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Chapter 18
Acid-Base Equilibria
18-1
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
18-2
Etching with acids
The inside surfaces of these light
bulbs are etched with HF.
18-3
Figure 18.1
Acids are used to wash away
oxides of silicon and metals during
the production of computer chips.
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Photo - JPEG decompressor
are needed to see this picture.
18-4
For reaction between a strong acid and strong base:
HCl(aq) + NaOH(aq)
H+(aq) + OH-(aq)
NaCl(aq) + H2O(l)
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.
18-5
The nature of the hydrated proton
Figure 18.2
18-6
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)
18-7
H2O(l)
∆Horxn = -55.9 kJ
Strong acids dissociate completely into ions in water.
H3O+ (aq) + A- (aq)
HA(g or l) + H2O(l)
Kc >> 1
Weak acids dissociate very slightly into ions in water.
H3O+ (aq) + A- (aq)
HA(aq) + H2O(l)
Kc << 1
Defining the acid dissociation constant
[H3O+][A-]
Kc =
[H2O][HA]
Kc[H2O] = Ka =
18-8
[H3O+][A-]
[HA]
stronger acid, higher [H3O+],
larger Ka
weaker acid, lower [H3O+],
smaller Ka
The extent of dissociation of strong acids
strong acid: HA(g or l) + H2O(l)
H3O+(aq) + A-(aq)
Figure 18.3
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
18-9
The extent of dissociation of weak acids
weak acid: HA(aq) + H2O(l)
Figure 18.3
18-10
H3O+(aq) + A-(aq)
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Reaction of zinc with a strong and a weak acid
Zn(s) + 2H3O+(aq)
Zn+2(aq)
+ 2H2O(l) + H2(g)
1 M HCl(aq)
Figure 18.4
18-11
1 M CH3COOH(aq)
Acid strength
decreases down
the table (smaller
Ka values)
18-12
SAMPLE PROBLEM 18.1
PROBLEM:
Classify each of the following compounds as a strong acid,
weak acid, strong base or weak base.
(a) H2SeO4
PLAN:
Classifying acid and base strength from the
chemical formula
(b) (CH3)2CHCOOH
(c) KOH
(d) (CH3)2CHNH2
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.
SOLUTION: (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.
18-13
The auto-ionization of water and the pH scale
+
H2O(l)
H2O(l)
+
18-14
H3O +(aq
)
OH-(aq)
H2O(l) + H2O(l)
Kc =
H3O+(aq) + OH-(aq)
[H3O+][OH-]
[H2O]2
Defining the ion-product constant of water
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0 x 10-14 at 25 oC
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-]
18-15
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions
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[H3O+]
divide into Kw
[OH-]
[H3O+] > [OH-]
[H3O+] = [OH-]
[H3O+] < [OH-]
acidic
solution
Figure 18.5
18-16
neutral
solution
basic
solution
SAMPLE PROBLEM 18.2
PROBLEM:
PLAN:
Calculating [H3O+] and [OH-] in an aqueous
solution
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?
Use the Kw at 25 oC and the [H3O+] to find the corresponding [OH-].
SOLUTION:
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.
18-17
The pH values of
some familiar
aqueous solutions
pH = -log [H3O+]
Related Expressions
pOH = -log [OH-]
pK = -log K
Figure 18.6
18-18
A low pK corresponds to a high K.
18-19
Relationships
between
[H3O+],
pH, [OH-] and
pOH
Figure 18.7
18-20
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SAMPLE PROBLEM 18.3
PROBLEM:
PLAN:
Calculating [H3O+], pH, [OH-], and pOH
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.
HNO3 is a strong acid so [H3O+] = [HNO3]. Use Kw to find the [OH-]
and then convert to pH and pOH.
SOLUTION:
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
For 0.3 M HNO3, [H3O+] = 0.30 M and -log [H3O+] = 0.52 = pH
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/0.30 = 3.3 x 10-14 M; pOH = 13.48
For 0.0063 M HNO3, [H3O+] = 0.0063 M and -log [H3O+] = 2.20 = pH
[OH-] = Kw/ [H3O+] = 1.0 x 10-14/6.3 x 10-3 = 1.6 x 10-12 M; pOH = 11.80
18-21
Methods for measuring the pH of an aqueous solution
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
pH (indicator) paper
Figure 18.8
18-22
pH meter
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.
18-23
Proton transfer is the essential feature of a BrønstedLowry acid-base reaction
lone pair
binds H+
+
+
HCl
(acid, H+ donor)
H 2O
Cl-
H 3 O+
(base, H+ acceptor)
lone pair
binds H+
+
+
NH3
H 2O
(base, H+ acceptor)
18-24
NH4+
OH-
(acid, H+ donor)
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Figure 18.9
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.
An acid reactant produces a base product and the two constitute
an acid-base conjugate pair.
Every acid has a conjugate base, and every base has a conjugate acid.
18-25
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
18-26
base1 + acid2
Table 18.4
Conjugate Pairs in Some Acid-Base Reactions
conjugate pair
acid1
+
base2
base1
+
acid2
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-
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18-27
SAMPLE PROBLEM 18.4
PROBLEM:
Identifying conjugate acid-base pairs
The following reactions are important environmental processes.
Identify the conjugate acid-base pairs.
(a) H2PO4-(aq) + CO32-(aq)
(b) H2O(l) + SO32-(aq)
PLAN:
OH-(aq) + HSO3-(aq)
Identify proton donors (acids) and proton acceptors (bases).
conjugate pair1
SOLUTION:
(a) H2PO4-(aq) + CO32-(aq)
proton
proton
donor
acceptor
conjugate pair1
(b) H2O(l) + SO32-(aq)
proton
proton
donor acceptor
18-28
HPO42-(aq) + HCO3-(aq)
conjugate pair2
HPO42-(aq) + HCO3-(aq)
proton
proton
acceptor
donor
conjugate pair2
OH-(aq) + HSO3-(aq)
proton
proton
acceptor
donor
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.
a
b
H2S + NH3
b
a
HS- + NH4+
Kc > 1
A competition for the proton between the two bases!
a
b
HF + H2O
18-29
b
a
F - + H3 O +
Kc < 1
Strengths of
conjugate acidbase pairs
A weaker acid has a stronger
conjugate base.
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.
Figure 18.10
18-30
SAMPLE PROBLEM 18.5
PROBLEM:
Predicting the net direction of an acid-base
reaction
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):
(a) H2PO4-(aq) + NH3(aq)
HPO42-(aq) + NH4+(aq)
(b) H2O(l) + HS-(aq)
PLAN:
OH-(aq) + H2S(aq)
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.
SOLUTION:
(a) H2PO4-(aq) + NH3(aq)
stronger acid
HPO42-(aq) + NH4+(aq)
stronger base
weaker base
weaker acid
Net direction is to the right; Kc > 1.
(b) H2O(l) + HS-(aq)
weaker acid
18-31
weaker base
OH-(aq) + H2S(aq)
stronger base
stronger acid
Net direction is to the left; Kc < 1.
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
18-32
SAMPLE PROBLEM 18.6
PROBLEM:
PLAN:
Find the Ka of a weak acid from the pH of its
solution
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?
Write the dissociation equation. Use pH and solution concentration to
find Ka.
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)
Ka =
H3O+(aq) + PAc-(aq)
[H3O+][PAc-]
[HPAc]
18-33
SAMPLE PROBLEM 18.6
concentration (M)
(continued)
HPAc(aq) + H2O(l)
H3O+(aq) + PAc-(aq)
initial
0.12
-
1 x 10-7
0
change
-x
-
+x
+x
0.12 - x
-
x +(<1 x 10-7)
x
equilibrium
[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-]
Ka =
(2.5 x 10-3) (2.5 x 10-3)
0.12
[HPAc]eq = 0.12 - x ≈ 0.12 M
= 5.2 x 10-5
[H3O+]water:
Testing the assumptions:
18-34
[HPAc]dissn:
1 x 10-7 M
10-3 M
x100 = 4 x 10-3 %
2.5 x
2.5 x 10-3 M
x 100 = 2.1 %
0.12 M
(the 5% rule is not violated in either case)
SAMPLE PROBLEM 18.7
PROBLEM:
PLAN:
Determining concentrations from Ka and
initial [HA]
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)?
Write the dissociation equation and Ka expression; make
assumptions about concentration that are valid; substitute.
For:
Assumptions:
HPr(aq) + H2O(l)
H3O+(aq) + Pr-(aq)
x = [HPr]diss = [H3O+]from HPr = [Pr-]
Ka =
[H3O+][Pr-]
[HPr]
SOLUTION:
concentration (M)
initial
change
equilibrium
HPr(aq) + H2O(l)
H3O+(aq) + Pr-(aq)
0.10
-
0
0
-x
-
+x
+x
0.10 - x
-
x
x
Since Ka is small, we assume that x << 0.10
18-35
SAMPLE PROBLEM 18.7
Ka = 1.3 x 10-5 =
(continued)
[H3O+][Pr-]
=
[HPr]
x  (0.10)(1.3x105 )
(x)2
0.10
= 1.1 x 10-3 M = [H3O+]
Checking assumptions:
[HPr]diss: 1.1 x 10-3 M / 0.10 M x 100 = 1.1%
18-36
As the initial concentration of a weak acid decreases, the
percent dissociation of the acid increases!
[HA]dissociated
percent HA dissociation =
In the prior problem:
x 100
[HA]initial
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)
18-37
Polyprotic
acids
Acids that contain more
than one ionizable proton
phosphoric acid, H3PO4
H3PO4(aq) + H2O(l)
H2PO4-(aq) + H3O+(aq)
Ka1 =
[H3O+] [H2PO4-]
[H3PO4]
= 7.2 x 10-3
H2PO4-(aq) + H2O(l)
HPO4
2-(aq)
+] [HPO 2-]
[H
O
+
3
4
+ H3O (aq) Ka2 =
[H2PO4-]
= 6.3 x 10-8
HPO42-(aq) + H2O(l)
PO43-(aq) + H3O+(aq)
Ka3 =
[H3O+] [PO43-]
[HPO42-]
= 4.2 x 10-13
Ka1 > Ka2 > Ka3
18-38
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18-39
Why are the Ka values successively smaller as more H+ ions
dissociate from a polyprotic acid like phosphoric acid?
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.
18-40
SAMPLE PROBLEM 18.8
PROBLEM:
PLAN:
Calculating equilibrium concentrations for a
polyprotic acid
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.
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.
[HAsc-][H3O +]
SOLUTION:
H2Asc(aq) + H2O(l)
HAsc-(aq) + H2O(l)
18-41
HAsc-(aq) + H3O+(aq)
Asc2-(aq)
+ H3O+(aq)
Ka1 =
[H2Asc]
= 1.0 x 10-5
[Asc2Ka2 = ][H3O +]
[HAsc-]
= 5 x 10-12
SAMPLE PROBLEM 18.8
concentration (M)
initial
change
equilibrium
(continued)
H2Asc(aq) + H2O(l)
HAsc-(aq) + H3O+(aq)
0.050
-
0
0
-x
-
+x
+x
0.050 - x
-
x
x
Ka1 = [HAsc-][H3O+]/[H2Asc] = 1.0 x 10-5 = (x)(x)/0.050 M
5
x  (0.050)(1.0x10 )
concentration (M)
initial
x = 7.1 x 10-4 M = [HAsc-]
HAsc-(aq) + H2O(l)
7.1 x 10-4
change
equilibrium
-x
7.1 x 10-4 - x
pH = -log(7.1 x 10-4) = 3.15
Asc2-(aq) + H3O+(aq)
-
0
-
+x
+x
-
x
x
+
x = [(7.1 x 10-4)(5 x 10-12)]0.5 = 6 x 10-8 M = [H3O ]
18-42
0
We can ignore the hydronium ion generated by second ionization.
[Asc-2] = (Ka2 x HAsc-)/[H3O+] = 5 x 10-12 M
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.
18-43
Weak Bases: Their Relationship to Weak Acids
B(aq) + H2O(l)
Kc =
BH+(aq) + OH-(aq)
[BH+][OH-]
[B][H2O]
Base-dissociation constant, Kb
Kb =
[BH+][OH-]
[B]
pKb decreases with increasing Kb (i.e., increasing base strength)
18-44
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
18-45
Abstraction of a proton from water by methylamine
lone pair
binds H+
+
CH3NH2
methylamine
H2O
+
Figure 18.11
18-46
CH3NH3+
methylammonium ion
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OH-
Base strength
decreases
down
the table
(smaller
Kb values)
18-47
SAMPLE PROBLEM 18.9
PROBLEM:
PLAN:
Determining pH from Kb and initial [B]
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?
Perform this calculation as done for acids. Keep in mind that you are
working with Kb and a base.
(CH3)2NH2+(aq) + OH-(aq)
(CH3)2NH(aq) + H2O(l)
Assumptions:
Kb >> Kw so [OH-]water is negligible
[(CH3)2NH2+] = [OH-] = x
SOLUTION:
concentration
initial
change
equilibrium
18-48
[(CH3)2NH]eq≈ [(CH3)2NH]initial
(CH3)2NH(aq) + H2O(l)
(CH3)2NH2+(aq) + OH-(aq)
1.50
-
0
0
-x
-
+x
+x
1.50 - x
-
x
x
SAMPLE PROBLEM 18.9 (continued)
Kb = 5.9 x 10-4 =
5.9 x
10-4
(x) (x)
=
[(CH3)2NH2+][OH-]
[(CH3)2NH]
x = 3.0 x 10-2 M = [OH-]
1.5 M
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
18-49
Anions of Weak Acids as Weak
Bases
A-(aq) + H2O(l)
F-(aq) + H2O(l)
HA(aq) + OH-(aq)
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.
For a conjugate acid-base pair:
Ka
18-50
x
Kb = Kw
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.
PLAN:
Sodium salts are soluble in water so [Ac-] = 0.25 M.
Use Ka to find Kb.
SOLUTION:
concentration
Ac-(aq) + H2O(l)
initial
change
equilibrium
Kb =
18-51
[HAc][OH-]
[Ac-]
=
HAc(aq) + OH-(aq)
0.25
-
0
0
-x
-
+x
+x
0.25 - x
-
x
x
Kw
Ka
Kb =
1.0 x 10-14
1.8 x 10-5
= 5.6 x 10-10
SAMPLE PROBLEM 18.10 (continued)
Kb =
[Ac-] = 0.25 M - x ≈ 0.25 M (since Kb is small)
[HAc][OH-]
5.6 x 10-10 ≈ x2/0.25 M
[Ac-]
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
18-52
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)
18-53
bond strength decreases,
acidity increases
The effect of atomic and molecular properties on
non-metal hydride acidity
6A(16)
7A(17)
H 2O
HF
H 2S
HCl
H2Se
HBr
H2Te
HI
electronegativity
increases, acidity
increases
Figure 18.12
18-54
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The relative strengths of oxoacids
Electronegativity increases, acidity increases
H
O


I
>
H
O


Br
>
H
O


O
H
O


Cl
<<
H

number of O atoms increases,
acidity increases
Figure 18.13
18-55
O

Cl
O
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O
Cl
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)xn+(aq) + H2O(l)
18-56
M(H2O)xn+(aq) + nNO3-(aq)
M(H2O)x-1OH(n-1)+(aq) + H3O+(aq)
The acidic behavior of the hydrated Al3+ ion
electron density
drawn toward Al3+
solvent H2O acts
as base
H 3 O+
H 2O
Al(H2O)63+
Figure 18.14
18-57
Al(H2O)5OH2+
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Photo - JPEG decompressor
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18-58
high charge
and small size
enhance acidity
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)
NaOH(s)
18-59
NO3-(aq) + H3O+(aq)
Na+(aq) + OH-(aq)
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.
NH4+ (aq) + Cl-(aq)
NH4Cl(s)
NH4+(aq) + H2O(l)
Fe(NO3)3(s) + 6H2O(l)
Fe(H2O)63+(aq) + H2O(l)
NaH2PO4(s)
Case 1
Fe(H2O)63+(aq) + 3NO3-(aq)
Case 2
Fe(H2O)5OH2+(aq) + H3O+(aq)
Na+(aq) + H2PO4- (aq)
H2PO4-(aq) + H2O(l)
18-60
NH3(aq) + H3O+ (aq)
HPO4
2-(aq)
+ H3O+(aq)
Case 3
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)
CH3COO-(aq) + H2O(l)
18-61
Na+(aq) + CH3COO-(aq)
CH3COOH(aq) + OH-(aq)
18-62
SAMPLE PROBLEM 18.11
PROBLEM:
Predicting the relative acidity of salt solutions
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
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.
SOLUTION: (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.
18-63
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.
18-64
SAMPLE PROBLEM 18.12
PROBLEM:
PLAN:
Predicting the relative acidity of salt solutions
from Ka and Kb of the ions
Determine whether an aqueous solution of zinc formate,
Zn(HCOO)2, is acidic, basic, or neutral.
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.
SOLUTION:
Zn(H2O)62+(aq) + H2O(l)
HCOO-(aq) + H2O(l)
Zn(H2O)5OH+(aq) + H3O+(aq)
HCOOH(aq) + OH-(aq)
Ka Zn(H2O)62+ = 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)62+ > Kb HCOO-; the solution is acidic.
18-65
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
18-66
Three Definitions of Acids and Bases
The Arrhenius Definition
The Brønsted-Lowry Definition
The Lewis Definition
18-67
The Lewis Acid-Base Definition
An acid is an electron-pair acceptor.
F
B
F
F
acid
A base is an electron-pair donor.
F
H
+
H
N
HH
base
B
F
F
N
The adduct
contains a
new covalent
bond.
HH
adduct
M(H2O)42+(aq)
M2 +
H2O(l)
18-68
adduct
Lewis Acids with Electron-Deficient Atoms
ROR’ + AlCl3
base
acid
R-O-R’
adduct
AlCl3
Lewis acids
contain (or can
generate) a
vacant orbital.
Lewis Acids with Polar Multiple Bonds
SO2 + H2O
acid
base
H2SO3
adduct
Metal Cations as Lewis Acids
M2+ + 4H2O
acid
18-69
base
M(H2O)42+
adduct
Metal ions act
as Lewis acids
when dissolved
in water.
Figure
18.15
Mg2+ ion as a Lewis acid in the
chlorophyll molecule
The ion accepts electron pairs from
four nitrogen atoms comprising part
of the chlorophyll molecule
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18-70
SAMPLE PROBLEM 18.13
PROBLEM:
Identifying Lewis acids and bases
Identify the Lewis acids and Lewis bases in the following reactions:
(a) H+ + OH-
H2O
(b) Cl- + BCl3
PLAN:
BCl4-
(c) K+ + 6H2O
K(H2O)6+
Look for electron pair acceptors (acids) and donors (bases).
SOLUTION:
acceptor
(a) H+ + OH-
H2O
donor
donor
(b) Cl- + BCl3
acceptor
BCl4-
acceptor
(c) K+ + 6H2O
donor
18-71
K(H2O)6+