A solution contains oxalic acid, H₂C₂O₄, and sodium hydrogen oxalate, NaHC₂O₄. A 25.0 mL portion needs 20.0 mL of 0.100 mol L⁻¹ NaOH to reach the phenolphthalein end point. A second 25.0 mL portion, in acid, needs 16.0 mL of 0.0400 mol L⁻¹ KMnO₄, where 2 MnO₄⁻ + 5 C₂O₄²⁻ + 16 H⁺ → 2 Mn²⁺ + 10 CO₂ + 8 H₂O. What is the molarity of NaHC₂O₄ in the solution?

ChemistryAdvancedHigh School (11–12)Atoms and Molecules › Stoichiometry and Concentration Olympiad

A solution contains oxalic acid, $\text{H}_2\text{C}_2\text{O}_4$ , and sodium hydrogen oxalate, $\text{NaHC}_2\text{O}_4$ . A $25.0\ \text{mL}$ portion needs $20.0\ \text{mL}$ of $0.100\ \text{mol L}^{-1}$ NaOH to reach the phenolphthalein end point. A second $25.0\ \text{mL}$ portion, in acid, needs $16.0\ \text{mL}$ of $0.0400\ \text{mol L}^{-1}$ $\text{KMnO}_4$ , where $2\,\text{MnO}_4^- + 5\,\text{C}_2\text{O}_4^{2-} + 16\,\text{H}^+ \rightarrow 2\,\text{Mn}^{2+} + 10\,\text{CO}_2 + 8\,\text{H}_2\text{O}$ . What is the molarity of $\text{NaHC}_2\text{O}_4$ in the solution?

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