A solution contains `Cu^(2+)` and `C_(2)O_(4)^(-)` ions which on titration with `1M KMnO_(4)` consumes 10 ml of the oxidizing agent for complete oxidation in acidic medium. The resulting solution is neutralized with `Na_(2) CO_(3)`, acidified with dilute `CH_(3)COOH` and is treated with excess of Kl. The liberated iodine requires 25 ml of 1 M of hypo solution, then what will be the molar ratio of `Cu^(2+)` to `C_(2)O_(4)^(2-)` ions in the solution?

Options

A. `5:2`
B. `1:2`
C. `2:1`
D. `1:1`

Answer

D

Explanation

Reactions are
`C_(2)O_(4)^(2-)to2CO_(2)+2e^(2-)`
`MnO_(4)^(-)+8H^(+)to5etoMn^(2+)+4H_(2)O`
`2Cu^(2+)+4KIto4K^(+)+Cu_(2)I_(2)+I_(2)`
`I_(2)+2Na_(2)S_(2)O_(3)toNa_(2)S_(4)O_(6)+2Nal`
mmol of `Na_(2)S_(2)O_(3)=25xx1` mmol of `Cu^(2+)` ions
mmol of `KMnO_(4)=10` mmol =50 meqs =meqs of `C_(2)O_(4)^(2-)`
`:.` mmol of `C_(2)O_(4)^(2-)=(50)/(2)=25` mmol
`("mmol of "Cu^(2+)ions)/("mmol of "C_(2)O_(4)^(2-)ions)=(25"mmol")/(25"mmol")=1:1`