Question

$KMnO_{4}$ can be prepared from $K_{2}MnO_{4}$ as per the reaction,
$3MnO_{4}^{2 -}+2H_{2}O\rightleftharpoons2MnO_{4}^{-}+MnO_{2}+4OH^{-}$
The reaction can go to completion by removing $OH^{-}$ ions by adding

• A. $CO_{2}$
• B. $SO_{2}$
• C. HCl
• D. KOH

Answer 1

Answer: (A)

To remove OH- ion a weak acid $\left(\right.H_{2}CO_{3}\left.\right)$ is required because a strong acid can reverse the reaction. So here HCl is a strong acid so it reverse the reaction.
KOH is strong base which will increase OH- concentration.
$SO_{2}$ will react with water to form a strong acid $H_{2}SO_{4}$ .
$CO_{2}$ combines with $OH^{-}$ ion to give carbonate which is easily removed.