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.
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^-).