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JEE Main & Advanced JEE Main Paper Phase-I (Held on 09-1-2020 Morning)

  • question_answer
    The \[{{K}_{sp}}\] for the following dissociation is \[1.6\text{ }\times \text{ }{{10}^{5}}\] \[PbC{{l}_{2\left( s \right)}}\rightleftharpoons Pb_{\left( aq \right)}^{2+}+2Cl_{\left( aq \right)}^{-}\] Which of the following choices is correct for a mixture of 300 mL 0.134 M \[Pb{{\left( N{{O}_{3}} \right)}_{2}}\]and 100 mL 0.4 M NaCl? [JEE MAIN Held on 09-01-2020 Morning]

    A) \[Q\text{ }<\text{ }{{K}_{sp}}\]

    B) \[Q\text{ }=\text{ }{{K}_{sp}}\]

    C) Not enough data provided

    D) \[Q\text{ }>\text{ }{{K}_{sp}}\]

    Correct Answer: D

    Solution :

    [d] \[PbC{{l}_{2}}\rightleftharpoons P{{b}^{2+}}+2C{{l}^{-}}\] \[{{K}_{sp}}\]of \[PbC{{l}_{2}}=1.6\times {{10}^{-5}}={{\left[ P{{b}^{2+}} \right]}_{eq}}\left[ C{{l}^{-}} \right]_{eq}^{2}\] \[Pb{{\left( N{{O}_{3}} \right)}_{2}}\to P{{b}^{2}}+2NO_{3}^{-}\] \[NaCl\to N{{a}^{+}}+C{{l}^{-}}\] \[\left[ P{{b}^{2+}} \right]=\frac{300\times 0.134}{400};\left[ C{{l}^{-}} \right]=\frac{100\times 0.4}{400}=0.1\] \[=0.1005\] \[\therefore \left[ P{{b}^{2+}} \right]{{\left[ C{{l}^{-}} \right]}^{2}}=0.1005\times {{\left( 0.1 \right)}^{2}}\] \[=1.005\times {{10}^{-3}}>{{K}_{sp}}\] \[i.e.\text{ }{{Q}_{IP}}\text{ }>\text{ }{{K}_{SP}}\]


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