| \[\text{KMn}{{\text{O}}_{\text{4}}}\] can be prepared from \[{{\text{K}}_{\text{2}}}\text{Mn}{{\text{O}}_{\text{4}}}\] as per reaction\[\text{3MnO}_{4}^{2-}+2{{H}_{2}}O2MnO_{4}^{-}\]\[+Mn{{O}_{2}}+4O{{H}^{-}}\][NEET 2013] |
| The reaction can go to completion by removing \[\text{O}{{\text{H}}^{\text{-}}}\] ions by adding |
A) HCI
B) KOH
C) \[C{{O}_{2}}\]
D) \[S{{O}_{2}}\]
Correct Answer: C
Solution :
| [c] Since, \[O{{H}^{-}}\] are generated from weak acid \[({{H}_{2}}O),\]a weak acid (like\[C{{O}_{2}}\]) should be used to remove it because of strong acid (HCI) reverse the reaction. KOH increases the concentration of \[O{{H}^{-}},\], thus again shifts the reaction in backward side. |
| \[C{{O}_{2}}\] combines with OH" to give carbonate which is easily removed. |
| \[S{{O}_{2}}\]reacts with water to give strong acid, so it cannot be used. |
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