Considering the following reaction, $3 MnO_4^{2-} + 2 H_2O \rightleftharpoons 2 MnO_4^{-} + MnO_2 + 4 OH^{-}$ The reagent that facilitates the removal of $OH^{-}$ ions to ensure the completion of the conversion of $K_2MnO_4$ to $KMnO_4$ is:

The conversion of manganate ion ($MnO_4^{2-}$) to permanganate ion ($MnO_4^-$) is a disproportionation reaction that takes place in a neutral or acidic solution . The given reaction represents an equilibrium where hydroxide ions ($OH^-$) are produced.

1. Le Chatelier's Principle: To drive the reaction to the right (completion), the concentration of the product $OH^-$ must be reduced. This requires an acidic reagent to neutralize the base.
2. Evaluating Reagents:

• KOH: Being a strong base, it increases $[OH^-]$, shifting the equilibrium to the left (reactants), thus preventing the conversion.
• HCl: While acidic, $HCl$ acts as a reducing agent. Permanganate oxidises $HCl$ to chlorine ($Cl_2$), consuming the desired product .
• SO₂: Sulphur dioxide is a strong reducing agent that would reduce the permanganate formed ($MnO_4^-$) to Manganese(II) ($Mn^{2+}$), destroying the product .
• CO₂: Carbon dioxide is a weakly acidic oxide. It reacts with $OH^-$ to form carbonate/bicarbonate, effectively removing $OH^-$ from the equilibrium and driving the reaction forward without acting as a reducing agent towards the permanganate.