1.

A solution containing Cu^(2+) and C_(2)O_(4)^(2-) ions is titrated with 20 " mL of " (M)/(4) KMnO_(4) solution in acidic medium. The resulting solution is treated with excess of KI after neutralisation. The evolved I_(2) is then absorbed is 25 " mL of " (M)/(10) hypo solution. Which of the following statements are correct?

Answer»

The difference of the number of m " mol of "`Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the SOLUTION is 10 m mol
The difference of the number of m " mol of "`Cu^(2+)` and `C_(2)O_(4)^(2-)` ions in the solution is 22.5 m mol.
The equivalent weight of `Cu^(2+)` ions in the titration with KI is equal to the atomic weight of `Cu^(2+)`
The equivalent weight of KI in the titratio is `(M)/(2)` `(M=` molesular weight of KI)

Solution :(i). `Cu^(2+)` does not react with `MnO_(4)^(ɵ)`
Only `C_(2)O_(4)^(2-)` reacts with `MnO_(4)^(2-)`
`MnO_(4)^(ɵ)=C_(2)O_(4)^(2-)`
`(n=5)(n=2)`
`mEq-=mEq`
`20xx(M)/(4)xx5-=mEq`
`thereforem" Eq of "C_(2)O_(4)^(2-)=25`
`mmoles of `C_(2)O_(4)^(2-)=(25)/(2)=12.5`
(II). `Cu^(2+)=KI-=I_(2)=S_(2)O_(3)^(2-)` (hypo)
`mEq-=mEq-=mEq-=mEq`
`(2Cu^(2+)+2I^(ɵ)toCu_(2)I_(2))(2S_(2)O_(3)^(2-)to_S_(4)O_(6)^(2-)+2e^(-))`
`(e+Cu^(2+)toCu^(1+))`
`m" Eq of "S_(2)O_(3)^(2-)-=25xx(M)/(10)xx1`
`=2.5m" Eq of "Cu^(2+)`
`m" Eq of "Cu^(2+)=2.5`
`mmoles of Cu^(2+)=(2.5)/(1)=2.5`
Difference in mmoles of `C_(2)O_(4)^(2-)` and `Cu^(2+)=12.5-2.5=10`
EW of `Cu^(2+)=("atomic weight" of Cu^(2+))/(1("n-factor"=1))`
Ew of `KI=(M)/("nfactor")=(M)/(1)=M{:(2I^(ɵ)toI_(2)+2e^(-)),(n=(2)/(2)=1):}`


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