Answer
The molar solubility is $6\times10^{-8} M$.
Work Step by Step
$Ksp=6.0\times10^{-16}$
Using the same procedure as 17.62a, we can find the molar solubility of nickel (II) hydroxide in this buffer.
$pH + pOH=14$
$pOH=14-pH=14-10=4.0$
$-> [OH^{-}]=10^{-pOH}=10^{-4} M$
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: $[Ni(OH)_{2}]=x M, [Ni^{2+}]= 0 M, [OH^{-}]=10^{-4} M$
Equilibrium: $[Ni(OH)_{2}]=0 M, [Ni^{2+}]= x M, [OH^{-}]=10^{-4} M$
As a saturated solution, $Ksp=[Ni^{2+}][OH^{-}]^2=6.0\times10^{-16}$
$-> x\times(10^{-4})^2=6.0\times10^{-16}$
$-> x=6.0\times10^{-8} M$ (use a calculator)
In conclusion, the molar solubility for nickel (II) hydroxide in this scenario is $6\times10^{-8} M$.
