1. CHAPTER 16 – ACIDS AND BASES
SVANTE ARRHENIUS
Proposed the first definitions of acids and bases
ACID – A compound that produces hydrogen ions in a water solution
HCl (g) → H+(aq) + Cl-(aq)
BASE – A compound that produces hydroxide ions in a water solution
NaOH (s) → Na+(aq) + OH-(aq)
8A-1 (of 51)

2. 1923
Expanded the definitions of acids and bases
THOMAS LOWRY
8A-2

3. 1923
Expanded the definitions of acids and bases
JOHANNES BRØNSTED
ACID – A hydrogen ion (or proton) donor
BASE – A hydrogen ion (or proton) acceptor
8A-3

4. HCl H2O →
Cl H3O+
acid
base
HYDRONIUM ION – H3O+, formed when a hydrogen ion attaches to a water - +
8A-4

5. NH H2O →
NH OH-
base
acid
AMPHOTERIC – A substance that can act as an acid or a base + -
8A-5

6. Acids turn into bases, and bases turn into acids
HCl H2O → Cl H3O+
acid
base - +
acid
base
8A-6

7. Acids turn into bases, and bases turn into acids
HCl H2O → Cl H3O+
acid
base
conjugate base of HCl
conjugate acid of H2O - +
acid
base
base
acid
Strong acids (or bases) have non-conjugates
8A-7

8. NH H2O → NH OH-
base
acid + -
base
acid
8A-8

9. NH H2O → NH OH-
base
acid
conjugate acid of NH3
conjugate base of H2O + -
base
acid
acid
base
Weak bases (or acids) have weak conjugates
8A-9

10. Conjugate base of HClO4
When HClO4 acts as an acid, it becomes:
ClO4-
Strong Acid
Non Base
Conjugate acid of CH3NH2
When CH3NH2 acts as a base, it becomes:
CH3NH3+
Weak Base
Non Acid
8A-10

11. POLYPROTIC ACID – An acid with more than one ionizable hydrogen ion
H2CO3
H3PO4
diprotic
triprotic
Hydrogen ions become successively more difficult to ionize
H2SO4 (aq) + H2O(l) → H3O+(aq) + HSO4-(aq)
Strong Acid
HSO4- (aq) + H2O(l) → H3O+(aq) + SO42-(aq)
Weaker Acid
8A-11

12. Conjugate base of H2CO3
When H2CO3 acts as an acid, it becomes:
HCO3-
Conjugate base of HCO3-
When HCO3- acts as an acid, it becomes:
CO32-
Conjugate acid of HCO3-
When HCO3- acts as an base, it becomes:
H2CO3
8A-12

13. WATER
Water ionizes to a small extent
8A-13

14. WATER
Water ionizes to a small extent + -
2H2O (l) → H3O+(aq) + OH-(aq)
Makes solutions acidic
Makes solutions basic
Equal amounts of H3O+ and OH- make a solution NEUTRAL
 pure water is neutral
8A-14

15. Ionization: 2H2O (l) → H3O+(aq) + OH-(aq)
Association: H3O+(aq) + OH-(aq) → 2H2O (l)
Equilibrium H2O (l) ⇆ H3O+(aq) + OH-(aq)
Because of this dynamic equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant
If we use [ ] to represent the molarity of a substance
K = [H3O+][OH-]
Where K is a constant
8A-15

16. Ionization: 2H2O (l) → H3O+(aq) + OH-(aq)
Association: H3O+(aq) + OH-(aq) → 2H2O (l)
Equilibrium H2O (l) ⇆ H3O+(aq) + OH-(aq)
Because of this dynamic equilibrium, the product of the molarity of the hydronium ions and the molarity of the hydroxide ions is a constant
If we use [ ] to represent the molarity of a substance
Kw = [H3O+][OH-]
ION-PRODUCT CONSTANT FOR WATER (Kw) – The product of the hydronium ion and hydroxide ion molarities in any water solution
8A-16

17. The Kw depends on temperature
Temp (ºC) Kw 10 25 40
0.29 x M2
1.00 x M2
2.92 x M2
8A-17

18. Calculate the hydronium ion and hydroxide ion molarities in pure water at 25ºC.
2H2O (l) ⇆ H3O+(aq) + OH-(aq)
Molarities before any Ionization:
Molarity Change:
Molarities at Equilibrium:
+x x
x x
Kw = [H3O+][OH-]
1.00 x M2 =
(x)(x)
= x2
1.00 x 10-7 M = x
In pure water, [H3O+] = [OH-] = x 10-7 M
8A-18

19. Calculate the hydroxide ion molarity in a water solution at 25ºC in which the [H3O+] = 1.00 x 10-5 M.
Kw = [H3O+][OH-]
1.00 x M2 =
(1.00 x 10-5 M)(x)
1.00 x M2
_________________
1.00 x 10-5 M
= x
1.00 x 10-9 M = x
= [OH-]
8A-19

20. Calculate the hydronium ion molarity in a water solution at 25ºC in which the [OH-] = 8.00 x 10-3 M.
Kw = [H3O+][OH-]
1.00 x M2 =
(x)(8.00 x 10-3 M)
1.00 x M2
_________________
8.00 x 10-3 M
= x
1.25 x M = x
= [H3O+]
8A-20

21. All water solutions contain both hydronium and hydroxide ions
[H3O+] = [OH-] :
[H3O+] > [OH-] :
[H3O+] < [OH-] :
the solution is NEUTRAL
the solution is ACIDIC
the solution is BASIC
For all water solutions: Kw = [H3O+][OH-]
8A-21

22. THE pH SCALE
pH – The negative logarithm of the hydronium ion molarity of a solution
The common logarithm of a number is:
the exponent to which 10 must be raised to equal the number
100
1000
0.001
200
100 = 102
1000 = 103
0.001 = 10-3
200 = 102.3
log 100 = 2
log 1000 = 3
log = -3
log 200 = 2.3
8A-22

23. [H3O+]
log [H3O+] pH M
0.01 M M
0.02 M
-1.0
-2.0
-3.0
-1.7
1.0
2.0
3.0
1.7
For logarithmic numbers, only the digits after the decimal point (called the MANTISSA) are significant figures, not the digits before (called the CHARACTERISTIC)
8A-23

24. Calculate the pH of pure water at 25ºC.
For pure water:
[H3O+] = x 10-7 M
pH = -log[H3O+]
= -log(1.00 x 10-7 M) =
8A-24

25. < 7 Acidic > 7 Basic [OH-]
10. M
1. M
0. 1 M
0.01 M
0.001 M M M M M M M M M M M M M M M M M M M M M M M M
0.001 M
0.01 M
0.1 M
1 M
< 7
Acidic
pH 7 = Neutral
> 7
Basic
[H3O+]
[OH-]
8A-25

26. < 7 Acidic > 7 Basic Common Substances
Battery Acid
Gastric Fluid
< 7
Acidic
Soda
Orange Juice
Tap Water
pH 7 = Neutral
“Pure” Water
Sea Water
Baking Soda
> 7
Basic
Ammonia
Bleach
Drain Cleaner
[H3O+]
Common Substances
8A-26

27. Calculate the pH of orange juice if it has a hydronium ion concentration of 1.6 x 10-3 M.
[H3O+] = 1.6 x 10-3 M
pH = -log[H3O+]
= -log(1.6 x 10-3 M)
= 2.80
8A-27

28. Calculate the pH of toothpaste that has a hydroxide ion concentration of 5.6 x 10-5 M.
[OH-] = 5.6 x 10-5 M
Kw = [H3O+][OH-]
1.00 x M2 =
(x)(5.6 x 10-5 M)
1.00 x M2 = x
_________________
5.6 x 10-5 M
= x M
= [H3O+]
pH = -log[H3O+]
= -log(1.79 x M)
= 9.75
8A-28

29. Calculate the pH of milk if it has a hydroxide ion concentration of 3
Calculate the pH of milk if it has a hydroxide ion concentration of 3.2 x 10-9 M.
[OH-] = 3.2 x 10-9 M
Kw = [H3O+][OH-]
1.00 x M2 =
(x)(3.2 x 10-9 M)
1.00 x M2 = x
_________________
3.2 x 10-9 M
= x 10-6 M
= [H3O+]
pH = -log[H3O+]
= -log(3.13 x 10-6 M)
= 5.51
8A-29

30. Calculate the hydronium ion concentration in blood, which has a pH of 7.4.
pH = -log [H3O+]
-pH = log [H3O+]
antilog (-pH) = [H3O+]
antilog (-7.4) = [H3O+]
= M
= x 10-8 M
= 4 x 10-8 M
For logarithmic numbers, only the digits after the decimal point are significant figures, not the digits before
8A-30

31. Calculate the hydroxide ion concentration in egg yolks, which have a pH of 5.65.
pH = -log [H3O+]
-pH = log [H3O+]
antilog (-pH) = [H3O+]
antilog (-5.65) = [H3O+]
= M
= x 10-6 M
Kw = [H3O+][OH-]
1.00 x M2 =
(2.24 x 10-6 M)(x)
1.00 x M2 = __________________
2.24 x 10-6 M x
= x 10-9 M
= [OH-]
8A-31

32. pH OF STRONG ACID SOLUTIONS
Strong acids completely ionize
HCl (g)
HCl (aq)
H+(aq) + Cl-(aq)
8A-32

33. Calculate the pH of a 0.015 M hydrochloric acid solution.
HCl completely ionizes, 
0.015 M H+ (H3O+)
0.015 M Cl-
pH = -log[H3O+]
= -log(0.015 M)
= 1.82
8A-33

34. Calculate the pH of a 0.0400 M sulfuric acid solution.
H2SO4 completely ionizes, 
M H+ (H3O+)
M SO42-
pH = -log[H3O+]
= -log( M) =
8A-34

35. pH OF STRONG BASE SOLUTIONS
Strong bases completely dissociate
NaOH (s)
NaOH (aq)
Na+(aq) + OH-(aq)
8A-35

36. Calculate the pH of a 0.40 M sodium hydroxide solution.
NaOH completely dissociates, 
0.40 M Na+
0.40 M OH-
not [H3O+]
Kw = [H3O+][OH-]
1.00 x M2 =
(x)(0.40 M)
1.00 x M2 = x
__________________
0.40 M
= x M
= [H3O+]
pH = -log[H3O+]
= -log(2.50 x M) =
8A-36

37. BUFFERS
BUFFER – A solution that resists a change in its pH even when a strong acid or strong base is added to it
7.00 pH
1.00 L H2O
0.10 moles HCl
8A-37

38. BUFFERS
BUFFER – A solution that resists a change in its pH even when a strong acid or strong base is added to it
1.00
3.45 pH pH
1.00 L H2O
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
0.10 moles HCl
8A-38

39. BUFFERS
BUFFER – A solution that resists a change in its pH even when a strong acid or strong base is added to it
1.00
3.37 pH pH
1.00 L H2O
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
0.10 moles HCl
8A-39

40. A solution is a buffer if is contains a weak acid and its conjugate baseHF F-
If a strong acid is added to the buffer:
1.00 L 1.0 M HF, 1.0 M F-
0.10 moles HCl
8A-40

41. A solution is a buffer if is contains a weak acid and its conjugate baseHF F-
If a strong acid is added to the buffer:
The strong acid (H+) is reacted away by the conjugate base (F-)
H+ + F- → HF
1.00 L 1.0 M HF, 1.0 M F-
 no more strong acid in the solution
0.10 moles HCl
8A-41

42. A solution is a buffer if is contains a weak acid and its conjugate baseHF F-
If a strong base is added to the buffer:
The strong base (OH-) is reacted away by the weak acid (HF)
OH- + HF → H2O + F-
1.00 L 1.0 M HF, 1.0 M F-
 no more strong base in the solution
0.10 moles OH-
8A-42