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Chemistry · EQUILIBRIUM
हिन्दीCalculate the percent dissociation of a \( 0.08\text{ M} \) hypochlorous acid (\( \text{HOCl} \)) solution if its ionization constant \( K_a \) is \( 2.5 \times 10^{-5} \).
Correct answer · Option 3
1.76%
Explanation
\( [\text{H}^+] = \sqrt{K_a \cdot c} = \sqrt{2.5 \times 10^{-5} \times 0.08} = \sqrt{2.0 \times 10^{-6}} = 1.41 \times 10^{-3}\text{ M} \). Percent dissociation = \( (1.41 \times 10^{-3} / 0.08) \times 100 = 1.76\% \).
