`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. KOH
B. `CO_(2)`
C. `SO_(2)`
D. HCI

Answer

B

Explanation

We can't use KOH as it increases the `OH^(-)` ions. To remove `OH^(-)` ions, we need an acidic substance which is not oxidized by `KMnO_(4)`. Both HCI and `SO_(2)` are oxidized by `KMnO_(4)` but `CO_(2)` (having C in the maximum oxidation state, +4) can not be oxidized by `KMnO_(4)`. Thus, we should use `CO_(2)` .