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What is the percent ionization of hypobromous acid in a 0.035 M HBrO solution? Ka =...

What is the percent ionization of hypobromous acid in a 0.035 M HBrO solution? Ka = 2.8 x 10-9

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Answer #1

[H+] = ka*C

         = 2.8*10^-9 *0.035

        = 98*10^-12

        = 9.9*10^-6 M

percent ionisation = [H+]*100/conc of acid

                                = 9.9*10^-6*100/0.035

                                = 0.028%

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