1. Aqueous Equilibria: Acids & Bases
McMurry and Fay ch. 14

2. Acid-base definition (Arrhenius)
Acid: increases H+ ions in solution
Base: increases OH- ions in solution

3. Arrhenius Acids & Bases
Acids: HCl  H+ + Cl- H2SO4  2 H+ + SO42- Bases: KOH  K+ + OH- NaOH  Na+ + OH-

4. Ammonia as a base
Ammonia (NH3) neutralizes acids. How can it make OH- in solution?

5. An important equilibrium constant
H2O H+ + OH-
(self-ionization of water)
[H+][OH-] = Kw

6. Ammonia as a base
NH3 + H2O  NH3 + H+ + OH-  NH4+ + OH-

7. Acid-base definition (Brønsted-Lowry)
Acid: proton donor
Base: proton acceptor
(the word proton refers to H+)

8. Brønsted-Lowry acid-base reactions are reversible
NH3 + H2O NH OH-
accepts
proton
BASE
donates
proton
ACID
donates
proton
ACID
accepts
proton
BASE

9. Brønsted-Lowry acid-base reactions are reversible
NH3 + H2O NH OH-
CONJUGATE
ACID
CONJUGATE
BASE
accepts
proton
BASE
donates
proton
ACID
donates
proton
ACID
accepts
proton
BASE

10. Conjugate acid-base pairs
Two substances that differ by only one proton
The acid has one more proton than the base
The base has one less proton than the acid

11. In the following reaction, which is the conjugate base of CH3COOH
In the following reaction, which is the conjugate base of CH3COOH? CH3COOH + H2O CH3COO- + H3O+
CH3COOH
H2O
CH3COO-
H3O+

12. Hydronium ion H+ is not very stable by itself
Charge on proton in solution is stabilized by IMFs
H+ ion is frequently written as H3O+ H+ O H
In this class we will use H+ and H3O+ interchangeably

13. Acid strength: old version
The stronger the acid,
the more H+ it generates
Acid strength: old version
HCl (aq) H+ (aq) + Cl- (aq)
(strong acid because no HCl remains; all turns into H+)
CH3COOH (aq) CH3COO-(aq) + H+ (aq)
(weak acid because little H+ is made; most remains as CH3COOH)

14. Acid strength: Equilibrium Version
Strength depends on equilibrium positions
HCl + H2O H3O+ + Cl-
CH3COOH + H2O CH3COO- + H3O+
In strong acids, the conjugate base form is favored at equilibrium
In weak acids, the acid form is favored at equilibrium

15. Characteristics of a Strong Acid
Weak bond (easy to break) between H & anion is weak
Anion is stable in solution

16. Acid strength: Binary acids in the same group of the periodic table
As valence shell increases, bond strength decreases
Easier to break bonds between H & larger atoms H F Cl H
Acid strength increases Br H I H

17. Acid strength: Binary acids in the same row of the periodic table- H H H H N N -
As you move to the right, electronegativity increases
More electronegative atoms are more happy as ions
More electronegative elements are stronger acids H H H H H H H H O - O - F H F

18. Acid strength: Oxoacids
Oxoacids are acids composed of hydrogen, oxygen, and some other element (examples: H2SO4, H3PO4, HNO3)
General rule: as central atom electronegativity increases,
strength of the O-H bond decreases
 Higher electronegativity = stronger acid strength

19. Acid strength: Oxoacids
For a given element, there can be oxoacids with different numbers of oxygen atoms bound to the central atom H2SO4 H2SO3 O O O S O O S O O

20. Acid strength: Oxoacids
More resonance structures = more stable conjugate base
General rule: The larger the number of lone oxygen atoms,
the stronger the acid

21. Lewis Acids and Bases
Explain acid-base chemistry that occurs outside of aqueous solutions Lewis acid: electron pair acceptor Lewis base: electron pair donor

22. Lewis Base examples:
H N H H O H H .. .. ..

23. Lewis acid-base example
NH3 + BF3

24. An important equilibrium constant
H2O H+ + OH-
(self-ionization of water)
[H+][OH-] = Kw = (at 25 oC)
What does the Kw value tell us about this equilibrium?

25. Self-ionization of water
H2O H+ + OH-
[H+][OH-] = Kw = (at 25 oC)
Any aqueous solution will have both H+ ions and OH- ions.
Kw will always be at 25oC.
You can’t have lots of acid AND lots of base

26. If you know [H+], you can calculate [OH-] (and vice versa)
[H+][OH-] = Kw so Kw
[OH-] [OH-]
[H+] = = Kw
[H+] [H+]
[OH-] = =

27. If [H+] = M, what is [OH-]? Kw
[OH-]
[H+] =

28. If [H+] = 6.24 x 10-13 M, what is [OH-]?

29. Definition of acidic & basic solutions
Neutral solution: [H+] = [OH-]
Acidic solution [H+] > [OH-]
Basic solution [H+] < [OH-]

30. p Notation
When talking about measurements that span several orders of magnitude, it is easiest to use a logarithmic scale
pX = -log X

31. Easily measured using pH electrode
pH = -log [H+]
Easily measured using pH electrode

32. pH of strong acid solutions
A strong acid is completely dissociated….
HCl  H+ + Cl-
(no HCl remains)
If you start with M HCl, you end with M H+
0.200 M Cl-

33. To find the pH of a strong acid solution:
1. Find molarity of acid solution
2. Assume [H+] = molarity
3. Use equation
pH = -log [H+]

34. What is the pH of a 0.0325 M solution of HNO3?
1. Find molarity of acid solution
2. Assume [H+] = molarity
3. Use equation
pH = -log [H+]

35. Converting from pH to [H+]
pH = -log [H+]
Divide each side by -1
- pH = log [H+]
Do the anti-log of both sides
(raise 10 to the power of each side)
10-pH = [H+]

36. Converting from pH to [H+]
10-pH = [H+]
pH 4.0 [H+] = 10-4 pH 8.0 [H+] = 10-8 pH 5.75 [H+] =
= 1.78 x 10-6

37. What is [H+] for an acid solution with pH 4.27?

38. pOH = -log [OH-] [H+][OH-] = 10-14 log[H+][OH-] = log (10-14)
Can’t be measured directly; calculate from pH
[H+][OH-] = 10-14
log[H+][OH-] = log (10-14)
log[H+] + log [OH-] = -14
-log [H+] – log [OH-] = 14
pH + pOH = 14

39. pH of strong base solutions
100% of strong base molecules generate OH- ions
pH + pOH = 14
pOH = 14 - pH

40. pH of strong base solutions
1. Find molarity of base solution
2. Assume [OH-] = molarity
3. Use equation
pOH = -log [OH-]
4. Use equation
pH = 14 - pOH

41. What is the pH of a 0.245 M solution of KOH?
1. Find molarity of base solution
2. Assume [OH-] = molarity
3. Use equation
pOH = -log [OH-]
4. Use equation
pH = 14 - pOH

42. What is the pH of 0.000581 M solution of NaOH?
1. Find molarity of base solution
2. Assume [OH-] = molarity
3. Use equation
pOH = -log [OH-]
4. Use equation
pH = 14 - pOH

43. Definition of weak acids and bases— not everything dissociates
pH is not a simple calculation

44. Acid equilibria HA H+ + A-
Generic equation for an acid dissociation reaction is:
HA H+ + A-
acid
(protonated)
conjugate base
(deprotonated)

45. Acid equilibrium constant, Ka
HA H+ + A-
[H+][A-]
[HA]
Ka =

46. pKa
pKa = - log Ka
Ka = 5.7 x pKa = 2.2
Ka = 9.8 x pKa = 6.0

47. Determining a Ka value from initial concentration & equilibrium pH
1. Make an ICE table
2. Use equation [H+] = 10-pH
3. Use step 2 to calculate change in [H+]
4. Use step 3 and stoichiometric ratios to calculate other concentration changes
5. Calculate equilibrium concentrations
6. Use final concentrations to calculate Ka

48. A student prepared a 0.10 M solution of formic acid (HCHO2) and measured a pH of Calculate Ka for formic acid.

49. 1. Make an ICE table HCHO2 H+ + CHO2-
[HCHO2] (M)
[H+] (M)
[CHO2-] (M)
Initial
Change
Equilibrium
A student prepared a 0.10 M solution of formic acid (HCHO2) and measured a pH of Calculate Ka for formic acid.

50. 2. Use equation [H+] = 10-pH HCHO2 H+ + CHO2-
[HCHO2] (M)
[H+] (M)
[CHO2-] (M)
Initial
Change
Equilibrium
A student prepared a 0.10 M solution of formic acid (HCHO2) and measured a pH of Calculate Ka for formic acid.

51. 3. Use step 2 to calculate change in [H+]
HCHO H+ + CHO2-
[HCHO2] (M)
[H+] (M)
[CHO2-] (M)
Initial
Change
Equilibrium

52. 4. Use step 3 and stoichiometric ratios to calculate other concentration changes
HCHO H+ + CHO2-
[HCHO2] (M)
[H+] (M)
[CHO2-] (M)
Initial
Change
Equilibrium

53. 5. Calculate equilibrium concentrations
HCHO H+ + CHO2-
[HCHO2] (M)
[H+] (M)
[CHO2-] (M)
Initial
Change
Equilibrium

54. 6. Use final concentrations to calculate Ka
HCHO H+ + CHO2-
[HCHO2] (M)
[H+] (M)
[CHO2-] (M)
Initial
Change
Equilibrium
[H+][CHO2-]
[HCHO2]
Ka =

55. Calculating pH using Ka
1. Make an ICE table; set change in [H+] to +x
2. Use stoichiometric ratios to calculate other concentration changes
3. Calculate equilibrium concentrations
4. Substitute equilibrium concentrations in to the Ka expression
5. Rewrite equation to form ax2 + bx + c
6. Solve for x using the quadratic equation
7. Use value of x to calculate pH

56. Calculate the pH of a 0.30 M solution of acetic acid (HC2H3O2), whose Ka = 1.8 x 10-5

57. 1. Make an ICE table; set change in [H+] to +x
C2H3O2H H+ + C2H3O2-
Calculate the pH of a 0.30 M of acetic acid (HC2H3O2H), whose Ka = 1.8 x 10-5

58. 2. Use stoichiometric ratios to calculate other concentration changes
C2H3O2H H+ + C2H3O2-

59. 3. Calculate equilibrium concentrations
C2H3O2H H+ + C2H3O2-

60. 4. Substitute equilibrium concentrations into the Ka expression
C2H3O2H H+ + C2H3O2-

61. 5. Rewrite equation to form ax2 + bx + c

62. 6. Solve for x using the quadratic equation
-b ± b2 – 4ac 2a
x =

63. 7. Use value of x to calculate pH

64. SHORTCUT!!
If Ka is much smaller than acid concentration (< 1/100th) …change (x) will be very small Subtracting x from [HA]0 will not substantially change [HA] APPROXIMATION: [HA]0 – x ≈ [HA]0

65. Calculate the pH of a 0.045 M solution of aniline (C6H5NH3), whose Ka = 2.51 x 10-5.
1. Make ICE table; set [H+] = +x
2. Calculate other changes
3. Calculate eq concs
4. Substitute eq concs into Ka
5. Solve for x
6. Use value of x to calculate pH

66. Base equilibria B + H2O BH+ + OH-
The generic base dissociation reaction is:
B + H2O BH+ + OH-

67. Base ionization constant, Kb
B + H2O BH+ + OH-
[BH+][OH-]
[B]
Kb =

68. pKb
pKb = - log Kb
Kb = 6.7 x pKb = 7.2

69. Relationship between pKa and pKb
pKa + pKb = pKw or
pKa + pKb = 14

70. Calculations for bases are very similar to those for acids
Use [OH-] calculated from
[OH-] = 10-14/[H+]

71. A 0. 10 M solution of for ethylamine, CH3CH2NH2, has a pH of 11. 87
A 0.10 M solution of for ethylamine, CH3CH2NH2, has a pH of Calculate the Kb and pKb.
1. Make an ICE table
2. Use pH + pOH = 14
3. Use step 2 to calculate change in [OH-]
4. Use step 3 & stoichiometric ratios to calc. other conc. changes
5. Calculate equilibrium concentrations
6. Use final concentrations to calculate Ka

72. Calculate the pH in a solution of a 0.40 M NH3. (Kb = 1.8 x 10-5)
1. Make an ICE table
2. Use pH + pOH = 14
3. Use step 2 to calculate change in [OH-]
4. Use step 3 & stoichiometric ratios to calc. other conc. changes
5. Calculate equilibrium concentrations
6. Use final concentrations to calculate Ka

73. Two ways to measure strength of a weak acid
The larger the Ka value, the more the acid dissociates and the more H+ it produces
Percentage ionization
The fraction of the acid that dissociates
[H+][A-]
[HA]
Ka =

74. Percentage ionization
% ionization = x 100%
Moles ionized per liter
Moles available per liter
Moles ionized = [A-] (or [BH+])
Moles available = [HA] + [A-] (or [B] + [BH+])

75. Enough benzoic acid (C7H6O2H) is dissolved in water to make an 0
Enough benzoic acid (C7H6O2H) is dissolved in water to make an M solution. The concentration of benzoate ions (C7H6O2-) was found to be 2.47 x 10-3 M. What is the percent ionization?
% ionization = x 100%
Moles ionized per liter
Moles available per liter

76. Enough benzoic acid (C7H6O2H) is dissolved in water to make an 0
Enough benzoic acid (C7H6O2H) is dissolved in water to make an M solution. The concentration of benzoate ions (C7H6O2-) was found to be 2.47 x 10-3 M. What is the percent ionization?
% ionization = x 100%
Moles ionized per liter
Moles available per liter

77. Using % ionization to calculate Ka or Kb
1. Make an ICE table and fill in known concentrations
2. Use % ionization & initial concentration to calculate change in [A-] or [BH+]
3. Calculate equilibrium concentrations
4. Plug equilibrium concentrations into expression for Ka or Kb

78. A 0. 0750 M solution of HF has a % ionization of 29. 6 %
A M solution of HF has a % ionization of 29.6 %. What is the Ka for HF?
1. Make an ICE table and fill in known concentrations
2. Use % ionization & initial concentration to calculate change in [A-] or [BH+]
3. Calculate equilibrium concentrations
4. Plug equilibrium concentrations into expression for Ka or Kb

79. A 0. 00392 M solution of imidazole (C3H3N2H2) has a % ionization of 0
A M solution of imidazole (C3H3N2H2) has a % ionization of 0.51%. What is the Ka for imidazole?
1. Make an ICE table and fill in known concentrations
2. Use % ionization & initial concentration to calculate change in [A-] or [BH+]
3. Calculate equilibrium concentrations
4. Plug equilibrium concentrations into expression for Ka or Kb

80. Using Ka to calculate pH and % ionization
1. Make an ICE table; set change in [H+] to +x
2. Use stoichiometric ratios to calculate other concentration changes
3. Calculate equilibrium concentrations
4. Substitute equilibrium concentrations in to the Ka expression
5. Solve for x
6. Use value of x to calculate pH
7. Use values in ICE table to calculate % ionization

81. Calculate the pH and % ionization of a 1
Calculate the pH and % ionization of a 1.0 M solution of chloroacetic acid (HC2H2ClO2), whose Ka = 1.36 x 10-3
1. Make ICE table; change in [H+] = +x
2. Calculate other conc changes
3. Calculate eq concs
4. Substitute eq concs into Ka
5. Solve for x
6. Use x to calculate pH
7. Use values in ICE table to calculate % ionization

82. Calculate the pH and % ionization of a 0
Calculate the pH and % ionization of a M solution of pyridine (C5H5NH), whose Ka = 6.3 x 10-6
1. Make ICE table; change in [H+] = +x
2. Calculate other conc changes
3. Calculate eq concs
4. Substitute eq concs into Ka
5. Solve for x
6. Use x to calculate pH
7. Use values in ICE table to calculate % ionization

83. Acid-Base Properties of Salt Solutions

84. Basic principle HA H+ + A- A- + H+ HA
A strong acid has a weak conjugate base A weak acid has a strong conjugate base
HA H+ + A-
A- + H HA
[H+][A-]
[HA]
[HA]
[H+][A-]
Ka =
Kb =

85. If salts… …have an anion that is the conjugate base of a weak acid…
If salts… …have an anion that is the conjugate base of a weak acid… OR …have a cation that is the conjugate acid of a weak base… They will affect the pH of a solution

86. 0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution?
Acetate (CH3CO2-) is the conjugate base of a fairly weak acid (acetic acid) and therefore is a fairly strong base
Sodium (Na+) is the conjugate acid of a very strong base (NaOH) and therefore is a very weak acid

87. pH calculated similarly to previous problems
1. Make an ICE table; set change in [H+] or [OH-] to +x
2. Fill out remainder of ICE table
3. Calculate Kb for conjugate base or Ka for conjugate acid
4. Substitute equilibrium concentrations in to the Ka or Kb expression
5. Solve for x
6. Use value of x to calculate pH (or convert pOH to pH)

88. 1. Make an ICE table; set change in [H+] or [OH-] to +x
H2O + C2H3O OH- + C2H3O2H
[C2H3O2-] (M)
[OH-] (M)
[C2H3O2H] (M) I C E
0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution?
Ka of CH3CO2H is 1.8 x 10-5.

89. 2. Fill out remainder of ICE table
H2O + C2H3O OH- + C2H3O2H
[C2H3O2-] (M)
[OH-] (M)
[C2H3O2H] (M) I C E
0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution?
Ka of CH3CO2H is 1.8 x 10-5.

90. 3. Calculate Kb for conjugate base or Ka for conjugate acid
H2O + C2H3O OH- + C2H3O2H
[C2H3O2-] (M)
[OH-] (M)
[C2H3O2H] (M) I C E
0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution?
Ka of CH3CO2H is 1.8 x 10-5.

91. 4. Substitute equilibrium concentrations in to the Ka or Kb expression
H2O + C2H3O OH- + C2H3O2H
[C2H3O2-] (M)
[OH-] (M)
[C2H3O2H] (M) I C E
0.10 mol sodium acetate (CH3CO2Na) is dissolved in enough water to make 1.00 L. What is the pH of this solution?
Ka of CH3CO2H is 1.8 x 10-5.

92. H2O + C2H3O2- OH- + C2H3O2H 5. Solve for x. [C2H3O2-] (M) [OH-] (M)
[C2H3O2H] (M) I C E

93. 6. Use value of x to calculate pH (or convert pOH to pH)
H2O + C2H3O OH- + C2H3O2H
[C2H3O2-] (M)
[OH-] (M)
[C2H3O2H] (M) I C E