1. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Acids and Bases
18-1

2. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
18-2

3. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Arrhenius Acid-Base Definition
An acid is a substance that increase H+ when dissolved in
water.
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
The ionization of acids actually produces H3O+ but
H3O+  H2O + H+
A base is a substance that increase OH- ions when dissolved in
water.
NaOH(aq) + H2O(l) Na+(aq) + OH-(aq)
Limitation: this theory cannot explain why NH3 is a base
18-3

4. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species which donates a H+.
A base is a proton acceptor, any species which accepts a H+.
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
An acid reactant will produce a base product and the two will
constitute an acid-base conjugate pair.
18-4

5. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Conjugate Acid-Base Pairs
An acid and base that differ only in the presence
or absence of a proton
18-5

6. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 18.4 The Conjugate Pairs in Some Acid-Base Reactions
Conjugate Pair
Acid + Base Base + Acid
Conjugate Pair
Reaction 1 HF + H2O F- + H3O+
Reaction 2 HCOOH + CN- HCOO- + HCN
Reaction 3 NH4+ + CO32- NH3 + HCO3-
Reaction 4 H2PO4- + OH- HPO42- + H2O
Reaction 5 H2SO4 + N2H5+ HSO4- + N2H62+
Reaction 6 HPO42- + SO32- PO43- + HSO3-
18-6

7. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM 18.4: Identifying Conjugate Acid-Base Pairs
PROBLEM: The following reactions are important environmental processes.
Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)
PLAN: Identify proton donors (acids) and proton acceptors (bases).
conjugate pair conjugate pair2
1
SOLUTION: (a) H2PO4-(aq) + CO32-(aq) HPO42-(aq) + HCO3-(aq)
proton proton proton proton
donor acceptor acceptor donor
conjugate pair1 conjugate pair2
(b) H2O(l) + SO32-(aq) OH-(aq) + HSO3-(aq)
proton proton proton proton
donor acceptor acceptor donor
18-7

8. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Amphoteric substances
A substance that can both accept and donate a proton–
i.e. act as an acid and as a base
Ex. H2O
Water acts as a base
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
Water acts as an acid
NH3 + H2O(l) NH4+(aq) + OH-(aq)
18-8

9. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 18.9
Strengths of
conjugate acid-
base pairs
18-9

10. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Relative Strengths of Acids and Bases
In every acid-base reaction, the position of
equilibrium favors the weaker acid
HCl(l) + H2O(l) H3O+(aq) + Cl-(aq)
Stronger acid Weaker acid
Since H3O+ is weaker, the forward reaction is favored over the
reverse reaction and the equilibrium lies to the right
18-10

11. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM 18.5: Predicting the Net Direction of an Acid-Base
Reaction
PROBLEM: Predict the net direction of equilibrium
(a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10
(button) to determine the relative strength of each. The stronger the
species, the more preponderant its conjugate.
SOLUTION: (a) H2PO4-(aq) + NH3(aq) HPO42-(aq) + NH4+(aq)
stronger acid weaker acid
Net direction is to the right with Kc > 1.
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
weaker acid stronger acid
Net direction is to the left with Kc < 1.
18-11

12. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Autoionization of Water
H2O(l) H3O+(aq) + OH-(aq)
Kc = [H3O ][OH ]
+ -
The Ion-Product Constant for Water
Kw = [H3O+][OH-] = 1.0 x 10-14 at 250C
A change in [H3O+] causes an inverse change in [OH-].
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]
18-12

13. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The relationship between [H3O+] and [OH-] and the
Figure 18.4
relative acidity of solutions.
[H3O+] Divide into Kw [OH-]
[H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-]
ACIDIC NEUTRAL BASIC
SOLUTION SOLUTION SOLUTION
18-13

14. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM 18.2: Calculating [H3O+] and [OH-] in an Aqueous
Solution
PROBLEM: A research chemist adds a measured amount of HCl gas to pure
water at 250C and obtains a solution with [H3O+] = 3.0x10-4M.
Calculate [OH-]. Is the solution neutral, acidic, or basic?
PLAN: Use the Kw at 250C and the [H3O+] to find the corresponding [OH-].
SOLUTION: K = 1.0x10-14 = [H O+] [OH-] so
w 3
[OH-] = Kw/ [H3O+] = 1.0x10-14/3.0x10-4 = 3.3x10-11M
[H3O+] is > [OH-] and the solution is acidic.
18-14

15. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Solve the following:
• [ H+] = 1.4 x 10-6 [OH-] =
b. [ OH-] = 1.0 x 10-7 [H+] =
18-15

16. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The pH Scale
-Measures the concentration of hydrogen ions
-the power/potential of the hydrogen ion
-a way of expressing H+ in a given solution
pH = - log [H+]
The higher the H+ concentration, the lower the pH
pOH = - log [OH-]
The higher the OH- concentration, the lower the pOH
pOH + pH = 14
18-16

17. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 18.5
The pH values of some
familiar aqueous
solutions
pH = -log [H3O+]
18-17

18. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 18.6 The relations among [H3O+], pH, [OH-], and pOH.
18-18

19. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM 18.3: Calculating [H3O+], pH, [OH-], and pOH
PROBLEM: Given [H3O+] concentrations: 2.0M, 0.30M, and 0.0063M,
Calculate pH, [OH-], and pOH of the three solutions at 250C.
PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-]
and then convert to pH and pOH.
SOLUTION: For [H3O+] = 2.0M and -log [H3O+] = -0.30 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/2.0 = 5.0x10-15M; pOH = 14.30
For [H3O+] = 0.30M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/0.30 = 3.3x10-14M; pOH = 13.48
For [H3O+] = 0.0063M and -log [H3O+] = 2.20 = pH
[OH-] = Kw/ [H3O+] = 1.0x10-14/6.3x10-3 = 1.6x10-12M; pOH = 11.80
18-19

20. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
STRONG ACIDS and BASES
Strong acids and bases dissociate completely into ions in water.
100% ionization
Strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-(aq)
18-20

21. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Strong Acids Strong Bases
HCl LiOH
HBr NaOH
HI KOH
HNO3 Ca(OH)2
H2 SO4 Sr(OH)2
HClO4 Ba(OH)2
All other acids and bases not listed here are considered weak.
18-21

22. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
For STRONG ACIDS,
H+ or OH- are equal to the original molarity of the solution
Problem: Calculate the pH of 0.5 M HCl and 0.10 M NaOH
solution:
HCl(g or l) + H2O(l) H3O+(aq) + Cl-(aq)
0.5 M 0.5 M 0.5 M
pH = - log (0.5 M) = 0.30
NaOH(aq) Na+(aq) + OH-(aq)
0.10 M 0.10 M 0.10 M
pOH = - log (0.10 M) = 1.00
pH = 14 – 1.00 = 13.00
18-22

23. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM 18.5:
Calculate the pH of a solution prepared by dissolving 5.00 g of KOH in
enough water to get 500.0 mL of solution
MW of KOH = 56.108 g/mol
Answer: 0.178 M of [OH-]
pOH = 0.74902
pH = 13.25
18-23

24. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
WEAK ACIDS and WEAK BASES
Weak acids and bases dissociate very slightly into ions in water.
- About 1% ionization, the equilibrium lies far to the left
Weak acid: HA(aq) + H2O(l) H2O+(aq) + A-(aq)
18-24

25. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Acid-Dissociation Constant, Ka
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
[H3O+][A-]
Kc =
[HA] stronger acid higher [H3O+]
larger Ka
[H3O+][A-]
Ka =
[HA]
smaller Ka lower [H3O+]
weaker acid
18-25

26. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM :
Write the equilibrium constant expression for the following weak acid:
a. NH4 +
NH4 + (aq) + H2O(l) H3O+(aq) + NH3
[H3O+][NH3]
Ka =
[NH4 + ]
18-26

27. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Table 18.3 The Relationship Between Ka and pKa
pKa = - log Ka
Acid Name (Formula) Ka at 250C pKa
Hydrogen sulfate ion (HSO4-) 1.02x10-2 1.991
7.1x10-4 3.15
Nitrous acid (HNO2)
1.8x10-5 4.74
Acetic acid (CH3COOH)
Hypobromous acid (HBrO) 2.3x10-9 8.64
Phenol (C6H5OH) 1.0x10-10 10.00
18-27

28. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Base-Dissociation Constant, Kb
A- (aq) + H2O(l) OH-(aq) + HA (aq)
[OH-][HA]
Kc =
[A-] stronger base higher [OH-]
larger Kb
[OH-][HA]
Kb =
[A-]
18-28

29. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM :
Write the equilibrium constant expression for the following weak base:
a. C5H5N
C5H5N (aq) + H2O(l) OH-(aq) + C5H5NH+
[C5H5NH+ ][OH-]
Kb =
[C5H5N]
18-29

30. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The Relationship Between Ka and Kb
Ka x Kb = Kw
Example: Calculate Kb if Ka = 1.5 x 10-5
1.0 x 10-14
Kb =
1.5 x 10-5
= 6.7 x 10-10
18-30

31. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Lewis Definition of Acids and Bases
An acid is an electron-pair acceptor.
A base is an electron-pair donor.
F F
H
H
B + N
B N
F HH F
F F HH
acid base adduct
M(H2O)42+(aq)
M2+
H2O(l) adduct
18-31

32. Copyright ©The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
SAMPLE PROBLEM 18.13: Identifying Lewis Acids and Bases
PROBLEM: Identify the Lewis acids and Lewis bases in the following reactions:
(a) H+ + OH- H2O
(b) Cl- + BCl3 BCl4-
(c) K+ + 6H2O K(H2O)6+
PLAN: Look for electron pair acceptors (acids) and donors (bases).
SOLUTION: acceptor
(a) H+ + OH- H2O
donor
donor
(b) Cl- + BCl3 BCl4-
acceptor
acceptor
(c) K+ + 6H2O K(H2O)6+
donor
18-32