JEE Main & Advanced Chemistry Equilibrium Question Bank Common ion effect, Isohydric solutions, Solubility product, Ionic product of water and salt hydrolysis

  • question_answer
    The solubility product of \[PbC{{l}_{2}}\] at \[{{20}^{o}}C\] is \[1.5\times {{10}^{-4}}.\] Calculate the solubility     [Bihar CEE 1995; BHU 2002]

    A)                 \[3.75\times {{10}^{-4}}\]            

    B)                 \[3.34\times {{10}^{-2}}\]

    C)                 \[3.34\times {{10}^{2}}\]             

    D)                 None of these

    Correct Answer: B

    Solution :

    \[\underset{S}{\mathop{PbC{{l}_{2}}}}\,\]⇌\[\underset{S\,\,\,}{\mathop{P{{b}^{++}}}}\,+\underset{{{(2S)}^{2}}}{\mathop{2C{{l}^{-}}}}\,\]                    \[{{K}_{sp}}=S\times {{(2S)}^{2}}=4{{S}^{3}}\]                                 \[S=\sqrt[3]{\frac{{{K}_{sp}}}{4}}=\sqrt[3]{\frac{1.5\times {{10}^{-4}}}{4}}=3.34\times {{10}^{-2}}\].


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