Titration problem

Question

If 27.21 ml of NaOH are required to titrate 50 ml of a solution that contains 1.2067g of KHP what is the molarity of the NaOH? You'll need to look up the molecular weight of KHP. Do not include units with your answer. You answer should have four significant figures.

Answer 1

I'll assume KHP is potassium hydrogen phthalate, which I'll write as HA (a weak monoprotic acid)

HA + NaOH ==> NaA + H2O

molar mass KHP = 204.23 g/mol

moles KHP present = 1.2067 g x 1mol/204.23 g = 0.0059085 moles KHP present

moles NaOH used = 0.0059085 moles since they react in a 1:1 mol ratio (see balanced equation)

molarity of NaOH = moles NaOH/Liters of solution = 0.0059085 moles/0.02721 L = 0.2171 M