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

1st case : Only `C_(2)O_(4)^(2-)` ions are oxidised by `KMnO_(4)` solution.
Normality of `KMnO_(4)` solution `=0.02xx5=0.1 N`
22.6 mL of 0.1 N `KMnO_(4)=22.6 mL " of" 0.1 N C_(2)O_(4)^(2-)` soln.
`underset("in the solution")("Mass of" C_(2)O_(4)^(2-))` ions `=(N xx E xx V)/(1000)=(N xx M xx V)/(1000xx2)`
No. of moles of `C_(2)O_(4)^(2-)` ions in the solution`=(N xx M xxV)/(1000xx2xxM)`
`=(N xx V)/(2000)`
`=(0.1xx22.6)/(2000)`
`=11.3xx10^(-4)`
2nd case : Only `Cu^(2+)` ions are reduced by KI and iodine liberated is neutralised by `Na_(2)S_(2)O_(3)` solution.
11.3 mL of `0.05 M Na_(2)S_(2)O_(3)-=11.3 mL of 0.05 N Na_(2)S_(2)O_(3)`
`=11.3 mL of 0.05 N I_(2)`
`=11.3 mL " of " 0.05 N Cu^(2+)`
Mass of `Cu^(2+)` ions in the solution`=(NxxExxV)/(1000)=(NxxMxxV)/(1000)`
No. of moles of `Cu^(2+)` ions in the solution `=(N xx M xx V)/(1000xxM)`
`=(N xx V)/(1000)`
`=(0.05xx11.3)/(1000)`
`=5.65xx10^(-4)`
Molar ratio of `(Cu^(2+))/(C_(2)O_(4)^(2-))=(5.65xx10^(-4))/(11.3xx10^(-4))=(1)/(2)`