Question

What volume of a 0.0225M solution of potassium permanganate are required to titrate 0.461g of a sample that is 70.6% sodium oxalate?

Answer 1

mass of sodium oxalate = 0.461 x 70.6 / 100 = 0.326 g

molar mass of sodium oxalate = 134 g/mol

moles of sodium oxalate = 0.326 / 134 = 2.43 x 10^-3

balanced equation:

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

2 moles MnO4- --------------------> 5 moles C2O4-2

x moles MnO4- ------------------> 2.43 x 10^-3 moles C2O4-2

x = 2 x 2.43 x 10^-3 / 5

x = 9.72 x 10^-4

volume = moles / molarity

            = 9.72 x 10^-4 / 0.0225

             = 0.0432 L

             = 43.2 mL

volume of potassium permanganate = 43.2 mL