JEE Main & Advanced Chemistry Equilibrium Question Bank Acids and Bases

  • question_answer \[1\times {{10}^{-12}}\] of \[1\,\,M\,\,{{H}_{2}}S{{O}_{4}}\] will completely neutralise

    A)                 \[10\,\,ml\] of \[1\,\,M\,\,NaOH\] solution

    B)                 \[10\,\,ml\] of \[2\,\,M\,\,NaOH\] solution

    C)                 \[5\,\,ml\] of \[2\,\,M\,\,KOH\] solution

    D)                 \[5\,\,ml\] of \[1\,\,M\,\,N{{a}_{2}}C{{O}_{3}}\] solution

    Correct Answer: B

    Solution :

               \[{{H}_{2}}S{{O}_{4}}+2{{H}_{2}}O\] ⇌ \[2{{H}_{3}}{{O}^{+}}+SO_{4}^{-\,-}\]                    NaOH ⇌ \[N{{a}^{+}}+O{{H}^{-}}\]                                 1 mole of \[{{H}_{2}}S{{O}_{4}}\]acid gives 2 moles of \[{{H}_{3}}{{O}^{+}}\]ions. So 2 moles of \[O{{H}^{-}}\] are required for complete neutralization.    


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