Solubility Product From Solubility Example Problem

This example problem demonstrates how to determine the solubility product of an ionic solid in water from a substances solubility. Problem The solubility of silver chloride, AgCl, is 1.26 x 10-5 M at 25 Â °C.The solubility of barium fluoride, BaF2, is 3.15 x 10-3 M at 25 Â °C.Calculate the solubility product, Ksp, of both compounds. Solution The key to solving solubility problems is to properly set up your dissociation reactions and define solubility. AgCl The dissociation reaction of AgCl in water isAgCl (s) ↔ Ag (aq) Cl- (aq)For this reaction, each mole of AgCl that dissolves produces 1 mole of both Ag and Cl-. The solubility would then equal the concentration of either the Ag or Cl ions.solubility [Ag] [Cl-]1.26 x 10-5 M [Ag] [Cl-]Ksp [Ag][Cl-]Ksp (1.26 x 10-5)(1.26 x 10-5)Ksp 1.6 x 10-10 BaF2 The dissociation reaction of BaF2 in water isBaF2 (s) ↔ Ba (aq) 2 F- (aq)This reaction shows that for every mole of BaF2 that dissolves, 1 mole of Ba and 2 moles of F- are formed. The solubility is equal to the concentration of the Ba ions in solution.solubility [Ba] 7.94 x 10-3 M[F-] 2 [Ba]Ksp [Ba][F-]2Ksp ([Ba])(2 [Ba])2Ksp 4[Ba]3Ksp 4(7.94 x 10-3 M)3Ksp 4(5 x 10-7)Ksp 2 x 10-6 Answer The solubility product of AgCl is 1.6 x 10-10.The solubility product of BaF2 is 2 x 10-6.