You have prepared a 1.50 M ZnCl2 solution. The pH must be below what value to avoid precipitation of zinc hydroxide in the solution? The solubility product of Zn(OH)2 is 3.0 x 10−16.

Answer 1

Zn(OH)2 ----> Zn^2+ + 2OH^-
Ksp = 3E-16 = (Zn^2+)(OH^-)^2
Plug in 1.5M = (Zn^2+) and solve for (OH^-) and convert that to pH. Post your work if you get stuck.