1. Acid – Base Titrations

2. Ca·Va·nº H = Cb·Vb·nº OH
EQUIVALENCE POINT
The equivalence point, or stoichiometric point, of a chemical reaction is the point at which an added titrant is stoichiometrically equal to the number of moles of substance (known as analyte) present in the sample: the smallest amount of titrant that is sufficient to fully neutralize or react with the analyte. In some cases there are multiple equivalence points, which are multiples of the first equivalence point, such as in the titration of a diprotic acid.
Acid-Base Equivalence Point - the point at which chemically equivalent quantities of acid and base have been mixed, can be found by means of an indicator
Ca·Va·nº H = Cb·Vb·nº OH

3. Titration Curve
A titration curve is a plot of pH vs. the amount of titrant added. Typically the titrant is a strong (completely) dissociated acid or base. Such curves are useful for determining endpoints and dissociation constants of weak acids or bases.

5. Features of the Strong Acid-Strong Base Titration Curve
The pH starts out low, reflecting the high [H3O+] of the strong acid and increases gradually as acid is neutralized by the added base.
Suddenly the pH rises steeply. This occurs in the immediate vicinity of the equivalence point. For this type of titration the pH is 7.0 at the equivalence point.
Beyond this steep portion, the pH increases slowly as more base is added.

6. Sample Calculation: Strong Acid-Strong Base Titration Curve
Problem Consider the titration of 40.0 mL of M HCl with M NaOH.
Region 1. Before the equivalence point, after adding 20.0 mL of M NaOH. (Half way to the equivalence point.)
Initial moles of H3O+ =
- Moles of OH- added =

7. Sample Calculation: Strong Acid-Strong Base Titration Curve (Cont. I)
Region 2. At the equivalence point, after adding 40.0 mL of M NaOH.
Initial moles of H3O+ = L x M = M H3O+
- Moles of OH- added = L x M = mol OH-

8. Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. II)
Region 3. After the equivalence point, after adding 50.0 mL of M NaOH. (Now calculate excess OH-)
Total moles of OH- = L x M = mol OH- -Moles of H3O+ consumed = L x M = mol

9. HPr = Propionic Acid

10. The four Major Differences Between a Strong Acid-Strong Base Titration Curve and a Weak Acid-Strong Base Titration Curve
The initial pH is higher.
A gradually rising portion of the curve, called the buffer region, appears before the steep rise to the equivalence point.
The pH at the equivalence point is greater than 7.00.
The steep rise interval is less pronounced.

11. Sample Calculation: Weak Acid-Strong Base Titration Curve
Problem Consider the titration of 40.0 mL of M HPr (Ka = 1.3 x 10-5) with M NaOH.
Region 1. The solution of weak acid to be titrated, before any base is added.
Solution:
Ans:

12. Sample Calculation: Weak Acid-Strong Base Titration Curve (Cont.I)
Problem Consider the titration of 40.0 mL of M HPr (Ka = 1.3 x 10-5) with M NaOH.
Region 2. After 30. mL of base (total) has been added. This is clearly in the buffer region of the titration curve.
Solution: Refer to Lecture 23.
Can use the calculator program, ‘Buf’ developed in lecture 23. But first must calculate the nominal amounts of acid and base forms of the weak acid created by addition of the strong base. These are: [HA]0 =
[A-]0 =
Ans: From buffer program: pH =

13. Sample Calculation: Weak Acid-Strong Base Titration Curve (Cont.ll)
Problem Consider the titration of 40.0 mL of M HPr (Ka = 1.3 x 10-5) with M NaOH.
Region 3. After 40. mL of base (total) has been added. This is clearly at the equivalence point of the titration curve.
Solution: Refer to Lecture 23.
Can use the calculator program developed in lecture 23. But first must calculate the nominal amounts of acid and base forms of the weak acid created by addition of the strong base. These are:
[HA]0 =
[A-]0 =
Ans: From buffer program: pH =

14. Sample Calculation: Strong Acid-Strong Base Titration Curve (cont. IIl)
Region 4. After the equivalence point, after adding 50.0 mL of M NaOH. (Now calculate excess OH-)
Total moles of OH- = Moles of weak acid consumed =
Moles of OH- remaining =

16. The four Major Differences Between a Weak Acid-Strong Base Titration Curve and a Weak Base-Strong Acid Titration Curve
The initial pH is above 7.00.
A gradually decreasing portion of the curve, called the buffer region, appears before a steep fall to the equivalence point.
The pH at the equivalence point is less than 7.00.
Thereafter, the pH decreases slowly as excess strong acid is added.

18. Features of the Titration of a Polyprotic Acid with a Strong Base
The loss of each mole of H+ shows up as separate equivalence point (but only if the two pKas are separated by more than 3 pK units).
The pH at the midpoint of the buffer region is equal to the pKa of that acid species.
The same volume of added base is required to remove each mole of H+.

20. Acid-Base Indicators and the Measurement of pH
Definition: A weak organic acid, HIn that has a different color than its conjugate base, In-, with the color change occurring over a specific and relatively narrow pH range.
Typically, one or both forms are intensely colored, so only a tiny amount of indicator is needed, far too little to perturb the pH of the solution.
Since the indicator molecule is a weak acid, the ratio of the two forms governed by the [H3O+] of the test solution:

21. Let us consider quantitatively, the case of titrating a weak acid with a strong base. If the weak acid has one dissociable proton, then the overall reaction is:
HA + OH- = A- + H20
We will assume that the strong base NaOH and the weak acid anion NaA are completely dissociated in solution.
Furthermore, we will not neglect the contribution of the dissociation of water.

22. Our Titration System is Governed by Four Equations

23. Such a system has 8 experimentally measurable variables: Ka, Kw, [HA]0 and [NaOH]0
If we assume that the first four (Ka, Kw, [HA]0 and [NaOH]0) are known, then we are left with 4 equations in 4 unknowns.
Of the 4 unknowns, the only one we can conveniently measure is [H+]. This suggests that we solve the four equations for [H+] by successive elimination.

24. The Exact Solution to the Titration Problem
We now proceed to solve the four equations for [H+] by successive elimination of variables. The result for [H+] is
Finding [H+] requires that we find the three roots of the above cubic equation and then selecting the one root that is consistent with physical reality, i.e. leads to all positive concentrations. Fortunately, in the above case only one root is positive and the other two are negative.
This positive root can be found by the ‘Solver’ function of your calculator. You can make a wild guess that is positive and the calculator will converge to the correct answer.

25. Answers
Region 1, pH = 1.477, Region 2, pH = 7.000, Region 3, pH =
Region 1, pH = 2.95 , Region 2, pH = 5.36 , Region 3,
pH = 8.79 , Region 4 = 12.05

26. Weak Acid Titration Calculations26

27. Strong and Weak acids Equilibrium and pH Strong Acids: Weak Acids:
HClO4 H2SO4
HNO3 HI
HBr HCl
HClO3
Weak Acids:
“The Rest” 27

28. Starting pH Equivalence point
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
Starting pH
Equivalence point
0.10 M NaOH 28

29. How do the two values compare?
Acetic acid is a weak acid. What would be the starting pH of the solution if it were a strong acid? (25.0 mL of 0.10M Acetic acid)
Do a quick calculation as if acetic acid were a strong acid and compare this value to that on the graph.
How do the two values compare?
Weak Acid: Graph starting pH = 2.87
Calculated as Strong acid: pH = -log(0.10) = 1.0
How and Why are they different?
“Let’s investigate!” 29

30. Weak
Strong 30

31. How do the two values compare?
Graph starting pH = 2.87
Strong acid calculation: pH = -log(0.10) = 1.0
What is the significance of this difference?
Since the two values differ by ~2 pH units and pH is a log scale, the concentration of H3O+ in the strong acid calculation is 1 x 102 or 100 times greater than that observed on the graph.
Let’s see WHY they are different? 31

32. 100% ionized (completely dissociated) in water. HCl + H2O  H3O+ + Cl-
Strong Acids:
100% ionized (completely dissociated) in water.
HCl + H2O  H3O+ + Cl-
Note the “one way arrow”.
Weak Acids:
Only a small % (dissociated) in water.
HC2H3O2 + H2O  H3O+ + C2H3O2-
Note the “2-way” arrow.
Why are they different? 32

33. ADD WATER to MOLECULAR ACID
Strong Acids:
HCl HCl
HCl
(H2O)
ADD WATER to MOLECULAR ACID 33

34. Strong Acids: Cl- H3O+ (H2O) Cl- H3O+ H3O+ Cl- Cl- H3O+ H3O+ Cl-
Note: No HCl molecules remain in solution, all have been ionized in water. 34

35. Add water to MOLECULES of WEAK Acid
Weak Acid Ionization:
HC2H3O2
HC2H3O2
(H2O)
HC2H3O2
HC2H3O2 
HC2H3O2
Add water to MOLECULES of WEAK Acid 35

36. This gives rise to an Equilbrium expression, Ka
Weak Acid Ionization:
HC2H3O2
HC2H3O2
(H2O)
HC2H3O2 
H C2H3O2-
HC2H3O2
HC2H3O2
Note: At any given time only a small portion of the acid molecules are ionized and since reactions are running in BOTH directions the mixture composition stays the same.
This gives rise to an Equilbrium expression, Ka 36

37. pH calculations:
To do pH calculations one must consider both the nature and condition (amount of ionization) of the species in the solution and then calculate the concentration of the hydronium ion (H+ or H3O+). 37

38. Strong acid: [H3O+] = concentration of acid
so: pH = -log [H3O+] = -log[acid]
Weak Acid: one must calculate the [H3O+] from an equilibrium ionization expression.
HA + H2O  H3O+ + A-
Ka = _ [H3O+][A-]
[HA]
These are equilibrium concentrations. 38

39. Starting pH, only acid and water.
25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 2 4 3 1.
Starting pH, only acid and water.
0.10 M NaOH 39

40. 4 2 3 1 Approximations: [H+][A-] pOH=-Log[XS OH-] [OH-]=[HA]40

41. Problem: Calculate the pH of 25.0 mL of 0.10M acetic acid (HOAc).
The Ka of HOAc = 1.8 x 10-5
Which region of a titration curve would this be? 41

42. Problem: Calculate the pH of 25.0 mL of 0.10M acetic acid (HOAc).
The Ka of HOAc = 1.8 x 10-5
Ka = [H+][OAc-]
[HOAc]
HOAc  H+ + OAC- I C E
-x x x
Ka = [x][x]
0.10-x
0.10-x x x
Small,drop, WHY?
Ka = x2
0.10
x = = [H+]
pH = -log =2.87 42

43. 2. 4 3 1. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
Some base has been added
This is also the “buffer region” 4 3
pKa = pH at this point (halfway point) 1.
0.10 M NaOH 43

44. Region 2 Theory and Calculations:
Base has been added and some of the acid has been neutralized.
What has changed and WHY?
1. The moles of acid is decreased.
2. The moles of acid anion (A-) is increased.
HA + OH-  H2O + A-
3. The total volume has increased.
HA + OH-  H2O + A- + XS HA
note: the 1:1 ratio 44

45. Why? Region 2 Theory and Calculations: Leads to: Ka= [H+][A-] [HA]
Base has been added and some of the acid has been neutralized.
intial RXN
HA + OH-  H2O + A-
Equilibrium XS HA + H2O  H3O+ + A-
2 sources
always
(calculate from equilibrium)
Leads to:
Ka= [H+][A-]
[HA]
Why?
and [H+][A-] 45

46. Solution: HOAc + OH-  H2O + OAc- 1:1 ratio
Region 2
Problem: Calculate the pH of a solution that is prepared by combining mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution: HOAc + OH-  H2O + OAc- 1:1 ratio
1. Find initial moles of Acid:
0.10mol HOAc
= mol HOAc L
0.0025 1L
2. find moles of OH-:
LNaOH
0.080molNaOH
1molOH-
_____________________________= mol OH- 1L
1molNaOH 46

47. 3. Subtraction gives what?
Problem: Calculate the pH of a solution that is prepared by combining mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution:
1. We found initial mol of Acid: HOAc + OH-  H2O + OAc- 1:1 ratio
0.10mol HOAc
_____ = mol HOAc L
0.0025 1L
2. We found mol of OH-: which =‘s initial moles of OAC- anion.
LNaOH
0.080molNaOH
1molOH-
_____________________________________= mol OH-
1molNaOH 1L
3. Subtraction gives what?
No, not the numerical answer!
moles of XS acid (initially present for OUR Eq. calculation). 47

48. 3. Subtraction gives moles of XS acid (initial for I.C.E.).
Problem: Calculate the pH of a solution that is prepared by combining mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution:
1. Find initial moles of Acid: HOAc + OH-  H2O + OAc- 1:1 ratio
0.10mol HOAc
_____ = mol HOAc L
0.0025 1L
2. find moles of OH-: which =‘s initial moles of OAC- anion.
LNaOH
0.080molNaOH
1molOH-
_____________________________________= mol OH- 1L
1molNaOH
3. Subtraction gives moles of XS acid (initial for I.C.E.).
= mole acid INITIAL
4. and..we have mole OAc- INITIAL
for I.C.E. 48

49. Solution: FROM PREVIOUS SLIDE.
Problem: Calculate the pH of a solution that is prepared by combining mL of 0.10M acetic acid (HOAc) with 14.00mL of 0.080M NaOH. (Ka of HOAc = 1.8 x 10-5).
Solution: FROM PREVIOUS SLIDE.
1. initial M of Acid: HOAc + OH-  H2O + OAc- 1:1 ratio
mole/0.039L = M acid after RXN =INITIAL
2. M of OH-: which =‘s initial M of OAC- anion.
mole/0.039L = M OAc- after RXN =INITIAL**
[HOAc]  [H+] + [OAc-]
I** C E
-x x x
x x x
Can we drop these?
x = =[H+]
pH = - Log [ ] = 4.65 49

50. 1. write balanced equation for initial weak/strong RXN.
Region 2 summary:
1. write balanced equation for initial weak/strong RXN.
2. find moles of acid
3. find moles of base
4. subtract to determine XS (and limiting reactant which equals initial moles of salt formed)
5. Change XS to M and limiting Reactant to M
6. Write balanced Chemical Equilibrium Equation:
HA  H A-
7. Use values from #5 a initial values in I.C.E. Chart
8. Work as Equilibrium problem. 50

51. 2. 4 3. 1. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5
Equivalence point 1.
0.10 M NaOH 51

52. º Region 3. Theory and Calculations.
Mole of acid = moles of OH- added. Equivalence pt.
HA + OH-  H2O + A- no XS of either
The titration curve indicates the pH to be above 7.
What causes this?
Hydrolysis of the A- anion: A- + H2O  HA + OH-
Since this equilibrium involves OH- being formed,
it is a Kb problem. 52

53. 1. Find moles of Acid (which =‘s moles of base at Eq pt.)
Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH
Ka(HOAc)=1.8 x 10-5.
1. Find moles of Acid (which =‘s moles of base at Eq pt.)
25.00mL = mol HOac
0.10mol HOAc
0.0025
1000mL
2. Find mol of OH-
31.25mL __ = mol OH-
0.0025
0.080mol NaOH
1molOH-
1000mL
1mol NaOH
3. Since the # of moles are the same, this is at the Eq. Pt. and therefore the initial moles of OAc- ion = mol 53

54. moles of Acid = moles of base = 0.0025 = Eq. Pt.
Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH
Ka(HOAc)=1.8 x 10-5.
moles of Acid = moles of base = = Eq. Pt.
New Volume = L..and [OAc-] = / =0.0444M 
Question: What equilibrium(ia) gives rise to the pH?
1. hydrolysis of the acetate anion:
OAc- + H2O 
HOAc + OH-
[OAc-] [HOAc] [OH-]
I C E
-x x x
x x x 54

55. moles of Acid = moles of base = 0.0025 = Eq. Pt.
Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5.
moles of Acid = moles of base = = Eq. Pt.
New Volume = L and [OAc-] =0.0444M
Question: What equilibrium(ia) gives rise to the pH?
1. hydrolysis of the acetate anion:
OAc- + H2O 
HOAc + OH-
OAc HOAc OH-
I C E
-x x x
x x x
Can we drop this x?
How do we find Kb?
Also: KaKb=Kw=1.0 x so Kb= x 10-10 55

56. moles of Acid = moles of base = 0.0025 = Eq. Pt.
Region 3 problem: Calculate the pH at the equivalence point of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 31.25mL of 0.080M NaOH Ka(HOAc)=1.8 x 10-5.
moles of Acid = moles of base = = Eq. Pt.
New Volume = L and [OAc-] =0.0444M
Question: What equilibrium(ia) gives rise to the pH?
1. hydrolysis of the acetate anion:
OAc- + H2O 
HOAc + OH-
OAc HOAc OH-
I C E
-x x x
x x x
pOH = 5.30
pH = = 8.70
x = 4.97 x 10-6 = [OH-] 56

57. 2. 4 3. 1. 25.0 mL of 0.10M Acetic acid Ka=1.8 x 10-5 XS base
0.10 M NaOH 57

58. Find moles of XS OH- and then use the new
Region 4 theory and Calculations:
All the acid has been neutralized and XS base has been added
HA + OH-  H2O + A- + XS OH-
What is the dominating factor that controls the pH?
The XS strong base (OH-)
Calculations:
Find moles of XS OH- and then use the new
volume to find [OH-] and then the pOH and pH. 58

59. 4. Find OH- concentration, pOH, and pH:
Region 4 problem: Calculate the pH of a solution formed by the titration of 25.00mL of 0.10M acetic acid with 40.00mL of 0.080M NaOH. Ka(HOAc)=1.8 x 10-5.
1. Find moles of acid:
0.10mol HOAc
25.00mL = mol HOAc
0.0025
1000mL
2. Find moles of OH-:
40.00mL = mol OH-
0.080molOH-
0.0032
1000mL
3. Subtract to find XS:
= mole XS OH-
4. Find OH- concentration, pOH, and pH:
pOH = -Log[0.0007/ ] = and pH = 12.03 59

60. STUDY FOR QUIZ 4 2 3 1 Approximations: [H+][A-] pOH=-Log[XS OH-]
[OH-]=[HA]
[H+]=[A-] 1 60

61. Titration Curves for Three Different Weak Acids
Ka = 1.4 x 10-7
Ka = 1.4 x 10-6 pH
Ka = 1.4 x 10-5
mL of Strong Base Added 61

62. 62

63. CO32- + H+  HCO3-
HCO3- + H+  HCO3 63

64. 64

65. acid + baseXS acid + salt HA+NaOH HA(XS)+NaA pOH=-Log[OH-]
Region 4:
Region 2:
acid + baseXS acid + salt
HA+NaOH HA(XS)+NaA
pOH=-Log[OH-]
Region 3: mol acid=mol base
hydrolysis of the salt anion
A- + H2OHA + OH-
Region 1: only acid (HA)
0.10 M NaOH 65

66. 66 Region 4: XS base pOH=-Log[OH-] Region 2: acid + baseXSacid + salt
HA+NaOH HA(XS)+NaA
Region 3: mol acid=mol base
hydrolysis of the salt anion
A- + H2OHA + OH-
Region 1: only acid (HA) 66