Calculate the percent ionization of a 0.394 M solution of hypochlorous acid. % Ionization =

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Expert Solution

the percent ionization of a 0.394 M solution of hypochlorous acid can be calculated using ICE table as

for HOCl ka = 3.5*10^-8

HOCl + H2O ==== H3O+ + OCl-

Initial--0.394----------------------0------------0

Change--- -x --------------------- +x ------ +x

Equil. 0,394-x -------------------- x -------- x

The acid dissociation constant Ka is givenby

Ka = [H3O+][OCl-]/[HOCl] = x^2/(0.394-x) = 3.5*10^-8

Solving for gives x = 1.117*10^-4

The [H3O+] = 1.117*10^-4 M

The percentage of dissociation is given by % diss. = (moles of acid dissociated/total no. of moles)*100

% diss. = (1.117*10^-4 /0.394)*100 = 0.029%

queen_honey_blossom answered 1 year ago

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