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NEET AIPMT SOLVED PAPER MAINS 2011

  • 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 be the concentration of \[A{{g}^{+}}\] and \[P{{b}^{2+}}\] be at equilibrium? (\[{{K}_{sp}}\]for \[AgCl=1.8\times {{10}^{-10}},\,\,{{K}_{sp}}\] for\[\text{PbC}{{\text{l}}_{\text{2}}}\text{=1}\text{.7 }\!\!\times\!\!\text{ 1}{{\text{0}}^{\text{-5}}}\])

    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 :

    \[{{K}_{sp}}\] for \[AgCl=[A{{g}^{+}}][C{{l}^{-}}]\] \[\therefore \]  \[[A{{g}^{+}}]=\frac{1.8\times {{10}^{-10}}}{{{10}^{-1}}}\]                 \[=1.8\times {{10}^{-9}}M.\] \[{{K}_{sp}}\]for\[PbC{{l}_{2}}=[P{{b}^{2}}^{+}]{{[C{{l}^{-}}]}^{2}}\] \[\therefore \]  \[[P{{b}^{2+}}]=\frac{1.7\times {{10}^{-5}}}{{{10}^{-1}}\times {{10}^{-1}}}\]                 \[=1.7\times {{10}^{-3}}M\]


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