1. Applications of Aqueous Equilibria

2. 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

3. Acid/Salt Buffering Pairs
The salt will contain the anion of the acid, and the cation of a strong base (NaOH, KOH)
Weak Acid
Formula
of the acid
Example of a salt of the weak acid
 Hydrofluoric 
HF 
 KF – Potassium fluoride 
 Formic 
 HCOOH 
 KHCOO – Potassium formate 
 Benzoic 
 C6H5COOH 
 NaC6H5COO – Sodium benzoate
 Acetic 
 CH3COOH 
 NaH3COO – Sodium acetate 
 Carbonic 
 H2CO3 
 NaHCO3 - Sodium bicarbonate
 Propanoic 
 HC3H5O2 
  NaC3H5O2  - Sodium propanoate
 Hydrocyanic 
 HCN 
 KCN - potassium cyanide 

4. Base/Salt Buffering Pairs
The salt will contain the cation of the base, and the anion of a strong acid (HCl, HNO3)
Base
Formula of the base
Example of a salt of the weak acid
Ammonia 
 NH3
 NH4Cl - ammonium chloride
 Methylamine
 CH3NH2
 CH3NH2Cl – methylammonium chloride
 Ethylamine
 C2H5NH2
 C2H5NH3NO3 -  ethylammonium nitrate
 Aniline
 C6H5NH2 
C6H5NH3Cl – aniline hydrochloride
 Pyridine
 C5H5N 
  C5H5NHCl – pyridine hydrochloride

5. Weak Acid/Strong Base Titration
Endpoint is above pH 7
A solution that is
0.10 M CH3COOH
is titrated with
0.10 M NaOH

6. Strong Acid/Strong Base Titration
Endpoint is at
pH 7
A solution that is
0.10 M HCl is titrated with
0.10 M NaOH

7. Strong Acid/Strong Base Titration
A solution that is
0.10 M NaOH is titrated with
0.10 M HCl
Endpoint is at
pH 7
It is important to recognize that titration curves are not always increasing from left to right.

8. Strong Acid/Weak Base Titration
A solution that is
0.10 M HCl is titrated with
0.10 M NH3
Endpoint is below
pH 7

9. Titration of an Unbuffered Solution
A solution that is
0.10 M HC2H3O2
is titrated with
0.10 M NaOH

10. Titration of a Buffered Solution
A solution that is
0.10 M HC2H3O2 and
0.10 M NaC2H3O2 is titrated with 0.10 M NaOH

11. Comparing Results Unbuffered Buffered
In what ways are the graphs different?
In what ways are the graphs similar?

12. Comparing Results
Buffered
Unbuffered

13. Buffer capacity The best buffers have a ratio [A-]/[HA] = 1
This is most resistant to change
True when [A-] = [HA]
Make pH = pKa (since log1=0)

14. General equation Ka = [H+] [A-] [HA] so [H+] = Ka [HA] [A-]
The [H+] depends on the ratio [HA]/[A-]
taking the negative log of both sides
pH = -log(Ka [HA]/[A-])
pH = -log(Ka)-log([HA]/[A-])
pH = pKa + log([A-]/[HA])

15. This is called the Henderson-Hasselbalch Equation

16. Using the Henderson-Hasselbalch Equation
pH = pKa + log([A-]/[HA])
pH = pKa + log(base/acid)
Calculate the pH of the following mixtures
0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
0.25 M NH3 and 0.40 M NH4Cl
(Kb = 1.8 x 10-5)

17. Prove they’re buffers
What would the pH be if mol of HCl is added to 1.0 L of both of the following solutions.
0.75 M lactic acid (HC3H5O3) and 0.25 M sodium lactate (Ka = 1.4 x 10-4)
0.25 M NH3 and 0.40 M NH4Cl (Kb = 1.8 x 10-5)
What would the pH be if mol of solid NaOH is added to each of the proceeding.
To do this I must introduce the BAAM table!

18. Buffer capacity
The pH of a buffered solution is determined by the ratio [A-]/[HA].
As long as this doesn’t change much the pH won’t change much.
The more concentrated these two are the more H+ and OH- the solution will be able to absorb.
Larger concentrations bigger buffer capacity.

19. Buffer Capacity
Calculate the change in pH that occurs when mol of HCl(g) is added to 1.0L of each of the following:
5.00 M HAc and 5.00 M NaAc
0.050 M HAc and M NaAc
Ka= 1.8x10-5

21. Summary Strong acid and base just stoichiometry.
Determine Ka, use for 0 mL base
Weak acid before equivalence point
Stoichiometry first
Then Henderson-Hasselbach
Weak acid at equivalence point Kb
Weak base after equivalence - leftover strong base.

22. Summary Determine Ka, use for 0 mL acid.
Weak base before equivalence point.
Stoichiometry first
Then Henderson-Hasselbach
Weak base at equivalence point Ka.
Weak base after equivalence - leftover strong acid.

23. Selection of Indicators

24. Solubility Equilibria
Will it all dissolve, and if not, how much?

25. All dissolving is an equilibrium.
If there is not much solid it will all dissolve.
As more solid is added the solution will become saturated.
Solid dissolved
The solid will precipitate as fast as it dissolves .
Equilibrium

26. Watch out Solubility is not the same as solubility product.
Solubility product is an equilibrium constant.
it doesn’t change except with temperature.
Solubility is an equilibrium position for how much can dissolve.
A common ion can change this.

27. Ksp Values for Some Salts at 25C
Name
Formula
Ksp
 Barium carbonate 
 BaCO3 
 2.6 x 10-9 
 Barium chromate 
 BaCrO4 
 1.2 x 10-10 
 Barium sulfate 
 BaSO4 
 1.1 x 10-10 
 Calcium carbonate 
 CaCO3 
 5.0 x 10-9 
 Calcium oxalate 
 CaC2O4 
 2.3 x 10-9 
 Calcium sulfate 
 CaSO4 
 7.1 x 10-5 
 Copper(I) iodide 
 CuI 
 1.3 x 10-12 
 Copper(II) iodate 
 Cu(IO3)2 
 6.9 x 10-8 
 Copper(II) sulfide 
 CuS 
 6.0 x 10-37 
 Iron(II) hydroxide 
 Fe(OH)2 
 4.9 x 10-17 
 Iron(II) sulfide 
 FeS 
 6.0 x 10-19 
 Iron(III) hydroxide 
 Fe(OH)3 
 2.6 x 10-39 
 Lead(II) bromide 
 PbBr2 
 6.6 x 10-6 
 Lead(II) chloride 
 PbCl2 
 1.2 x 10-5 
 Lead(II) iodate 
 Pb(IO3)2 
 3.7 x 10-13 
 Lead(II) iodide 
 PbI2 
 8.5 x 10-9 
 Lead(II) sulfate 
 PbSO4 
 1.8 x 10-8 
Name
Formula
Ksp
 Lead(II) bromide 
 PbBr2 
 6.6 x 10-6 
 Lead(II) chloride 
 PbCl2 
 1.2 x 10-5 
 Lead(II) iodate 
 Pb(IO3)2 
 3.7 x 10-13 
 Lead(II) iodide 
 PbI2 
 8.5 x 10-9 
 Lead(II) sulfate 
 PbSO4 
 1.8 x 10-8 
 Magnesium carbonate 
 MgCO3 
 6.8 x 10-6 
 Magnesium hydroxide 
 Mg(OH)2 
 5.6 x 10-12 
 Silver bromate 
 AgBrO3 
 5.3 x 10-5 
 Silver bromide 
 AgBr 
 5.4 x 10-13 
 Silver carbonate 
 Ag2CO3 
 8.5 x 10-12 
 Silver chloride 
 AgCl 
 1.8 x 10-10 
 Silver chromate 
 Ag2CrO4 
 1.1 x 10-12 
 Silver iodate 
 AgIO3 
 3.2 x 10-8 
 Silver iodide 
 AgI 
 8.5 x 10-17 
 Strontium carbonate 
 SrCO3 
 5.6 x 10-10 
 Strontium fluoride 
 SrF2 
 4.3 x 10-9 
 Strontium sulfate 
 SrSO4 
 3.4 x 10-7 
 Zinc sulfide 
 ZnS 
 2.0 x 10-25 

28. 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

29. 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

30. Relative solubilities
Ksp will only allow us to compare the solubility of solids the that fall apart into the same number of ions.
The bigger the Ksp of those the more soluble.
If they fall apart into different number of pieces you have to do the math.

31. The Common Ion Effect
When the salt with the anion of a weak acid is added to that acid,
It reverses the dissociation of the acid.
Lowers the percent dissociation of the acid.
The same principle applies to salts with the cation of a weak base..
The calculations are the same as last chapter.

32. 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

33. Precipitation and Qualitative Analysis

34. pH and solubility OH- can be a common ion. More soluble in acid.
For other anions if they come from a weak acid they are more soluble in a acidic solution than in water.
CaC2O4 Ca+2 + C2O4-2
H+ + C2O HC2O4-
Reduces C2O4-2 in acidic solution.

35. Precipitation Ion Product, Q =[M+]a[Nm-]b
If Q>Ksp a precipitate forms.
If Q<Ksp No precipitate.
If Q = Ksp equilibrium.
A solution of mL of 4.00 x 10-3M Ce(NO3)3 is added to mL of x 10-2M KIO3. Will Ce(IO3)3 (Ksp= 1.9 x 10-10M)precipitate and if so, what is the concentration of the ions?

36. Selective Precipitations
Used to separate mixtures of metal ions in solutions.
Add anions that will only precipitate certain metals at a time.
Used to purify mixtures.
Often use H2S because in acidic solution Hg+2, Cd+2, Bi+3, Cu+2, Sn+4 will precipitate.

37. Selective Precipitation
In Basic adding OH-solution S-2 will increase so more soluble sulfides will precipitate.
Co+2, Zn+2, Mn+2, Ni+2, Fe+2, Cr(OH)3, Al(OH)3

38. Selective precipitation
Follow the steps first with insoluble chlorides (Ag, Pb, Ba)
Then sulfides in Acid.
Then sulfides in base.
Then insoluble carbonate (Ca, Ba, Mg)
Alkali metals and NH4+ remain in solution.

39. Complex Ions A Complex ion is a charged species composed of:
1. A metallic cation
2. Ligands – Lewis bases that have a lone electron pair that can form a covalent bond with an empty orbital belonging to the metallic cation

40. NH3, CN-, and H2O are Common Ligands

41. The addition of each ligand has its own equilibrium
Usually the ligand is in large excess.
And the individual K’s will be large so we can treat them as if they go to equilibrium.
The complex ion will be the biggest ion in solution.

42. Coordination Number
Coordination number refers to the number of ligands attached to the cation
2, 4, and 6 are the most common coordination numbers
Coordination
number
Example(s) 2
Ag(NH3)2+ 4
CoCl Cu(NH3)42+ 6
Co(H2O) Ni(NH3)62+

43. Complex Ions and Solubility
AgCl(s)  Ag+ + Cl- Ksp = 1.6 x 10-10
Ag+ + NH3  Ag(NH3) K1 = 1.6 x 10-10
Ag(NH3)+ NH3  Ag(NH3) K2 = 1.6 x 10-10
AgCl + 2NH3  Ag(NH3)2+ + Cl-
K = KspK1K2