The solubility product constant ( K_ sp ) of Mg(OH) ₂ is 1.0 10⁻¹¹ . The molar solubility of Mg(OH) ₂ in a buffer solution of pH 12 is:

Step-by-step solution

Given pH = 12 , the pOH can be calculated as: pOH = 14 - pH = 14 - 12 = 2 Therefore, the concentration of hydroxide ions is [ OH ^-] = 10⁻² M . Since the solution is buffered, the [ OH ^-] remains fixed at 10⁻² M . The dissociation of Mg(OH) ₂ is: Mg(OH) ₂ Mg ²⁺ + 2 OH ^- Let s be the molar solubility of Mg(OH) ₂ . Then [ Mg ²⁺] = s . The expression for the solubility product is: K_ sp = [ Mg ²⁺][ OH ^-]^2 Substituting the given values: 1.0 10⁻¹¹ = (s)(10⁻²)^2 1.0 10⁻¹¹ = s 10⁻⁴ s = 1.0 10⁻¹¹ 10⁻⁴ = 1.0 10⁻⁷ M The