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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Critical Thinking

  • question_answer
    \[{{K}_{sp}}\] for \[Cr{{(OH)}_{3}}\]is \[2.7\times {{10}^{-31}}\]. What is its solubility in moles / litre.        [JEE Orissa 2004]

    A)                 \[1\times {{10}^{-8}}\]  

    B)                 \[8\times {{10}^{-8}}\]

    C)                 \[1.1\times {{10}^{-8}}\]              

    D)                 \[0.18\times {{10}^{-8}}\]

    Correct Answer: A

    Solution :

               \[Cr{{(OH)}_{3}}\ \to \ \underset{x}{\mathop{C{{r}^{+3}}}}\,+\underset{3x}{\mathop{3O{{H}^{-}}}}\,\]                    \[{{K}_{sp}}=x.{{(3x)}^{3}}=27{{x}^{4}}\]                    \[x=\sqrt[4]{\frac{{{K}_{sp}}}{27}}\] ; \[x=\sqrt[4]{\frac{2.7\times {{10}^{-31}}}{27}}\]                                 \[x=1\times {{10}^{-8}}\]mole/litre.


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