`KMnO_4` can be prepared from`K_2MnO_4` as per the reaction:
The reaction can go the completion by removing `OH^(ɵ)` ions by adding.

Options

A. `CO_2`
B. `SO_2`
C. `HCl`
D. `KOH`

Answer

A

Explanation

The `overset(ɵ)(O)H` can be removed by adding HCl or `CO_2` since `MnO_4^(ɵ)` is formed in the product, which will oxidise `HCl` or `Cl^(ɵ)` ion to `Cl_2` hence `HCl` will not react with `overset(ɵ)(O)H` ions Therefore `CO_2` is used to remove `overset(ɵ)(O)H` ions.