1. pH and Buffers
The Whole Story

3. Acid & Base and pH 0 7 14 1 M 10-7 M 10-14 M Very Neutral Very
This should be a review but we will go over it to help refresh your memory
1 M M M
Very Neutral Very
Acidic Basic

4. Trivia time:
Why was the concept of pH developed?

5. Trivia time: Why was the concept of pH developed?
By the Danish Biochemist Sorensen to test the acidity of the beer he was making
Who said biochemistry isn’t cool?

6. pH - pouvoir hydrogene (the power of hydrogen)
Water undergoes ionization
Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions
Water can act as both an acid and base
The equilibrium constant for the ionization of water is:
The concentration of pure water
1 liter = 1000g MW of water is
the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9
Very little water actually dissociates
So Keq is very small – not easily measured or easy to use

7. pH - pouvoir hydrogene (the power of hydrogen)
Water undergoes ionization
Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions
Water can act as both an acid and base
The equilibrium constant for the ionization of water is:
The concentration of pure water
1 liter = 1000g MW of water is
the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9
Very little water actually dissociates
So Keq is very small – not easily measured or easy to use H 2 O 3 + OH _

8. pH - pouvoir hydrogene (the power of hydrogen)
Water undergoes ionization
Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions
Water can act as both an acid and base
The equilibrium constant for the ionization of water is:
The concentration of pure water
1 liter = 1000g MW of water is
the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9
Very little water actually dissociates
So Keq is very small – not easily measured or easy to use H 2 O 3 + OH _ K eq =
[products]
[reactants]
[H+ ]
[OH- ]
[H2O]2

9. pH - pouvoir hydrogene (the power of hydrogen)
Water undergoes ionization
Water ionizes to form the hydronium (hydroxyl) ion and hydroxide ions
Water can act as both an acid and base
The equilibrium constant for the ionization of water is:
The concentration of pure water
1 liter = 1000g MW of water is
the final concentration of water is 55M and H+ concentration is about 1.8 x 10-9
Very little water actually dissociates
So Keq is very small – not easily measured or easy to use H 2 O 3 + OH _
[H2O] is
effectively
constant.
[products]
[H+ ]
[OH- ] K = = eq
[reactants]
[H2O]2

10. Instead a different constant is used where the denominator is ignored
Kw= 1.0 X
pH is a measure of the acidity and basicity of a solution
when [H+] = [OH-] the solution is neutral and pH is 7
when [H+] > [OH-] the solution is acidic and pH is less than 7
when [H+] < [OH-] the solution is basic and pH is more than 7
a change in 1 pH units is = a ten fold change in hydrogen ion concentration

11. Instead a different constant is used where the denominator is ignored
Kw= 1.0 X
pH is a measure of the acidity and basicity of a solution
when [H+] = [OH-] the solution is neutral and pH is 7
when [H+] > [OH-] the solution is acidic and pH is less than 7
when [H+] < [OH-] the solution is basic and pH is more than 7
a change in 1 pH units is = a ten fold change in hydrogen ion concentration _ +
Therefore: K =
[H ]
[OH ] w
And as measured in pure water
[H ] = +
[OH ] = 1 X 10 _ - 7

12. Instead a different constant is used where the denominator is ignored
Kw= 1.0 X
pH is a measure of the acidity and basicity of a solution
pH = -log[H+]
when [H+] = [OH-] the solution is neutral and pH is 7
when [H+] > [OH-] the solution is acidic and pH is less than 7
when [H+] < [OH-] the solution is basic and pH is more than 7
a change in 1 pH units is = a ten fold change in hydrogen ion concentration _ +
Therefore: K =
[H ]
[OH ] w
And as measured in pure water
[H ] = +
[OH ] = 1 X 10 _ - 7

13. The extent of ionization of a weak acid is a function of its acid dissociation constant pKa
Bronsted and Lowry acid and bases
acid donates protons
bases accepts protons
Strong acids dissociate nearly fully
Weak acids only partially dissociate
Acids with Ka < 1 are considered weak acids
Ka for acetic acid is 1.76 x > difficult to work with so instead use log scale:
pKa = -log Ka
So the pKa of acetic acid is = -log 1.76 x = 4.75
The pH is a measure of acidity and the pKa is a measure of acid strength

14. The extent of ionization of a weak acid is a function of its acid dissociation constant pKa
Bronsted and Lowry acid and bases
acid donates protons
bases accepts protons
Strong acids dissociate nearly fully
Weak acids only partially dissociate
Acids with Ka < 1 are considered weak acids
Ka for acetic acid is 1.76 x > difficult to work with so instead use log scale:
pKa = -log Ka
So the pKa of acetic acid is = -log 1.76 x = 4.75
The pH is a measure of acidity and the pKa is a measure of acid strength
HA H O H O A 2 3 + _
acid base conjugate conjugate
acid base

15. [H+] = [ acid] and thus pH = -log [acid]
The extent of ionization of a weak acid is a function of its acid dissociation constant pKa
Bronsted and Lowry acid and bases
acid donates protons
bases accepts protons
Strong acids dissociate nearly fully
[H+] = [ acid] and thus pH = -log [acid]
Weak acids only partially dissociate
Acids with Ka < 1 are considered weak acids
Ka for acetic acid is 1.76 x > difficult to work with so instead use log scale:
pKa = -log Ka
So the pKa of acetic acid is = -log 1.76 x = 4.75
The pH is a measure of acidity and the pKa is a measure of acid strength
HA H O H O A 2 3 + _
acid base conjugate conjugate
acid base

16. [H+] = [ acid] and thus pH = -log [acid]
The extent of ionization of a weak acid is a function of its acid dissociation constant pKa
Bronsted and Lowry acid and bases
acid donates protons
bases accepts protons
Strong acids dissociate nearly fully
[H+] = [ acid] and thus pH = -log [acid]
Weak acids only partially dissociate
Acids with Ka < 1 are considered weak acids
Ka for acetic acid is 1.76 x > difficult to work with so instead use log scale:
pKa = -log Ka
So the pKa of acetic acid is = -log 1.76 x = 4.75
The pH is a measure of acidity and the pKa is a measure of acid strength + _
HA H O H O A 2 3
acid base conjugate conjugate
acid base K =
[H ] +
[A ] _
[HA] a
pKa and Ka are used for all acids.

17. Dissociation constants and pKa values
More
Acidic C
hemi c al K a ( M ) p K a F o r
mic
aci d -4 3 . 8 1 . 77 x 10 C a rb
oni c a ci d -7 6 . 4 4 . 30 x 10 B ic a r b o na te
-11 10 . 2 5 . 61 x 10 Am m o n
ium
-10 9 . 2 5 . 62 x 10
Note relationship between pKA, Ka and acid strength.

18. Titrations and pKa
pKa of a weak acid is determined experimentally by titration.
pKa is when the concentration of acid and base is equal in a titration (ask yourself if that then equal to pH 7).
There is a point in a titration of a weak acid where the change in pH is very little. This is the buffer action of the acid.
Many acids have more than one ionizable group (polyprotic)

19. The relationship between pH and pKa FOR A WEAK ACID is described by the Henderson-Hasselbalch equation
HA H+ + A-

20. The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid

21. The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a

22. The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get

23. The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
[A-]

24. Multiply each side by log
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]

25. Multiply each side by log
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]

26. Multiply each side by log
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1

27. Multiply each side by log
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1
[HA]
-Log[H+] = -Log K - Log a
[A-]

28. Multiply each side by log
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1
[HA]
Substitute
pH = -Log[H+]
pKa = Log Ka
-Log[H+] = -Log K - Log a
[A-]

29. Multiply each side by log
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1
[HA]
Substitute
pH = -Log[H+]
pKa = Log Ka
-Log[H+] = -Log K - Log a
[A-]
[HA]
pH = pKa - Log
[A-]

30. Multiply each side by log Remove (-) and invert last term
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1
[HA]
Substitute
pH = -Log[H+]
pKa = Log Ka
-Log[H+] = -Log K - Log a
[A-]
[HA]
pH = pKa - Log
[A-]
Remove (-) and invert last term

31. Multiply each side by log Remove (-) and invert last term
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1
[HA]
Substitute
pH = -Log[H+]
pKa = Log Ka
-Log[H+] = -Log K - Log a
[A-]
[HA]
pH = pKa - Log
[A-]
Remove (-) and invert last term
[A-]
pH = pKa + Log
[HA]

32. Multiply each side by log Remove (-) and invert last term
The relationship between pH and pKa is described by the Henderson-Hasselbalch equation
HA H+ + A-
Start with a weak acid K = [
H+] [A-] HA ] a
Arrange to get
[HA]
[H+] K = a
Multiply each side by log
[A-]
[HA]
Log[H+] = Log K + Log a
[A-]
Multiply by -1
[HA]
Substitute
pH = -Log[H+]
pKa = Log Ka
-Log[H+] = -Log K - Log a
[A-]
[HA]
pH = pKa - Log
[A-]
Remove (-) and invert last term
[A-]
pH = pKa + Log
[HA]
Henderson -Hasselbalch

33. What is the H-H Equation Used For?
This is used to determine the concentration of acid and base at a given pH. It is Also used to determine the pH of a known solution. These concepts are used to calculate buffer strength and understand the pH of a biological solution.
Remember that buffers are mixtures of weak acids and their conjugate bases that resist pH by shifting the equilibrium between the acid and base in response to the pH of a solution.

34. Case 1) when the concentration of base equals the acid._
[A ]
pH = pKa + Log
[HA]
[ x ]
= pKa + Log
[ x ]
pH = pKa
When pH = pKa 50% of the acid is dissociated

35. Case 2) when the pH is above or below 1 pH unit of the pKa_
[A ]
5.00 = Log
[HA] _
[A ]
1.00 = Log
[HA] _
[A ]
10.0 =
[HA]
Then 90% of the buffer is in the conjugate base for
If pH is 2 units different then 99% of buffer is in the conjugate base form
What does this mean about the buffering ability if more acid or base is added?

36. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.

37. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.
Start with the HH equation
pH = pKa + Log [A-]/[HA]

38. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.
Start with the HH equation
pH = pKa + Log [A-]/[HA]
Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms

39. Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.
Start with the HH equation
pH = pKa + Log [A-]/[HA]
Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms
pH = Log (0.1/0.25)

40. Remember [X] indicates molar concentration
Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.
Start with the HH equation
pH = pKa + Log [A-]/[HA]
Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms
pH = Log (0.1/0.25)
Remember [X] indicates molar concentration

41. Remember [X] indicates molar concentration
Calculate the pH of a mixture of 250 mM acetic acid and 100 mM Na acetate. The pKa of acetic acid is 4.75.
Start with the HH equation
pH = pKa + Log [A-]/[HA]
Recognize that a compound that ends with -ate is the base form of a weak acid and plug in the known terms
pH = Log (0.1/0.25)
Remember [X] indicates molar concentration
pH =4.35

42. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?

43. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation

44. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]

45. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms

46. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]

47. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]
Arrange

48. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]
Arrange
1.14 = Log [A-]/[HA]

49. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]
Arrange
1.14 = Log [A-]/[HA]
Inv Log on both sides

50. What is the ratio of lactic acid to lactate in a buffer at pH of 5. 00
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]
Arrange
1.14 = Log [A-]/[HA]
Inv Log on both sides
Ratio = 13.80

51. That means that for each mole of acid there are 13.8 moles of base
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]
Arrange
1.14 = Log [A-]/[HA]
Inv Log on both sides
Ratio = 13.80
That means that for each mole of acid there are 13.8 moles of base

52. That means that for each mole of acid there are 13.8 moles of base
What is the ratio of lactic acid to lactate in a buffer at pH of The pKa of lactic acid is 3.86?
Again we recognize this is a weak acid and start with the HH equation
pH = pKa + Log [A-]/[HA]
Plug in the known terms
5.00 = Log [A-]/[HA]
Arrange
1.14 = Log [A-]/[HA]
Inv Log on both sides
Ratio = 13.80
That means that for each mole of acid there are 13.8 moles of base

53. What is the concentration of base and acid you need to add to make a 50 mM solution of lactate buffer at pH 4.0? The MW of Lactic acid is 91 amu and sodium lactate is 102 amu. This is for you to take home try and we will calculate an answer with Dr Provost - Also start to do the pH Homework!