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

  • question_answer
    On addition of increasing amount of \[AgN{{O}_{3}}\] to 0.1 M each of \[NaCl\] and \[NaBr\] in a solution, what % of \[B{{r}^{-}}\] ion get precipitated when \[C{{l}^{-}}\]ion starts precipitating.\[{{K}_{sp}}\left( AgCl \right)=1.0\times {{10}^{-10}}\], \[{{K}_{sp}}(AgBr)=1\times {{10}^{-13}}\]

    A) 0.11                 

    B) 99.9 

    C) 0.01                 

    D) 9.99

    Correct Answer: B

    Solution :

    [b] To precipitate the \[AgCl\] \[[A{{g}^{+}}]\] required \[=\frac{{{K}_{sp}}(AgCl)}{[C{{l}^{-}}]}=\frac{1.0\times {{10}^{-10}}}{0.1}=1.0\times {{10}^{-9}}M\] \[\left[ B{{r}^{-}} \right]\] left at this stage \[=\frac{{{K}_{sp}}(AgBr)}{[A{{g}^{+}}]}\] \[=\frac{1.0\times {{10}^{-13}}}{1.0\times {{10}^{-9}}}=1.0\times {{10}^{-4}}M\] % of remaining \[[B{{r}^{-}}]=\frac{1.0\times {{10}^{-4}}}{0.1}\times 100=0.1\] % of \[B{{r}^{-}}\] to be precipitated \[=100-0.1=99.9\]


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