Calculate the Ph of an aqueous solution that is 3.00% KOH, by mass, and has a density of 1.0242 g/ml.

Answer 1

Assume the mass of the solution is 1024.2 grams, or one liter.

Then three percent (by mass) is KOH. How many moles is that?

Assuming total dissociation, then

[OH}=MolarityKOH= moles/1.0000liter

Then [H]=14-[OH]

and pH= -log[H]

Answer 2

I think bobpursley meant that:
pOH = -log[OH-]
and
pH = 14 - pOH

Use these relationships to get the pH after finding the molarity of KOH = [OH-] as bobpursley suggested.