# College Chemistry

what is the approximate pH of a solution labeled 0.050 M HClO? I realized my mistake on my last post but this isnt a strong acid nor a strong base so how do i solve this?

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1. HClO ==> H^+ + ClO^-

Set up ICE chart, substitute into Ka expression, and solve for (H^+). Convert to pH.

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2. i still don't get it.
(H+)=x/0.050M

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3. Absolutely not. Where's the ICE chart?
HClO ==> H^+ + ClO^-

initial:
HClO = 0.05 M
H^+ = 0
ClO = 0

change:
HClO = -x
H^+ = +x
ClO^- = +x

equilibrium:
HClO = 0.05-x
H^+ = x
ClO^- = x

Ka = (H^+)(ClO^-)/(HClO)
Ka = 3.0 x 10^-8 in my book. Yours may be different.
(x)^2/(0.05 = 3.0 x 10^-8
x^2 = 1.5 x 10^-9
x = 3.87 x 10^-5 for pH = 4.41

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