Posted by S on Tuesday, February 21, 2012 at 9:23pm.
pH + pOH = pKw = 14. Solve for pOH, then pOH = -log(OH^-).
Solve for OH^-
OR you can use
pH = -log(H^+) and solve for )H^+), then
(H^+)(OH^-) = Kw = 1E-14
and solve for OH^-
Is this the correct way to set it up & solve for the problem?
[OH-] = 10^-11.63
[OH-] = 2.34E-12
I wasn't sure if it was the same set up as it is for the hydronium ion.
No. The pH is 11.63, your set up calculates (H^+). I would do
14-11.63 = 2.37 = pOH, then
(OH^-) = 10^-2.37 = 4.27E-3.
Okay, thanks so much for your help!! :)
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