The solubility product ( K_ sp ) of zinc hydroxide, Zn(OH) ₂ , is 3.2 10⁻¹⁴ at 298 K . What is the pH of its saturated aqueous solution at this temperature? (Given : 2 = 0.3010 )

Step-by-step solution

Let the solubility of Zn(OH) ₂ be s . Zn(OH) ₂(s) Zn ²⁺(aq) + 2 OH ^-(aq) K_ sp = [ Zn ²⁺][ OH ^-]^2 = (s)(2s)^2 = 4s^3 4s^3 = 3.2 10⁻¹⁴ s^3 = 0.8 10⁻¹⁴ = 8 10⁻¹⁵ s = 2 10⁻⁵ M [ OH ^-] = 2s = 4 10⁻⁵ M pOH = - [ OH ^-] = - (4 10⁻⁵) = 5 - 2 2 pOH = 5 - 2(0.3010) = 5 - 0.6020 = 4.398 pH = 14 - pOH = 14 - 4.398 = 9.602 Answer: 9.602