# chem 12

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Calculate [H3O] an [OH] for a sodium hydroxide solution having a pH of 12.5.

I know this is a strong base, and i got the pH 12.5. But i don't know how to go from there.

• chem 12 -

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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