The degree of ionization of 0.05 M acetic acid (\( K_a = 1.8 \times 10^{-5} \)) is \( \alpha_0 \). When sodium acetate is added to make the solution 0.05 M in acetate ions, the new degree of ionization becomes \( \alpha' \). The ratio \( \frac{\alpha_0}{\alpha'} \) is:

Initially, \( \alpha_0 = \sqrt{K_a/C} = \sqrt{1.8 \times 10^{-5}/0.05} = \sqrt{3.6 \times 10^{-4}} = 0.01897 \). In presence of common ion, \( K_a = \frac{[\text{H}^+][\text{Ac}^-]}{[\text{HAc}]} \approx \frac{(C\alpha')(C)}{C} \implies \alpha' = \frac{K_a}{C} = \frac{1.8 \times 10^{-5}}{0.05} = 3.6 \times 10^{-4} \). Ratio \( = \frac{0.01897}{3.6 \times 10^{-4}} \approx 52.7 \).