JEE MainChemistryIonic Equilibrium
The solubility product constant ( K_ sp ) of Mg(OH) ₂ at 298 K is 4 10⁻⁹ . The solubility of Mg(OH) ₂ in pure water is ________ mg/L . (Nearest integer) (Given molar mass of Mg(OH) ₂ = 58 g mol ⁻¹ )
Correct answer
58
Step-by-step solution
Let the molar solubility of Mg(OH) ₂ be S mol/L . The dissociation reaction is: Mg(OH) ₂(s) Mg ²⁺(aq) + 2 OH ^-(aq) The solubility product expression is: K_ sp = [ Mg ²⁺][ OH ^-]^2 = (S)(2S)^2 = 4S^3 Substitute the given value of K_ sp : 4S^3 = 4 10⁻⁹ S^3 = 10⁻⁹ S = 10⁻³ mol/L To find the solubility in g/L , multiply by the molar mass: Solubility = 10⁻³ mol/L 58 g/mol = 0.058 g/L To convert to mg/L , multiply by 1000 : Solubility = 0.058 1000 mg/L = 58 mg/L Answer: 58