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3 September, 00:33

At a certain temperature, the percent dissociation (ionization) of chlorous acid, HClO2, in a 1.43 M solution water is 8.0%. Calculate the value of Ka for chlorous acid at this temperature.

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  1. 3 September, 00:51
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    Kₐ = 9.9 * 10⁻³

    Explanation:

    HClO₂ + H₂O ⇌ H₃O⁺ + ClO₂⁻

    0.080 * 1.43 = 0.1144

    So, at equilibrium,

    [H₃O⁺] = [ClO₂⁻] = 0.1144

    [HClO₂] = 1.43 - 0.1144 = 1.316

    Kₐ = {[H₃O⁺][ClO₂⁻]}/[HClO₂]; Substitute values

    Kₐ = (0.1144 * 0.1144) / 1.316 Do the operations

    Kₐ = 9.9 * 10⁻³
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