Answer
See proof below.
Work Step by Step
$Zn(OH)_2(s)\leftrightarrow Zn^{2+}(aq)+2OH^-(aq)$
$Ksp=[Zn^{2+}][OH^-]^2=3\dot{}10^{-17}$
$Zn^{2+}(aq)+4\ OH^-(aq)\leftrightarrow Zn(OH)_4^{2-}(aq)$
$Kf=[Zn(OH)_4^{2-}]/[Zn^{2+}][OH^-]=4.6\dot{}10^{17}$
$Zn(OH)_2(s)+2\ OH^-\leftrightarrow Zn(OH)_4^{2-}(aq)$
$K=Kf\dot{}Ksp=13.8$
Since K>1, this reaction is product favored.
