Question

A solution of NaOH is standardized with potassium acid phthalate (KHP), KHC₈H₈O₄, molar mass = 204 g/mol. If 2.550 g of KHP is titrated with 54.50 mL of the NaOH solution, what is the molarity of NaOH?

Answer 1

Given : mass KHC₈H₈O₄ = 2.550 g

moles KHC₈H₈O₄ = (mass KHC₈H₈O₄) / (molar mass KHC₈H₈O₄)

moles KHC₈H₈O₄ = (2.550 g) / (204 g/mol)

moles KHC₈H₈O₄ = 0.0125 mol

moles NaOH consumed = moles KHC₈H₈O₄ present

moles NaOH consumed = 0.0125 mol

molarity NaOH = (moles NaOH consumed) / (volume NaOH in Liter)

molarity NaOH = (0.0125 mol) / (54.50 x 10^-3 L)

molarity NaOH = 0.2294 M