Answer
The molar solubility is $6×10^{−12}M$.
Work Step by Step
$Ksp=6.0×10^{−16}$
Using the same procedure as 17.62a and 17.62b, we can find the molar solubility of nickel (II) hydroxide in this buffer.
$pH+pOH=14$
$pOH=14−pH=14−12=2.0$
$−>[OH−]=10^{−pOH}=10^{−2}M$
Suppose $x $ M is the amount of Ni(OH)2 solid needed to dissolve and form a saturated solution. We can construct the following table:
$Ni(OH)_{2}(s)Ni^{2+}(aq)+2OH^{−}(aq)$
Initial: $[Ni(OH)_{2}]=xM,[Ni^{2+}]=0M,[OH^{−}]=10^{−2}M$
Equilibrium: $[Ni(O)_{2}]=0M,[Ni^{2+}]=xM,[OH^{−}]=10^{-2}M$
As a saturated solution, $Ksp=[Ni^{2+}][OH^{−}]^2=6.0×10^{−16}$
$−>x×(10^{−2})^2=6.0×10^{−16}$
$−>x=6.0×10^{-12}M$ (use a calculator)
In conclusion, the molar solubility for nickel (II) hydroxide in this scenario is $6×10^{−12}M$.
