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

Options

A. 11.4 mL
B. 12 mL
C. 33.3 mL
D. 35 mL

Answer

C

Explanation

`MnO_(4)^(-)+5Fe^(2+)+8H^(+) to Mn^(2+)+5Fe^(3+)+4H_(2)O`
`(M_(1)V_(1))/(1)=(M_(2)V_(2))/(5)`
`(M_(1)xx20)/(1)=(M_(2)xx25)/(5)`
`M_(1)=(M_(2))/(4)`
Neutral medium
`MnO_(4)^(-)+3Fe^(2+) +4H^(+) to MnO_(2)+3Fe^(3+)+2H_(2)O`
`(M_(1)V_(1))/(1)=(M_(2)V_(2))/(3)`
`((M_(2))/(4))xxV_(1)=(M_(2)xx25)/(3)`
`V_(1)=33.3 mL`