1. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Contents
16-1 The Arrhenius Theory: A Brief Review
16-2 Brønsted-Lowry Theory of Acids and Bases
16-3 The Self-Ionization of Water and the pH Scale
16-4 Strong Acids and Strong Bases
16-5 Weak Acids and Weak Bases
16-6 Polyprotic Acids
16-7 Ions as Acids and Bases
16-8 Molecular Structure and Acid-Base Behavior
16-8 Lewis Acids and Bases
Focus On Acid Rain
General Chemistry: Chapter 16

2. 5-3 Acid-Base Reactions Latin acidus (sour)
Sour taste
Arabic al-qali (ashes of certain plants)
Bitter taste
Svante Arrhenius 1884 Acid-Base theory.

3. Some Definitions Arrhenius
An acid is a substance that, when dissolved in water, ionizes and increases the concentration of hydrogen ions, H+.
HCl → H+ + Cl-
A base is a substance that, when dissolved in water, increases the concentration of hydroxide ions, OH-. But, Not all bases contain OH-
NaOH → Na+ + OH-

4. 16-1 The Arrhenius Theory: A Brief Review
Chemistry 140 Fall 2002
16-1 The Arrhenius Theory: A Brief Review
H2O
HCl(g) → H+(aq) + Cl-(aq)
NaOH(s) → Na+(aq) + OH-(aq)
H2O
Na+(aq) + OH-(aq) + H+(aq) + Cl-(aq) → H2O(l) + Na+(aq) + Cl-(aq)
H+(aq) + OH-(aq) → H2O(l)
Arrhenius theory did not handle non OH- bases such as ammonia very well.
General Chemistry: Chapter 16

5. Brønsted-Lowry An acid is a proton donor. A base is a proton acceptor.
e.g. ammonia, NH3, is a base.

6. A Brønsted-Lowry acid…
…must have an ionizable or removable (acidic) proton.
A Brønsted-Lowry base…
…must have a pair of nonbonding electrons.

7. In an acid – base reaction, the acid donates a proton (H+) to the base.

8. What Happens When an Acid Dissolves in Water?
Water acts as a Brønsted-Lowry base and abstracts a proton (H+) from the acid.
As a result, the conjugate base of the acid and a hydronium ion are formed.

9. If a substance can be either…
…it is amphiprotic.
H2O
HCO3-

10. Conjugate Acids and Bases
The term conjugate comes from the Latin word “conjugare,” meaning “to join together.”
Reactions between acids and bases always yield their conjugate bases and acids.

11. What is the conjugate base of each of the following acids?
HClO4
H2S
HCO3-
PH4+
All the above substances are acid; simply remove H+ to obtain the conjugate base.

12. What is the conjugate acid of each of the following?
CN-
SO22-
H2O
HCO3-
CN H+ →
SO H+ →
H2O H+ →
HCO3- + H+ →

13. Acid and Base Strength
Strong acids are completely dissociated in water.
Weak acids only dissociate partially in water.

14. General Chemistry: Chapter 16

15. General Chemistry: Chapter 16
The Solvated Proton
General Chemistry: Chapter 16

16. General Chemistry: Chapter 16
In an acid-base reaction, the favored direction of the reaction is from the stronger to the weaker member of a conjugate acid-base pair.
General Chemistry: Chapter 16

17. General Chemistry: Chapter 16
A Strong Acid
General Chemistry: Chapter 16

18. Predict the favored direction of the following reactions:
HCl + OH-
H2O + I-
H2O + H2O
CH3COOH + H2O
NH3 + H2O

19. General Chemistry: Chapter 16

20. A Weak Acid

21. Acid Ionization Constant
conjugate
base
conjugate
acid
acid
base
CH3CO2H + H2O CH3CO2- + H3O+
Kc=
[CH3CO2H][H2O]
[CH3CO2-][H3O+]
Ka= Kc[H2O] =
= 1.810-5
[CH3CO2H]
[CH3CO2-][H3O+]
General Chemistry: Chapter 16

22. General Chemistry: Chapter 16
A Weak Base
General Chemistry: Chapter 16

23. Base Ionization Constant
conjugate
acid
conjugate
base
base
acid
NH3 + H2O NH4+ + OH-
Kc=
[NH3][H2O]
[NH4+][OH-]
Kb= Kc[H2O] =
[NH3]
[NH4+][OH-]
= 1.810-5
General Chemistry: Chapter 16

24. Autoionization of Water
As we have seen, water is amphoteric.
In pure water, a few molecules act as bases and a few act as acids.
This is referred to as autoionization.
H2O (l) + H2O (l)
H3O+ (aq) + OH- (aq)

25. 16-3 The Self-Ionization of Water and the pH Scale
General Chemistry: Chapter 16

26. General Chemistry: Chapter 16
The equilibrium condition of the self-ionization of water is called the Ion Product of Water
conjugate
acid
conjugate
base
base
acid
H2O + H2O H3O+ + OH-
Kc=
[H2O][H2O]
[H3O+][OH-]
KW= Kc[H2O][H2O] =
= 1.010-14
[H3O+][OH-]
General Chemistry: Chapter 16

27. For ALL aqueous solutions:
Kc = [H3O+] [OH-]
At 25C, Kw = [H3O+]·[OH-] = 1.0· = Kw (at 25 ºC)
[H3O+] = [OH-] the solution is….
[H3O+] > [OH-] the solution is….
[H3O+] < [OH-] the solution is….

28. pH
pH is defined as the negative base-10 logarithm of the concentration of hydronium ion.
pH = -log [H3O+]

29. pH In pure water, Kw = [H3O+] [OH-] = 1.0  10-14
Since in pure water [H3O+] = [OH-],
[H3O+] =  = 1.0  10-7

30. pH Therefore, in pure water, pH = -log (1.0  10-7) = 7.00
An acid has a higher [H3O+] than pure water, so its pH is <7.
A base has a lower [H3O+] than pure water, so its pH is >7.

31. pH
These are the pH values for several common substances.

32. Other “p” Scales
The “p” in pH tells us to take the negative base-10 logarithm of the quantity (in this case, hydronium ions).
Some similar examples are
pOH: -log [OH-]
pKw: -log Kw

33. -log [H3O+] + -log [OH-] = -log Kw = 14.00
Watch This!
Because
[H3O+] [OH-] = Kw = 1.0  10-14,
we know that
-log [H3O+] + -log [OH-] = -log Kw = 14.00
or, in other words,
pH + pOH = pKw = 14.00

34. How Do We Measure pH? For less accurate measurements, one can use
Litmus paper
“Red” paper turns blue above ~pH = 8
“Blue” paper turns red below ~pH = 5
Or an indicator.

35. How Do We Measure pH?
For more accurate measurements, one uses a pH meter, which measures the voltage in the solution.

36. Strong Acids
You will recall that the seven strong acids are HCl, HBr, HI, HNO3, H2SO4, HClO3, and HClO4.
These are, by definition, strong electrolytes and exist totally as ions in aqueous solution.
For the monoprotic strong acids,
[H3O+] = [acid].
What is the pH of a solution of 0.3 M hydrochloric acid?

37. Dissociation Constants
For a generalized acid dissociation,
the equilibrium expression would be
This equilibrium constant is called the acid-dissociation constant, Ka.
HA (aq) + H2O (l)
A- (aq) + H3O+ (aq)
[H3O+] [A-]
[HA]
Kc =

38. Dissociation Constants
The greater the value of Ka, the stronger is the acid.

39. HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)
Calculating pH from Ka
Calculate the pH of a 0.30 M solution of acetic acid, HC2H3O2, at 25C.
HC2H3O2 (aq) + H2O (l) H3O+ (aq) + C2H3O2- (aq)
Ka for acetic acid at 25C is 1.8  10-5.

40. Calculating pH from Ka [H3O+] [C2H3O2-] Ka = [HC2H3O2] [C2H3O2], M
Initially
Change
At Equilibrium
[H3O+] [C2H3O2-]
[HC2H3O2]
Ka =
We are assuming that x will be very small compared to 0.30 and can, therefore, be ignored.

41. General Chemistry: Chapter 16
EXAMPLE 16-5
Determining a Value of KA from the pH of a Solution of a Weak Acid. Butyric acid, HC4H7O2 (or CH3CH2CH2CO2H) is used to make compounds employed in artificial flavorings and syrups. A M aqueous solution of HC4H7O2 is found to have a pH of Determine Ka for butyric acid.
HC4H7O2 + H2O C4H7O2 + H3O+
Ka = ?
General Chemistry: Chapter 16

42. Chemistry 140 Fall 2002
EXAMPLE 16-5
Solution:
For HC4H7O2 KA is likely to be much larger than KW. Therefore assume self-ionization of water is unimportant.
HC4H7O2 + H2O C4H7O2 + H3O+
Initial conc M 0 0
Changes -x M +x M +x M
Equilibrium (0.250-x) M x M x M
Concentration

43. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
EXAMPLE 16-5
HC4H7O2 + H2O C4H7O2 + H3O+
Log[H3O+] = -pH = -2.72
[H3O+] = = 1.910-3 = x
[H3O+] [C4H7O2-]
[HC4H7O2]
Ka=
1.910-3 · 1.910-3
(0.250 – 1.910-3) =
Ka= 1.510-5
Check assumption: Ka >> KW.
General Chemistry: Chapter 16

44. General Chemistry: Chapter 16
Percent Ionization
HA + H2O H3O+ + A-
Degree of ionization =
[H3O+] from HA
[HA] originally
Percent ionization =
[H3O+] from HA
[HA] originally
 100%
General Chemistry: Chapter 16

45. Strong Bases
Strong bases are the soluble hydroxides, which are the alkali metal and heavier alkaline earth metal hydroxides (Ca2+, Sr2+, and Ba2+).
Again, these substances dissociate completely in aqueous solution.

46. Bases react with water to produce hydroxide ion.
Weak Bases
Bases react with water to produce hydroxide ion.

47. General Chemistry: Chapter 16
A Weak Base
General Chemistry: Chapter 16

48. Weak Bases [CH3NH3+][HO-] Kb= = 4.310-4 [CH3NH2]
pKb= -log(4.210-4) = 3.37

49. Weak Bases [HB] [OH-] [B-] Kb =
The equilibrium constant expression for this reaction is
[HB] [OH-]
[B-]
Kb =
where Kb is the base-dissociation constant.

50. Kb can be used to find [OH-] and, through it, pH.
Weak Bases
Kb can be used to find [OH-] and, through it, pH.

51. pH of Basic Solutions [NH4+] [OH-] [NH3] Kb = = 1.8  10-5
What is the pH of a 0.15 M solution of NH3?
NH3 (aq) + H2O (l)
NH4+ (aq) + OH- (aq)
[NH4+] [OH-]
[NH3]
Kb =
= 1.8  10-5

52. [NH3], M
[NH4+], M
[OH-], M
Initially
0.15
At Equilibrium
x  0.15 x

53. Ka and Kb Ka and Kb are related in this way: Ka  Kb = Kw
Therefore, if you know one of them, you can calculate the other.

54. 16-7 Ions as Acids and Bases
CH3CO2- + H2O CH3CO2H + OH-
base
acid
[NH3] [H3O+]
Ka=
[NH4+]
= ?
NH4+ + H2O NH3 + H3O+
acid
base
[NH3] [H3O+] [OH-]
Ka=
[NH4+] [OH-] = KW Kb =
1.010-14
1.810-5
= 5.610-10
Ka Kb = Kw
General Chemistry: Chapter 16

56. 16-6 Polyprotic Acids Phosphoric acid: A triprotic acid.
H3PO4 + H2O H3O+ + H2PO4-
Ka = 7.110-3
H2PO4- + H2O H3O+ + HPO42-
Ka = 6.310-8
HPO42- + H2O H3O+ + PO43-
Ka = 4.210-13

57. Phosphoric Acid Ka1 >> Ka2
All H3O+ is formed in the first ionization step.
H2PO4- essentially does not ionize further.
Assume [H2PO4-] = [H3O+].
[HPO42-]  Ka2 regardless of solution molarity.

58. EXAMPLE 16-9
Calculating Ion Concentrations in a Polyprotic Acid Solution. For a 3.0 M H3PO4 solution, calculate:
(a) [H3O+]; (b) [H2PO4-]; (c) [HPO42-] (d) [PO43-]
H3PO H2O H2PO H3O+
Initial conc. 3.0 M 0 0
Changes -x M +x M +x M
Equilibrium (3.0-x) M x M x M
Concentration

59. General Chemistry: Chapter 16
EXAMPLE 16-9
H3PO H2O H2PO H3O+
[H3O+] [H2PO4-]
x · x
Ka= =
= 7.110-3
[H3PO4]
(3.0 – x)
Assume that x << 3.0
x2 = (3.0)(7.110-3) x = 0.14 M
[H2PO4-] = [H3O+] = 0.14 M
General Chemistry: Chapter 16

60. General Chemistry: Chapter 16
EXAMPLE 16-9
H2PO H2O HPO H3O+
Initial conc M M
Changes -y M +y M +y M
Equilibrium ( y) M y M (0.14 +y) M
Concentration
[H3O+] [HPO42-]
[H2PO4-]
Ka=
y · ( y)
( y) =
= 6.310-8
y << 0.14 M y = [HPO42-] = 6.310-8
General Chemistry: Chapter 16

61. EXAMPLE 16-9 HPO4- + H2O PO43- + H3O+ [H3O+] [HPO42-] (0.14)[PO43-]
Ka= =
[H2PO4-]
6.310-8
[PO43-] = 1.910-19 M

62. Sulfuric Acid Sulfuric acid: A diprotic acid. H2SO4 + H2O H3O+ + HSO4-
Ka = very large
HSO4- + H2O H3O+ + SO42-
pKa = 1.96

63. General Chemistry: Chapter 16
pH of Salt Solutions
Salt solutions can be neutral, acidic or basic in nature.
The pH of a salt solution depends on the behavior of its corresponding Cations and Anions towards water.
General Chemistry: Chapter 16

64. General Chemistry: Chapter 16
Hydrolysis
The reaction between an ion and water is called hydrolysis.
Na+ + H2O → Na+ + H2O
No reaction
Cl- + H2O → Cl- + H2O
No reaction
NH4+ + H2O → NH3 + H3O+
Hydrolysis
General Chemistry: Chapter 16

65. General Chemistry: Chapter 16
In general:
Cations of strong bases do not undergo hydrolysis; they don’t react with water.
 Cations of weak bases undergo hydrolysis.
 Anions of strong acids do not react with water.
 Anions of weak acids undergo hydrolysis.
General Chemistry: Chapter 16

66. 16-8 Molecular Structure and Acid-Base Behavior
Why is HCl a strong acid, but HF is a weak one?
Why is CH3CO2H a stronger acid than CH3CH2OH?
There is a relationship between molecular structure and acid strength.
Bond dissociation energies are measured in the gas phase and not in solution.
General Chemistry: Chapter 16

67. Factors Affecting Acid Strength
The more polar the H-X bond and/or the weaker the
H-X bond, the more acidic the compound.
So acidity increases from left to right across a row and from top to bottom down a group.

68. Strengths of Binary Acids
Chemistry 140 Fall 2002
Strengths of Binary Acids
HI HBr HCl HF
Bond length
160.9 > > > pm
Bond energy
297 < 368 < 431 < kJ/mol
Acid strength
109 > 108 > 1.3106 >> 6.610-4
HF + H2O → [F-·····H3O+] F- + H3O+
ion pair
H-bonding
free ions

69. General Chemistry: Chapter 16
Strengths of Oxoacids
Factors promoting electron withdrawal from the OH bond to the oxygen atom:
High electronegativity (EN) of the central atom.
A large number of terminal O atoms in the molecule.
H-O-Cl H-O-Br
ENCl = 3.0 ENBr= 2.8
Ka = 2.910-8 Ka = 2.110-9
General Chemistry: Chapter 16

70. Factors Affecting Acid Strength
In oxoacids, in which an -OH is bonded to another atom, Y, the more electronegative Y is, the more acidic the acid.

71. General Chemistry: Chapter 16
Strengths of Oxoacids S O H ·· S O H ··
Ka 103 Ka =1.310-2 S O H ·· - 2+ S O H ·· - +
General Chemistry: Chapter 16

72. Strengths of Organic Acids·· O H H H ·· ·· H C C O H H C C O H ·· ·· H H H
acetic acid ethanol
Ka = 1.810-5 Ka =1.310-16
General Chemistry: Chapter 16

73. Focus on the Anions Formed
Ethoxide H H ·· - H C C O ·· ·· H H
Acetate C O H - ·· C O H - ··
General Chemistry: Chapter 16

74. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Structural Effects ·· H O
Ka = 1.810-5
Ka = 1.310-5 H C C ·· - O H ·· H C H C H C H C H C H C H C O - ·· H C H
General Chemistry: Chapter 16

75. General Chemistry: Chapter 16
Chemistry 140 Fall 2002
Structural Effects ·· H O
Ka = 1.810-5
Ka = 1.410-3 H C C ·· - O H ·· ·· Cl O H C C ·· - O H ··
General Chemistry: Chapter 16

76. Strengths of Amines as Bases·· Br N ·· H H
ammonia bromamine
pKb = 4.74 pKb = 7.61
General Chemistry: Chapter 16

77. Strengths of Amines as Bases
Chemistry 140 Fall 2002
Strengths of Amines as Bases H H H H H H H C
NH2 H C C
NH2 H C C C
NH2 H H H H H H
methylamine ethylamine propylamine
pKb = 4.74 pKb = 3.38 pKb = 3.37
General Chemistry: Chapter 16

78. General Chemistry: Chapter 16
Resonance Effects
General Chemistry: Chapter 16

79. General Chemistry: Chapter 16
16-9 Lewis Acids and Bases
Lewis Acid
A species (atom, ion or molecule) that is an electron pair acceptor.
Lewis Base
A species that is an electron pair donor.
base
acid
adduct
General Chemistry: Chapter 16

80. Showing Electron Movement