chemistry
posted by Miley .
give the [OH] of a vinegar solution that has a pH of 3.2. i got 6.3 *10^4. is that right?

I don't get your number. I think you did this.
pH = log(H^+)
3.2 = log(H^+)
3.2 = log(H^+)
(H^+) = 5.86 x 10^4
The problem asks for OH, so
(H^+)(OH^) = Kw.
Kw = 1 x 10^14
You know (H^+). Calculate (OH^)
A far easier way to do it is this.
pH = 3.2; therefore, pOH = 14  3.2 = 10.8.
Then pOH = 10.8 = log)OH^)
10.8 = log(OH^)
(OH^) = ??