$KMnO_4$ can be prepared from $K_2MnO_4$ as per the reaction : <br> $3MnO_4^{2-} + 2H_2O \rightleftharpoons 2MnO_4^- + MnO_2 + 4 OH^-$. <br> The reaction can go to completion by removing $OH^-$ ions by adding -

Question

( A ) KOH
( B ) $SO_2$
( C ) HCl
( D ) $CO_2$