Question again...

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

pH + pOH = pK_w = 14 @25degC
So, if pOH = 6.113, pH = 7.887
[H⁺] = 10^-pH = 10^-7.887 = 1.30x10^-8 M
[H⁺] [OH^-] = K_w = 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