1. Acid-Base Equilibria
AP Chem Unit 15

2. Unit 15 Content Solutions of Acids or Bases Containing a Common Ion
Buffered Solutions
Buffering Capacity
Titrations and pH Curves
Acid-Base Indicators

3. Introduction
Most chemistry in the natural world takes place in an aqueous solution. One of the most significant aqueous reactions is one with acids and bases. Most living systems are very sensitive to pH and yet are subjected to acids and bases.
The main idea of this unit is to understand the chemistry of a buffered system. Buffered systems contain chemical components that enable a solution to be resistant to change in pH.

4. Solutions of Acids or Bases Containing a Common Ion
15.1

5. Common Ions
Solutions that contain a weak acid and the salt of it conjugate base create a common ion system.
A common ion system can also be a weak base and the salt of its conjugate acid.
Example: HF and NaF solution or
NH3 and NH4Cl solution.

6. Common Ions
The soluble salts of conjugate acids and conjugate bases are typically strong electrolytes and therefore dissociate 100%.
This dissociation increases the concentration of the conjugate ion and a shift in equilibrium occurs according to Le Chatelier.
Example:
Shift Left!

7. Common Ion Effect
The shift left in the second equation is known as the common ion effect.
This effect makes a solution of NH3 and NH4Cl less basic than NH3 alone

8. Common Ion Effect
The shift left in the second equation is known as the common ion effect. This effect makes a solution of NaF and HF less acidic than a solution of HF alone.

9. Common Ion Effect
Calculations for common ion equilibria is similar to weak acid calculations except that the initial concentrations of the anion are not 0.
Initial concentrations for the equilibrium calculation must include the concentration of ions from the complete dissociation of the salt.

10. Practice Problem #1
The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF. Major species? ICE: I C E HF F- H+

11. Practice Problem #1
The equilibrium concentration of H+ in a 1.0 M HF solution is 2.7x10-2M, and the percent dissociation of HF is 2.7%. Calculate [H+] and the percent dissociation of HF in a solution containing 1.0 M HF (Ka=7.2x10-4) and 1.0 M NaF. ANS: .072%

12. Buffered Solutions
15.2

13. Buffered Solutions
The most important application of a common ion system is buffering.
A buffered solution resists a change in its pH when either hydroxide ions or protons are added.
Blood is a practical example of a buffered solution; it is buffered with carbonic acid and the bicarbonate ion (among others).

14. Buffered Solutions
A solution can be buffered at any pH by choosing the appropriate components.
Smaller amounts of H+ or OH- can be added with little pH effect.

15. Practice Problem #2
A buffered solution contains 0.50 M acetic acid (Ka= 1.8 x 10-5) and 0.50 M sodium acetate. Calculate the pH of this solution. pH = 4.74

16. How Does Buffering Work?
When OH- ions are added to a solution of a weak acid, the OH- ions react with the best source of H+ (weak acid) to make water:
The net result is that OH- ions are replaced by A- ions and water until they are all consumed.

17. How Does Buffering Work?
When H+ ions are added to a solution of a weak acid, the H+ ions react with the strong conjugate base.
The net result is that more weak acid is produced, but free H+ do not accumulate to make large changes in the pH.

18. Buffered Solution Summary
Buffered solutions are simply solutions of weak acids or bases containing a common ion. The pH calculations on buffered solutions require exactly the same procedures introduced in Chapter 14.
When a strong acid or base is added to a buffered solution, it is best to deal with the stoichiometry of the resulting reaction first. After the stoichiometric calculations are completed, then consider the equilbrium calculations.

19. Practice Problem #3
Calculate the change in pH that occurs when mol solid NaOH is added to the buffered solution in #2 (0.50 M acetic and 0.50 M sodium acetate). Compare this pH change with the original (pH=4.74) and the pH that occurs with mol of solid NaOH is added to 1.0 L of water.

20. Practice Problem #3 Major species: OH- +HC2H3O2  C2H3O2- + H2O
mole table:
HC2H3O2
OH-
C2H3O2-
H2O
Before
After

21. Practice Problem #3 ICE table: pH= 4.76 (+.02 off original)
pH of just OH-=12.00 I C E
HC2H3O2
OH-
C2H3O2-

22. Henderson-Hasselbalch Equation
Buffered solutions work well as long as the concentration of the salts and weak acids/bases in solution are much larger than the amount of OH- and H+ being added.
The amount of [H+] being in a buffered solution is often solved using a rearrangement of the equilibrium expression:

23. Henderson-Hasselbalch Equation
This equation, , can be changed into another
useful form by taking the negative log of both sides:
This log form of the expression is called the Henderson-Hasselbalch Equation.

24. Henderson-Hasselbalch Equation
Henderson-Hasselbalch Equation: When using this equation it is often assumed that the the equilibrium concentrations of A_ and HA are equal to their initial concentrations due to the validity of most approximations.
Since the initial concentrations of HA and A- are relatively large in a buffered solution, this assumption is generally acceptable.

25. Practice Problem #4
Calculate the pH of a solution containing 0.75 M lactic acid (Ka=1.4x10-4) and 0.25 M sodium lactate. Lactic acid (HC3H5O3) is a common constituent of biologic systems. For example, it is found in milk and is present in human muscle tissue during exertion. pH = 3.38

26. Reminder
Buffered solutions can be formed from a weak base and the corresponding conjugate acid. In these solutions, the weak base B reacts with any H+ added:
The conjugate acid BH+ reacts with any added OH-:

27. Practice Problem #5
A buffered solution contains 0.25 M NH3 (Kb=1.8x10-5) and 0.40 M NH4Cl. Calculate the pH of this solution. pH=9.05

28. Practice Problem #6
Calculate the pH of the solution that results when 0.10 mol gaseous HCl is added to 1.0 L of the buffered solution from #5 (.25M NH3 and .40 NH4Cl). pH= 8.73

29. Summary
Buffered solutions contain relatively large concentrations of a weak acid and the corresponding weak base. They can involve a weak acid HA and the conjugate base A- or a weak base B and the conjugate acid BH+.
When H+ is added to a buffered solution, it reacts essentially to completion with the weak base:

30. Summary
When OH- is added to a buffered solution, it reacts essentially to completion with the weak acid present:
The pH in the buffered solution is determined by the ratio of the concentrations of the weak acid and weak base. As long as this ratio remains virtually constant, the pH will remain virtually constant. This will be the case as long as the concentration of the buffering materials (HA and A- or B and BH+) are large compared with the amounts of H+ or OH- added.

31. Buffering Capacity
15.3

32. Buffering Capacity
The buffering capacity of a buffered solution represents the amount of protons or hydroxide ions the buffer can absorb without a significant change in pH.
The pH of a buffered solution is determined by the ratio [A-] / [HA]. The capacity of a buffered solution is determined by the magnitudes of [HA] and [A-].

33. Practice Problem #7
Calculate the change in pH that occurs when mol of gaseous HCl is added to 1.0 L of each of the following solutions:
Soln A: 5.00 M HC2H3O2 and 5.00 M NaC2H3O2
Soln B: M HC2H3O2 and M NaC2H3O2
Ka=1.8x10-5.
Soln A: pH= Soln B: pH=4.56

34. Optimal Buffering
The optimal buffering occurs when [HA] is equal to [A-].
This ratio of 1 resists the most amount of pH change.
The pKa of the weak acid to be used in the buffer should be as close as possible to the desired pH.
For example, if a buffered solution is needed with a pH of The most effect buffering will occur when [HA] is equal to [A-] and the pKa of the acid is close to 4.0 or Ka=1.0x10-4.

35. Practice Problem #8
A chemist needs a solution buffered at pH=4.30 and can choose from the following acids and their sodium salts. Calculate the ratio [HA]/[A-] required for each system to yield a pH of Which system will work best?
chloroacetic acid (Ka=1.35x10-3)
propanoic acid (Ka=1.3x10-5)
benzoic acid (Ka=6.4x10-5)
hypochlorous acid (Ka=3.5x10-8)
benzoic acid

36. Titrations and pH Curves
15.4

37. Titration Curve An acid-base titration is often graphed by plotting
the pH of the solution
(y-axis) vs. the amount of
titrant added (x-axis).
Point (s) is the equivalence
point/stoichiometric point.
This is where [OH-]=[H+].

38. Strong Acid-Strong Base Titrations
The net ionic reaction for a strong acid- strong base titration is:
To compute [H+] at any point in a titration, the moles of H+ that remains at a given point must be divided by the total volume of the solution.
Titrations often include small amounts. The mole is usually very large in comparison.
Millimole (mmol) is often used for titration amounts.

39. Molarity with mmols

40. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
NaOH has not been added.
Since HNO3 is a strong acid (complete dissociation), pH = 0.699

41. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
10.0 ml of M NaOH has been added.
H+ = 10 mmol – 1.0 mmol of OH- = 9.0mmol
9.0 mmol H+/ 60 ml = 0.15 M
pH=0.82

42. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
20.0 mL of M NaOH has been added.
pH=0.942

43. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
50.0 mL of M NaOH has been added.
pH=1.301

44. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
100.0 mL of M NaOH has been added.
The stoichiometric point/equivalence point is reached. pH=7

45. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
150.0 mL of M NaOH has been added.
pH = 12.40

46. Strong Acid-Base Titration Example Calculation
50.0 ml of M HNO3 is titrated with M NaOH. Calculate the pH of the solution at the following points during the titration.
200.0 mL of M NaOH has been added.
pH=12.40

47. Strong Acid-Base Titration Example Calculation
The results of these calculations are summarized by this graph: The pH changes very gradually until the titration is close to the equivalence point, then dramatic change occurs. Near the E.P., small changes produces large changes in the OH-/H+ ratio.

48. Strong Acid-Base Summary
Before the equivalence point, [H+] can be calculated by dividing the number of millimoles of H+ remaining by the total volume of the solution in mL.
At the equivalence point, pH=7.0
After the equivalence point, [OH-] can be calculated by dividing the number of millimoles of excess OH- by the total volume of the solution. [H+] is calculated from Kw.

49. Titrations of Weak Acids with Strong Bases
When acids don’t completely dissociate, calculations used previously need adjusted.
To calculate [H+] after a certain amount of strong base has been added, the weak acid dissociation equilibrium must be used.
Remember: that a strong base reacts to completion with a weak acid.

50. Calculating the pH Curve for a Weak Acid-Strong Base Titration
Calculating the pH Curve for a Weak Acid- Strong Base Titration is broken down into two steps:
A stoichiometric problem. The reaction of hydroxide ion with the weak acid is assumed to run to completion, and the concentrations of the acid remaining and the conjugate base formed are determined.
An equilibrium problem: The position of the weak acid equilibrium is determined, and the pH is calculated.

51. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
NaOH has not been added.
pH = 2.87

52. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
10.0 mL of 0.10 M NaOH has been added.
pH= 4.14

53. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
25.0 mL of 0.10 M NaOH has been added.
pH=4.74

54. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
40.0 mL of 0.10 M of NaOH has been added.
pH=5.35

55. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
50.0 mL of NaOH is added.
pH=8.72

56. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
60.0 mL of 0.10 M NaOH is added.
pH = 11.96

57. Weak Acid-Strong Base Titration Example Calculation
50.0 mL of 0.10 M acetic acid (HC2H3O2, Ka=1.8x10-5) is titrated with 0.10 M NaOH.
75.0 mL of 0.10 M NaOH is added.
pH = 12.30

58. Weak Acid-Strong Base Titration
Some of the differences in the the titration curves:
The pH increases more rapidly in the beginning of the weak acid titration.

59. Weak Acid-Strong Base Titration
Some of the differences in the the titration curves:
The titration curve levels off near the halfway point due to buffering effects.
Buffering happens when [HA]=[A-]. This is exactly the halfway point of the titration.

60. Weak Acid-Strong Base Titration
Some of the differences in the the titration curves:
The value of the pH at the equivalence point is not 7. The value of the e.p. is greater than 7.

61. Practice Problem #9
Hydrogen cyanide gas (HCN), a powerful respiratory inhibitor, is highly toxic. It is a very weak acid (Ka=6.2x10-10) when dissolved in water. If a 50.0 mL sample of M HCN is titrated with M NaOH, calculate the pH of the solution.
after 8.00 mL of M NaOH has been added.
at the halfway point of the titration
at the equivalence point of the titration
a) pH= b) pH= c)pH=10.96

62. Important Conclusions
When comparing the example calculation and the practice problem with weak acids, two major conclusions can be made:
The same amount of 0.10 M NaOH is required to reach the equivalence point even though HCN is a much weaker acid.
It is the amount of acid, not its strength that determines the equivalence point.

63. Important Conclusions
When comparing the example calculation and the practice problem with weak acids, two major conclusions can be made:
The pH value at the equivalence point is affected by the acid stregth. The pH at the e.p. for acetic acid is The pH at the e.p. for hydrocyanic acid is
The CN- ion is a much stronger base than C2H3O2-.
The smaller acid strength and stronger conjugate base produces a higher pH.

64. Weak Acid Titration Curves
The strength of a weak acid has a significant effect on the shape of its pH curve. The e.p. occurs at the same point, but the shapes of the curve is dramatically different.

65. Practice Problem #10
A chemist has synthesized a monoprotic weak acid and wants to determine its Ka value. To do so, the chemist dissolves 2.00 mmol of the solid acid in mL water and titrates the resulting solution with M NaOH. After 20.0 mL NaOH has been added, the pH is What is the Ka value for the acid? Ka=1.0x10-6

66. Titration of Weak Bases with Strong Acids Example Calculation
Find the pH of a solution of mL of M NH3 is titrated with 0.10 M HCl (Kb=1.8x10- 5).
HCl has not been added.
pH is found with Kb equilibrium. pH = 10.96

67. Titration of Weak Bases with Strong Acids Example Calculation
Find the pH of a solution of mL of M NH3 is titrated with 0.10 M HCl (Kb=1.8x10- 5).
10.0 mL of HCl is added.
H+ ions react to completion. pH=9.85

68. Titration of Weak Bases with Strong Acids Example Calculation
Find the pH of a solution of mL of M NH3 is titrated with 0.10 M HCl (Kb=1.8x10- 5).
25.0 mL of HCl is added.
pH = 9.25

69. Titration of Weak Bases with Strong Acids Example Calculation
Find the pH of a solution of mL of M NH3 is titrated with 0.10 M HCl (Kb=1.8x10- 5).
50.0 mL of HCl is added.
pH = 5.36

70. Titration of Weak Bases with Strong Acids Example Calculation
Find the pH of a solution of mL of M NH3 is titrated with 0.10 M HCl (Kb=1.8x10- 5).
60.0 mL of HCl is added.
pH= 2.21

71. The Titration Curve for a Weak Base with a Strong Acid

72. Acid-Base Indicators
15.5

73. Acid-Base Indicators
There are two common methods for determining the equivalence point of a titration:
Use a pH meter to monitor the pH and then plot the titration curve. The center of the vertical region of the pH curve indicates the equivalence point.
Use an acid-base indicator, which marks the end point of a titration by changing color.

74. Acid-Base Indicators
The equivalence point of a titration is defined by the stoichiometry, but it is not necessarily the same as the end point where the indicator changes color).
Selection of the right indicator for the titration is very important.

75. Indicators
The most common acid-base indicators are complex molecules that are weak acids (HIn).
Most exhibit one color when the proton is attached to the molecule and a different color when the proton is absent.
Example: Phenolphthalein is colorless in
its HIn form and pink in its In-, or basic
form.

76. Indicator Example (RED) (BLUE)
The color will depend on the ratio of [In-] to [HIn]. For most indicators, approximately 1/10th of the initial form must be converted to the final form before a color is perceived by the human eye.
the color change will occur at a pH when

77. Practice Problem #11
Bromthymol blue, an indicator with Ka=1.0x10- 7, is yellow in its HIn form and blue in its In- form. Suppose we put a few drops of this indicator in a strongly acidic solution. If the solution is then titrated with NaOH, at what pH will the indicator color change first be visible? pH=6.0

78. Henderson-Hasselbalch and Indicators
The H-H equation is very useful in determining the pH at which an indicator changes color.
Example:Bromthymol blue (Ka=1x10-7, or pKa=7), the pH at the color change is pH=7- 1=6

79. Indicator Titrations
When a basic solution is titrated, the indicator will initially exist as In- in solution, but as acid is added more HIn is formed. Color change will occur at:
Substituting this reciprocal into the H-H equation gives us:
Bromthymol blue=pH 7+1=8;
The useful range of bromthymol blue is pH(6-8)

80. Indicator Ranges

81. THE END