KMnO_4​ can be prepared from K₂​MnO₄​ as per the reaction:

$3Mn{O}_4^{2-}+2H_{2}O\rightleftharpoons 2Mn{O}_4^{-}+Mn{O}_2+4O{H}^{-}$
The reaction can go to completion by removing OH⁻ ions by adding:

$3{\mathrm{MnO}}_4^{2-}+2{\mathrm{H}}_2\mathrm{O}\rightleftharpoons 2{\mathrm{MnO}}_4^{-}+{\mathrm{MnO}}_2+4^{-}\mathrm{OH}$
Removal of ^–OH is favourable for the formation of KMnO₄ , i.e., forward reaction. To remove ^–OH ion, a weak acid is required because a strong acid reverses the reaction. So here, among all, CO₂ is the weak acid that combines with ^–OH ion to give carbonate, which is easily removed.
(2) $\text{ (O) }+{\mathrm{SO}}_2+{\mathrm{H}}_2\mathrm{O}\to {\mathrm{H}}_2{\mathrm{SO}}_4$
${\mathrm{H}}_2{\mathrm{SO}}_4$  is a strong acid, which will reverse equilibrium reaction.
(3) HCl is a strong acid, which favors equilibrium in backward reaction.