$$KMnO_{4}$$ can be prepared from $$K_{2}MnO_{4}$$ as per the reaction $$3MnO_{4}^{2-} + 2H_{2}O \rightleftharpoons 2MnO_{4}^{-} + MnO_{2} + 4OH^{-}$$ The reaction can go to completion by removing $${OH}^{-}$$ ions by adding :

A

$$SO_{2}$$

B

$$HCl$$

C

$$KOH$$

D

$$CO_{2}$$