Concentration Titration question:

Question

Potassium oxalate was added to a 25.00 mL solution containing an unknown amount of Th4+, which resulted in the formation of the precipitate Th(C2O4)2 The precipitate was washed and then dissolved in acid. The resulting oxalic acid was titrated with 44.79mL of 0.0521 M KMnO4

5H2C2O4+2MnO−4+6H+⟶2Mn2++10CO2+8H2O

Calculate the concentration of Th4+ in the unknown solution.

I tried to calculate and got 0.467M and that was labeled wrong and I'm not sure why.

Answer 1

Let's write the balanced redox equation so it can be more easily read:

5H₂C₂O₄ + 2MnO₄^- + 6H⁺ ==> 2Mn²⁺ + 10CO₂ + 8H₂O

moles MnO₄^- used = 44.79 ml x 1 L / 1000 ml x 0.0521 mol / L = 0.002334 mols MnO₄^-

moles H₂C₂O₄ present = 0.002334 mol MnO₄^- x 5 mol H₂C₂O₄ / 2 mol MnO₄^- = 0.005834 mol H₂C₂O₄

To find mols Th⁴⁺ that was present, we must look at the stoichiometry between the thorium (Th) and the oxalic acid: 2C₂O₄^2- + Th⁴⁺ ==>Th(C₂O₄)₂(s)

mols Th⁴⁺ present = 0.005834 mol C₂O₄^2- x 1 mol Th⁴⁺ / 2 mol C₂O₄^2- = 0.00292 mol Th⁴⁺

[Th⁴⁺] = 0.00292 mol / 0.025 L = 0.117 M

(be sure to check the math)