1. TOPIC 18 ACIDS AND BASES
18.3
pH Curves

2. ESSENTIAL IDEA NATURE OF SCIENCE (3.7)
pH curves can be investigated experimentally but are mathematically determined by the dissociation constants of the acid and base. An indicator with an appropriate end point can be used to determine the equivalence point of the reaction.
NATURE OF SCIENCE (3.7)
Increased power of instrumentation and advances in available techniques – development in pH meter technology has allowed for more reliable and ready measurement of pH.

3. THEORY OF KNOWLEDGE
Is a pH curve an accurate description of reality or an artificial representation? Does science offer a representation of reality?

4. UNDERSTANDING/KEY IDEA 18.3.A
The composition and action of a buffer solution.

5. What is a Buffer?
A buffer solution is resistant to changes in pH on the addition of small amounts of acid or alkali.
The most practical example is our blood which can absorb the acids and bases produced in biologic reactions without changing its pH.

6. Composition of buffer solutions
A buffer solution may contain a weak acid and its salt or a weak base and its salt.
Ex: HF and NaF
Ex: NH3 and NH4Cl
Acidic buffers maintain a pH below 7 and basic buffers maintain a pH above 7.

7. Buffer solutions are a mixture containing both an acid and a base of a weak conjugate pair.
The buffer’s acid neutralizes added alkali or OH- and the buffer’s base neutralizes added acid or H+.

8. Example 1 (Basic weak acid)
Calculate the [H+] and percent dissociation of a mol/dm3 solution of HF (Ka = 7.2x10-4).
What are the major species?
HF and H2O
Set up your ICE table.
HF ↔ H+ + F- I
C -x x x
E x x x
Assume 1.0-x = 1.0 because Ka is so small.

9. Set up your Ka expression and solve for x.
Ka = [H+][F-] = 7.2x10-4 = x x = [H+]=.027
[HF]
Calculate the % dissociation.
% dissoc = x = x 100% = 2.7%
[conc]ini

10. Example 2 (Addition of common ion)
Calculate the [H+] and percent dissociation of a 1.0 mol/dm3 solution of HF (Ka = 7.2x10-4) and 1.0 mol/dm3 NaF.
What are the major species?
HF , H2O, Na+ and F-
Set up your ICE table.
HF ↔ H+ + F- I
C -x x x
E x x x
Assume 1.0-x = 1.0 and 1.0+x = 1.0 because Ka is so small.

11. Set up your Ka expression and solve for x.
Ka = [H+][F-] = 7.2x10-4 = x (1.0) x = [H+]= 7.2x10-4
[HF]
Calculate the % dissociation.
% dissoc = x = 7.2x10-4 x 100% = .072%
[conc]ini
Notice that with the addition of the common ion, equilibrium was pushed back to the reactant side due to LeChatelier’s principle and less of the acid dissociated.

12. Example 3 (Buffered solution)
A buffered solution contains 0.50 mol/dm3 HC2H3O2
(Ka = 1.8x10-5) and 0.50 mol/dm3 NaC2H3O2. Calculate the pH of this solution.
What are the major species?
HC2H3O2, H2O, Na+ and C2H3O2-
The Na+ has no effect on the equilibrium.
HC2H3O2 ↔ H+ + C2H3O2- I
C x x x
E x x x
Assume .50-x = .50 and .50+x = .50 because Ka is so small.

13. Set up your Ka expression and solve for x.
Ka = [H+][C2H3O2-] = 1.8x10-5 = x (.50) x = [H+]= 1.8x10-5
[HC2H3O2]
Solve for pH pH = -log[H+] = -log 1.8x10-5 = 4.74

14. Example 4 (Adding OH- to buffer)
Calculate the pH when mol NaOH is added to dm3 of the solution in Ex 3.
The major species in solution before any reaction occurs are: HC2H3O2, H2O, Na+, OH- and C2H3O2-
The strong base (OH-) will react with the best source of protons until there are no OH- ions left in solution.
The acetic acid is a better source for H+ than water so the reaction is as follows:
OH- + HC2H3O2 ↔ C2H3O H2O
Notice that adding OH- will use up the acid and produce more of the conj base.

15. There are always 2 steps when adding acids and bases to buffer solutions.
A stoichiometry problem where you deal with the addition of the OH- or the H+ added.
Followed by an equilibrium problem with the new concentrations of the acid and conj base.

16. Back to Example 4
Stoichiometry problem (Before and After) Always use moles in this type of problem, not concentration.
1.0dm3 x .50mol/dm dm3 x .50mol/dm3
OH- + HC2H3O2 ↔ C2H3O H2O
Before
After
Notice that all the OH- has been used up and no longer exists.
Before we proceed to the equilibrium problem, we must convert back to concentration.
[HC2H3O2] = .49mol/1dm3 and [C2H3O2-] = .51mol/1dm3

17. Equilibrium problem (ICE table with new values)
HC2H3O2 ↔ H+ + C2H3O2- I
C x x x
E x x x
Assume .49-x = .49 and .51+x = .51 because Ka is so small.
Ka = [H+][C2H3O2-] = 1.8x10-5 = x (.51) x = [H+]= 1.7x10-5
[HC2H3O2]
The new pH = -log1.7x10-5 = 4.76

18. Parctice

21. Henderson Hasselbalch Equation
Ka = [H+][A-]
[HA]
Rearranging to solve for [H+] gives: [H+] = Ka[HA]
[A-]
Take the negative logarithms of both sides to give:
pH = pKa + log [salt] or pOH = pKb + log [salt]
[acid] [base]

22. More on buffers
When the concentrations of the salt and the acid or the salt and the base are equal, then pH = pKa and pOH = pKb.
The pH of a buffer solution depends upon the pKa or pKb of the acid or base and the ratio of the initial concentrations of the salt/acid or salt/base pairs.

23. Factors influencing buffers
Dilution – does not change Ka or Kb and it does not change the ratio of acid or base to salt concentration. Dilution does not change its pH. It does lower the buffering capacity because dilution affects how much strong acid or base can be absorbed.
Temperature – changes Ka or Kb so pH is affected. Temp needs to be held constant.

24. APPLICATION/SKILLS
Know that the nature of acid-base buffers remains the same, but they can be prepared in two different ways:
Half neutralization with strong acid or base
Equal parts weak acid/conj salt

25. Making Buffer Solutions
Start with an acid or a base with a pKa or pKb as close as possible to the required pH of the buffer needed.
Then react the acid or base with enough OH- or H+ to exactly half neutralize it.
Then you can use pH of buffer = pKa or pKb.

26. Example 5
How would you prepare a buffer solution of pH 3.75 starting with methanoic acid, HCOOH?
From the IB data booklet we have pKa of HCOOH = 3.75, so a buffer with equal amounts of the acid and its salt NaCOOH will have a pH of 3.75.
The solution is prepared by reacting the acid with enough NaOH so that half of it is converted into its salt so [HCOOH]=[HCOO-].

27. When acids and bases react, they form salt and water.
These salts can be neutral, basic or acidic.
A strong acid with a strong base produces a neutral salt.
A strong acid with a weak base produces an acidic salt. (strong acid overrules)
A strong base with a weak acid produces a basic salt. (strong base overrules)

28. Salt is another name for ionic compound.
When they dissolve in water, the ions can behave as acids or bases.
Remember that strong acids and bases dissolve completely in water so their ions have no affinity for the H+ or OH- ions.

29. The anions of strong acids such as Cl-, NO3-, I-, ClO4- will not combine with H+ so they have no affect on pH and are therefore true spectator ions.
The cations of strong bases such as Na+, Li+, Ba2+ will not combine with OH- so they are also true spectator ions and will have no affect on pH.
Therefore, mixtures of strong acids and strong bases produce neutral salt solutions in water.

30. When given the salt, you can easily determine which acid and base reacted to form it.
NaCl
NaCl breaks down into Na+ and Cl- ions. The base has to be NaOH and the acid has to be HCl. Do you know why?

31. Hydrolysis
When ions react with water, they can hydrolyze the water by either releasing H+ or OH- ions.

32. Anion Hydrolysis
An anion can be the conjugate base of the parent acid.
When the acid is weak, the conj base is strong enough to hydrolyze water.
A- + H2O(l) ↔ HA + OH-
The release of OH- ions causes the pH of the solution to increase.

33. Strong base with weak acid
NaOH + HC2H3O2 ↔ NaC2H3O2 + H2O
The NaC2H3O2 is soluble so it breaks into Na+ and C2H3O2- ions. The strong cong base C2H3O2- reacts with water in the solution as follows to produce OH-.
C2H3O2- + H2O ↔ HC2H3O2 + OH-
This is the real reason that a strong base with a weak acid produces a basic salt.

34. Example 1
Calculate the pH of a 0.30M solution of NaF. The Ka for HF is 7.2x10-4.
Step 1: Write the equation.
NaF ↔ Na+ + F-
Step 2: Which species reacts with water?
F- + H2O ↔ HF + OH-
Step 3: This is actually a Kb problem, but we only have
Ka. We can solve for Kb by KaKb=Kw.
Kb = (1.00x10-14)/(7.2x10-4)= 1.4x10-11

35. F- + H2O ↔ HF + OH- I
C -x x x
E x x
Kb = [HF][OH-] = 1.4x10-11
[F-] so x2 = 1.4x10-11
.30
x = [OH-] = 2.0x pOH = 5.69, pH = = 8.31

36. Cation Hydrolysis
A cation can be the conjugate acid of the parent base.
When the base is weak and the conj acid a non-metal (NH4+), it is able to hydrolyze water.
M+ + H2O(l) ↔ MOH + H+
The release of H+ into solution causes the pH to decrease.

37. If the cation is a metal, the situation is more complex and depends upon the charge density.
Metal ions from Group 1 and Group 2 (except Be2+) do not have enough charge density to hydrolyze water to generate H+ so they do not show acidic behavior in water.

38. Strong acid with weak base
HCl + NH3 ↔ NH4+ + Cl-
The NH4+ then reacts with water in the solution as follows to produce H+:
NH4+ ↔ NH3 + H+
This is the real reason that a strong acid with a weak base produces an acidic salt.

39. Example 2
Calculate the pH of a 0.10 M NH4Cl solution. The Kb value for NH3 is 1.8x10-5.
Step 1: Write the equation.
NH4Cl ↔ NH4+ + Cl-
Step 2: Which species reacts with water?
NH4+ ↔ NH3 + H+
Step 3: This is actually a Ka problem, but we only have
Kb. We can solve for Ka by KaKb=Kw.
Ka = (1.00x10-14)/(1.8x10-5)= 5.6x10-10

40. NH4+ ↔ NH3 + H+ I
C -x x x
E x x
Ka = [H+][NH3-] = 5.6x10-10
[NH4+] so x2 = 5.6x10-10
.10
x = [H+] = 7.5x pH = 5.12

41. Metal ions that are small and have 2 or 3 positive charges such as Be2+, Al3+ and transition metals most notably Fe3+ can hydrolyze water by pulling enough electron density away from the O-H bond to release the H+ to cause a decrease the pH.

42. UNDERSTANDING/KEY IDEA 18.3.B
The characteristics of the pH curves produced by the different combinations of strong and weak acids and bases.

43. APPLICATION/SKILLS
Know the general shapes of graphs of pH against volume for titrations involving strong and weak acids and bases with an explanation of their important features.

44. What is titration?
Technique in which one solution is used to analyze another.
The neutralization reactions between acids and bases can be carried out in a controlled way using titration.
Standardization is the calculation of the exact concentration of one solution when the other is known.

45. The equivalence point is the point when the acid and base exactly neutralize each other. At this point the major species present are the salt and water.
The titrant is the solution of known concentration delivered from a burette.
The titration or pH curve is the plot of pH vs volume of titrant delivered.

46. NaOH + HCl → NaCl + H2O
Initially, there is only the strong acid present, so to get the initial pH, you just take the negative log of the initial concentration.
You have 50.0 cm3 of 0.10 mol/dm3 HCl
pH = -log(0.1) = 1 Major species: H+ and Cl-
The next step is to add some base, so we are going to add 25.0cm3 of 0.10mol/dm3 NaOH.
Major species before reaction: H+, Cl-, Na+, OH-, H2O
All the OH- will react with available H+ until all OH- ions are consumed.

47. Adding 25.0 cm3 base (con’t) H+ + OH- → H2O
Remember to convert cm3 to dm3 by dividing by 1000:
50cm3 x 1dm3/1000cm3=0.050dm3 & 25cm3 x 1dm3/1000cm3 = 0.025dm3
We now need a before and after table using moles, not concentration.
H OH- → H2O
Before: dm3x.10mol/dm3= x .10=
.005mol H mol OH-
-.0025mol mol
After: mol H mol OH-
To find pH, we need to convert back to concentration of H+ using the new volume of dm dm3 = 0.075dm3
So [H+] = .0025mol/0.075dm3 = pH= -log(.0333) = 1.48

48. Next add 49.0 cm3 of the base Follow the same procedure as before:
H OH- → H2O
Before: dm3x.10mol/dm3= x .10=
.005mol H mol OH-
-.0049mol mol
After: mol H mol OH-
To find pH, we need to convert back to concentration of H+ using the new volume of dm dm3 = dm3
So [H+] = .0001mol/0.0990dm3 = pH= -log(.00101) = 2.99

49. Add 50.0 cm3 of strong base H+ + OH- → H2O
Follow the same procedure as before:
H OH- → H2O
Before: dm3x.10mol/dm3= x .10=
.005mol H mol OH-
-.0050mol mol
After: mol H mol OH-
At this point, all of the OH- ions have consumed all the H+ ions so we are at the equivalence point where only salt and water exist.
The pH = 7 at the equivalence point of a strong acid with a strong base.

50. Add 51.0 cm3 of the strong base
Follow the same procedure as before:
H OH- → H2O
Before: dm3x.10mol/dm3= x .10=
.005mol H mol OH-
-.005mol mol
After: mol H mol OH- in excess
We are now past the equivalence point so we use up all the H+ and then have extra OH- left in solution. We can calculate pH from pOH
So [OH-] = .0001mol/0.101dm3 = pOH= -log(.00099) = 3
pH = 14 – 3 = 11

51. Add 75.0 cm3 of the strong base
Follow the same procedure as before:
H OH- → H2O
Before: dm3x.10mol/dm3= x .10=
.005mol H mol OH-
-.005mol mol
After: mol H mol OH- in excess
We are now past the equivalence point so we use up all the H+ and then have extra OH- left in solution. We can calculate pH from pOH
So [OH-] = .0025mol/0.125dm3 = pOH= -log(.02) = 1.7
pH = 14 – 1.7 = 12.3

52. Titration curve for strong acid and strong base.
Be sure to be able to draw and label a titration curve for a strong acid with a strong base and the opposite conditions.

54. Points to deduce from graph
1. The initial pH = 1 (pH of strong acid)
2. pH changes gradually until equivalence.
3. There is a very sharp jump in pH at equivalence from pH 3 to pH 11.
Remember these were the points right before and right after the equivalence pt.
4. After equivalence, the curve flattens out at a high value (pH of a strong base)
5. pH at equivalence = 7

55. Note
All example problems start with 50 cm3 of 0.10 mol/dm3 acid. The titrant also has the concentration of 0.10 mol/dm3.

56. Weak acid (Ac = acetic acid) with strong base HAc + NaOH ↔ NaAc + H2O
Initially, there is only the weak acid present, so to get the initial pH, you must use an ICE table.
What are the major species?
HAc and H2O (Ka = 1.8x10-5)
Set up your ICE table.
HAc ↔ H+ + Ac- I
C -x x x
E x x x
Ka = [H+][Ac-] = 1.8x10-5= x x = [H+]= pH = 3
[HAc]

57. Add 25.0cm3 of strong base Major species before reaction:
HAc, Na+, OH- and H2O
All the OH- will react with available H+ until all OH- ions are consumed.
Each addition of OH- in the titration requires a “Before and After Table” using moles, followed by a new ICE Table using concentration.
***This is the halfway to the equivalence point where HAc and Ac- will be equal.
So pH = pKa.***

58. HAc + OH- → Ac- + H2O Before: 0.050dm3x.10mol/dm3= .025 x .10=
.005mol H mol OH
-.0025mol mol mol
After: mol H mol OH mol Ac-
Next, change back to concentrations to use the ICE table.
.0025mol/.075dm3 = .033mol/dm3
HAc ↔ H+ + Ac- I
C -x x x
E x x x
Ka = [H+][Ac-] = 1.8x10-5= x(.033) x = [H+]= 1.8x10-5 pH = 4.74
[HAc]

59. Add 50cm3 of strong base (Equiv pt) HAc + OH- → Ac- + H2O
Before: dm3x.10mol/dm3= x .10=
.0050mol H mol OH
-.0050mol mol mol
After: mol H mol OH mol Ac-
You now have different major species: Ac-, Na+, H2O
Ac- + H2O ↔ HAc + OH-
I Kb = Kw/Ka
C -x x x =1.0x10-14
E .05-x x x x10-5
Kb = [HAc][OH-] = 5.6x10-10= x x = [OH-]= 5.3x10-6 pOH = 5.28
[Ac-] pH = 8.72

60. Add 75cm3 of strong base HAc + OH- → Ac- + H2O
Before: dm3x.10mol/dm3= x .10=
.0050mol H mol OH
-.0050mol mol mol
After: mol H mol OH mol Ac-
The pH is now determined by the excess OH- in solution because it overrules the Ac-.
[OH-] = mol/.125dm3 = .02mol/dm3
pOH = -log(.02) = 1.70
pH = = 12.3

61. Buffered Weak Acid-Strong Base Titration Curve
The initial pH is higher than the unbuffered acid
As with a weak acid and a strong base, the equivalence point for a buffered weak acid is higher than pH =7
The conjugate base is strong enough to affect the pH

62. Weak Acid VS
Strong Base
Strong Acid VS
Weak Base

63. UNDERSTANDING/KEY IDEA 18.3.C
The buffer region on the pH curve represents the region where small additions of acids or base result in little or no change in pH.

65. Important points about graph for weak acid with a strong base
1. The initial pH is higher than 1. (pH of a weak acid)
2. pH stays fairly consistent until equivalence (buffer region).
3. There is a smaller jump in pH at the equivalence from about 7.0 – (This is not as large a jump as with strong with strong.)
4. After equivalence, the pH curve flattens out at a high value (pH of a strong base).
5. pH at equivalence is greater than 7 (anion hydrolysis releases OH-)

66. Titration curve for weak acid with a strong base
Be able to draw this titration curve and that of a weak acid with a strong base.

67. Important to note
All of the above graphs can be reversed so that you are adding the acid to the base.
You are expected to be able to draw and label both types of graphs.

68. Important points about graph for a strong acid with a weak base
1. The initial pH is 1. (pH of a strong acid)
2. pH stays fairly consistent until equivalence (buffer region).
3. There is a smaller jump in pH at the equivalence from about 3.0 – 7.0. (This is not as large a jump as with strong with strong.)
4. After equivalence, the pH curve flattens out at a lower value (pH of a weak base).
5. pH at equivalence is less than 7 (cation hydrolysis releases H+)

69. Important points about graph for weak acid with a weak base
1. The initial pH is fairly high. (pH of a weak acid)
2. Addition of a base causes pH to rise steadily.
3. The change in pH at equivalence is much less sharp than the other titrations.
4. After equivalence, the pH curve flattens out at a lower value (pH of a weak base).
5. This is not a good technique to determine the equivalence point.

70. Indicators are substances that change color reversibly according to the pH of the solution.
An indicator can be a weak acid or a weak base in which the undissociated and dissociated forms have different colors.

71. HIn ↔ H+ + In- The HIn is one color and the In- ion is another color.
Remember that the H+ is the factor that affects the pH.
Using Le Chatelier’s principle, you can predict how the equilibrium will respond to a change in the pH by the addition or deletion of H+ ions.
Increasing [H+]: equilibrium shifts to HIn
Decreasing [H+]: equilibrium shifts to In-

72. UNDERSTANDING/KEY IDEA 18.3.D
An acid-base indicator is a weak acid or a weak base where the components of the conjugate acid-base pair have different colors.

73. Equivalence point vs End point
Do not confuse the equivalence point with the end point.
The equivalence point is where stoichiometrically equal amounts of acids and bases have neutralized each other.
The end point is the pH at which the indicator changes color. This occurs at the pH equal to its pKa.

74. HIn ↔ H+ + In- Ka = [H+][In-] [HIn]
When [In-] = [HIn], the indicator is exactly in the middle of its color change and Ka=[H+] so pKa = pH. This is the end point.
Different indicators change colors at different pH values.

75. BOH ↔ B+ + OH- color a color b
Kb = [B+][OH-]
[BOH]
When [B+] = [BOH], the indicator is exactly in the middle of its color change and Kb=[OH-] so pKb = pH. This is the end point.
Different indicators change colors at different pH values.

76. UNDERSTANDING/KEY IDEA 18.3.E
The relationship between the pH range of an acid-base indicator, which is a weak acid, and its pKa value.

77. GUIDANCE
Be able to select an appropriate indicator for a titration, given the equivalence point of the titration and the end point of the indicator.

78. Steps for choosing an indicator
1. Determine what type of titration you have. (ex: strong acid with weak base, etc.)
2. Deduce the pH of the salt solution at the equivalence.
3. Choose an indicator with an end point in the range of the equivalence point from a table.
4. The end point range should be within +/- 1 pH unit on either side of the pKa.

79. Strong acid with strong base
pH range at equivalence is 3-11
Indicators: phenolphtalein
pKa = 9.50
End point range of pth =
Colorless to pink
methyl orange
pKa = 3.46
End point range of methyl orange =
Red to yellow

80. Weak acid with strong base
pH range at equivalence is 7-11
Indicators: phenolphtalein
pKa = 9.50
End point range of pth =
Colorless to pink
phenol red
pKa = 8.00
End point range of phenol red =
Yellow to red

81. Strong acid with weak base
pH range at equivalence is 3-7
Indicators: methyl orange
pKa = 3.46
End point range of methyl orange =
Red to yellow
bromophenol blue
pKa = 4.10
End point range of bromophenol blue =
Yellow to blue

89. Citations
International Baccalaureate Organization. Chemistry Guide, First assessment Updated Brown, Catrin, and Mike Ford. Higher Level Chemistry. 2nd ed. N.p.: Pearson Baccalaureate, Print. Most of the information found in this power point comes directly from this textbook. The power point has been made to directly complement the Higher Level Chemistry textbook by Catrin and Brown and is used for direct instructional purposes only.