1. Buffers and Titrations
Chapter 18
Buffers and Titrations

2. Problem Calculate the pH of 1.0 L of a 0.25 M acetic acid solution?
Ka = 1.8 x 10-5
CH3COOH + H2O CH3COO H3O+

3. CH3COOH + H2O CH3COO H3O+
(I)
(C) x x x
(E) –x x x

4. [CH3COO-] [H3O+]
Ka =
[CH3COOH]
(x) (x)
1.8 x = (ignore x)
( x)
(x)2
1.8 x 10-5 =
(0.25)
X (H3O+) =

5. [H3O+] =
pH = -log [H3O+]
pH = -log (0.0021)
pH = 2.67

6. Usually prepared from a conjugate acid-base pair.
Buffers: A buffer causes a solution to be resistant to a change in pH when a strong acid or base is added.
Usually prepared from a conjugate acid-base pair.
1) Weak acid and conjugate base
CH3COOH / CH3COO-
2) Weak base and conjugate acid
NH3 / NH4+

7. How a buffer works CH3COOH + OH- CH3COO- + H2O
CH3COO- + H3O CH3COOH + H2O

10. Buffer Problem Example, page 807
If 1.0 L of a 0.25 M acetic acid solution has a pH of 2.67, what is the pH after adding 0.1 moles of sodium acetate, NaCH3CO2?
Ka = 1.8 x 10-5
CH3COOH + H2O CH3COO H3O+

11. CH3COOH + H2O CH3COO H3O+
(I)
(C) x x x
(E) –x x x

12. [CH3COO-] [H3O+]
Ka =
[CH3COOH]
(0.1 + x) (x)
1.8 x = (ignore x)
( x)
(0.1) (x)
1.8 x 10-5 =
(0.25)
X (H3O+) = 4.5 x 10-5

13. [H3O+] = 4.5 x 10-5
pH = -log [H3O+]
pH = -log (4.5 x 10-5)
pH = 4.35

14. Buffer Problem Example 18.2, page 812
What is the pH of an acetic acid/sodium acetate buffer solution with [CH3COOH] = 0.70 M, and [NaCH3CO2] = 0.60 M?
Ka = 1.8 x 10-5
CH3COOH + H2O CH3COO H3O+

15. CH3COOH + H2O CH3COO H3O+
(I)
(C) x x x
(E) –x x x

16. [CH3COO-] [H3O+]
Ka =
[CH3COOH]
(0.6 + x) (x)
1.8 x = (ignore x)
(0.7 - x)
(0.6) (x)
1.8 x 10-5 =
(0.7)
X (H3O+) = 2.1 x 10-5

17. [H3O+] = 2.1 x 10-5
pH = -log [H3O+]
pH = -log (2.1 x 10-5)
pH = 4.68

18. Henderson-Hasselbalch Equation
[CH3COO-] [H3O+]
Ka =
[CH3COOH]
[conj. base] [H3O+]
Ka =
[acid]
[acid]
[H3O+] =
x Ka
[conj. base]

19. and re-write the equation:
[acid]
[H3O+] =
x Ka
[conj. base]
If you take the –log of both sides of the equation
and re-write the equation:
[conj. base] pH
= pKa + log
[acid]
Henderson-Hasselbalch Equation

20. Henderson-Hasselbalch Equation
(for bases)
[conj. acid]
pOH
= pKb + log
[base]
Also:
Kw = Ka x Kb

21. The pH of a buffer solution is controlled by:
The strength of the acid (Ka or pKa)
Concentration of the acid and conjugate base

22. If the concentrations of the acid and conjugate base are the same:
[conj. base]
= 1
[acid]
The log of 1 = 0, so pH = pKa

23. The Ka for benzoic acid is 6.3 x 10-5 So the pKa = 4.20
Example 18.3, page 813
Benzoic acid (C6H5COOH, 2.00 grams) and sodium benzoate (C6H5COONa, 2.00 grams)
are dissolved in enough water to make 1.0 L of solution. Calculate the pH of the solution.
C6H5COOH + H2O C6H5COO- Na+ + H3O+
The Ka for benzoic acid is 6.3 x 10-5
So the pKa = 4.20

24. Benzoic acid (C6H5COOH): 2.00 grams x = 0.0164 moles (in 1.0 L)
Sodium benzoate (C6H5COONa):
2.00 grams x = mole (in 1.0 L)
1mole
144.1 g

25. (0.0139) pH
= log
(0.0164) pH
= log (0.848)
pH = 4.13

26. To be useful. A buffer solution must:
Control the pH of a solution at a desired value.
Keep the pH relatively constant after the addition of some acid or base.
If too much acid or base is added any buffer will loose it’s capacity to control the pH of a solution.
3) When choosing a buffer, select an acid / conjugate base close to the desire pH.

27. Acid Conjugate base Ka pKa
You wish to prepare a buffer solution with a pH of Which acid / conjugate base pair would you choose:
Acid Conjugate base Ka pKa
HSO4- SO x
CH3CO2H CH3CO x
HCO CO x

28. What is the pH of a buffer solution containing
Example
What is the pH of a buffer solution containing
2.0 moles of NH3 (ammonia) and 3.0 moles of NH4Cl in a volume of 1.0 liter of water.
NH H2O NH4Cl + OH-
The Kb for ammonia is 1.8 x 10-5
So the pKb = 4.74

29. [conj. acid]
pOH
= pKb + log
[base]
(3 M)
pOH
= log
(2 M)
pOH
= log (1.5)
pOH = 4.92
pH = 14 – 4.92 = 9.08

30. How does a buffer maintain pH?
What is the pH of a solution containing 2.0 moles of NH3 (ammonia) and 3.0 moles of NH4Cl after 0.50 moles of HCl is added. pKb = 4.74
NH HCl NH4Cl
moles moles moles
HCl is the limiting reagent (smallest number of moles) and will be completely consumed in the reaction.

31. NH HCl NH4Cl
(I)
2.0 –
(E)

32. [conj. acid]
pOH
= pKb + log
[base]
(3.50 M)
pOH
= log
(1.50 M)
pOH
= log (2.33)
pOH = 5.10
pH = 14 – 5.10 = (pH is slightly lower than previous problem, 9.08)

33. Example 18.5, page 816
What is the pH of a solution when 1.00 mL of 1.00 M HCl is added to 1.0 L of an acetic acid / sodium acetate buffer where [CH3COOH] = 0.70 M, and [NaCH3CO2] = 0.60 M?
Ka = 1.8 x so the pKa = 4.74
CH3COO- + H CH3COOH
moles moles moles

34. CH3COO HCl
CH3COOH
(I)
0.60 –
(E)

35. (0.598) pH
= log
(0.701) pH
= log (0.853)
pH = 4.67
The original buffer solution before adding HCl has a
pH of So adding a small amount of HCl does not change the pH that much.

36. Acid-Base Titrations: A titration can be used to determine the quantity of an acid or base or determining the purity of a substance.
Equivalence Point: The point where the amount of base = amount of acid.

37. Strong acid strong base titration:
the pH at the equivalence point is 7
Weak acid – strong base titration:
the pH at the equivalence point is > 7
Strong acid – weak base:
the pH at the equivalence point is < 7

38. There are four distinct regions in a titration curve of an acid and base:
The initial pH
The pH before the equivalence point.
The pH at the equivalence point.
The pH after the equivalence point.

39. Strong Acid and Strong Base

40. Problem – page 818
Find the pH of a solution after 10.0 mL of M NaOH has been added to 50 mL of M HCl.
To solve the problem, set up an ICE table using moles as the initial amounts.
Moles = c x v
To determine the final pH you will then need to divide moles by the final volume to get [H+]

41. H3O OH H2O
(I)
moles
(NaOH is consumed)
(E)
After the reaction:
.004 moles / 0.06 L = M solution of HCl

42. Initial pH:
pH = -log [H+]
pH = -log (0.10) = 1.0
Final pH after NaOH addition:
pH = -log (0.0667) = 1.18

43. Problem – page 818
Find the pH of a solution after 49.5 mL of M NaOH has been added to 50 mL of M HCl.
To solve the problem, set up an ICE table using moles as the initial amounts.
Moles = c x v
To determine the final pH you will then need to divide moles by the final volume to get [H+]

44. H3O OH H2O
(I)
moles
(NaOH is consumed)
(E)
After the reaction:
moles / L = M solution of HCl

45. Initial pH:
pH = -log [H+]
pH = -log (0.10) = 1.0
Final pH after NaOH addition:
pH = -log (0.0005) = 3.3

46. Strong Acid and Strong Base
The pH at the equivalence point is the steepest part of the vertical line

47. Problem – page 820
Find the pH of a solution after 55.0 mL of M NaOH has been added to 50 mL of M HCl.
After the equivalence point, the HCl will be completely consumed in the reaction and therefore the limiting reagent.

48. H3O OH H2O
(I)
moles
(HCl is consumed)
(E)
After the reaction:
moles / L = M solution of NaOH

49. Initial pH:
pH = -log [H+]
pH = -log (0.10) = 1.0
Final pH after NaOH addition:
pOH = -log [OH-]
pOH = -log (0.0048) = 2.32
pH: 14 – 2.32 = (solution is basic)

50. Weak Acid and Strong Base
Let’s look at the titration of mL of 0.10 M acetic acid with 0.10 M NaOH.
Calculate the pH before adding NaOH
The pH before the equivalence point. (buffer region)
The pH at the equivalence point.

51. Weak Acid and Strong Base

52. CH3COOH + NaOH Na+CH3COO- + H2O
At the beginning of the titration (before NaOH is added) the pH can be calculated:
CH3COOH + H2O CH3COO H3O+
[CH3COO-] [H3O+]
Ka =
[CH3COOH]
pH = -log [H3O+]

53. At the beginning of the titration (before NaOH is added) the pH can be calculated:
CH3COOH + H2O CH3COO H3O+
(X) (X)
Ka = = 1.8 x 10-5
(0.1)
pH = -log ( ) = 2.87

54. After the NaOH is added, the pH of the solution can be calculated using the Henderson-Hasselbalch equation since this is a buffer problem:
[conj. base] pH
= pKa + log
[acid]
Since this is a buffer problem, the pH only rises slowly according to the chart.

55. This point is at halfway to the equivalence point.
If the concentrations of the acid and conjugate base are the same, [CH3COOH] = [CH3COO-] :
[conj. base]
= 1
[acid]
The log of 1 = 0, so pH = pKa
This point is at halfway to the equivalence point.

56. Problem 18.6 – page 822
Consider the titration of mL of M acetic acid with M NaOH.
CH3COOH + NaOH Na+CH3COO H2O
What is the pH of the solution when 90.0 mL of M NaOH has been added to mL of M acetic acid?

57. CH3COOH + NaOH Na+CH3COO- + H2O
(I)
moles
(E)
NaOH is used up in the reaction because there are less
moles of NaOH versus acetic acid

58. (0.0090) pH
= log
(0.0010) pH
= log (9.0)
pH = 5.70

59. What is the pH at the equivalence point?
Problem 18.6 – page 822
Consider the titration of mL of M acetic acid with M NaOH.
CH3COOH + NaOH Na+CH3COO H2O
What is the pH at the equivalence point?

60. CH3COOH + NaOH Na+CH3COO- + H2O
(I)
moles
(E) (moles)
To reach the equivalence point the moles of acetic acid and
NaOH are equal. Therefore, for every mole of NaOH reacted an equal number of moles of acetate ion are formed

61. CH3COO H2O CH3COOH + OH-
(I)
molarity
- x x x
(E) x x x
The concentration of acetate ion formed is
(0.01 mol / 0.20 L) = M. Use this in the above ICE table to calculate [OH-] and then pH. Kb = 5.6 x 10-10

62. (x) (x)
Kb = = 5.6 x 10-10
( x)
x2 = 2.8 x 10-11
x = 5.3 x 10-6
pOH = -log [OH-]
pOH = -log (5.3 x 10-6) = 5.28
pH = 14 – 5.28 = 8.72

63. Titration of Weak Polyprotic Acids
H2C2O4 + NaOH HC2O H2O
HC2O NaOH C2O H2O

65. pH Indicators

66. An indicator is usually a weak acid that changes color as it is deprotonated.
HInd + H2O Ind H3O+
one color different color