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JEE Main & Advanced Chemistry Equilibrium / साम्यावस्था Question Bank Common ion effect, Isohydric solutions, Solubility product, Ionic product of water and salt hydrolysis

  • question_answer
    The solubility in water of a sparingly soluble salt \[A{{B}_{2}}\] is \[1.0\times {{10}^{-5}}mol\,{{l}^{-1}}\]. Its solubility product number will be [AIEEE 2003]

    A)                 \[4\times {{10}^{-15}}\]               

    B)                 \[4\times {{10}^{-10}}\]

    C)                 \[1\times {{10}^{-15}}\]               

    D)                 \[1\times {{10}^{-10}}\]

    Correct Answer: A

    Solution :

               \[A{{B}_{2}}\] ⇌ \[\underset{1\times {{10}^{-5}}}{\mathop{{{A}^{+}}\,\,\,}}\,\,\,+\,\,\,\,\,\underset{2\times {{10}^{-5}}}{\mathop{2{{B}^{-}}}}\,\]                                 \[{{K}_{sp}}=[1\times {{10}^{-5}}]\,\,{{[2\times {{10}^{-5}}]}^{2}}=4\times {{10}^{-15}}\]


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