Question

1.   A solution of 0.05M KMnO₄ was titrated with a solution of 0.268g Na₂C₂O₄ 100mL. What is the volume of KMnO₄ required to the end point?

2.   Calculate the % C₂O₄^2- in each of the following :-

a.   Na₂C₂O₄

b.   K₂[Cu (C₂O₄)₂].2H₂O

Answer 1

5C2O4^2- + 2MnO4^- + 16H⁺ ---------> 2Mn⁺² +10Co2 + 8H2O

no of moles of Na2C2O4 = W/G.M.Wt

                                          = 0.268/134 = 0.02moles

5 moles of Na2C2O4   = 2 moles of KMnO4

0.02 moles of Na2C2O4 = 2*0.02/5   = 0.008 moles of KMnO4

molarity of KMnO4   = no of moles /volume in L

0.05                           = 0.008/volume in L

volume in L             = 0.008/0.05 = 0.16 L = 160ml

2. % C₂O₄^2-          = 78*100/134   = 58.2%

% C₂O₄^2-           = 2*78*100/575.4   = 27.11%