If the end point in the titration of oxalic acid with NaOH solution is super-passed (too pink), will the molar concentration of NaOH be hiegher or lower than the actual value ? Explain ..

Answer 1

H2C2O4 + 2NaOH ==> Na2C2O4 + 2H2O

eqn 1. moles H2C2O4 = grams/molar mass
eqn 2. moles NaOH = 2 x moles H2C2O4
eqn 3. M NaOH = moles NaOH/L NaOH

eqn 1 doesn't change.
eqn 2 doesn't change.
eqn 3 changes. If L NaOH is too high, then M NaOH is too low.