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4 April, 12:04

Calculate the ph of a 0.10 m solution of hypochlorous acid, hocl. ka of hocl is 3.5Ã10â8 at 25 âc. express your answer numerically using two decimal places.

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  1. 4 April, 14:49
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    we know that: Ka = [H+][OCl-] / [HOCl] and HOCl dissociates into equal number of ions H + = OCl - We substitute [OCl-] for [H+} Hence Ka = [H+][H+] / {HOCl] Ka = [H+]^2 / [HOCl] [H+]^2 = Ka X [HOCl] [H+] = sqrt{Ka X [HOCl]} [H+] = sqrt{ 3.5 x 10^-8 x 0.10} [H+] = sqrt[ 3.5 x 10^-9 [H+] = 5.916 x 10^-5 you know: pH = - log (10) [H+] pH = - log (10) 5.916 x 10^-5 pH = - (-4.227965) pH = 4.227965 pH = 4.23.
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