$\mathrm{KMnO}_4$ can be prepared from $\mathrm{K}_2 \mathrm{MnO}_4$ as per reaction \(3…
$\mathrm{KMnO}_4$ can be prepared from $\mathrm{K}_2 \mathrm{MnO}_4$ as per reaction \(3 \mathrm{MnO}_2^{2-}+2 \mathrm{H}_2 \mathrm{O} \rightleftharpoons 2 \mathrm{MnO}_4^{-}+\mathrm{MnO}_2+4 \mathrm{OH}^{-}\). The reaction can go to completion by removing $\mathrm{OH}^{-}$ions by adding
$\mathrm{HCl}$
$\mathrm{KOH}$
$\mathrm{CO}_2$
$\mathrm{SO}_2$
Solution
Since, $\mathrm{OH}^{-}$are generated from weak acid $\left(\mathrm{H}_2 \mathrm{O}\right)$, a weak acid $\left(\right.$ like $\left.\mathrm{CO}_2\right)$ should be used to remove it because of strong acid $(\mathrm{HCl})$ reverse the reaction. $\mathrm{KOH}$ increases the concentration of $\mathrm{OH}^{-}$, thus again shifts the reaction in backward side.
$\mathrm{CO}_2$ combines with $\mathrm{OH}^{-}$to give carbonate which is easily removed.
$\mathrm{SO}_2$ reacts with water to give strong acid, so it cannot be used.