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.

Explanation

At the first stage , `C_(2)O_(4)^(2-) ` is oxidised to `CO_(2)` by the oxidant `KMnO_(4)`
`{:(5C_(2)O_(4)^(2-)+2MnO_(4)^(-)+16H^(+)to,10CO_(2)+2Mn^(2+)+3H_(2)O),(" +7"," +2"):}`
` :. ` normality of `KMnO_(4)` solution = `0.02 xx (7-2) = 0.1 N `
At the second stage `Cu^(2+)` liberates `I_(2)` from KI and this liberated `I_(2)` requires
11.3 mL of `0.05 " M " Na_(2)S_(2)O_(3)` solution for complete reaction .
`{:(2Cu^(2+)+2I^(-)to,2Cu^(+)+I_(2)),(2S_(2)O_(3)^(2-)+I_(2) to ,S_(4)O_(6)^(2-)+2I^(-)),(" +4"," +5"):}`
Normality of `Na_(2)S_(2)O_(3)` solution ` = 0.05 (5-4) = 0.05 N `
Now , `(" m.e of "Cu^(2+))/( "m.e of " C_(2)O_(4)^(2-))= (" m.e of"Na_(2)S_(2)O_(3))/("m.e of"KMnO_(4)) = (0.05 xx 11.3 )/(0.1 xx 22.3) = 1/4 `
As `{:(Cu^(2+)to,Cu^(+)and,C_(2)O_(4)^(2-) to,2CO_(2)),(+2,+1,+6,+8):}`
` :. ("mmol of "Cu^(2+)xx1)/("mmol of"C_(2)O_(4)^(2-)xx2) =1/4`
or `("mole of "Cu^(2+))/(" mole of " C_(2)O_(4)^(2-)) = 1/2 `