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

  • A
    $CO_2$
  • B
    $SO_2$
  • C
    $HCl$
  • D
    $KOH$

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