Question

The solubility product of $N i( OH )_{2}$ at $298 K$ is $2 \times 10^{-15} mol ^{3} \cdot d m^{-9}$. The $pH$ value if its aqueous and saturated solution is

• A. 5
• B. 7.5
• C. 9
• D. 13

Answer 1

Answer: (C)

$N i(O H)_{2} \rightleftharpoons \underset{ S\,mol / L}{N i^{2+}}+\underset{2S\,mol/L}{2 OH ^{-}}$
$\Rightarrow K_{s p}=\left[N i^{2+}\right]\left[O H^{-}\right]^{2}=S \times(2 S)^{2}=4 S^{3}$
$=2 \times 10^{-15}( mol / L)^{3}$
$=2 \times 10^{-15} mol ^{3} d m^{-9}$
$\Rightarrow S=\left(\frac{2}{4} \times 10^{-15}\right)^{1 / 3}=8 \times 10^{-6} mol / L$
$\left[O H^{-}\right]=2 S=2 \times 8 \times 10^{-6} mol / L$
$p O H=6-\log 16=4.983=5$
$\therefore p H=14-p O H=14-5=9$ at $25^{\circ} C$