{{\mathrm{K}}_{\mathrm{a}}}{\mathrm{C}}}=\sqrt{\frac{10^{-4}}{1.2\times 10^{-3}}}=\sqrt{\frac{1}{12}}$
$\mathrm{\%}\mathrm{ }\mathrm{\alpha }=28.86\mathrm{\%}\mathrm{ }$, as value of$\mathrm{\alpha }$is greater than 5% so we can not take assumptions and will have to use the exact formula.
So$\mathrm{\alpha }$will be given by:
$\mathrm{\alpha }=\frac{-10^{-4}+\sqrt{10^{-8}+4\times 10^{-3}\times 1.2\times 10^{-4}}}{2\times 1.2\times 10^{-3}}$
So the correct value of$\mathrm{\alpha }=25\mathrm{ }\mathrm{\%}$