Question

The Ionic product of $Ni{{(OH)}_{2}}$ is $2.0\times {{10}^{-15}}$. Molar solubility of $Ni{{(OH)}_{2}}$ in 0.10M NaOH will be ________.
(A)- $1.0\times {{10}^{-13}}M$
(B)- $2.0\times {{10}^{-13}}M$
(C)- $4.0\times {{10}^{-13}}M$
(D)- $8.0\times {{10}^{-13M}}$

Answer 1

Hint: Ionic product of a salt or electrolyte is equal to the product of molar concentration of its ions where each concentration term is raised to the number of ions produced on dissociation of one molecule of the salt or electrolyte.
If salt of the type ${{M}_{x}}{{N}_{y}}$ dissociates in solution.
     \[{{M}_{x}}{{N}_{y}}\rightleftarrows x{{M}^{y-}}+y{{N}^{x-}}\]

Then, it ionic product will be given as: ${{\left[ {{M}^{y+}} \right]}^{x}}{{\left[ {{N}^{x-}} \right]}^{y}}$
Mathematical expression for ionic product and solubility product is the same.

Complete step by step answer:
Nickel hydroxide $Ni{{(OH)}_{2}}$ dissociates in solution as:
     \[Ni{{(OH)}_{2}}\rightleftarrows N{{i}^{2+}}+2O{{H}^{-}}\]
We can write the ionic product constant for $Ni{{(OH)}_{2}}$ as: $\left[ N{{i}^{2+}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$

Now the concentration of $N{{i}^{2+}}$ and $O{{H}^{-}}$ ions (in mol${{L}^{-1}}$) produced on dissociation is equal to the amount of $Ni{{(OH)}_{2}}$ that is dissolved.
Let the molar solubility be s mol.
     \[\begin{matrix}
   Ni{{(OH)}_{2}} & \rightleftarrows & N{{i}^{2+}} & + & 2O{{H}^{-}} \\
   s & {} & s & {} & 2s \\
\end{matrix}\]

One mole of $Ni{{(OH)}_{2}}$ gives one mole of $N{{i}^{2+}}$ and two moles $O{{H}^{-}}$, it means that s mol${{L}^{-1}}$ of $Ni{{(OH)}_{2}}$ will give s mol${{L}^{-1}}$of $N{{i}^{2+}}$and 2s mol${{L}^{-1}}$ of $O{{H}^{-}}$ ions.

However, the solution has 0.10M NaOH which affects the solubility of $Ni{{(OH)}_{2}}$. 0.10M NaOH completely dissociates to give $N{{a}^{+}}$ and $O{{H}^{-}}$ ions.
     \[\begin{matrix}
   NaOH & \rightleftarrows & N{{a}^{+}} & + & O{{H}^{-}} \\
   0.10 & {} & 0.10 & {} & 0.10 \\
\end{matrix}\]

Since the solution also contains 0.10 moles of $O{{H}^{-}}$ ions per liter of solution produced from NaOH, the total concentration of $O{{H}^{-}}$ ions will be (2s + 0.10) mol${{L}^{-1}}$.
We know that the mathematical expression for the solubility product is the same as that of the ionic product of $Ni{{(OH)}_{2}}$.

The value of ionic product is equal to the solubility product, ${{K}_{sp}}=2.0\times {{10}^{-15}}$.
Therefore, the solubility product of $Ni{{(OH)}_{2}}$ is given as: ${{K}_{sp}}=\left[ N{{i}^{2+}} \right]{{\left[ O{{H}^{-}} \right]}^{2}}$.
Substituting the $\left[ N{{i}^{2+}} \right]$= s mol${{L}^{-1}}$, $\left[ O{{H}^{-}} \right]$= 2s + 0.10 mol${{L}^{-1}}$ and ${{K}_{sp}}=2.0\times {{10}^{-15}}$ in the above equation, we get
     \[2.0\times {{10}^{-15}}=s\times {{(2s+0.10)}^{2}}\]

From the above equation, we can infer that s is very small as compared to 0.10. So it is reasonable to assume that $2s+0.10\approx 0.10$. Then, the above equation is simplified as
     \[\begin{align}
  & 2.0\times {{10}^{-15}}=s\times {{(0.10)}^{2}} \\
 & 2.0\times {{10}^{-15}}=s\times {{10}^{-2}} \\
\end{align}\]
 Solving for s, we get
     \[s=2\times {{10}^{-13}}\]

Therefore, the molar solubility of $Ni{{(OH)}_{2}}$ in 0.10M NaOH is $2\times {{10}^{-13}}$ mol${{L}^{-1}}$ or $2\times {{10}^{-13}}$ M.
So, the correct answer is “Option B”.

Note: Do not get confused between solubility product and ionic product. Both represent the product of concentration of ions. Ionic product is given for all types of solutions saturated and unsaturated at all concentrations. But solubility products are only applicable to saturated solutions. In other words, the ionic product of the saturated solution is equal to the solubility product.