What is the percent ionization of 0.025 M chlorous acid, HClO2 solution?

Ka HClO2 = 1.1*10^(-2)

Answer: 48%

What I did:

HClO2(aq) + H2O(l) -><-
ClO2(aq) + H3O(aq)

initial-change-end table results:

x^(2)/(.025-x)=1.1*10^(-2) x=1.6*10^(-2)

% ionization = (1.1*10^(-2))/(1.6*10^(-2))*100 = 69%

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  1. Two problems with what you did.
    1. %ion = [(H3O^+)/0.025]*100 = ??
    2. Your equation is set up correctly; i.e., x^2/(0.025-x) = Ka.
    However, with Ka so large and (HClO2) so small, you MUST solve the quadratic equation. That is, you may not assume 0.025-x = 0.025.
    3. I get 47.9% which rounds to 48% if you solve the quadratic and substitute into #1 above correctly.

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  2. thank you; I was pretty sure I couldn't assume x negligible for Ka but I thought that the test for negligibility was the same as the test for % ionization.

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