`MnO_(4)^(-)` ions are reduced in acidic condition to `Mn^(2+)` ions whereas they are reduced in neutral condition to `MnO_(2)`. The oxidataion of 25 mL of a solution X containing `Fe^(2+)` ions required in acidic medium 20 mL of a solution Y containing `MnO_(4)^(-)` ions. What volume of solution Y would be required to oxidise 25 mL of solution Y containing `Fe^(2+)` ions in neutral condition ?

Options

A. 11.4 mL
B. 12.0 mL
C. 33.3 mL
D. 25.0 mL

Answer

C

Explanation

In acidic medium,
`overset(+7)MnO_(4)^(-) to overset(+2)Mn^(2+)`
Change in oxidation number = 5
`underset(Fe^(2+))underbrace(N_(1)V_(1))= underset(MnO_(4)^(-))underbrace(N_(2)V_(2))`
`N xx 25 = 5 M xx 20 " " …(i)`
25 N = 100 M
IN neutral medium,
`overset(+7)MnO_(4)^(-) to overset(+4)MnO_(2)`
Change in oxidation number = 3
`underset(Fe^(2+))underbrace(N_(1)V_(1))= underset(MnO_(4)^(-))underbrace(N_(2)V_(2))`
`25 xxN = 3 M xx V`
25N = 3MV
Equating (i) and (ii)
100 M = 3 MV
or `V = (100)/(3) = 33.3` mL