KMnO4 can be prepared from K2MnO4 as per the reaction, 3MnO42 -+2H2Oleftharpoons2MnO4-+MnO2+4OH- The reaction can go to completion by removing OH- ions by adding

Question

KMnO4 can be prepared from K2MnO4 as per the reaction, 3MnO42 -+2H2Oleftharpoons2MnO4-+MnO2+4OH- The reaction can go to completion by removing OH- ions by adding (A) CO2 (B) SO2 (C) HCl (D) KOH

Tardigrade Admin · Accepted Answer

To remove OH- ion a weak acid (.H2CO3.) 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. SO2 will react with water to form a strong acid H2SO4 . CO2 combines with OH- ion to give carbonate which is easily removed.

Q. $K M n O_{4}$ can be prepared from $K_{2} M n O_{4}$ as per the reaction,
$3 M n O_{4}^{2 -} + 2 H_{2} O ⇌ 2 M n O_{4}^{-} + M n O_{2} + 4 O H^{-}$
The reaction can go to completion by removing $O H^{-}$ ions by adding

$C O_{2}$

$S O_{2}$

HCl

KOH

Q. KMnO4​ can be prepared from K2​MnO4​ as per the reaction, 3MnO42−​+2H2​O⇌2MnO4−​+MnO2​+4OH− The reaction can go to completion by removing OH− ions by adding

CO2​

SO2​

HCl

KOH

Q. $K M n O_{4}$ can be prepared from $K_{2} M n O_{4}$ as per the reaction,
$3 M n O_{4}^{2 -} + 2 H_{2} O ⇌ 2 M n O_{4}^{-} + M n O_{2} + 4 O H^{-}$
The reaction can go to completion by removing $O H^{-}$ ions by adding

$C O_{2}$

$S O_{2}$