Answer
$1.4\times10^3$ g/L
Work Step by Step
$Ksp=[Mn^{2+}][OH^{-}]^2=1.6\times10^{-13}$ (from table D.3 Appendix D).
In a buffered solution, its pH value (or pOH value) hardly changes over time. For the sake of simplicity, let's consider the pH is held constant (also because we don't know the buffer's capacity).
First, we need to find the initial concentration of hydroxide ions before $Mn(OH)_{2}$.
$pH=7 -> pOH=14-7=7$
$-> [OH^{-}]=10^{-7}$ M
The solubility of manganese (II) hydroxide is calculated based on the amount it dissolves to form a saturated solution. So, let $x$ (mol/L) be the amount of $Mn(OH)_{2}$ needed to make that saturated solution.
Because pH is held constant, $[OH^{-}]$ would be always $10^{-7}$ M, regardless of how many $Mn(OH)_{2}$ is added.
$Mn(OH)_{2}(s) Mn^{2+} (aq) + 2OH^{-} (aq)$
initial: $[Mn(OH)_{2}]=x$ M, $[Mn^{2+}]=0$ M, $[OH^{-}]=10^{-7}$ M
equilibrium: $[Mn(OH)_{2}]=0$ M, $[Mn^{2+}]=x$ M, $[OH^{-}]=10^{-7}$ M
Therefore:
$Ksp=[Mn^{2+}][[OH^{-}]^2=x\times(10^{-7})^2=1.6\times10^{-13}$
$-> x=[Mn^{2+}]=16$ M
The question asks us to find the solubility in g/L. Then, we need to convert the unit from M (mol/L) to g/L.
$16mol/L=\frac{16 mol}{1 L}\times\frac{88.95g}{1mol}=1.4\times10^3g/L$
