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JEE Main & Advanced AIEEE Solved Paper-2003

  • 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  Solved  Paper-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 :

    \[A{{B}_{2}}\,\underset{S}{\mathop{{{A}^{2+}}}}\,\,\,+2\,\underset{2S}{\mathop{{{B}^{-}}}}\,\] \[{{K}_{sp}}=[{{A}^{2+}}]\,{{[{{B}^{-}}]}^{2}}\]     \[=(S)\,{{(2S)}^{2}}=4{{S}^{3}}\]     \[=4{{(1\times {{10}^{-5}})}^{3}}\]              \[=4\times {{10}^{-15}}\]


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