Punjab Medical Punjab - MET Solved Paper-2006

  • question_answer
    The solubility product of \[A{{g}_{2}}CrO\] in water a\[298K\]is\[3.2\times {{10}^{-11}}\]. What will be the concentration of \[CrO_{4}^{2-}\] ions in the saturate solution of\[A{{g}_{2}}Cr{{O}_{4}}\]?

    A) \[2\times {{10}^{-4}}M\]                             

    B) \[5.7\times {{10}^{-5}}M\]

    C) \[5.7\times {{10}^{-6}}M\]                          

    D) \[3.2\times {{10}^{-11}}\]

    Correct Answer: A

    Solution :

    For saturated solution of\[A{{g}_{2}}Cr{{O}_{4}}\], if solubility is \[s\]\[mo{{l}^{-1}}\,\,{{L}^{-1}}\].Then               \[A{{g}_{2}}Cr{{O}_{4}}2\underset{2s}{\mathop{A{{g}^{+}}}}\,(aq)+CrO_{4}^{2-}(aq)\]                 \[{{K}_{sp}}={{(2s)}^{2}}(s)=4{{s}^{3}}\]                 \[{{K}_{sp}}=3.2\times {{10}^{-11}}\](given)                 \[\therefore \]\[3.2\times {{10}^{-11}}=4{{s}^{3}}\]                 \[{{s}^{3}}=\frac{3.2\times {{10}^{-11}}}{4}=8\times {{10}^{-12}}\] \[\therefore \]  \[s=\sqrt{8\times {{10}^{-12}}}=2\times {{10}^{-4}}M\]


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