1. AP Chem Get Equilibrium Practice Stamped
Today: Solubility Equilibrium (Ksp)
Unit 6 Test: Th 1/31 (MC) and Fri 2/1(FRQ)

2. Solubility & Equilibrium
Initially, when a solid is placed in water:
XY(s)  X+ (aq) + Y- (aq)
As time continues, more of the solid dissolves and the [ions] begins to increase.

3. Solubility & Equilibrium
When the concentration of ions increases, eventually there is a greater the chance that they will collide and reform the solid
XY(s)  X+ (aq) + Y- (aq)
When it becomes saturated it reaches equilibrium.
XY(s) ⇌ X+ (aq) + Y- (aq)

4. XY(s) ⇌ X+ (aq) + Y- (aq) Equilibrium Constant for Solubility-Ksp
Ksp = [X+]x [Y-]y
The concentrations of the ions at equilibrium is known as the solubility or molar solubility of the substance.

5. Meaning of Ksp
The value of Ksp tells you how soluble a substance is (how much it will dissolve and dissociate into ions).
A really small Ksp means the substance is relatively not that soluble in water (not much will dissolve and dissociate into ions)
We can still use ICE tables to solve Ksp problems.

6. AgI(s) ⇆ Ag+(aq)+ I-(aq)
Write the Ksp expression for this solubility equilibrium.
Ksp= [Ag+][I-]

7. AgI(s) ⇆ Ag+(aq)+ I-(aq)C E
Consider the solubility equilibrium: AgI (s) ⇌ Ag+(aq) + I-(aq) (Ksp = 8.52 x 10-17). Calculate the solubility of AgI in pure water. +x +x x x
Ksp= [Ag+][I-]
8.52 x = x2
x = 9 x 10-9 M  molar solubility of AgI in pure water

8. Common Ion Effect
Instead of adding your solute to water, you’re adding it to a solution that contains a common ion as your solute
This means the starting concentration of one of your ions will NOT be 0.

9. AgI (s) ⇆ Ag+(aq) + I-(aq)C E
0.001
Then calculate the new solubility of AgI in a solution containing x 10-3 M NaI. +x +x x x
Ksp= [Ag+][I-]
8.52 x = x(.001+x)
x = 8.52 x M  solubility in a solution containing NaI