Question

A solution of NaOH needed to be standardized before use in other titrations.   If 0.356 g of KHP (MW = 204.23 g/mol ) required 30.89 mL of the base to reach the equivalence point, what is the concentration of NaOH? Write a balanced equation to verify titrant: analyte ratio.

Answer 1

moles = mass / molar mass

so

moles of KHP = 0.356 g / (204.23 g/mol) = 1.743 x 10-3 mol

now

KHP + NaOH ==> KNaP + H2O

so

moles of NaOH required = 1.743 x 10-3 mol KHP x (1 mol NaOH / 1 mol KHP)

moles of NaOH required = 1.743 x 10-3 mol NaOH

now

concentration = moles of NaOH / volume of base required (L)

concentration of NaOH = 1.743 x 10-3 mol / 0.03089 L

concentration of NaOH = 0.05643 mol / L

concentration of NaOH = 0.05643 M