In a titration of 42.0 mL of a 0.500 M solution of a diprotic acid H₂C₃H₂O₄ (malonic acid) with 0.255 M NaOH, how many grams of NaOH are required to reach the second equivalence point?

Question

In a titration of 42.0 mL of a 0.500 M solution of a diprotic acid H₂C₃H₂O₄ (malonic acid) with 0.255 M NaOH, how many grams of NaOH are required to reach the second equivalence point?

Answer 1

H₂C₃H₂O₄ + 2NaOH ==> Na₂C₃H₂O₄ + 2H₂O ... balanced equation

molar mass H₂C₃H₂O₄ = 104.06 g / mol

moles H₂C₃H₂O₄ present = 42.0 ml x 1 L / 1000 ml x 0.500 mol/L = 0.021 mols

moles NaOH needed = 0.021 mols H₂C₃H₂O₄ x 2 mol NaOH / mol H₂C₃H₂O₄ = 0.042 mols NaOH

molar mass NaOH = 40.0 g / mol

Grams NaOH needed = 0.042 mols NaOH x 40.0 g / mol = 1.68 g NaOH to reach 2nd equivalence

In case the question meant to ask for the volume of 0.255 M NaOH, we would solve that as follows:

0.042 mols NaOH x 1 L / 0.255 mols = 0.165 L = 165 mls