JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Self Evaluation Test - Equilibrium

  • question_answer
    The solubility of \[Pb{{I}_{2}}\] at \[25{}^\circ C\] is\[0.7\text{ }g\text{ }{{L}^{-1}}\]. The solubility product of \[Pb{{I}_{2}}\] at this temperature is (molar mass of\[Pb{{I}_{2}}=461.2gmo{{l}^{-1}}\])

    A) \[1.40\times {{10}^{-9}}\]        

    B) \[0.14\times {{10}^{-9}}\]

    C) \[140\times {{10}^{-9}}\]         

    D) \[14.0\times {{10}^{-9}}\]

    Correct Answer: D

    Solution :

    [d] \[Pb{{I}_{2}}\rightleftharpoons \underset{s}{\mathop{P{{b}^{++}}}}\,+\underset{2s}{\mathop{2{{I}^{-}}}}\,\] \[{{K}_{sp}}=s\times {{(2s)}^{2}}=4{{s}^{3}}\]  \[=4\times {{\left( \frac{0.7}{461.2} \right)}^{3}}=14.0\times {{10}^{-9}}\]


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