Redox and Equivalent ConceptMedium
Question
KMnO4 reacts with oxalic acid according to the equation :
2MnO4- + 5C2O4- + 16H+ → 2Mn+2 + 10CO2 + 8H2O
Here 20 ml of 0.1 M KMnO4 is equivalent to :
2MnO4- + 5C2O4- + 16H+ → 2Mn+2 + 10CO2 + 8H2O
Here 20 ml of 0.1 M KMnO4 is equivalent to :
Options
A.20 mL of 0.5 M H2C2O4
B.50 mL of 0.5 M H2C2O4
C.50 mL of 0.1 M H2C2O4
D.20 mL of 0.1M H2C2O4
Solution
Meq of A = Meq of B.
Meq of KMnO4 = 20×0.5 = 10
Meq of 50 ml of 0.1 M
H2C2O4 = 50× 0.2 = 10
(0.1 M H2C2O4 = 0.2 N H2C2O4)
Meq of KMnO4 = 20×0.5 = 10
Meq of 50 ml of 0.1 M
H2C2O4 = 50× 0.2 = 10
(0.1 M H2C2O4 = 0.2 N H2C2O4)
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