KHP titration help

Question

answer the following questions based on the reaction below: NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

A 1.864 g sample of impure KHP was titrated with a 0.0978 M solution of NaOH. To completely react the KHP in the sample, 72.14 mL of base was needed. KHP (potassium hydrogen phthalate, 204.23 g/mol)

a.) How many grams of KHP were in the unknown sample?

b.) What is the percentage of KHP in the unknown sample?

Answer 1

NaOH(aq) + KHP(s) --> NaKP(aq)+H2O(I)

moles NaOH needed = 72.14 ml x 1 L/1000 ml x 0.0978 mol/L = 0.007055 moles NaOH

a). grams KHP present = 0.007055 mol NaOH x 1 mol KHP/mol NaOH x 204.23 g/mol = 1.44 g KHP

b). % KHP = 1.44 g/1.864 g (x100%) = 77.3% KHP