Question

$K_{sp}$ of a salt $Ni(OH{)}_2$ is $2\times {10}^{-15}$ then molar solubility of $Ni(OH{)}_2$ in $0.01MNaOH$ is:

• A. $2\times {10}^{-15}M$
• B. $2^{1/3}\times {10}^{-5}M$
• C. $2\times {10}^{-11}M$
• D. ${10}^{-7}M$

Answer 1

The correct option is C $2\times {10}^{-11}M$

Solubility is represented as:

$Ni{\left(OH\right)}_2\rightleftharpoons {Ni}^{2+}+2{OH}^{-}$

Also, $NaOH\to N{a}^{+}+{OH}^{-}$

If molar solubility of $Ni(OH{)}_2$ is $S$, then at equilibrium it provides $S$ mol/L of $N{i}^{2+}$ ions and $2Smol/L$ of $O{H}^{-}$ ions.

Therefore solubility product is defined as

$K_{sp}=\left[N{i}^{2+}\right]{\left[O{H}^{-}\right]}^2$=$2\times {10}^{-15}$

where $\left[N{i}^{2+}\right]=Sand$

$\left[O{H}^{-}\right]=2S+0.01$ due to the presence of strong base $NaOH$ that gives 0.01 M of $O{H}^{-}$ ions as well.

Hence, $2\times {10}^{-15}=S\times (2S+0.01{)}^2$

since $K_{sp}$ is very small therefore 2S<<0.01

So,$2S+0.01\approx 0.01$

upon substitution we get:

$2\times {10}^{-15}=0.0001S$ and

$S=2\times {10}^{-11}M$

Therefore option C is correct.