`KMnO_4` can be prepared from `K_2MnO_4` as per the reaction,
`3MnO_(4)^(2-)+2H_(2)O Leftrightarrow 2MnO_(4)^(-)+MnO_(2)+4OH^(-)`
The reaction can go to completion by removing `OH^-` ions by adding

Options

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

Answer

A

Explanation

HCl and `SO_2` are reducing agents, So they can reduce `MnO_4^-`.
`CO_2` is neither oxidising nor reducing agent, it will provide only acidic medium. The `OH^-` will be reduced in acidic medium as it will react with `H^+` ions. Thus, it can shift the reaction in forward direction and the reaction can go to completion.