`KMnO_(4)` can be prepared from `K_(2)MnO_(4)` as per the reaction
`3MnO_(4)^(2-) + 2H_(2)O hArr 2MnO_(4)^(2-) + MnO_2 + 4OH^-`
The reaction can go to completion by removing `OH^(-)` ions by adding.

Options

A. HCl
B. KOH
C. `CO_(2)`
D. `SO_(2)`

Answer

C

Explanation

`3MnO_(4)^(2-) + 2H_(2)O hArr 2MnO_(4)^(-) + MnO_(2) + 4OH^(-)`
The reaction can goto completion by removing `OH^(-)` ions by adding `CO_(2)` as it reacts with `OH^(-)` ion to form `CO_(3)^(2-)` ions
`4OH^(-) + 2CO_(2) rarr 2CO_(3)^(2-) + 2H_(2)O`