1. Chapter 17 Additional Aspects of Acid-Base Equilibria
Dr. Peter Warburton

2. The common-ion effect Say we make two acid solutions:
0.100 M HCl (a strong acid) and
0.100 M CH3COOH (a weak acid).
A M HCl solution by itself would have a pH  1.0 ([H3O+] = M) since the reaction goes to completion.

3. x = [H3O+] = [CH3COO-] = 1.3 x 10-3 M
The common-ion effect
A M CH3COOH solution
(Ka = 1.8 x 10-5) by itself would have a
pH  2.8 since an equilibrium is established where the equilibrium concentration of acetic acid  M
x = [H3O+] = [CH3COO-] = 1.3 x 10-3 M

4. The common-ion effect

5. The common-ion effect Say we put together a solution that is BOTH
0.100 M HCl (a strong acid) and
0.100 M CH3COOH (a weak acid).
The two reactions of
the acids with water take place
in the same container
at the same time!

6. which appears to be true
The common-ion effect
Both reactions are a source of H3O+ and so we could expect that
[H3O+] = M x 10-3 M
[H3O+] = M
and pH  0.99
which appears to be true

7. Le Chatalier’s Principle!
The common-ion effect
However, both reactions share the common ion H3O+, and so they cannot be treated as independent reactions. Something that affects one reaction must also affect the other reaction.
Le Chatalier’s Principle!

8. The common-ion effect
Imagine we start with the M CH3COOH and then we add M H3O+ by adding M HCl.
Le Chatalier’ Principle tells us our reaction will shift back towards reactants!

9. The value of x has decreased because of the added H3O+!
The common-ion effect
If we assume x is much smaller than M we will quickly find that x = Ka
x = [CH3COO-] = 1.8 x 10-5 M
The value of x has decreased because of the added H3O+!

10. H3O+ as a common-ion

11. OH- as a common-ion

12. Salts as source of a basic common-ion

13. Salts as source of a basic common-ion
On the left is a solution of M CH3COOH while on the right is a solution that is both M CH3COOH and M CH3COONa which is a source of the basic common ion CH3COO-.
The reaction has shifted back towards reactants!

14. Salts as source of an acidic common-ion

15. Salts as source of an acidic common-ion
On the left is a solution of M NH3 while on the right is a solution that is both M NH3 and M NH4Cl which is a source of the acidic common ion NH4+.
The reaction has shifted back towards reactants!

16. These solutions are resistant to changes in pH.
Buffer solutions
Solutions that contain both a weak acid and its conjugate base are buffer solutions.
These solutions are resistant to changes in pH.

17. very close in composition to the original equilibrium mixture.
Buffer solutions
The system has enough of the
original acid and conjugate base molecules
in the solution to react with
added acid or added base,
so the new equilibrium mixture will be
very close in composition to the original equilibrium mixture.

18. CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)
Buffer solutions
A 0.10 molL-1 acetic acid – 0.10 molL-1 acetate mixture has a pH of 4.74 and is a buffer solution!
CH3COOH (aq) + H2O (l)  H3O+ (aq) + CH3COO- (aq)

19. Buffer solutions If we rearrange the Ka expression
and make the assumption that x is much less than 0.10, we see
If [CH3COOH] = [CH3COO-],
then [H3O+] = 1.8 x 10-5 M = Ka
and pH = pKa = 4.74

20. Buffer solutions

21. Buffer solutions
What happens if we add 0.01 mol of NaOH (strong base) to 1.00 L of the acetic acid – acetate buffer solution?
CH3COOH (aq) + OH- (aq) → H2O (l) + CH3COO- (aq)
This goes to completion and keeps occurring until we run out of the limiting reagent OH-
New [CH3COOH] = 0.09 M and new [CH3COO-] = 0.11 M

22. Note we’ve made the assumption that x << 0.09!
Buffer solutions
With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check after calculations are done!), we find
Note we’ve made the assumption that x << 0.09!
pH = - log [H3O+]
pH = - log 1.5 x 10-5
pH = 4.82

23. Buffer solutions
Adding 0.01 mol of OH- to 1.00 L of water would have given us a pH of 12.0!
There is no significant amount of acid in water for the base to react with.

24. Buffer solutions
What happens if we add 0.01 mol of HCl (strong acid) to 1.00 L of the acetic acid – acetate buffer solution?
CH3COO- (aq) + H3O+ (aq) → H2O (l) + CH3COOH (aq)
This goes to completion and keeps occurring until we run out of the limiting reagent H3O+
New [CH3COOH] = 0.11 M and new [CH3COO-] = 0.09 M

25. Note we’ve made the assumption that x << 0.09!
Buffer solutions
With the assumption that x is much smaller than 0.09 mol (an assumption we always need to check after calculations are done!), we find
Note we’ve made the assumption that x << 0.09!
pH = - log [H3O+]
pH = - log 2.2 x 10-5
pH = 4.66

26. Buffer solutions
Adding 0.01 mol of H3O+ to 1.00 L of water would have given us a pH of 2.0!
There is no significant amount of base in water for the acid to react with.

27. Adding acid or base to a buffer

28. Problem
Calculate the pH of a L buffer solution that is 0.25 mol/L in HF and 0.50 mol/L in NaF.
With the assumption that x is much smaller than 0.25 mol (an assumption we always need to check after calculations are done!), we find

29. Problem
pH = - log [H3O+]
pH = - log 1.75 x 10-4
pH = 3.76

30. New [HF] = 0.27 M and new [F-] = 0.48 M
Problem
a) What is the change in pH on addition of mol of HNO3?
New [HF] = 0.27 M and new [F-] = 0.48 M

31. Problem
Notice we’ve made the assumption that x << We should check this!
pH = - log [H3O+]
pH = - log 1.97 x 10-4
pH = 3.71

32. New [HF] = 0.21 M and new [F-] = 0.54 M
Problem
b) What is the change in pH on addition of mol of KOH?
New [HF] = 0.21 M and new [F-] = 0.54 M

33. Problem
Notice we’ve made the assumption that x << We should check this!
pH = - log [H3O+]
pH = - log 1.36 x 10-4
pH = 3.87

34. Predicting whether a solution is a buffer
Any solution that becomes a mixture of a conjugate acid-base pair will be a buffer.
1) Weak acid-base conjugate pairs like CH3COOH and CH3COO- or NH4+ and NH3.
2) Weak acid reacting with small amounts of strong base like CH3COOH and NaOH.
3) Weak base reacting with small amounts of strong acid like NH3 and HCl.

35. Problem
Describe how a mixture of a strong acid such as HCl and a salt of a weak acid such as CH3COONa can be a buffer solution.

36. Problem
What is the pH of a buffer solution prepared by dissolving 23.1 g of NaCHO2 (molar mass is gmol-1) in a sufficient volume of M HCHO2 to make mL of the buffer?
Ka of formic acid is 1.8 x 10-4
Answer: pH = 3.94

37. The Henderson-Hasselbalch equation
We’ve seen that, for solutions with both members of a conjugate acid-base pair, that
pH = pKa + log [base] / [acid]
This is called the Henderson-Hasselbalch Equation.

38. The Henderson-Hasselbalch Equation
If we have a buffer solution of a conjugate acid-base pair, then the pH of the solution will be close to the pKa of the acid.
This pKa value is modified by the logarithm of ratio of the concentrations of the base and acid in the solution to give the actual pH.

39. The Henderson-Hasselbalch equation
pH = pKa + log [base] / [acid]
ALWAYS remember when you use the H-H eqn that there is the assumption that the equilibrium concentrations of acid and base are relatively unchanged from the initial concentrations.
That is we have assumed x is very small compared to the initial concentrations!

40. The Henderson-Hasselbalch equation
pH = pKa + log [base] / [acid]
Generally this assumption is valid as long as we know
0.10 < [base] / [acid] < 10
AND
[base] / Ka > 100 AND [acid] / Ka > 100

41. Alternate Henderson-Hasselbalch equation
We can always look at a buffer solution as a base combined with its conjugate acid
B (aq) + H2O (l)  OH- (aq) + BH+ (aq)
pOH = pKb + log [acid] / [base]

42. Alternate Henderson-Hasselbalch equation
If we have a buffer solution of a conjugate acid-base pair, then the pOH of the solution will be close to the pKb of the base.
This pKb value is modified by the logarithm of ratio of the concentrations of the acid and base in the solution to give the actual pH.

43. We should also check the validity of using H-H at the end to be sure!
Problem
Use the Henderson-Hasselbalch equation to calculate the pH of a buffer solution prepared by mixing equal volumes of 0.20 mol/L NaHCO3 and 0.10 mol/L Na2CO3.
Ka of HCO3- = 4.7 x 10-11
(see Table 16.4)
We should also check the validity of using H-H at the end to be sure!

44. pH = (-log 4.7 x 10-11) + log (0.05) / (0.10)
Problem answer
If we mix equal volumes, the total volume is TWICE the volume for the original acid or base solutions.
Since the number of moles of acid or base DON’T CHANGE on mixing,
the concentrations will be
half the given values.
pH = pKa + log [base] / [acid]
pH = (-log 4.7 x 10-11) + log (0.05) / (0.10)
pH = – 0.30
pH = 10.03

45. Preparing buffer solutions
If we want to create a buffer solution of a specific pH, the H-H equation tells us we need to pick a conjugate acid-base pair with a pKa for the acid close to the pH we want, and then we adjust the amounts of the conjugate acid and base.

46. How do we adjust the concentrations?

47. Problem
How many grams of (NH4)2SO4 (molar mass is gmol-1) must be dissolved in L of 0.35 M NH3 to produce a solution with pH = 9.00? Assume that the solution volume remains at L.
Kb for ammonia is 1.8 x 10-5.
Answer: 21 g

48. Buffer capacity
Buffer capacity is the measure of the ability of a buffer to absorb acid or base without significant change in pH.
Larger volumes of buffer solutions have a higher buffer capacity, and buffer solutions of higher initial concentrations of the conjugate acid-base pair have a larger buffer capacity.

49. (x << [base] and x << [acid])
Buffer range
We’ve seen that as long as
0.10 < [base] / [acid] < 10
then the assumption that the Henderson-Hasselbalch equation is based upon
(x << [base] and x << [acid])
is likely to be valid.

50. pH = pKa + log [base] / [acid]
Buffer range
pH = pKa + log [base] / [acid]
pH = pKa + log 10
pH = pKa + 1.0 OR
pH = pKa + log 0.10
pH = pKa - 1.0

51. useful range of pH that is pKa  1.0
Buffer range
In general buffer solutions have a
useful range of pH that is pKa  1.0
For instance an acetic acid - acetate buffer has a useful pH range of about
3.7 to 5.7
since pKa is 4.7 (Ka = 1.8 x 10-5)

52. Application of buffers
Many biological processes can only occur at very specific pH values (usually between pH = 6 and pH = 8). The reactions often take place in buffered environments (e.g. human blood is buffered to pH = 7.4 – see text pg. 734).

53. HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
Acid-base indicators
We often measure the pH of a solution with a chemical acid-base indicator. Such indicators are weak acids in their own right (symbolized HIn) and indicate pH because the acid form has a different colour than the conjugate base form (In-)
HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
colour A colour B

54. Diprotic indicator!

55. HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
Acid-base indicators
HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
colour A colour B
If we increase the [H3O+] we shift this reaction towards reactants. The colour will be that of HIn.
If we decrease the [H3O+] we shift this reaction towards products. The colour will be that of In-.
A [H3O+] in between these two extremes will give a colour that is a mixture of the two colours because both HIn and In- are present in significant amounts!

56. HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
Acid-base indicators
HIn (aq) + H2O (l)  H3O+ (aq) + In- (aq)
colour A colour B
More specifically, the HIn and In- form a buffer so the indicator works in a pH range of about 1 around the pKHIn of HIn.

58. Since indicators work in a range of about 2 pH units we often put several indicators that have different ranges in a single solution to give a universal indicator for pH range of about

59. Universal indicator

60. Applications of indicators
Indicators are useful when we want a general idea of the value of the pH without using a pH meter. Pool chlorination (with Cl2 or NaOCl) is done to avoid algae growth. This works best at pH = 7.4, so we might need to add some acid or base…

61. Neutralization reactions
A reaction of an acid and a base often produces water and an aqueous salt as products. Such reactions are called neutralization reactions, and can be categorized by the strengths of the acid and base involved.

62. Strong acid – strong base neutralization
The reaction of a strong acid such as HCl and a strong base such as NaOH becomes a reaction of H3O+ and OH-,
Therefore while the overall reaction is
HCl (aq) + NaOH (aq)  H2O (l) + NaCl (aq)
the actual net ionic equation is
H3O+ (aq) + OH- (aq)  2 H2O (l)
where the Na+ and Cl- ions are not involved (they are neutral spectator ions which don’t react with water!)
See slides 102, 103 and 106 of last chapter!

63. Strong acid – strong base neutralization
If we mix equal numbers of moles of HCl and NaOH, we will create twice as much water, while leaving an excess of NEITHER ion.
In this case the [H3O+] = [OH-] = 1.0 x 10-7 M
at 25 °C (reaction goes to completion).
H3O+ (aq) + OH- (aq)  2 H2O (l)
is the reverse of the autoionization of water reaction so

64. Strong acid – strong base neutralization
Since the equilibrium constant is very large, we see the reaction goes to completion, and at the end, the equilibrium mixture will consist of water and an aqueous salt of ions that are neutral in character because they don’t react with water.
The pH after the reaction will be 7
(a neutral solution).

65. Strong acid titration by strong base
Say we add the strong base to the strong acid solution drop by drop. Each drop of base (the titrant) will react with some of the acid to completion until the added base (the limiting reagent!) is all gone. If we measure the pH after we add each drop of base, we can plot a pH versus total volume of added base graph.
This is called a titration curve.

66. Titration curve – strong acid titrated by strong base

67. End point versus equivalence point
How do we know when we’ve reached the end of our titration? We often will use an indicator to tell us when the titrated solution reaches a specific pH. This is the end point of the titration.
The end point DEPENDS on the indicator we use!

68. End point versus equivalence point
The equivalence point is that point in the titration where we have added equal numbers of MOLES of acid and base. For strong acid – strong base neutralizations
the equivalence point occurs when the pH is 7
(a neutral solution)

69. Titration curve – strong acid titrated by strong base
Different end points depending on indicator

70. End point versus equivalence point
Bromothymol blue is a good indicator for strong acid-strong base titrations because the end point is very close to the equivalence point.
Methyl red and phenolphtalein are pretty good choices too because the titration curve is very steep in their effective pH ranges.

71. Titration curve – strong base titrated by strong acid
Equivalence point
The curve looks exactly the same, just flipped vertically!

72. Because of this, we often do calculations in millimoles
Millimoles (mmol)
Since we often set up titrations where our titrant concentration is less than 1 M and total titrant volume used is less than 50 mL this means we are adding about 5.00 x 10-3 moles of acid or base in our titration.
Because of this, we often do calculations in millimoles
(1 mmol = 1 x 10-3 mol)

73. Millimoles (mmol)
M = mol / L
M = mmol / mL

74. Problem
For the titration of mL of M HCl with M NaOH, calculate
a) the initial pH;
b) the pH when the neutralization is 50.0% complete;
c) the pH when the neutralization is 100% complete; and
d) the pH when 1.00 mL of NaOH is added beyond the equivalence point.

75. Problem answers a) pH = 0.824 (Vtotal = 25.00 mL)
b) pH = (Vtotal = mL)
c) pH = 7.00 (Vtotal = mL)
d) pH = (Vtotal = mL)

76. HA (aq) + OH- (aq)  H2O (l) + A- (aq)
Weak acid – strong base
Because a weak acid HA is largely undissociated in water, while the strong base is completely dissociated in water (becoming a source of OH-), the neutralization reaction of a weak acid and strong base becomes that of
HA (aq) + OH- (aq)  H2O (l) + A- (aq)
Acetic acid is a weak acid, so when it reacts with NaOH, the net ionic equation is
CH3COOH (aq) + OH- (aq)  H2O (l) + CH3COO- (aq)
Now remember that
Knet = K1 x K2 x K3 x … x Kn

77. Weak Acid – Strong Base
The equilibrium constant shows us the reaction goes to completion, so for equal numbers of moles of weak acid and strong base, we expect only water and the aqueous salt in the equilibrium mixture.

78. See slide 107 of last chapter!
Weak acid – strong base
However, the salt comprises of a neutral cation (Na+ in this case) and a weakly basic anion (CH3COO-), meaning the equilibrium mixture will be BASIC and have a pH greater than 7.
See slide 107 of last chapter!
In a weak acid titration by a strong base this means the equivalence point is NOT at pH = 7 but rather at pH > 7!

79. Titration curve – weak acid titrated by strong base
6. Added strong base dominates weak base and determines pH
3. pH = pKa when moles added OH- = ½ initial moles weak acid
5. pH > 7 because of water hydrolysis by conjugate base
4. buffer breaks!
2. creation of a buffer
1. higher initial pH for a weak acid
Titration of CH3COOH with NaOH.

80. Problem
For the titration of mL of M HF with M NaOH, calculate
a) the initial pH; Ka = 6.6 x 10-4
b) the pH when the neutralization is 25.0% complete;
c) the pH when the neutralization is 50.0% complete; and
d) the pH when the neutralization is 100% complete.

81. Problem answers a) pH = 2.00 (Vtotal = 20.00 mL)
b) pH = 2.70 (Vtotal = mL)
c) pH = 3.18 (Vtotal = mL)
d) pH = 8.08 (Vtotal = mL)

82. H3O+ (aq) + B (aq)  H2O (l) + BH+ (aq)
Weak base – strong acid
Because a weak base B is largely undissociated in water, while the strong acid is completely dissociated in water (becoming a source of H3O+), the neutralization reaction of a strong acid and weak base becomes that of
H3O+ (aq) + B (aq)  H2O (l) + BH+ (aq)
Ammonia is a weak base, so when it reacts with HCl, the net ionic equation is
H3O+ (aq) + NH3 (aq)  H2O (l) + NH4+ (aq)
Again
Knet = K1 x K2 x K3 x … x Kn

83. Weak base – strong acid
Again the equilibrium constant shows us the reaction goes nearly to completion, so for equal number of reactant moles, we expect only water and the aqueous salt in the equilibrium mixture.

84. Weak base – strong acid
However, the salt comprises of a neutral anion (Cl- in this case) and a weakly acidic cation (NH4+), meaning the equilibrium mixture will be ACIDIC and have a pH less than 7.
In a weak base titration by a strong acid this means the equivalence point is NOT at pH = 7 but rather at pH < 7!

85. Titration curve – weak base titrated by strong acid
pH = pKb when moles added H3O+ = ½ initial moles weak base

86. Problem
For the titration of mL of M NH3 with M HCl, calculate
a) the initial pH; Kb = 1.8 x 10-5
b) the pH when the neutralization is 25.0% complete;
c) the pH when the neutralization is 50.0% complete; and
d) the pH when the neutralization is 100% complete.

87. Problem answers a) pH = 11.14 (Vtotal = 50.00 mL)
b) pH = 9.74 (Vtotal = mL)
c) pH = 9.26 (Vtotal = mL)
d) pH = 5.20 (Vtotal = mL)