Question

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

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

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Answer #1

5C2O42- + 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 C2O42- = 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 C2O42-= (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

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