Question

If the ionic product of ${Ni\left(OH\right)}_2$ is $1.9\times {10}^{-15}$, the molar solubility of ${Ni\left(OH\right)}_2$ in $1.0M$ $NaOH$ is:

• A. $1.9\times {10}^{-18}M$
• B. $1.9\times {10}^{-13}M$
• C. $1.9\times {10}^{-15}M$
• D. $1.9\times {10}^{-14}M$

Answer 1

Answer: (C)

$1.9\times {10}^{-15}M$

Let, the solubility be $SM$.

$\left[Ni\right(OH{)}_2]=[N{i}^{2+}]=SM$

$\left[O{H}^{-}\right]=1.0M$

Now,

$K_{sp}=\left[N{i}^{2+}\right][O{H}^{-}{]}^2$

$1.9\times {10}^{-15}=S\times (1.0{)}^2$

$\therefore S=1.9\times {10}^{-15}M$