1. Chapter 17 Additional Aspects of Aqueous Equilibria
Chemistry, The Central Science, 10th edition
Theodore L. Brown; H. Eugene LeMay, Jr.; and Bruce E. Bursten
Chapter 17 Additional Aspects of Aqueous Equilibria
John D. Bookstaver
St. Charles Community College
St. Peters, MO
 2006, Prentice Hall, Inc.

2. March 11 COMMON ION EFFECT 17.11, 13, 15
BUFFER PROBLEMS 17, 19, 23, 25
TITRATIONS 31 to 39 odd, 43

3. The Common-Ion Effect Consider a solution of acetic acid:
If acetate ion is added to the solution, Le Châtelier says the equilibrium will shift to the left.
HC2H3O2(aq) + H2O(l)
H3O+(aq) + C2H3O2−(aq)

4. The Common-Ion Effect
“The extent of ionization of a weak electrolyte is decreased by adding to the solution a strong electrolyte that has an ion in common with the weak electrolyte.”

5. The Common-Ion Effect
Calculate the fluoride ion concentration and pH of a solution that is 0.20 M in HF and 0.10 M in HCl.
Ka for HF is 6.8  10−4.
[H3O+] [F−]
[HF]
Ka =
= 6.8  10-4

6. The Common-Ion Effect HF(aq) + H2O(l) H3O+(aq) + F−(aq)
Because HCl, a strong acid, is also present, the initial [H3O+] is not 0, but rather 0.10 M.
[HF], M
[H3O+], M
[F−], M
Initially
0.20
0.10
Change −x +x
At Equilibrium
0.20 − x  0.20
x  0.10 x

7. The Common-Ion Effect (0.10) (x) 6.8  10−4 = (0.20)
(0.20) (6.8  10−4)
(0.10)
= x
1.4  10−3 = x

8. The Common-Ion Effect Therefore, [F−] = x = 1.4  10−3
[H3O+] = x =  10−3 = 0.10 M
So, pH = −log (0.10)
pH = 1.00

9. Buffers: Solutions of a weak conjugate acid-base pair.
They are particularly resistant to pH changes, even when strong acid or base is added.

10. Example
Write the reaction that occurs in a buffer solution containing HF/F- when:
H+ is added H+ + F-  HF
OH- is added OH- + HF  F- + H2O

11. Buffers
If a small amount of hydroxide is added to an equimolar solution of HF in NaF, for example, the HF reacts with the OH− to make F− and water.

12. Buffers
If acid is added, the F− reacts to form HF and water.

13. Buffer Calculations
Consider the equilibrium constant expression for the dissociation of a generic acid, HA:
HA + H2O
H3O+ + A−
[H3O+] [A−]
[HA]
Ka =

14. Buffer Calculations Rearranging slightly, this becomes
[HA]
Ka =
[H3O+]
Taking the negative log of both side, we get
base
[A−]
[HA]
−log Ka =
−log [H3O+] + −log
pKa pH
acid

15. Buffer Calculations So pKa = pH − log [base] [acid]
Rearranging, this becomes
pH = pKa + log
[base]
[acid]
This is the Henderson–Hasselbalch equation.

16. MARCH 14 BUFFER PROBLEMS Calculation of the ph of a buffer
Buffer capacity
Addition of strong acid or bases to buffers
BUFFER PROBLEMS 17, 19, 23, 25
TITRATIONS a)Strong acid and Strong base
b ) Weak acid – Strong Base
Titration of polyprotic Acids
TITRATIONS problems 31 to 39 odd, 43

17. Henderson–Hasselbalch Equation
What is the pH of a buffer that is 0.12 M in lactic acid, HC3H5O3, and 0.10 M in sodium lactate? Ka for lactic acid is
1.4  10−4.

18. Henderson–Hasselbalch Equation
pH = pKa + log
[base]
[acid]
pH = −log (1.4  10−4) + log
(0.10)
(0.12)
pH = (−0.08)
pH = 3.77

19. Buffer Capacity
The amount of acid of base the buffer can neutralize before the pH begins to change to an appreciable degree. It depends on the amount of acid and base from which the buffer is made.
The pH of the buffer depends on the Ka of the acid, and on the relative concentrations of acid and base.

20. Example A solution 1M in HA, 1M in NaA will have its pH=Ka.
A solution .1M in HA and .1M in NaA will have SAME pH (Ka) but the first one will have greater buffer capacity.

21. pH Range
The pH range is the range of pH values over which a buffer system works effectively.
It is best to choose an acid with a pKa close to the desired pH.
In practice if the concentrations of one component of the buffer is 10 times the concentration of the other component the buffering action is poor.

22. Addition of Strong Acids or Bases to Buffers
The amount of strong acid or base added results in a neutralization reaction:
X- + H3O+  HX + H2O
HX + OH-  X- + H2O.
By knowing how much H3O+ or OH- was added (stoichiometry) we know how much HX or X- is formed.
With the concentrations of HX and X- (note the change in volume of solution) we can calculate the pH from the Henderson-Hasselbalch equation or the original equilibrium expression (ICE chart).

23. When Strong Acids or Bases Are Added to a Buffer…
…it is safe to assume that all of the strong acid or base is consumed in the reaction.

24. Calculating pH Changes in Buffers
A buffer is made by adding mol HC2H3O2 and mol NaC2H3O2 to enough water to make 1.00 L of solution. The pH of the buffer is Calculate the pH of this solution after mol of NaOH is added.

25. Calculating pH Changes in Buffers
Before the reaction, since
mol HC2H3O2 = mol C2H3O2−
pH = pKa = −log (1.8  10−5) = 4.74

26. Calculating pH Changes in Buffers
The mol NaOH will react with mol of the acetic acid:
HC2H3O2(aq) + OH−(aq)  C2H3O2−(aq) + H2O(l)
HC2H3O2
C2H3O2−
OH−
Before reaction
0.300 mol
0.020 mol
After reaction
0.280 mol
0.320 mol
0.000 mol

27. Calculating pH Changes in Buffers
Now use the Henderson–Hasselbalch equation to calculate the new pH:
pH = log
(0.320)
(0. 200)
pH =
pH = 4.80

28. Addition of Strong Acid or Base to a Buffer
Determine how the neutralization reaction affects the amounts of the weak acid and its conjugate base in solution.
Use the Henderson–Hasselbalch equation to determine the new pH of the solution.

29. Examples
Consider a buffer prepared by placing 0.60 moles of HF (Ka = 7.2 x 10-4) and 0.48 moles of NaF in a 1.00 L solution.
a. Calculate the pH of the buffer
b. Calculte the pH after the addition of 0.08 moles of HCl
c. Calculate the pH after the addition of a total of 0.16 moles of HCl
d. Calculate the pH after the addition of 0.10 moles of NaOH.

30. Consider a buffer containing 0. 78 moles of HC2H3O2 (Ka = 1
Consider a buffer containing 0.78 moles of HC2H3O2 (Ka = 1.8 x 10-5) and 0.67 moles of NaC2H3O2 in 1.00 L of solution.
a. Calculate the pH of the buffer solution.
b. Calculate the pH after the addition of 0.10 moles of HNO3.
c. Calculate the pH after the addtiion of a total of 0.20 moles of HNO3.
d. Calculate the pH after the addition of 0.10 moles of NaOH.

31. Titration
A known concentration of base (or acid) is slowly added to a solution of acid (or base).

32. Titration
A pH meter or indicators are used to determine when the solution has reached the equivalence point, at which the stoichiometric amount of acid equals that of base.

33. Titration of a Strong Acid with a Strong Base
From the start of the titration to near the equivalence point, the pH goes up slowly.

34. Titration of a Strong Acid with a Strong Base
Just before and after the equivalence point, the pH increases rapidly.

35. Titration of a Strong Acid with a Strong Base
At the
equivalence point,
mol acid = mol base, the solution contains only water and the salt from the cation of the base and the anion of the acid.

36. Titration of a Strong Acid with a Strong Base
As more base is added, the increase in pH again levels off.

37. Titration of a Weak Acid with a Strong Base
Unlike in the previous case, the conjugate base of the acid affects the pH when it is formed.
The pH at the equivalence point will be >7.
Phenolphthalein is commonly used as an indicator in these titrations.

38. Titration of a Weak Acid with a Strong Base
At each point below the equivalence point, the pH of the solution during titration is determined from the amounts of the acid and its conjugate base present at that particular time.

39. Titration of a Weak Acid with a Strong Base
With weaker acids, the initial pH is higher and pH changes near the equivalence point are more subtle.

40. Titration of a Weak Base with a Strong Acid
The pH at the equivalence point in these titrations is < 7.
Methyl red is the indicator of choice.

41. Titrations of Polyprotic Acids
In these cases there is an equivalence point for each dissociation.

42. Solubility Products
Consider the equilibrium that exists in a saturated solution of BaSO4 in water:
BaSO4(s)
Ba2+(aq) + SO42−(aq)

43. Solubility Products
The equilibrium constant expression for this equilibrium is
Ksp = [Ba2+] [SO42−]
where the equilibrium constant, Ksp, is called the solubility product.

44. Solubility Products Ksp is not the same as solubility.
Solubility is generally expressed as the mass of solute dissolved in 1 L (g/L) or 100 mL (g/mL) of solution, or in mol/L (M).

45. Factors Affecting Solubility
The Common-Ion Effect
If one of the ions in a solution equilibrium is already dissolved in the solution, the equilibrium will shift to the left and the solubility of the salt will decrease.
BaSO4(s)
Ba2+(aq) + SO42−(aq)

46. Factors Affecting SolubilitypH
If a substance has a basic anion, it will be more soluble in an acidic solution.
Substances with acidic cations are more soluble in basic solutions.

47. Factors Affecting Solubility
Complex Ions
Metal ions can act as Lewis acids and form complex ions with Lewis bases in the solvent.

48. Factors Affecting Solubility
Complex Ions
The formation of these complex ions increases the solubility of these salts.

49. Factors Affecting Solubility
Amphoterism
Amphoteric metal oxides and hydroxides are soluble in strong acid or base, because they can act either as acids or bases.
Examples of such cations are Al3+, Zn2+, and Sn2+.

50. Will a Precipitate Form?
In a solution,
If Q = Ksp, the system is at equilibrium and the solution is saturated.
If Q < Ksp, more solid will dissolve until Q = Ksp.
If Q > Ksp, the salt will precipitate until Q = Ksp.

51. Selective Precipitation of Ions
One can use differences in solubilities of salts to separate ions in a mixture.