NEET Sample Paper NEET Sample Test Paper-14

  • question_answer 52) When solid lead iodide is added to water, the  equilibrium concentration of \[{{\text{I}}^{-}}\]becomes \[2.6\times {{10}^{-3}}\,M.\] What is the \[{{K}_{sp}}\]for \[Pb{{I}_{2}}?\]

    A) \[2.2\times {{10}^{-9}}\]                

    B) \[8.8\times {{10}^{-9}}\]

    C) \[1.8\times {{10}^{-8}}\]                

    D) \[3.5\times {{10}^{-8}}\]    

    Correct Answer: B

    Solution :

    The equilibrium is \[Pb{{l}_{2}}\Leftrightarrow P{{b}^{2}}+2{{I}^{-}}\] On the basis of this equation, the concentration of \[P{{b}^{2+}}\]ions will be half of the concentration of \[{{\text{I}}^{-}}\]ions. Thus, \[[{{I}^{-}}]=2.6\times {{10}^{-3}}M\]and\[[P{{b}^{2+}}]=1.3\times {{10}^{-3}}M.\]  \[{{K}_{sp}}=[P{{b}^{2+}}]{{[{{I}^{-}}]}^{2}}=[1.3\times {{10}^{-3}}M]{{[2.6\times {{10}^{-3}}M]}^{2}}\] \[{{K}_{sp}}=8.8\times {{10}^{-9}}.\] Hence, the correct option is [b].

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