When HCl is added to the solution, the concentration of H+ ions will increase. Therefore, the equilibrium will shift in the backward direction i.e., dissociation of acetic acid will decrease

Case I: When 0.01 M HCl is taken.

Let x be the amount of acetic acid dissociated after the addition of HCl.

${\mathrm{CH}}_3\mathrm{COOH}\leftrightarrow {\mathrm{H}}^{+}+{\mathrm{CH}}_3{\mathrm{COO}}^{-}$
$\mathrm{Initial} \mathrm{conc}. 0.05 \mathrm{M} 0 0$
$\mathrm{After} \mathrm{dissociation} 0.05-5 0.1+\mathrm{x} \mathrm{x}$

As the dissociation of a very small amount of acetic acid will take place, the values i.e., 0.05 – x and 0.01 + x can be taken as 0.05 and 0.01 respectively.

${\mathrm{K}}_{\mathrm{\alpha }}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$
$\therefore {\mathrm{K}}_{\mathrm{\alpha }}=\frac{(0.01)\mathrm{x}}{0.05}$
$\mathrm{x}=\frac{1.82\times {10}^{-5}\times 0.05}{0.01}$
$\mathrm{x}=1.82\times {10}^{-3}\times 0.05\mathrm{M}$
$\hspace{0.17em}\mathrm{Now},$
$\mathrm{\alpha }=\frac{\mathrm{Ammount} \mathrm{of} \mathrm{acid} \mathrm{dissociated}}{\mathrm{Amount} \mathrm{of} \mathrm{acid} \mathrm{taken}}$
$=\frac{1.82\times {10}^{-3}\times 0.05}{0.05}$
$=1.82\times {10}^{-3}$

Case II: When 0.1 M HCl is taken. Let the amount of acetic acid dissociated, in this case, be X. As we have done in the first case, the concentrations of various species involved in the reaction are:

$\left[{\mathrm{CH}}_3\mathrm{COOH}\right]=0.05-\mathrm{X};0.05\mathrm{M}$
$\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]=\mathrm{X}$
$\left[{\mathrm{H}}^{+}\right]=0.1+\mathrm{X} ; 0.1\mathrm{M}$
${\mathrm{K}}_{\mathrm{\alpha }}=\frac{\left[{\mathrm{CH}}_3{\mathrm{COO}}^{-}\right]\left[{\mathrm{H}}^{+}\right]}{\left[{\mathrm{CH}}_3\mathrm{COOH}\right]}$
$\therefore {\mathrm{K}}_{\mathrm{\alpha }}=\frac{(0.1)\mathrm{X}}{0.05}$
$\mathrm{x}=\frac{1.82\times {10}^{-5}\times 0.05}{0.1}$
$\mathrm{x}=1.82\times {10}^{-4}\times 0.05\mathrm{M}$
$\mathrm{Now},$
$\mathrm{\alpha }=\frac{\mathrm{Amount} \mathrm{of} \mathrm{acid} \mathrm{dissociated}}{\mathrm{Amount} \mathrm{of} \mathrm{acid} \mathrm{taken}}$
$=\frac{1.82\times {10}^{-4}\times 0.05}{0.05}$
$=1.82\times {10}^{-4}$