question_answer 28)

Calculate the molar solubility of \[Ni{{(OH)}_{2}}\]in 0.1 M\[NaOH\]. The ionic product of\[Ni{{(OH)}_{2}}\]is\[2\times {{10}^{-15}}\].  

Answer:

Let solubility of\[Ni{{(OH)}_{2}}\] in \[NaOH\]is s. \[\underset{s}{\mathop{Ni{{(OH)}_{2}}}}\,\to \underset{s}{\mathop{N{{i}^{2+}}}}\,+\underset{(2s+0.1)\approx 0.1}{\mathop{2O{{H}^{-}}}}\,\] \[{{K}_{sp}}=[N{{i}^{2+}}]{{[O{{H}^{-}}]}^{2}}\] \[2\times {{10}^{-15}}=s\times {{(0.1)}^{2}}\] \[s=2\times {{10}^{-13}}M\]