Posted by Sarah on Tuesday, October 11, 2011 at 2:16pm.
pH = -log(H^+)
12.5 = 0log.....
-12.5 = log(H^+).
Take antilog both sides to arrive at (H^+). Then
(H^+)(OH^-) = Kw = 1E-14.
You know (H^+), solve for (OH^-).
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