Equivalent weight of `MnO_(4)^(-)` in acidic, neutral and basic media are in ratio of:

Options

A. `3:5:15`
B. `5:3:1`
C. `5:1:3`
D. `3:5:5`

Answer

D

Explanation

`KMnO_(4)` can act as an O.A in acidic, basic and neutral mediums as follows :
In acidic medium,
`MnO_(4)^(-)+8H^(+)+5e^(-)rarrMn^(2+)+4H_(2)O`
In basic medium
`MnO_(4)^(-)+2H_(2)O+3Fe^(-)rarrMnO_(2)+4OH^(-)`
In neutral medium
`MnO_(4)^(-)+2H_(2)O+3Fe^(-)rarrMnO_(2)+4OH^(-)`
Let molecular mass of `KMnO_(4)=M`
`:.` Ratio of Eq mass of `KMnO_(4)` in acidic, basic and neutral mediums are `(M)/(5):(M)/(3):(M)/(3)or3:5:5`.