`MnO_(4)^(-)` ions are reduced in acidic conditions 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 value of solution y would be required to oxidize 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

`overset(+7)(MnO_(4)^(-))+5e^(-)overset(H+)(rarr)overset(+2)(Mn^(2+))`
`overset(+7)(MnO_(4)^(-))+3e^(-)overset(("neutral"))(rarr)overset(+4)(MnO_(2))`
`:.` 20 mL of `MnO_(4)^(-)(H^(+))`
`-=(20xx5)/(3)"mL of " MnO_(4)^(-)` (neural)
`:. y=33.33 mL`