PYQ NEET- Chemical Equilibrium-1 L-4

Question: $\mathrm{KMnO}_4$ can be prepared from $\mathrm{K}_2 \mathrm{MnO}_4$ as per reaction,#

$$ \begin{alignedat} 3 \mathrm{MnO}_4^{2-}+2 \mathrm{H}_2 \mathrm{O} & \rightleftharpoons \mathrm{MnO}_4^- + 4 \mathrm{OH}^- 2 \mathrm{MnO}_4^{-} \rightarrow \mathrm{MnO}_2+4 \mathrm{OH}^{-} \end{aligned} $$

The reaction can go to completion by removing $\mathrm{OH}^{-}$ ions by adding

A) $\mathrm{HCl}$

B) $\mathrm{KOH}$

C) $\mathrm{CO}_2$

D) $\mathrm{SO}_2$

Answer: $\mathrm{CO}_2$#

Solution:#

Since, $\mathrm{OH}^{-}$ are generated from weak base $\left(\mathrm{H}_2 \mathrm{O}\right)$, and a weak base (like $\mathrm{CO}_3^{2-}$ ) should be used to remove it. Because if we add strong acid like $\mathrm{HCl}$ it reverses 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.