Warm Up #5 __ AgCl(s) ⇌ __ Ag+1(aq) + __ Cl-1(aq) Balance the following and construct a K equation. What is the ratio between the solid and the two products? Given that the concentration of the silver ion was 4.50x10-5, and the solid broke down completely and evenly…calculate the K for this reaction. Which are favored, the products or reactants? Why, given the states of matter, does this make sense?

Chapter 18.3 Solubility & KsP

AgCl(s) ⇌ Ag+1(aq) + Cl-1(aq) Remember… Keq = [products]/[reactants] NO LIQUIDS OR SOLIDS Keq > 1…products favored Keq < 1…reactants favored Keq = 1…equilibrium WHAT IF…. AgCl(s) ⇌ Ag+1(aq) + Cl-1(aq) Keq = [Ag+1] [Cl-1]

Solubility Solubility – ability to be dissolved If not…insoluble (PRECIPITATE!)…(s) AgCl(s) = precipitate Ksp = RATE of solubility Decomposition reactions Similar to Keq For precipitates…Ksp = NEGATIVE TRY THIS ONE: MgCl2(s) = Mg+2(aq) + 2 Cl-1(aq)

Finding Ksp…Two Scenarios AgCl(s) ⇌ Ag(aq) + Cl(aq) MgCl2(s) ⇌ Mg(aq) + 2 Cl(aq) Ksp = [Ag+1] [Cl-1] Molar Solubility = 1.4x10-4 [Ag+1] = 1:1 ratio…1.4x10-4 [Cl-1] = 1:1 ratio…1.4x10-4 Ksp = [1.4x10-4] [1.4x10-4] Ksp = [Mg+2] [Cl-1]2 Molar Solubility = 1.4x10-4 [Mg+2] = 1:1 ratio…1.4x10-4 [Cl-1] = 1:2 ratio…2.8x10-4 Ksp = [1.4x10-4] [2.8x10-4]2

Try This One! __Mg3(PO4)2(s)  __ Mg+2(aq) + __ PO4-3(aq)

Finding Molar Solubility, given Ksp

Warm Up #6 With the molar solubility of 4.7x10-6, calculate the Ksp for Ca(NO3)2 (s) If the Ksp for Na3(PO4) is 9.1x10-19, calculate the molar solubility. Looking at a chemical reaction, how do you know it is endothermic? (two possible answers)