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.

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 .

Alpha particles have positive charge, beta particles have negative charge, and answer 4 just doesn’t make sense. thus, you are left with only one possible answer.

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