The solubility of Zn(OH)2 in 1.00 M NaOH is calculated below:The reaction is: Zn(OH)2(s) ⇌ Zn2+(aq) + 2OH-(aq)Initial: 0M 0M 1.00 MChange: -S + S + 2S Equilibrium: -S S 1.00M + 2SThe Ksp for Zn(OH)2 is 3.0 x 10^-16; hence, [Zn2+] [OH-]^2 = 3.0 x 10^-16 …(1)The Kf for Zn(OH)4^2- is 3.0 x10^15; hence,Zn(OH)2(s) + 4OH-(aq) ⇌ Zn(OH)4^2-(aq)Kf = ([Zn(OH)4^2-]/([Zn2+][OH-]^4) 3.0 x10^15 = ([Zn(OH)4^2-]/([Zn2+][OH-]^4) [Zn(OH)4^2-] = 3.0 x 10^15 [Zn2+][OH-]^4 …(2)From (1), [Zn2+] = (3.0 x 10^-16)/[OH-]^2Substituting [Zn2+] into (2) gives:[Zn(OH)4^2-] = 3.0 x 10^15 [(3.0 x 10^-16)/[OH-]^2][OH-]^4[Zn(OH)4^2-] = 9.0 x 10^-1 [OH-]^2[Zn(OH)4^2-] = [OH-]^2 = 9.49 x 10^-10 MThe solubility of Zn(OH)2 is 2[OH-] = 1.90 x 10^-9 M.This is 150 words.