` 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. 12.0 mL
C. 33.3 mL
D. 35.0 mL

Answer

C

Explanation

in acidic medium
5 vol of `Fe^(2+)` requires 1 vol of `MnO_(4)^(-) ` in acidic medium
` therefore 25 Vol ` of ` Fe^(2+)` requires ` 1/5 xx 25 vol ` of ` MnO_4^(- ) ` in acidic medium
`1/5 xx 25 vol ` of `MnO_4 ^(- ) ~= 20 vol or 20 mL`
In neutral medium
3 mol of ` Fe^(2+) ` requires 1 vol of `MnO_(4)^(-)` in neutral medium then 25 vol of ` Fe^(2+)` requires `1/3 xx 25 vol` of ` MnO_4` in neutral medium
`1/5 xx 25 vol ` of `MnO_(4)^(- ) ~= 20 vol`
`therefore 1/3 x 25 vol ` of ` MnO_4^(-) =(20) /(5) xx (25)/( 3 ) vol` of `MnO_(4)^(-)`
`= 33.3 vol or mL`