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Chemistry · EQUILIBRIUM
हिन्दीA \( 0.08\text{ M} \) solution of hypochlorous acid (\( \text{HOCl} \)) has an ionization constant \( K_a = 2.5 \times 10^{-5} \). What is the percent dissociation of \( \text{HOCl} \)?
Correct answer · Option 1
\( 1.76\% \)
Explanation
\( \alpha = \sqrt{\frac{K_a}{c}} = \sqrt{\frac{2.5 \times 10^{-5}}{0.08}} = \sqrt{3.125 \times 10^{-4}} \approx 0.0176 \). Percent dissociation \( = 0.0176 \times 100 = 1.76\% \).
