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Manipal Engineering Manipal Engineering Solved Paper-2008

  • question_answer
    Solubility product of a salt\[AB\]is\[1\times {{10}^{-8}}{{M}^{2}}\]in a solution in which the concentration of\[{{A}^{+}}\]ions is\[{{10}^{-3}}M\]. The salt will precipitate when the concentration of\[{{B}^{-}}\]ions is kept

    A)  between\[{{10}^{-8}}M\]to\[{{10}^{-7}}M\]

    B)  between\[{{10}^{-7}}M\]to\[{{10}^{-8}}M\]

    C) \[>{{10}^{-5}}M\]

    D) \[<{{10}^{-8}}M\]

    Correct Answer: C

    Solution :

    A salt is precipitated only when the product of ionic concentration is more than its solubility product.                 \[{{K}_{sp}}=1\times {{10}^{-8}}\]                 \[[{{A}^{+}}]={{10}^{-3}}M\]                 \[[{{B}^{-}}]=\frac{1\times {{10}^{-8}}}{{{10}^{-3}}}={{10}^{-5}}M\] So, AB will be precipitated only when the concentration of\[{{B}^{-}}\]is more than\[{{10}^{-5}}M\].


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