Question

Find out the solubility of $Ni{\left(OH\right)}_2$ in$0.1\mathrm{M}$ $\mathrm{NaOH}.$

The ionic product of $Ni{\left(OH\right)}_2$ is $2\times {10}^{-15}$

• A. $1\times {10}^{-13}M$
• B. $1\times {10}^{-8}M$
• C. $2\times {10}^{-13}M$
• D. $1\times {10}^{8}M$

Answer 1

The correct option is C

$2\times {10}^{-13}M$

Correct Answer is Option ( c) $2\times {10}^{-13}M$

Writing Dissociation reaction of $Ni{(OH)}_2$

Nickel hydroxide $Ni{(OH)}_2$ dissociates in solution as:
$\mathrm{Ni}{\left(\mathrm{OH}\right)}_2\rightleftarrows {\mathrm{Ni}}^{2+}+2{\mathrm{OH}}^{-}$
We can write the ionic product constant for $Ni{(OH)}_2$ as: $\left[{\mathrm{Ni}}^{2+}\right]{\left[{\mathrm{OH}}^{-}\right]}^2$

Calculating the concentration of $N{i}^{2+}$ and $O{H}^{-}$ ions (in ${molL}^{-1}$) produced on the dissociation of $Ni{(OH)}_2$.
$$\begin{gathered} \mathrm{Ni}{\left(\mathrm{OH}\right)}_2\left(\mathrm{aq}\right)\rightleftharpoons {\mathrm{Ni}}^{2+}\left(\mathrm{aq}\right)+2{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ \mathrm{s}\mathrm{s}2\mathrm{s} \end{gathered}$$

s is solubility/molar concentration

• One mole of $Ni{(OH)}_2$ gives one mole of $N{i}^{2+}$ and two moles $O{H}^{-}$,
• The concentration of $N{i}^{2+}$ and $O{H}^{-}$ ions (in ${molL}^{-1}$) produced on dissociation is equal to the amount of $Ni{(OH)}_2$
• Therefore, $smol{L}^{-1}$of $Ni{(OH)}_2$ will gives $smol{L}^{-1}$of $N{i}^{2+}$and $2smol{L}^{-1}$ of $O{H}^{-}$ ions.

Calculating $O{H}^{-}$concentration through $\mathrm{NaOH}$ dissociation reaction

$0.1\mathrm{M}$$\mathrm{NaOH}$ completely dissociates to give$N{a}^{+}$ and $O{H}^{-}$ ions.

$$\begin{gathered} \mathrm{NaOH}\left(\mathrm{aq}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{OH}}^{-}\left(\mathrm{aq}\right) \\ 0.10.10.1 \end{gathered}$$

Total Concentration of $O{H}^{-}$= $0.1\mathrm{M}+2\mathrm{s}$

Calculating the Ionic Product

• Ionic Product =${\mathrm{K}}_{\mathrm{sp}}=\mathrm{s}\times {\left(0.1+2\mathrm{s}\right)}^2$
• From the above equation, we can infer that $\mathrm{s}$ is very small as compared to $0.1\mathrm{M}$ .
• So we can assume that $2s+0.10\approx 0.10$.
• Then the equation is simplified as

$\begin{array}{rl}\Rightarrow & 2.0\times {10}^{-15}=s\times {\left(0.10\right)}^2\\ \Rightarrow & 2.0\times {10}^{-15}=s\times {10}^{-2}\end{array}$
Solving for $\mathrm{s}$, we get
$\mathrm{s}=2\times {10}^{-13}{\mathrm{molL}}^{-1}\mathrm{or}\mathrm{M}$

Hence, the molar solubility of $Ni{(OH)}_2$ in $0.1\mathrm{M}$ $\mathrm{NaOH}$ is $2\times {10}^{-13}$ ${molL}^{-1}$ or $2\times {10}^{-13}M$