Consider the following reaction that describes the solubility equilibria of solid Zn(OH)2 in aqueous solution (Ksp of Zn(OH)2 4.0x10^-17. Calculate molar solubility of Zn(OH)s in an acidic solution of pH5.0.

Answer 1

Zn(OH)2 ==> Zn^2+ + 2OH^-
Ksp = 4.0E-17 = (Zn^+)(OH^-)^2
pH = 5.0; use
pH + pOH = pKw = 14 to solve for pOH, then -log(OH^-) to solve for (OH^-). Substitute OH into Ksp expression and solve for Zn^2+. That is the solubility of Zn(OH)2.

Answer 2

I followed your direction and I calculated the solubility to be 40. Am I correct?