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JEE Main & Advanced JEE Main Paper (Held on 12-4-2019 Morning)

  • question_answer
              What is the molar solubility of \[Al{{(OH)}_{3}}\]in 0.2 M NaOH solution? Given that, solubility product of \[Al{{(OH)}_{3}}=2.4\times {{10}^{-24}}\]:                               [JEE Main Held on 12-4-2019 Morning]

    A) \[12\times {{10}^{-23}}\]               

    B) \[12\times {{10}^{-21}}\]

    C) \[3\times {{10}^{-19}}\]                 

    D) \[3\times {{10}^{-22}}\]

    Correct Answer: D

    Solution :

    \[\underset{S'}{\mathop{Al{{(OH)}_{3}}A{{l}^{+3}}}}\,\underset{0.2+3(S')\simeq 0.2}{\mathop{+3O{{H}^{-}}}}\,\]           \[S'\times {{(0.2)}^{2}}={{k}_{sp}}=2.4\times {{10}^{-24}}\]           \[(S')=3\times {{10}^{-22}}M\]


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