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WB JEE Medical WB JEE Medical Solved Paper-2013

  • question_answer
    \[KMn{{O}_{4}}\]can be prepared from \[{{K}_{2}}Mn{{O}_{4}}\]as per  reaction \[3MnO_{4}^{2-}+2{{H}_{2}}O\rightleftharpoons 2MnO_{4}^{-}\] \[+Mn{{O}_{2}}+4O{{H}^{-}}\] The reaction can go to completion by removing \[O{{H}^{-}}\]ions by adding

    A)  HCI

    B)  KOH

    C) \[C{{O}_{2}}\]

    D)  \[S{{O}_{2}}\]

    Correct Answer: D

    Solution :

     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 \[O{{H}^{-}}\]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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