## Chemistry and Chemical Reactivity (9th Edition)

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