The solubility product ( K_ sp ) of Mg(OH)₂ at 298 K is 3.2 10⁻¹¹ . The pH of its saturated aqueous solution at this temperature is: (Given: 2 = 0.3 )

Step-by-step solution

Let the molar solubility of Mg(OH)₂ be s . The dissociation reaction is: Mg(OH)₂ Mg²⁺ + 2OH⁻ The solubility product expression is: K_ sp = [Mg²⁺][OH⁻]^2 = (s)(2s)^2 = 4s^3 Given K_ sp = 3.2 10⁻¹¹ = 32 10⁻¹² 4s^3 = 32 10⁻¹² s^3 = 8 10⁻¹² s = 2 10⁻⁴ M The concentration of hydroxide ions is: [OH⁻] = 2s = 2 (2 10⁻⁴) = 4 10⁻⁴ M Now, calculate the pOH : pOH = - [OH⁻] = - (4 10⁻⁴) pOH = 4 - 4 = 4 - 2 2 pOH = 4 - 2(0.3) = 4 - 0.6 = 3.4 At 298 K , pH + pOH = 14 pH = 14 - 3.4 = 10.6 Answer: 10.6