Posted by JOHN on Saturday, November 17, 2012 at 12:04am.
I would convert 1.94E-10 to pOH.
pOH = -log(OH^-), then
pH + pOH = pKw = 14.
You know pOH and pKw, solve for pH.
If you prefer you can do it this way(H^+)(OH^-) = Kw = 1E-14
You have OH^-, solve for (H^+), then
pH = -log(H^+)
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