The solubility product of a sparingly soluble salt BA is 6.4 10⁻¹³ . Calculate it's solubility in g dm ⁻³ . Molar mass of salt is 190 g mol ⁻¹ .

Step-by-step solution

For a sparingly soluble salt BA , the dissociation equilibrium is given by BA B^+ + A^- . Let the solubility of the salt be s in mol dm ⁻³ . The solubility product K_ sp is given by: K_ sp = [B^+][A^-] = s s = s^2 Substituting the given value of K_ sp : s^2 = 6.4 10⁻¹³ = 64 10⁻¹⁴ s = 64 10⁻¹⁴ = 8 10⁻⁷ mol dm ⁻³ To convert the solubility from mol dm ⁻³ to g dm ⁻³ , multiply by the molar mass of the salt: Solubility in g dm ⁻³ = s Molar mass Solubility = 8 10⁻⁷ mol dm ⁻³ 190 g mol ⁻¹ Solubility = 1520 10⁻⁷ g dm ⁻³ =