1. Aqueous Equilibria Chapter 15
DE Chemistry
Dr. Walker

2. Common Ion Effect
When multiple compounds with a common ion are in solution, equilibrium is shifted and the pH is affected.
HF (aq) and NaF (aq)
HF (aq) H+ (aq) + F-(aq)
NaF (aq) Na+ (aq) + F-(aq)
Presence of flouride ion shifts equilibrium to the left,
which lowers [H+], raising the pH

3. Common Ion Effect
When multiple compounds with a common ion are in solution, equilibrium is shifted and the pH is affected.
NH4OH and NH4Cl (aq)
NH3 (aq) + H2O NH4+ + OH- (aq)
NH4Cl(aq) NH4+ (aq) + Cl- (aq)
Presence of ammonium ion shifts equilibrium to the left, which lowers [OH-], lowering the pH

4. Example
Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.

5. Example
Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.
Notice that we have a source of F- from two different places, the
dissociation of NaF and the dissociation of HF. The presence of F-
from NaF will keep some HF from dissociating.
NaF F- + Na+
HF F- + H+
Add F-, shift equilibrium left, lower [H+]

6. Example
Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.
HF F- + H+
From 1.0 M NaF! HF F- H+ I
1.0 C -x +x E
1.0 - x
1.0 + x
+ x

7. Example
Calculate the pH of a solution containing 1.0 M HF (Ka = 7.2 x 10-4) and 1.0 M NaF.
pH = -log (7.2 x 10-4) = 3.14
Compare to a 1.0 M solution of only HF, the pH = 1.57
Pretty big difference!!

8. Buffered Solutions
A solution that resists a change in pH when either hydroxide ions or protons are added.
Buffered solutions contain either:
A weak acid and its salt
A weak base and its salt

9. Buffered Solutions
HC2H3O C2H3O2- + H+

10. Buffered Solutions HC2H3O2 C2H3O2- + H+ HC2H3O2 C2H3O2- H+ I 0.5 C -xC -x +x E
0.5 - x
0.5 + x
+ x

11. Buffered Solutions
HC2H3O C2H3O2- + H+

12. A Different Scenario
The previous example used a buffer system where the concentrations were equal. This doesn’t always happen
The Henderson-Hasselbalch equation helps us solve for this scenario as long as our given concentrations are significantly larger than the pKa.
pKa = - log Ka

13. Unequal Buffering Example

14. Unequal Buffering Example
We can still use the standard method (ICE) to determine our pH, but Henderson-Hasselbalch is actually less work here. Lactic acid is our “acid” and sodium lactate is our “base”

15. Solubility Equilibria
Many ionic compounds dissociate completely in aqueous solutions
Nitrates, Group 1 cations, most sulfates, halides
Others are technically “insoluble”
Some of the material still dissolves, just a small fraction of it
Similar to the dissociation of strong vs. weak acids

16. Solubility Equilibria
“Soluble” compounds
Similar to strong acids
Dissociate completely
Large K value, not really an equilibrium process
“Insoluble” compounds
Similar to weak acids
Dissociate partially
Mostly reactant at equilibrium, small K value
Reactant NOT in equilibrium expression, since it’s a solid

17. Solubility Equilibria
When ionic compounds are dissolved in solution, they dissociate to some degree
Ksp is the solubility constant
Higher Ksp = more soluble
Remember, pure liquids and solids are NOT included in the law of mass action
These problems work the same as problems for equilibrium, weak acids/bases, and buffers

19. Solving Solubility Problems
What is the solubility for the salt AgI at 25C, Ksp = 1.5 x 10-16
AgI(s)  Ag+(aq) + I-(aq) I C E O O +x +x x x

20. Solving Solubility Problems
For the salt AgI at 25C, Ksp = 1.5 x 10-16
AgI(s)  Ag+(aq) + I-(aq) I C E O O +x +x x x
1.5 x = x2
x = solubility of AgI in mol/L = 1.2 x 10-8 M

21. Solving Solubility Problems
For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5
PbCl2(s)  Pb2+(aq) + 2Cl-(aq) I C E O O +x
+2x x 2x
1.6 x 10-5 = (x)(2x)2 = 4x3
x = solubility of PbCl2 in mol/L = 1.6 x 10-2 M

22. Solving Solubility Problems
What is the solubility for the salt For the salt PbCl2 at 25C, Ksp = 1.6 x 10-5
PbCl2(s)  Pb2+(aq) + 2Cl-(aq) I C E O O +x
+2x x 2x

23. Solving Solubility with a Common Ion
For the salt AgI at 25C, Ksp = 1.5 x 10-16
What is its solubility in 0.05 M NaI?
AgI(s)  Ag+(aq) + I-(aq) I C E O
0.05 +x
0.05+x x
0.05+x
1.5 x = (x)(0.05+x)  (x)(0.05)
x = solubility of AgI in mol/L = 3.0 x M