Question

What volume of 0.300 M KMnO4 would be required to titrate 0.41 g of K2[Cu(C2O4)2]⋅2H2O?

Answer 1

5C₂O₄^2- + 2MnO₄^- + 16 H⁺ -----> 10CO₂ + 2Mn²⁺ + 8H₂O

Moles of K₂[Cu(C₂O₄)₂]2H₂O = Mass/ Molar mass = 0.41g/ (353.81 g/mol) = 1.15 x 10^-3 moles

Moles of C₂O₄^2- = 2 x Moles of K₂[Cu(C₂O₄)₂]2H₂O = 2 x (1.15 x 10^-3 moles) = 2.31 x 10^-3 moles

Moles of MnO₄^- = (2/5) x Moles of C₂O₄^2-= (2/5) x (2.31 x 10^-3 moles) = 9.27 x 10^-4 moles

Volume of KMnO₄= Moles of KMnO₄/Molarity = (9.27 x 10^-4 moles)/0.3 M = 0.00309 L = 3.09 mL

volume of 0.300 M KMnO₄ required = 3.09 mL