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NEET Sample Paper NEET Sample Test Paper-89

  • question_answer
    In qualitative analysis, the metals of Group I can be separated from other ions by precipitating them as chloride salts. A solution initially contains \[A{{g}^{+}}\] and \[P{{b}^{2+}}\] at a concentration of 0.10 M. Aqueous HCl is added to this solution until the \[C{{l}^{-}}\]concentration is 0.10 M. What will the concentrations of \[A{{g}^{+}}\] and \[P{{b}^{2+}}\] be at equilibrium? \[({{K}_{sp}}\,for\,AgCl=1.8\times {{10}^{-10}},\]\[{{K}_{sp}}\,for\,PbC{{l}_{2}}=1.7\times {{10}^{-5}})\]

    A) \[[A{{g}^{+}}]=1.8\times {{10}^{-7}}M;\] \[[P{{b}^{2}}^{+}]=1.7\times {{10}^{-6}}M\]

    B) \[[A{{g}^{+}}]=1.8\times {{10}^{-11}}M;\] \[[P{{b}^{2}}^{+}]=8.5\times {{10}^{-5}}M\]

    C) \[[A{{g}^{+}}]=1.8\times {{10}^{-9}}M;\] \[[P{{b}^{2}}^{+}]=1.7\times {{10}^{-3}}M\]

    D) \[[A{{g}^{+}}]=1.8\times {{10}^{-11}}M;\] \[[P{{b}^{2}}^{+}]=8.5\times {{10}^{-4}}M\]

    Correct Answer: C

    Solution :

    [c] \[{{K}_{sp}}=[A{{g}^{+}}][C{{l}^{-}}]\] \[1.8\times {{10}^{-10}}=[A{{g}^{+}}][0.1]\] \[[A{{g}^{+}}]=1.8\times {{10}^{-9}}M\] \[{{K}_{sp}}=[P{{b}^{+2}}]{{[C{{l}^{-}}]}^{2}}\] \[1.7\times {{10}^{-5}}=[P{{b}^{+2}}]{{[0.1]}^{2}}\] \[[P{{b}^{+2}}]=1.7\times {{10}^{-3}}M\]


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