2009/07/22

Question again...

In exam #3 question #1 how do you find the H+ or OH- from pOH of 6.113?

pH + pOH = pKw = 14 @25degC
So, if pOH = 6.113, pH = 7.887
[H+] = 10-pH = 10-7.887 = 1.30x10-8 M
[H+] [OH-] = Kw = 10-14 @25degC so
(1.30x10-8)[OH-] = 10-14
[OH-] = 7.71x10-7 M
As a check,
pOH = -log[OH-] = -log(7.71x10-7) = 6.113

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