Tutorial 10-Solutions

Chemistry 12 Tutorial 10—Solutions Chemistry 12 Tutorial 10—Ksp Calculations Solutions 1. Calculate the solubility of SrF in moles/Litre in water. 2 -s +s +2s 2+ -Equilibrium Equation: SrF Sr + 2F (where s = molar solubility) 2(s) (aq) (aq) 2+ - 2Ksp = [Sr ] [F ] 2Ksp = s (2s) 2Ksp = s x 4s 3Ksp = 4s 3s = Ksp 4 -9 -3s = K s p = 4 . 3 x 1 0 = 1.0244 x 10 M 3 3 4 4 -3Because the Ksp was 2 SD’s, the answer will be: Molar Solubility = 1.0 x 10 M ********************************************************** 2. Calculate the solubility of Ag CO in grams/Litre. 2 3 -s +2s +s + 2-Equilibrium Equation: Ag CO 2Ag + CO (where s = molar solubility) 2 3(s) (aq) 3 (aq) + 2 2-Ksp = [Ag ] [CO ] 3 2Ksp = (2s) x s 2Ksp = 4s x s 3Ksp = 4s 3s = Ksp 4 –12 -4Ksp s = = 8 . 5 x 1 0 = 1.286 x 10 M 3 3 4 4 Chemistry 12—Tutorial 10 Solutions Page 1 of 5 pages Chemistry 12 Tutorial 10—Solutions -4Molar Solubility = 1.286 x 10 M To get solubility in g/L: -4 -21.286 x 10 mol x 275.8 g Ag CO = 3.5 x 10 g/L (or 0.035 g/L) 2 3 L ...