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JEE Main & Advanced JEE Main Online Paper (Held on 9 April 2013)

  • question_answer
                                    Sodium Carbonate cannot be used in place of   \[{{({{\operatorname{NH}}_{4}})}_{2}}{{\operatorname{CO}}_{3}}\] for the identification of \[C{{a}^{2+}},B{{a}^{2+}}\] and \[\operatorname{S}{{r}^{2+}}\] ions (in group V) during mixture analysis because:                   JEE Main Online Paper (Held On 09 April 2013)

    A)                 \[M{{g}^{2+}}\]ions will also be precipitated                

    B)                 Concentration of \[CO_{3}^{2-}\] ions is very low                

    C)                 Sodium ions will acid radicals                

    D)                 \[{{\operatorname{Na}}^{+}}\] ions will interfere with the detection of \[C{{a}^{2+}},\operatorname{B}{{a}^{2+}},{{\operatorname{Sr}}^{2+}}\] ions                

    Correct Answer: A

    Solution :

                    In fifth group \[{{(N{{H}_{4}})}_{2}}C{{O}_{3}}\] is used in presence of \[N{{H}_{4}}Cl\]. In the presence of \[N{{H}_{4}}Cl\] dissociation of \[{{(N{{H}_{4}})}_{2}}C{{O}_{3}}\] decreases and due to low concentration of \[CO_{3}^{2-}\] ions only V group radical are precipitated out. If \[N{{a}_{2}}C{{O}_{3}}\] is used concentration of \[CO_{3}^{2-}\] ions will increase and \[M{{g}^{2+}}\] ions will also be precipitated.                


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