Zinc hydroxide, Zn(OH)2 (Ksp = 4.1 x 10^-17), is nearly insoluble in water but is more soluble in strong base because Zn2+ forms the soluble complex ion [Zn(OH)4]2- (Kf = 3 x 10^15). (a) What is the molar solubility of Zn(OH)2 in pure water? (You may ignore the OH- from the self-dissociation of water.) (b) What is the pH of the solution in part (a)? (c) What is the molar solubility of Zn(OH)2 in 0.10 M NaOH?

Verified step by step guidance

1. Step 1: For part (a), write the dissolution equation for Zn(OH)_2 in water: Zn(OH)_2 (s) ⇌ Zn^{2+} (aq) + 2OH^{-} (aq).
2. Step 2: Express the solubility product constant (Ksp) for Zn(OH)_2: Ksp = [Zn^{2+}][OH^{-}]^2. Given Ksp = 4.1 x 10^{-17}, let the molar solubility of Zn(OH)_2 be 's'. Then, [Zn^{2+}] = s and [OH^{-}] = 2s.
3. Step 3: Substitute the expressions for [Zn^{2+}] and [OH^{-}] into the Ksp expression: Ksp = s(2s)^2 = 4s^3. Solve for 's' to find the molar solubility of Zn(OH)_2 in pure water.
4. Step 4: For part (b), calculate the pH of the solution using the concentration of OH^{-} ions. Use the relation pOH = -log[OH^{-}] and pH + pOH = 14 to find the pH.
5. Step 5: For part (c), consider the effect of the common ion (OH^{-}) from NaOH. Write the equilibrium expression for the formation of the complex ion: Zn^{2+} + 4OH^{-} ⇌ [Zn(OH)_4]^{2-}. Use the formation constant (Kf) and the initial concentration of OH^{-} from NaOH to find the new solubility of Zn(OH)_2.