NEET Chemistry Equilibrium / साम्यावस्था NEET PYQ-Ionic Equilibrium

  • question_answer
    Find out the solubility of \[Ni{{(OH)}_{2}}\] in 0.1 M\[NaOH\]. Given that the ionic product of \[Ni{{(OH)}_{2}}\] is \[2\times {{10}^{15}}\]                                 [NEET 2020]

    A)  \[2\times {{10}^{8}}M\]         

    B)       \[1\times {{10}^{13}}M\]

    C)  \[1\times {{10}^{8}}M\]         

    D)       \[2\times {{10}^{13}}M\]

    Correct Answer: D

    Solution :

    [d] \[\underset{s}{\mathop{Ni{{(OH)}_{2}}}}\,\underset{s}{\mathop{N{{i}^{2+}}}}\,+\underset{2s}{\overset{\Theta }{\mathop{2OH}}}\,\]
                \[\underset{0.1}{\mathop{NaOH}}\,\xrightarrow{{}}\underset{0.1}{\mathop{N{{a}^{+}}}}\,+\underset{0.1}{\overset{\Theta }{\mathop{OH}}}\,\]
                Total \[[\overset{\Theta }{\mathop{OH}}\,]=2s+0.1\approx 0.1\]
                Ionic product \[=[N{{i}^{2+}}]{{[\overset{-}{\mathop{O}}\,H]}^{2}}\]
                \[2\times {{10}^{-15}}=s{{(0.1)}^{2}}\]
                \[s=2\times {{10}^{-13}}\]
                Solubility of \[Ni{{(OH)}_{2}}=2\times {{10}^{-13}}M\]


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