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A solution of 0.2 g of a compound containing Cu^(2+) "and" C_(2)O_(4)^(2-) ions on titration with 0.02 M KMnO_(4) in presence of H_(2)SO_(4) consumes 22.6 mL of the oxidant. The resultant solution is neutralized with Na_(2)CO_(3) acidified with dil. acetic acid and treated with excess KI. The liberated iodine requires 11.3 mL of 0.5 M Na_(2)S_(2)O_(3) solution for complete reduction. Find out the mole ratio of Cu^(2+) "to" C_(2)O_(4)^(2-)in the compound. Write down the balanced redox reactions involved in the above titrations. |
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Answer» Solution :Let 'a' moles of `Cu^(2+)` and 'b' moles of `C_(2)O_(4)^(2-)` be PRESENT in solution (i) The solution is oxidised by `KMnO_(4)` with only `C_(2)O_(4)^(2-)` `Mn^(7+) + 5 E^(-) rarr Mn^(2+)` `C_(2)^(6+) rarr 2C^(4+) + 2e^(-)` Meq. of `C_(2)O_(4)^(2-)` = Meq. of `KMnO_(4)` `b XX 2 xx 1000 = 0.02 xx 5 xx 2.26 ` `b = 1.13 xx 10^(-3)` (ii) After oxidation of `C_(2)O_(4)^(2-)` the resulting solution is neutralized by `Na_(2)CO_(3)` acidified with dilute `CH_(3)COOH`and then treated with excess of KI. The liberated `I_(2)` REQUIRED `Na_(2)S_(2)O_(3)` for itsneutralization, i.e. `Cu^(2+) overset("KI")to I_(2) overset(Na_(2)S_(2)O_(3))to Na_(2)S_(4)O_(6) + I^(-)` Meq. of `Cu^(2+)` = Meq. of `I_(2)` liberated = Meq. of `Na_(2)S_(2)O_(3)` used Meq. of `Cu^(2+)` = Meq. of `Na_(2)S_(2)O_(3)` used `a xx 1 xx 1000 = 11.3 xx 0.5 xx 1` `a = 5.65 xx 10^(-3)` Molar ratio of `Cu^(2+)//C_(2)O_(4)^(2-) = a/bimplies 5.65 xx 10^(-3)/(1.13 xx 10^(-3)) = 5/1 ` Hence, molar ratio of `Cu^(2+) , C_(2)O_(4)^(2-) = 5 : 1` Involved reactions are : `2 MnO_(4)^(-) + 5 C_(2)O_(4)^(2-) + 16H^(+) rarr 2MN^(2+) + 10 CO_(2) + 8H_(2)O` `2Cu^(2+) + 4I^(-) rarr Cu_(2)I_(2) + I_(2)` `I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) + S_(4)O_(6)^(2-)` |
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