A solution of `0.2 g` of a compound containing `Cu^(2+)` and `C_(2)O_(4)^(2-)` ions on titration with `0.02M KMnO_(4)` in presence of `H_(2)SO_(4)` consumes `22.6mL` oxidant. The resulting solution is neutralized by `Na_(2)CO_(3)`, acidified with dilute `CH_(3)COOH` and titrated with excess of `KI`. The liberated `I_(2)` required `11.3 mL "of" 0.05M Na_(2)S_(2)O_(3)` for complete reduction. Find out mole ratio of `Cu^(2+)` and `C_(2)O_(4)^(2+)` in compound.

Answer

`(1:2)`

Explanation

With `KMnO_(4)`, oxalate ion is oxidised only as:
`5C_(2)O_(4)^(2-) + 2MnO_(4)^(-) + 16H^(+) rarr 2Mn^(2+) +10 CO_(2)+8H_(2)O`
Let in the given mass of compounf, x millimol of `C_(2)O_(4)^(2-)` ion is present, then
Meq of `C_(2)O_(4)^(2-) = "Meq of" MnO_(4)^(-)`
`rArr 2x = 0.02 xx 5 xx 22.6 rArr x = 1.13`
At the later stage, with `I^(-), Cu^(2+)` is reduced as :
`2CU^(2+) + 4I^(-) rarr 2CuI + I_(2)`
and `I_(2) + 2S_(2)O_(3)^(2-) rarr 2I^(-) +S_(4)O_(6)^(2-)`
Let there be x millimol of `Cu^(2+)`.
`rArr` Meq of `Cu^(2+)` = Meq of `I_(2)` = meq of hypo
`rArr x = 11.3 xx 0.05 = 0.565`
`rArr` Moles of `Cu^(2+)` : moles of `C_(2)O_(4)^(2-) = 0.565 : 1.13 = 1:2`