A chemist mixes HClO with water. The concentration is originally 0.24 M HClO, but after reacting...?

A chemist mixes HClO with water. The concentration is originally 0.24 M HClO, but after reacting with the water, it is something less than 0.24 M. The Ka = 1.1 x 10–8. Find the [H+].

Answer:
The equilibrium is :
HClO + H2O <> H3O+ + ClO-
initial concentration
0.24 M
at equilibrium
0.24-x . . . . . . . . . . .x . . . . . .x
Ka = 1.1 x 10^-8 = (x)(x) / 0.24 - x

x = 5.14 x 10^-5 M = [H3O+] or [H+]
Ka= ((H+)(ClO-))/(HClO)


for every H+ there is a ClO- so the top is just x^2

1.1x10-8= x^2/0.24

x= 0.000051381, thus [H+] is 0.000051381 as is the [ClO-]

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