1. 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?

2. Chapter 18.3 Solubility & KsP

3. 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]

4. 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)

5. 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

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

7. Finding Molar Solubility, given Ksp

8. 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)