This document provides an overview of acid-base equilibria, including: - Definitions of acids and bases according to Arrhenius, Brønsted-Lowry, and Lewis theories. - A discussion of auto-ionization of water and the pH scale. Weak acids only partially dissociate in water according to their acid dissociation constant, Ka. - Explanations of acid-base terminology like conjugate acid-base pairs and the effect of relative acid strength on the direction of acid-base reactions. - Examples of calculating quantities like [H3O+], pH, pKa, and determining the direction of acid-base reactions. The document

1. 18-1
Chapter 18
Acid-Base Equilibria

2. 18-2
Acid-Base Equilibria
18.1 Acids and bases in water
18.2 Auto-ionization of water and the pH scale
18.3 Proton transfer and the Brønsted-Lowry acid-base definition
18.4 Solving problems involving weak acid equilibria
18.5 Weak bases and their relationships to weak acids
18.6 Molecular properties and acid strength
18.7 Acid-base properties of salt solutions
18.8 Generalizing the Brønsted-Lowry concept: The Leveling Effect
18.9 Electron-pair donations and the Lewis acid-base definition

3. 18-3 Figure 18.1
Etching with acids
The inside surfaces of these light
bulbs are etched with HF.
Acids are used to wash away
oxides of silicon and metals during
the production of computer chips.

4. 18-4

5. 18-5
For reaction between a strong acid and strong base:
HCl(aq) + NaOH(aq) NaCl(aq) + H2O(l)
H+(aq) + OH-(aq) H2O(l)
For acid dissociation:
HA(g or l) + H2O(l) A-(aq) + H3O+(aq)
In aqueous solution, the H+ ions bind covalently to
water to form solvated hydronium ions.

6. 18-6
Figure 18.2
The nature of the hydrated proton

7. 18-7
The Classical (Arrhenius) Definition of Acids and Bases
An acid is a substance that has H in its formula and dissociates in water
to yield H3O+.
A base is a substance that has OH in its formula and dissociates in water
to yield OH-.
Arrhenius acids contain covalently bonded H atoms
that ionize in water.
Neutralization occurs when the H+ ion from the acid and the OH- ion
from the base combine to form water.
H+(aq) + OH- (aq) H2O(l) ∆Ho
rxn = -55.9 kJ

8. 18-8
Defining the acid dissociation constant
Weak acids dissociate very slightly into ions in water.
Strong acids dissociate completely into ions in water.
HA(g or l) + H2O(l) H3O+ (aq) + A- (aq)
HA(aq) + H2O(l) H3O+ (aq) + A- (aq)
Kc >> 1
Kc << 1
Kc =
[H3O+][A-]
[H2O][HA]
Kc[H2O] = Ka =
[H3O+][A-]
[HA]
weaker acid, lower [H3O+],
smaller Ka
stronger acid, higher [H3O+],
larger Ka

9. 18-9
Figure 18.3
strong acid: HA(g or l) + H2O(l) H3O+(aq) + A-(aq)
The extent of dissociation of strong acids
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

10. 18-10
Figure 18.3
The extent of dissociation of weak acids
weak acid: HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

11. 18-11
Figure 18.4
Reaction of zinc with a strong and a weak acid
1 M HCl(aq) 1 M CH3COOH(aq)
Zn(s) + 2H3O+(aq) Zn+2(aq)
+ 2H2O(l) + H2(g)

12. 18-12
Acid strength
decreases down
the table (smaller
Ka values)

13. 18-13
SAMPLE PROBLEM 18.1
SOLUTION:
Classifying acid and base strength from the
chemical formula
PROBLEM: Classify each of the following compounds as a strong acid,
weak acid, strong base or weak base.
(a) H2SeO4
(b) (CH3)2CHCOOH (c) KOH (d) (CH3)2CHNH2
PLAN: Pay attention to the text definitions of acids and bases. Look at O for
acids and for the -COOH group; watch for amine groups and cations
in bases.
(a) strong acid - H2SeO4 - the number of oxygen atoms
exceeds the number of ionizable protons by a factor of 2.
(b) weak acid - (CH3)2CHCOOH is an organic acid having a -COOH group
(a carboxylic acid).
(c) strong base - KOH is a Group 1A hydroxide.
(d) weak base - (CH3)2CHNH2 has a lone pair of electrons on the
nitrogen and is an amine.

14. 18-14
The auto-ionization of water and the pH scale
H2O(l) H2O(l)
H3O +(aq
)
OH-(aq)
+
+

15. 18-15
Kc =
[H3O+][OH-]
[H2O]2
Kc[H2O]2 = [H3O+][OH-
]
Defining the ion-product constant of water
Kw =
A change in [H3O+] causes an inverse change in [OH-].
= 1.0 x 10-14 at 25 oC
H2O(l) + H2O(l) H3O+(aq) + OH-(aq)
In an acidic solution, [H3O+] > [OH-]
In a basic solution, [H3O+] < [OH-]
In a neutral solution, [H3O+] = [OH-]

16. 18-16
Figure 18.5
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions
[H3O+] [OH-]
divide into Kw
acidic
solution
basic
solution
[H3O+] > [OH-] [H3O+] = [OH-] [H3O+] < [OH-]
neutral
solution
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

17. 18-17
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 25 oC and obtains a solution with [H3O+] = 3.0 x 10-4 M.
Calculate [OH-]. Is the solution neutral, acidic or basic?
SOLUTION:
PLAN: Use the Kw at 25 oC and the [H3O+] to find the corresponding [OH-].
Kw = 1.0 x 10-14 = [H3O+] [OH-]
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/3.0 x 10-4 = 3.3 x 10-11 M
[H3O+] > [OH-]; the solution is acidic.

18. 18-18
Figure 18.6
The pH values of
some familiar
aqueous solutions
pH = -log [H3O+]
pOH = -log [OH-]
pK = -log K
Related Expressions

19. 18-19
A low pK corresponds to a high K.

20. 18-20
Relationships
between
[H3O+],
pH, [OH-] and
pOH
Figure 18.7

21. 18-21
SAMPLE PROBLEM 18.3 Calculating [H3O+], pH, [OH-], and pOH
PROBLEM: In a restoration project, a conservator prepares copper-plate etching
solutions by diluting concentrated nitric acid, HNO3, to 2.0 M, 0.30 M,
and 0.0063 M HNO3. Calculate [H3O+], pH, [OH-], and pOH of the
three solutions at 25 oC.
SOLUTION:
PLAN: HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-]
and then convert to pH and pOH.
For 2.0 M HNO3, [H3O+] = 2.0 M and -log [H3O+] = -0.30 = pH
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/2.0 = 5.0 x 10-15 M; pOH =
14.30
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/0.30 = 3.3 x 10-14 M; pOH = 13.48
For 0.3 M HNO3, [H3O+] = 0.30 M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/6.3 x 10-3 = 1.6 x 10-12 M; pOH = 11.80
For 0.0063 M HNO3, [H3O+] = 0.0063 M and -log [H3O+] = 2.20 = pH

22. 18-22
Figure 18.8
Methods for measuring the pH of an aqueous solution
pH meter
pH (indicator) paper
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

23. 18-23
The Brønsted-Lowry Definition of Acids and Bases
An acid is a proton donor, that is, any species that donates an H+ ion.
All Arrhenius acids are Brønsted-Lowry acids.
A base is a proton acceptor, that is, any species
that accepts an H+ ion.
In the Brønsted-Lowry definition, an acid-base reaction is a proton
transfer process.

24. 18-24
Figure 18.9
Proton transfer is the essential feature of a Brønsted-
Lowry acid-base reaction
(acid, H+ donor) (base, H+ acceptor)
HCl H2O
+
Cl- H3O+
+
lone pair
binds H+
(base, H+ acceptor) (acid, H+ donor)
NH3 H2O
+
NH4
+ OH-
+
lone pair
binds H+
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

25. 18-25
An acid reactant produces a base product and the two constitute
an acid-base conjugate pair.
The Conjugate Acid-Base Pair
H2S + NH3 HS- + NH4
+
H2S and HS- are a conjugate acid-base pair.
HS- is the conjugate base of the acid H2S.
NH3 and NH4
+ are a conjugate acid-base pair.
NH4
+ is the conjugate acid of the base NH3.
Every acid has a conjugate base, and every base has a conjugate acid.

26. 18-26
The conjugate base of the pair has one fewer H and one more
negative charge than the acid.
The conjugate acid of the pair has one more H and one less
negative charge than the base.
A Bronsted-Lowry acid-base reaction occurs when an acid and a base
react to form their conjugate base and conjugate acid, respectively.
acid1 + base2 base1 + acid2

27. 18-27
Table 18.4 Conjugate Pairs in Some Acid-Base Reactions
base1 acid2
+
acid1 base2
+
conjugate pair
conjugate pair
reaction 4 H2PO4
- OH-
+
reaction 5 H2SO4 N2H5
+
+
reaction 6 HPO4
2- SO3
2-
+
reaction 1 HF H2O
+ F- H3O+
+
reaction 3 NH4
+ CO3
2-
+
reaction 2 HCOOH CN-
+ HCOO- HCN
+
NH3 HCO3
-
+
HPO4
2- H2O
+
HSO4
- N2H6
2+
+
PO4
3- HSO3
-
+
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

28. 18-28
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) + CO3
2-(aq) HPO4
2-(aq) + HCO3
-(aq)
(b) H2O(l) + SO3
2-(aq) OH-(aq) + HSO3
-(aq)
SOLUTION:
PLAN: Identify proton donors (acids) and proton acceptors (bases).
(a) H2PO4
-(aq) + CO3
2-(aq) HPO4
2-(aq) + HCO3
-(aq)
proton
donor
proton
acceptor
proton
acceptor
proton
donor
conjugate pair1
conjugate pair2
(b) H2O(l) + SO3
2-(aq) OH-(aq) + HSO3
-(aq)
conjugate pair2
conjugate pair1
proton
donor
proton
acceptor
proton
acceptor
proton
donor

29. 18-29
Relative Acid-Base Strength and Reaction Direction
General Rule: An acid-base reaction proceeds to the greater extent
in the direction in which a stronger acid and stronger base form a
weaker acid and a weaker base.
H2S + NH3 HS- + NH4
+ Kc > 1
A competition for the proton between the two bases!
HF + H2O F- + H3O+ Kc < 1
a b b a
a b b a

30. 18-30
Figure 18.10
Strengths of
conjugate acid-
base pairs
An acid-base reaction proceeds to
the right if the acid reacts with a
base that is lower on the list because
this combination produces a weaker
conjugate base and a weaker
conjugate acid.
A weaker acid has a stronger
conjugate base.

31. 18-31
SAMPLE PROBLEM 18.5 Predicting the net direction of an acid-base
reaction
PROBLEM: Predict the net direction and whether Ka is greater or less than 1
for each of the following reactions (assume equal initial
concentrations of all species):
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
(a) H2PO4
-(aq) + NH3(aq) HPO4
2-(aq) + NH4
+(aq)
SOLUTION:
PLAN: Identify the conjugate acid-base pairs and then consult Figure 18.10
to determine the relative strength of each. The stronger the species,
the more preponderant will be its conjugate.
(a) H2PO4
-(aq) + NH3(aq) HPO4
2-(aq) + NH4
+(aq)
stronger acid weaker acid
stronger base weaker base
Net direction is to the right; Kc > 1.
(b) H2O(l) + HS-(aq) OH-(aq) + H2S(aq)
stronger base
weaker base stronger acid
weaker acid
Net direction is to the left; Kc < 1.

32. 18-32
Weak Acid Equilibrium Problems
Follow the general procedures described in Chapter 17
Two Key Assumptions
[H3O+] from the auto-ionization of water is negligible.
[HA]eq = [HA]init - [HA]dissoc ≈ [HA]init
(because a weak acid has a small Ka)
[HA]dissoc is very small

33. 18-33
SAMPLE PROBLEM 18.6
Find the Ka of a weak acid from the pH of its
solution
PROBLEM: Phenylacetic acid (C6H5CH2COOH, denoted as HPAc) builds up
in the blood of people afflicted with phenylketonuria, an inherited
genetic disorder that, if left untreated, causes mental retardation
and death. A study of the acid shows that the pH of a 0.12 M
solution of HPAc is 2.60. What is the Ka of phenylacetic acid?
PLAN: Write the dissociation equation. Use pH and solution concentration to
find Ka.
Ka = [H3O+][PAc-]
[HPAc]
Assumptions: At a pH of 2.60, [H3O+]HPAc >> [H3O+]water
[PAc-]eq ≈ [H3O+]eq, and since HPAc is weak, [HPAc]eq ≈
[HPAc]initial = [HPAc]initial - [HPAc]dissociation
SOLUTION: HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)

34. 18-34
SAMPLE PROBLEM 18.6 (continued)
concentration (M) HPAc(aq) + H2O(l) H3O+(aq) + PAc-(aq)
initial 0.12 - 1 x 10-7 0
change -
-x +x +x
equilibrium -
0.12 - x x
x +(<1 x 10-7)
[H3O+] = 10-pH = 2.5 x 10-3 M, which is >> 10-7 ([H3O+] from water)
x ≈ 2.5 x 10-3 M ≈ [H3O+] ≈ [PAc-] [HPAc]eq = 0.12 - x ≈ 0.12 M
Ka =
(2.5 x 10-3) (2.5 x 10-3)
0.12
= 5.2 x 10-5
Testing the assumptions:
= 4 x 10-3 %
x 100
[HPAc]dissn: 2.5 x 10-3 M
0.12 M
[H3O+]water:
1 x 10-7 M
2.5 x 10-3 M
x100
= 2.1 %
(the 5% rule is not violated in either case)

35. 18-35
SAMPLE PROBLEM 18.7 Determining concentrations from Ka and
initial [HA]
PROBLEM: Propanoic acid (CH3CH2COOH, simplified as HPr) is an organic
acid whose salts are used to retard mold growth in foods. What is
the [H3O+] of a 0.10 M aqueous solution of HPr (Ka = 1.3 x 10-5)?
SOLUTION:
PLAN: Write the dissociation equation and Ka expression; make
assumptions about concentration that are valid; substitute.
x = [HPr]diss = [H3O+]from HPr = [Pr-]
Assumptions: For: HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)
Ka = [H3O+][Pr-]
[HPr]
HPr(aq) + H2O(l) H3O+(aq) + Pr-(aq)
concentration (M)
initial 0.10 - 0 0
change -
-x +x +x
equilibrium -
0.10 - x x
x
Since Ka is small, we assume that x << 0.10

36. 18-36
SAMPLE PROBLEM 18.7 (continued)
(x)2
0.10
1.3 x 10-5 =
[H3O+][Pr-]
[HPr]
=
x  (0.10)(1.3x105
) = 1.1 x 10-3 M = [H3O+]
Checking assumptions:
[HPr]diss: 1.1 x 10-3 M / 0.10 M x 100 = 1.1%
Ka =

37. 18-37
percent HA dissociation =
[HA]dissociated
[HA]initial
x 100
As the initial concentration of a weak acid decreases, the
percent dissociation of the acid increases!
In the prior problem: for 0.10 M HPr, 1.1% dissociation
for 0.010 M HPr, 3.6% dissociation
Rationale: larger solution volume accommodates more ions
(analogous to pressure effects on gas equilibria when ∆ngas is non-zero)

38. 18-38
Polyprotic
acids
Acids that contain more
than one ionizable proton
H3PO4(aq) + H2O(l) H2PO4
-(aq) + H3O+(aq)
H2PO4
-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
HPO4
2-(aq) + H2O(l) PO4
3-(aq) + H3O+(aq)
Ka1 =
[H3O+] [H2PO4
-]
[H3PO4]
Ka2 =
[H3O+] [HPO4
2-]
[H2PO4
-]
Ka3 =
[H3O+] [PO4
3-]
[HPO4
2-]
Ka1 > Ka2 > Ka3
= 7.2 x 10-3
= 6.3 x 10-8
= 4.2 x 10-13
phosphoric acid, H3PO4

39. 18-39
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40. 18-40
Since successive acid dissociation constants typically differ
by several orders of magnitude, pH calculations for aqueous
solutions of the free acids can be simplified
by neglecting H3O+ generated by subsequent dissociations.
Why are the Ka values successively smaller as more H+ ions
dissociate from a polyprotic acid like phosphoric acid?

41. 18-41
SAMPLE PROBLEM 18.8 Calculating equilibrium concentrations for a
polyprotic acid
PROBLEM: Ascorbic acid (H2C6H6O6; abbreviated H2Asc), known as
vitamin C, is a diprotic acid (Ka1 = 1.0 x 10-5 and Ka2 = 5 x 10-12)
found in citrus fruit. Calculate [H2Asc], [HAsc-], [Asc2-], and the
pH of a 0.050 M aqueous solution of H2Asc.
SOLUTION:
PLAN: Write out expressions for both dissociations and make assumptions.
Ka1 >> Ka2 so the first dissociation produces virtually all of the H3O+.
Ka1 is small, thus [H2Asc]initial ≈ [H2Asc]eq
After finding concentrations of the various species for the first dissociation,
use them as initial concentrations for the second dissociation.
H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)
Ka1 =
[HAsc-][H3O +]
[H2Asc]
= 1.0 x 10-5
HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq) Ka2 =
[Asc2-
][H3O +]
[HAsc-]
= 5 x 10-12

42. 18-42
SAMPLE PROBLEM 18.8 (continued)
H2Asc(aq) + H2O(l) HAsc-(aq) + H3O+(aq)
concentration (M)
initial 0.050 - 0 0
change - x - + x + x
equilibrium 0.050 - x - x x
Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0 x 10-5 = (x)(x)/0.050 M
pH = -log(7.1 x 10-4) = 3.15
HAsc-(aq) + H2O(l) Asc2-(aq) + H3O+(aq)
7.1 x 10-4 - 0 0
change - x - + x + x
equilibrium 7.1 x 10-4 - x - x x
initial
 (0.050)(1.0x105
)
x x = 7.1 x 10-4 M = [HAsc-]
concentration (M)
x = [(7.1 x 10-4)(5 x 10-12)]0.5 = 6 x 10-8 M = [H3O+]

43. 18-43
We can ignore the hydronium ion generated by second ionization.
Also: 7.1 x 10-4/0.050 x 100 = 1.4%
This value is < 5%, thus the assumption made in the analysis of the first
dissociation reaction is justified.
[Asc-2] = (Ka2 x HAsc-)/[H3O+] = 5 x 10-12 M

44. 18-44
Weak Bases: Their Relationship to Weak Acids
B(aq) + H2O(l) BH+(aq) + OH-(aq)
[BH+][OH-]
[B][H2O]
Base-dissociation constant, Kb
[BH+][OH-]
[B]
pKb decreases with increasing Kb (i.e., increasing base strength)
Kc =
Kb =

45. 18-45
Main Classes of Weak Bases: nitrogen-containing molecules (ammonia
and amines) and anions of weak acids.
Ammonia:
NH3(aq) + H2O(l) NH4
+(aq) + OH-(aq) Kb = 1.76 x 10-5 (25 oC)
Amines:
RNH2, R2NH and R3N: all have a lone pair of electrons that
can bind a proton donated by an acid

46. 18-46
+
Figure 18.11
Abstraction of a proton from water by methylamine
+
lone pair
binds H+
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
CH3NH2
methylamine
H2O
CH3NH3
+
methylammonium ion
OH-

47. 18-47
Base strength
decreases
down
the table
(smaller
Kb values)

48. 18-48
SAMPLE PROBLEM 18.9 Determining pH from Kb and initial [B]
PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in detergent
manufacture, has a Kb = 5.9 x 10-4. What is the pH of a 1.5 M
aqueous solution of (CH3)2NH?
SOLUTION:
PLAN: Perform this calculation as done for acids. Keep in mind that you are
working with Kb and a base.
(CH3)2NH(aq) + H2O(l) (CH3)2NH2
+(aq) + OH-(aq)
Assumptions:
[(CH3)2NH2
+] = [OH-] = x [(CH3)2NH]eq≈ [(CH3)2NH]initial
Kb >> Kw so [OH-]water is negligible
initial 1.50 0 0
-
change - x - + x + x
equilibrium 1.50 - x - x x
(CH3)2NH(aq) + H2O(l) (CH3)2NH2
+(aq) + OH-(aq)
concentration

49. 18-49
SAMPLE PROBLEM 18.9 (continued)
Kb = 5.9 x 10-4 =
[(CH3)2NH2
+][OH-]
[(CH3)2NH]
5.9 x 10-4 =
(x) (x)
1.5 M
x = 3.0 x 10-2 M = [OH-]
Check assumption: [3.0 x 10-2 M / 1.5 M] x 100 = 2% (error is
< 5%; thus, assumption is justified)
[H3O+] = Kw/[OH-] = 1.0 x 10-14/3.0 x 10-2 = 3.3 x 10-13 M
pH = -log (3.3 x 10-13) = 12.48

50. 18-50
Anions of Weak Acids as Weak
Bases
A-(aq) + H2O(l) HA(aq) + OH-(aq)
F-(aq) + H2O(l) HF(aq) + OH-(aq)
Kb = ([HF][OH-]) / [F-]
Rationale for the basicity of A-(aq): e.g., 1 M NaF
Acidity is determined by [OH-] generated from the F- reaction and
[H3O+] generated from the auto-ionization of water; since
[OH-] >> [H3O+], the solution is basic.
Ka x Kb = Kw
For a conjugate acid-base pair:

51. 18-51
SAMPLE PROBLEM 18.10 Determining the pH of a solution of A-
PROBLEM: Sodium acetate (CH3COONa, abbreviated NaAc) has
applications in photographic development and textile dyeing.
What is the pH of a 0.25 M aqueous solution of NaAc? Ka of
acetic acid (HAc) is 1.8 x 10-5.
SOLUTION:
PLAN: Sodium salts are soluble in water so [Ac-] = 0.25 M.
Use Ka to find Kb.
initial 0.25 - 0 0
change -x +x +x
-
equilibrium -
0.25 - x x x
Ac-(aq) + H2O(l) HAc(aq) + OH-(aq)
concentration
Kb =
[HAc][OH-]
[Ac-]
=
Kw
Ka
= 5.6 x 10-10
Kb =
1.0 x 10-14
1.8 x 10-5

52. 18-52
SAMPLE PROBLEM 18.10 (continued)
Kb =
[HAc][OH-]
[Ac-]
[Ac-] = 0.25 M - x ≈ 0.25 M (since Kb is small)
5.6 x 10-10 ≈ x2/0.25 M
x ≈ 1.2 x 10-5 M = [OH-]
Check assumption: [1.2 x 10-5 M / 0.25 M] x 100 = 4.8 x 10-3 %
[H3O+] = Kw/[OH-] = 1.0 x 10-14/1.2 x 10-5 = 8.3 x 10-10 M
pH = -log (8.3 x 10-10 M) = 9.08

53. 18-53
Molecular Properties and Acid Strength
Non-metal hydrides: two factors determine acid strength, namely,
the electronegativity of the central non-metal atom E and the strength
of the E-H bond
Non-metal hydride acid strength increases across a period
(E electronegativity effect)
Non-metal hydride acid strength increases down a group
(E-H bond strength effect)

54. 18-54
Figure 18.12
The effect of atomic and molecular properties on
non-metal hydride acidity
6A(16)
H2O
H2S
H2Se
H2Te
7A(17)
HF
HCl
HBr
HI
electronegativity
increases, acidity
increases
bond
strength
decreases,
acidity
increases
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55. 18-55
H O I H O Br H O Cl
> >
H O Cl
O
O
O
<<
Figure 18.13
The relative strengths of oxoacids
     
H O Cl
 
 
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Electronegativity increases, acidity increases
number of O atoms increases,
acidity increases

56. 18-56
Acidity of Hydrated Metal Ions
Aqueous solutions of certain metal ions are acidic because the hydrated
metal ion transfers an H+ ion to water.
Generalized Reactions
M(NO3)n(s) + xH2O(l) M(H2O)x
n+(aq) + nNO3
-(aq)
M(H2O)x
n+(aq) + H2O(l) M(H2O)x-1OH(n-1)+(aq) + H3O+(aq)

57. 18-57
Al(H2O)5OH2+
Al(H2O)6
3+
Figure 18.14
The acidic behavior of the hydrated Al3+ ion
H2O H3O+
electron density
drawn toward Al3+
solvent H2O acts
as base
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58. 18-58
high charge
and small size
enhance acidity

59. 18-59
Acid-Base Properties of Salt Solutions
A. Salts That Yield Neutral Solutions: the anion of a strong acid and
the cation of a strong base (the ions do not react with water)
The anion of a strong acid is a much weaker base than water (HNO3).
The cation of a strong base only becomes hydrated (NaOH).
HNO3(l) + H2O(l) NO3
-(aq) + H3O+(aq)
NaOH(s) Na+(aq) + OH-(aq)

60. 18-60
B. Salts That Yield Acidic Solutions: (1) the anion of a strong acid
and the cation of a weak base (the cation acts as a weak acid); (2) small,
highly charged metal ions; (3) cations of strong bases and anions of
polyprotic acids with another ionizable proton.
NH4Cl(s) NH4
+ (aq) + Cl-(aq)
NH4
+(aq) + H2O(l) NH3(aq) + H3O+ (aq)
Case 1
Fe(NO3)3(s) + 6H2O(l) Fe(H2O)6
3+(aq) + 3NO3
-(aq)
Fe(H2O)6
3+(aq) + H2O(l) Fe(H2O)5OH2+(aq) + H3O+(aq)
Case 2
NaH2PO4(s) Na+(aq) + H2PO4
- (aq)
H2PO4
-(aq) + H2O(l) HPO4
2-(aq) + H3O+(aq)
Case 3

61. 18-61
C. Salts That Yield Basic Solutions: the anion of a weak acid and the
cation of a strong base (the anion acts as a weak base)
CH3COONa(s) Na+(aq) + CH3COO-(aq)
CH3COO-(aq) + H2O(l) CH3COOH(aq) + OH-(aq)

62. 18-62

63. 18-63
SAMPLE PROBLEM 18.11 Predicting the relative acidity of salt solutions
PROBLEM: Predict whether aqueous solutions of the following compounds
are acidic, basic, or neutral (write an equation for the reaction of
the appropriate ion with water to explain pH effect).
(a) potassium perchlorate, KClO4 (b) sodium benzoate, C6H5COONa
(c) chromium trichloride, CrCl3 (d) sodium hydrogen sulfate, NaHSO4
SOLUTION:
PLAN: Consider the acid-base nature of the anions and cations. Strong
acid-strong base combinations produce a neutral solution; strong
acid-weak base, acidic; weak acid-strong base, basic.
(a) The ions are K+ and ClO4
-, which come from a strong base
(KOH) and a strong acid (HClO4). The salt solution will be neutral.
(b) Na+ comes from the strong base NaOH while C6H5COO- is the anion of a
weak organic acid. The salt solution will be basic.
(c) Cr3+ is a small cation with a large + charge, so its hydrated form will react
with water to produce H3O+. Cl- comes from the strong acid HCl. The
salt solution will be acidic.
(d) Na+ comes from a strong base. HSO4
- can react with water to form H3O+.
The salt solution will be acidic.

64. 18-64
Salts of Weakly Acidic Cations and Weakly Basic Anions
Both components react with water!
Acidity is determined by the relative acid strength
and base strength of the separated ions.
example: NH4HS
NH4
+(aq) + H2O(l) NH3(aq) + H3O+(aq)
HS-(aq) + H2O(l) H2S(aq) + OH-(aq)
Ka(NH4
+) = 5.7 x 10-10 Kb(HS-) = 1 x 10-7
Since Kb > Ka, the solution is basic.

65. 18-65
SAMPLE PROBLEM 18.12 Predicting the relative acidity of salt solutions
from Ka and Kb of the ions
PROBLEM: Determine whether an aqueous solution of zinc formate,
Zn(HCOO)2, is acidic, basic, or neutral.
SOLUTION:
PLAN: Both Zn2+ and HCOO- come from weak conjugates. In order to find
the relatively acidity, write the dissociation reactions and use the
information in Tables 18.2 and 18.7.
Ka Zn(H2O)6
2+ = 1 x 10-9
Ka HCOOH = 1.8 x 10-4 ; Kb = Kw/Ka = 1.0 x 10-14/1.8 x 10-4 = 5.6 x 10-11
Ka for Zn(H2O)6
2+ > Kb HCOO-; the solution is acidic.
Zn(H2O)6
2+(aq) + H2O(l) Zn(H2O)5OH+(aq) + H3O+(aq)
HCOO-(aq) + H2O(l) HCOOH(aq) + OH-(aq)

66. 18-66
The Leveling Effect
In water, the strongest acid possible is H3O+ and the
strongest base possible is OH-.
Any acid stronger than H3O+ donates its proton to H2O, and
any base stronger than OH- accepts a proton from H2O; thus,
water exerts a leveling effect (levels the strengths of all strong
acids and bases).
To rank strong acids: must dissolve in a solvent that is a weaker
base than water (i.e., one that accepts their protons less readily).
HCl(g) + CH3COOH(l) Cl-(acet) + CH3COOH2
+(acet)
HBr(g) + CH3COOH(l) Br-(acet) + CH3COOH2
+(acet)
HI(g) + CH3COOH(l) I-(acet) + CH3COOH2
+(acet)
KHI > KHBr > KHCl

67. 18-67
Three Definitions of Acids and Bases
The Arrhenius Definition
The Brønsted-Lowry Definition
The Lewis Definition

68. 18-68
F
B
F F
H
N
H H
+
F
B
F F
H
N
H H
acid base adduct
An acid is an electron-pair acceptor. A base is an electron-pair donor.
M2+
H2O(l)
M(H2O)4
2+(aq)
adduct
The Lewis Acid-Base Definition
The adduct
contains a
new covalent
bond.

69. 18-69
Lewis Acids with Electron-Deficient Atoms
Lewis Acids with Polar Multiple Bonds
Metal Cations as Lewis Acids
ROR’ + AlCl3 R-O-R’
AlCl3
base acid
adduct
SO2 + H2O H2SO3
M2+ + 4H2O M(H2O)4
2+
base
acid adduct
adduct
base
acid
Metal ions act
as Lewis acids
when dissolved
in water.
Lewis acids
contain (or can
generate) a
vacant orbital.

70. 18-70
Figure
18.15
Mg2+ ion as a Lewis acid in the
chlorophyll molecule
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The ion accepts electron pairs from
four nitrogen atoms comprising part
of the chlorophyll molecule

71. 18-71
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
+
SOLUTION:
PLAN: Look for electron pair acceptors (acids) and donors (bases).
(a) H+ + OH- H2O
acceptor
donor
(b) Cl- + BCl3 BCl4
-
donor
acceptor
(c) K+ + 6H2O K(H2O)6
+
acceptor
donor