2. Solved Papers
  3. JEE Main & Advanced

JEE Main & Advanced AIEEE Solved Paper-2010

  • question_answer
    At\[25{}^\circ C,\] the solubility product of\[Mg{{(OH)}_{2}}\]is\[1.0\times {{10}^{-11}}\]. At which pH, will\[M{{g}^{2+}}\]ions start precipitating in the form of\[Mg{{(OH)}_{2}}\]from a solution of\[0.001\text{ }M\text{ }M{{g}^{2+}}\]ions ?       AIEEE  Solved  Paper-2010

    A) 8                             

    B)        9                             

    C)                        10                          

    D)        11

    Correct Answer: C

    Solution :

    \[{{K}_{sp}}=[M{{g}^{+2}}]{{[O{{H}^{-}}]}^{2}}\] \[1\times {{10}^{-11}}=[0.001]{{[O{{H}^{-}}]}^{2}}\] \[[o{{h}^{-}}]={{10}^{-4}}\] \[P\text{ }on=4;\text{ }pn=10\]


You need to login to perform this action.
You will be redirected in 3 sec spinner