posted by susan .
Calculate the pH of an aqueous solution
of 11.5 mg of KOH dissolved in 56 mL of
mols KOH = grams/molar mass
Then (KOH) = mols KOH/L solution
Then (OH^-) = (KOH) since KOH is a strong electrolyte and dissociates completely.
I would then do pOH = -log(OH^-) and follow with pH + pOH = pKw = 14 and solve for pH.