Question

In: Chemistry

1.   A solution of 0.05M KMnO4 was titrated with a solution of 0.268g Na2C2O4 100mL. What...

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 % C2O42- in each of the following :-

a.   Na2C2O4

b.   K2[Cu (C2O4)2].2H2O

Solutions

Expert Solution

5C2O42- + 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. % C2O42-          = 78*100/134   = 58.2%

% C2O42-           = 2*78*100/575.4   = 27.11%

  

queen_honey_blossom answered 2 years ago
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