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Chemistry · EQUILIBRIUM
हिन्दीThe ionization constant of aqueous ammonia is \( K_b = 1.77 \times 10^{-5} \). If the concentration of ammonia is diluted from 0.05 M to 0.005 M at 298 K, the percentage increase in its degree of ionization (\( \alpha \)) is approximately:
Correct answer · Option 4
\( 216\% \)
Explanation
By Ostwald's dilution law, \( \alpha = \sqrt{K_b/C} \). \( \alpha_1 = \sqrt{\frac{K_b}{0.05}} \), \( \alpha_2 = \sqrt{\frac{K_b}{0.005}} = \sqrt{10}\alpha_1 \approx 3.162\alpha_1 \). Percentage increase \( = \frac{\alpha_2 - \alpha_1}{\alpha_1} \times 100\% = (3.162 - 1) \times 100\% \approx 216\% \).
