titration question

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these are the previous questions i dont need the answer but it might help in answering question C
Ethanedioic acid acts, initially, as a monoprotic acid.
(b) A buffer solution is made by adding 6.00 × 10–2 mol of sodium hydroxide to a solution containing 1.00 × 10–1 mol of ethanedioic acid (H2C2O4).
Assume that the sodium hydroxide reacts as shown in the following equation and that in this buffer solution, the ethanedioic acid behaves as a monoprotic acid.
H2C2O4(aq) + OH–(aq) HC2O4−(aq) + H2O(l)
The dissociation constant Ka for ethanedioic acid is 5.89 × 10–2 mol dm–3.
(c) In a titration, the end point was reached when 25.0 cm3 of an acidified solution containing ethanedioic acid reacted with 20.20 cm3 of 2.00 ×10–2 mol dm–3 potassium manganate(VII) solution.
Deduce an equation for the reaction that occurs and use it to calculate the original concentration of the ethanedioic acid solution.
the markschem answer for c is
5H2C2O4 + 6H+ + 2MnO4– 2Mn2+ + 10CO2 + 8H2O
OR 5C2O42– + 16H+ + 2MnO4– 2Mn2+ + 10CO2 + 8H2O
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Moles of KMnO4 = 20.2 × 2.00 × 10–2 / 1000 = 4.04 × 10–4
1
Moles of H2C2O4 = 5 / 2 × 4.04 × 10–4 = 1.01 × 10–3
1
Concentration = moles / volume (in dm3)
= 1.01 × 10–3 × 1000 / 25 = 4.04 × 10–2 (mol dm–3)