`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

Redox equation for oxidation of `Fe^(2+)` ions in acidic and neutral media respectviely are
`MnO_(4)^(-)+8H^(+)+5Fe^(2+)rarrMn^(2+)+5Fe^(3+)+4H_(2)O`
`MnO_(4)^(-)+2H_(2)O+3Fe^(2)rarrMnbO_(2)+3Fe^(3+)+4OH`
since under acidic condition 1 mole of `MnO_(4)^(-)` oxidises 5 moles of `Fe^(2+)` ions but only 3 moles under neutral conditions therefore volume of `MnO_(4)^(-)` used oxidation of 25 ml of same `Fe^(2+)` solution under neutral condition
`(20xx5)/(3)=33.3mL`