Posted by Miley on Thursday, May 15, 2008 at 3:15pm.
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^-) = ??
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