Acid-Base Equilibria
18-1
Table 18.1
18-2
Some Common Acids and Bases and their
Household Uses.
Arrhenius Acid-Base Definition
This is the earliest acid-base definition, which classifies
these substances in terms of their behavior in water.
An acid is a substance with H in its formula that
dissociates to yield H3O+.
A base is a substance with OH in its formula that
dissociates to yield OH-.
When an acid reacts with a base, they undergo
neutralization:
H+(aq) + OH-(aq) → H2O(l) DH°rxn = -55.9 kJ
18-3
Strong and Weak Acids
A strong acid dissociates completely into ions in water:
HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)
A dilute solution of a strong acid contains no HA molecules.
A weak acid dissociates slightly to form ions in water:
HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
In a dilute solution of a weak acid, most HA molecules are
undissociated.
Kc =
18-4
[H3O+][A-]
[HA][H2O]
has a very small value.
Figure 18.1A
The extent of dissociation for strong acids.
Strong acid: HA(g or l) + H2O(l) → H3O+(aq) + A-(aq)
There are no HA molecules in solution.
18-5
Figure 18.1B
The extent of dissociation for weak acids.
Weak acid: HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
Most HA molecules are undissociated.
18-6
Figure 18.2
Reaction of zinc with a strong acid (left) and a
weak acid (right).
Zinc reacts rapidly with
the strong acid, since
[H3O+] is much higher.
1 M HCl(aq)
18-7
1 M CH3COOH(aq)
The Acid Dissociation Constant, Ka
HA(aq) + H2O(l)
[H3O+][A–]
Kc =
[HA][H2O]
H3O+(aq) + A–(aq)
[H3O+][A–]
Kc[H2O] = Ka =
[HA]
The value of Ka is an indication of acid strength.
Stronger acid
higher [H3O+]
Weaker acid
lower % dissociation of HA
18-8
larger Ka
smaller Ka
Table 18.2
18-9
Ka Values for some Monoprotic Acids at 25°C
Classifying the Relative Strengths of Acids
• Strong acids include
– the hydrohalic acids (HCl, HBr, and HI) and
– oxoacids in which the number of O atoms exceeds the number
of ionizable protons by two or more (eg., HNO3, H2SO4, HClO4.)
• Weak acids include
– the hydrohalic acid HF,
– acids in which H is not bonded to O or to a halogen (eg., HCN),
– oxoacids in which the number of O atoms equals or exceeds
the number of ionizable protons by one (eg., HClO, HNO2), and
– carboxylic acids, which have the general formula RCOOH (eg.,
CH3COOH and C6H5COOH.)
18-10
Classifying the Relative Strengths of Bases
• Strong bases include
– water-soluble compounds containing O2- or OH- ions.
– The cations are usually those of the most active metals:
• M2O or MOH, where M = Group 1A(1) metal (Li, Na, K, Rb, Cs)
• MO or M(OH)2 where M = group 2A(2) metal (Ca, Sr, Ba).
• Weak bases include
– ammonia (NH3),
– amines, which have the general formula
– The common structural feature is an N atom with a lone
electron pair.
18-11
Sample Problem 18.1
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) KOH
(b) (CH3)2CHCOOH
(c) H2SeO4
(d) (CH3)2CHNH2
PLAN: We examine the formula and classify each acid or base, using
the text descriptions. Particular points to note for acids are the
numbers of O atoms relative to ionizable H atoms and the
presence of the –COOH group. For bases, note the nature of
the cation or the presence of an N atom that has a lone pair.
SOLUTION:
(a) Strong base: KOH is one of the group 1A(1) hydroxides.
18-12
Sample Problem 18.1
(b) Weak acid: (CH3)2CHCOOH is a carboxylic acid, as indicated by
the –COOH group. The –COOH proton is the only ionizable
proton in this compound.
(c) Strong acid: He2SO4 is an oxoacid in which the number of O
atoms exceeds the number of ionizable protons by two.
(d) Weak base: (CH3)2CHNH2 has a lone pair of electrons on the N
and is an amine.
18-13
Autoionization of Water
Water dissociates very slightly into ions in an equilibrium
process known as autoionization or self-ionization.
2H2O (l)
18-14
H3O+ (aq) + OH- (aq)
The Ion-Product Constant for Water (Kw)
2H2O (l)
H3O+ (aq) + OH- (aq)
[H3O+][OH-]
Kc =
[H2O]2
Kc[H2O]2 = Kw = [H3O+][OH-] = 1.0x10-14 (at 25°C)
In pure water,
[H3O+] = [OH-] =
= 1.0x10-7 (at 25°C)
Both ions are present in all aqueous systems.
18-15
A change in [H3O+] causes an inverse change in [OH-],
and vice versa.
Higher [H3O+]
lower [OH-]
Higher [OH-]
lower [H3O+]
We can define the terms “acidic” and “basic” in terms
of the relative concentrations of H3O+ and OH– ions:
In an acidic solution,
In a neutral solution,
In a basic solution,
18-16
[H3O+] > [OH–]
[H3O+] = [OH–]
[H3O+] < [OH–]
Figure 18.3
18-17
The relationship between [H3O+] and [OH-] and the
relative acidity of solutions.
Sample Problem 18.2
Calculating [H3O+] or [OH–] in an
Aqueous Solution
PROBLEM: A research chemist adds a measured amount of HCl
gas to pure water at 25°C and obtains a solution with
[H3O+] = 3.0x10–4 M. Calculate [OH–]. Is the solution
neutral, acidic, or basic?
PLAN: We use the known value of Kw at 25°C (1.0x10–14) and the
given [H3O+] to solve for [OH–]. We can then compare [H3O+]
with [OH–] to determine whether the solution is acidic, basic,
or neutral.
SOLUTION:
Kw = 1.0 x 10–14 = [H3O+] [OH–] so
Kw
–
[OH ] =
[H3O+]
1.0 x 10–14
=
3.0 x 10–4
= 3.3x10–11 M
[H3O+] is > [OH–] and the solution is acidic.
18-18
The pH Scale
pH = -log[H3O+]
The pH of a solution indicates its relative acidity:
In an acidic solution,
In a neutral solution,
In a basic solution,
pH < 7.00
pH = 7.00
pH > 7.00
The higher the pH, the lower the [H3O+] and the less
acidic the solution.
18-19
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Figure 18.4
The pH values of some
familiar aqueous
solutions.
pH = –log [H3O+]
18-20
Table 18.3
The Relationship between Ka and pKa
Acid Name (Formula)
Ka at 25°C
pKa
Hydrogen sulfate ion (HSO4-)
1.0x10–2
1.99
Nitrous acid (HNO2)
7.1x10–4
3.15
Acetic acid (CH3COOH)
1.8x10–5
4.75
Hypobromous acid (HBrO)
2.3x10–9
8.64
Phenol (C6H5OH)
1.0x10–10
10.00
pKa = –logKa
A low pKa corresponds to a high Ka.
18-21
pH, pOH, and pKw
Kw = [H3O+][OH–] = 1.0x10–14 at 25°C
pH = –log[H3O+]
pOH = –log[OH–]
pKw = pH + pOH = 14.00 at 25°C
pH + pOH = pKw for any aqueous solution at any temperature.
Since Kw is a constant, the values of pH, pOH, [H3O+],
and [OH–] are interrelated:
• If [H3O+] increases, [OH–] decreases (and vice versa).
• If pH increases, pOH decreases (and vice versa).
18-22
Figure 18.5
The relations among [H3O+], pH, [OH-], and pOH.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
18-23
Sample Problem 18.3
Calculating [H3O+], pH, [OH-], and pOH for
Strong Acids and Bases
PROBLEM: Calculate [H3O+], pH, [OH–], and pOH for each solution at
25°C:
(a) 0.30 M HNO3, used for etching copper metal; and
(b) 0.0042 M Ca(OH)2, used in leather tanning to remove hair from hides.
PLAN:
HNO3 is a strong acid; so it dissociates completely, and [H3O+] = [HNO3]init.
Likewise, Ca(OH)2 is a strong base that dissociates completely, and
[OH–] = 2[Ca(OH)2]init.
We use the given concentrations and the value of Kw at 25°C to find [OH–]
and [H3O+]. We can then calculate pH and pOH.
18-24
Sample Problem 18.3
SOLUTION:
(a) Calculating the values for 0.30 M HNO3:
[H3O+] = 0.30 M
pH = –log[H3O+] = -log(0.30) = 0.52
–14
K
1.0
x
10
w
-14 M
[OH–] =
=
=
3.3
x
10
[H3O+]
0.30
pOH = –log[OH–] = –log(3.3 x 10–14) = 13.48
(b) Calculating the values for 0.0042 M Ca(OH)2:
Ca(OH)2 is a strong electrolyte: Ca(OH)2(aq) → Ca2+(aq) + 2 OH–(aq)
[OH–] = 2 (0.0042 M) = 0.0084 M
pOH = –log[OH–] = -log 0.0084 = 2.08
Kw
1.0 x 10-14
+
[H3O ] =
=
= 1.2 x 10–12 M
[OH–]
0.0084
pH = –log[H3O+] = –log(1.2 x 10–12) = 11.92
18-25
Figure 18.6
Methods for measuring the pH of an aqueous solution.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
pH paper
pH meter
18-26
Brønsted-Lowry Acid-Base Definition
An acid is a proton donor, any species that donates an
H+ ion.
• An acid must contain H in its formula.
A base is a proton acceptor, any species that accepts
an H+ ion.
• A base must contain a lone pair of electrons to bond
to H+.
An acid-base reaction is a proton-transfer process.
18-27
Figure 18.7
Dissolving of an acid or base in water as a BrønstedLowry acid-base reaction.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Lone pair
binds H+
(acid, H+ donor)
(base, H+ acceptor)
Lone pair
binds H+
(base, H+ acceptor)
18-28
(acid, H+ donor)
Conjugate Acid-Base Pairs
In the forward reaction:
NH3 accepts an H+ to form NH4+.
H2S + NH3
HS– + NH4+
H2S donates an H+ to form HS-.
In the reverse reaction:
NH4+ donates an H+ to form NH3.
H2S + NH3
HS– + NH4+
HS– accepts an H+ to form H2S.
18-29
Conjugate Acid-Base Pairs
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.
A Brønsted-Lowry acid-base reaction occurs when an
acid and a base react to form their conjugate base
and conjugate acid, respectively.
acid1 + base2
18-30
base1 + acid2
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–
18-31
+
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)
(b) H2O(l) + SO32–(aq)
HPO42-–(aq) + HCO3–(aq)
OH–(aq) + HSO3–(aq)
PLAN: To find the conjugate pairs, we find the species that donated
an H+ (acid) and the species that accepted it (base). The acid
donates an H+ to become its conjugate base, and the base
accepts an H+ to become its conjugate acid.
SOLUTION:
(a) H2PO4–(aq) + CO32–(aq)
acid1
base2
HPO42–(aq) + HCO3–(aq)
base1
acid2
The conjugate acid-base pairs are H2PO4–/HPO42– and CO32–/HCO3–.
18-32
Sample Problem 18.4
(b) H2O(l) + SO32–(aq)
acid1
base2
OH–(aq) + HSO3–(aq)
base1
acid2
The conjugate acid-base pairs are H2O/OH– and SO32–/HSO3–.
18-33
Net Direction of Reaction
The net direction of an acid-base reaction depends on
the relative strength of the acids and bases involved.
A reaction will favor the formation of the weaker acid
and base.
H2S
stronger acid
+
NH3
HS–
+
NH4+
weaker base
stronger base
weaker acid
This reaction favors the formation of the products.
18-34
Figure 18.8
Strengths of conjugate acid-base pairs.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
The stronger the acid is, the
weaker its conjugate base.
When an acid reacts with a
base that is farther down the
list, the reaction proceeds to
the right (Kc > 1).
18-35
Sample Problem 18.5
Predicting the Net Direction of an AcidBase Reaction
PROBLEM: Predict the net direction and whether Kc 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)
OH–(aq) + H2S(aq)
PLAN: We identify the conjugate acid-base pairs and consult figure
18.8 to see which acid and base are stronger. The reaction
favors the formation of the weaker acid and base.
SOLUTION:
(a) H2PO4–(aq) + NH3(aq)
stronger acid stronger base
HPO42–(aq) + NH4+(aq)
weaker base weaker acid
The net direction for this reaction is to the right, so Kc > 1.
18-36
Sample Problem 18.5
(b) H2O(l) + HS–(aq)
OH–(aq) + H2S(aq)
weaker acid weaker base stronger base stronger acid
The net direction for this reaction is to the left, so Kc < 1.
18-37
Sample Problem 18.6
Using Molecular Scenes to Predict the Net
Direction of an Acid-Base Reaction
PROBLEM: Given that 0.10 M of HX (blue and green) has a pH of 2.88,
and 0.10 M HY (blue and orange) has a pH of 3.52, which
scene best represents the final mixture after equimolar
solutions of HX and Y- are mixed?
PLAN:
18-38
A stronger acid and base yield a weaker acid and base, so we
have to determine the relative acid strengths of HX and HY to
choose the correct molecular scene. The concentrations of the
acid solutions are equal, so we can recognize the stronger acid
by comparing the pH values of the two solutions.
Sample Problem 18.6
SOLUTION:
The HX solution has a lower pH than the HY solution, so HX is the
stronger acid and Y– is the stronger base. Therefore the reaction of HX
and Y– has a Kc > 1, which means the equilibrium mixture will contain
more HY than HX.
Scene 1 has equal numbers of HX and HY, which could occur if the
acids were of equal strength. Scene 2 shows fewer HY than HX,
which would occur if HY were the stronger acid.
Scene 3 is consistent with the relative acid strengths, because it
contains more HY than HX.
18-39
Solving Problems Involving
Weak-Acid Equilibria
Problem-solving approach
1. Write a balanced equation.
2. Write an expression for Ka.
3. Define x as the change in concentration that
occurs during the reaction.
4. Construct a reaction table in terms of x.
5. Make assumptions that simplify the calculation.
6. Substitute values into the Ka expression and
solve for x.
7. Check that the assumptions are justified.
18-40
Solving Problems Involving
Weak-Acid Equilibria
The notation system
• Molar concentrations are indicated by [ ].
• A bracketed formula with no subscript
indicates an equilibrium concentration.
The assumptions
• [H3O+] from the autoionization of H2O is
negligible.
• A weak acid has a small Ka and its
dissociation is negligible. [HA] ≈ [HA]init.
18-41
Sample Problem 18.7
Finding Ka of a Weak Acid from the
Solution pH
PROBLEM: Phenylacetic acid (C6H5CH2COOH, simplified here as
HPAc) builds up in the blood of persons with
phenylketonuria, an inherited disorder that, if untreated,
causes mental retardation and death. A study of the acid
shows that the pH of 0.12 M HPAc is 2.62. What is the Ka
of phenylacetic acid?
PLAN:
We start with the balanced dissociation equation and write
the expression for Ka. We assume that [H3O+] from H2O is
negligible and use the given pH to find [H3O+], which equals
[PAc–] and [HPAc]dissoc. We assume that [HPAc] ≈ [HPAc]init
because HPAc is a weak acid.
SOLUTION: HPAc(aq) + H2O(l)
Ka =
[H3O+][PAc–]
[HPAc]
18-42
H3O+(aq) + PAc–(aq)
Sample Problem 18.7
Concentration (M) HPAc(aq) +
Initial
Change
Equilibrium
0.12
–x
0.12 – x
H2O(l)
-
H3O+(aq) + PAc–(aq)
0
+x
x
0
+x
x
[H3O+] = 10–pH = 2.4 x 10–3 M which is >> 10–7 (the [H3O+] from water)
x ≈ 2.4 x 10–3 M ≈ [H3O+] ≈ [PAc–]
[HPAc] = 0.12 - x ≈ 0.12 M
(2.4 x 10–3) (2.4 x 10–3)
So Ka =
= 4.8 x 10–5
0.12
Checking the assumptions by finding the percent error in concentration:
1 x 10–7 M
+
[H3O ]from H2O =
x 100 = 4 x 10–3 % (< 5%; assumption is justified).
2.4 x 10–3 M
–3
[HPAc]dissoc = 2.4 x 10 M x 100 = 2.0 % (< 5%; assumption is justified).
0.12 M
18-43
Sample Problem 18.8
Determining Concentration from Ka and
Initial [HA]
PROBLEM: Propanoic acid (CH3CH2COOH, which we simplify as
HPr) is a carboxylic acid whose salts are used to retard
mold growth in foods. What is the [H3O+] of 0.10 M HPr
(Ka = 1.3x10−5)?
PLAN: We write a balanced equation and the expression for Ka. We
know [HPr]init but not [HPr] (i.e., the concentration at
equilibrium). We define x as [HPr]dissoc and set up a reaction
table. We assume that, since HPr has a small Ka value, it
dissociates very little and therefore [HPr] ≈ [HPr]init.
SOLUTION:
HPr(aq) + H2O(l)
H3O+(aq) + Pr−(aq)
Ka = [H3O+][Pr−]
[HPr]
18-44
Sample Problem 18.8
Concentration (M)
Initial
Change
Equilibrium
HPr(aq) + H2O(l)
H3O+(aq) + Pr–(aq)
0.10
-x
-
0
+x
0
+x
0.10 - x
-
x
x
Since Ka is small, we will assume that x << 0.10 and [HPr] ≈ 0.10 M.
Ka =
1.3x10–5
=
[H3O+][Pr–]
[HPr]
≈
x2
0.10
x ≈ √(0.10)(1.3 x 10–5) = 1.1 x 10–3 M = [H3O+]
1.1 x 10–3 M
Check: [HPr]diss =
x 100 = 1.1% (< 5%; assumption is justified.)
0.10 M
18-45
Concentration and Extent of Dissociation
Percent HA dissociated =
[HA]dissoc
x 100
[HA]init
As the initial acid concentration decreases, the percent
dissociation of the acid increases.
HA(aq) + H2O(l)
H3O+(aq) + A–(aq)
A decrease in [HA]init means a
decrease in [HA]dissoc = [H3O+] = [A–],
causing a shift toward the products.
The fraction of ions present increases, even though the
actual [HA]dissoc decreases.
18-46
Sample Problem 18.9
Finding the Percent Dissociation of a Weak
Acid
PROBLEM: In 2011, researchers showed that hypochlorous acid (HClO)
generated by white blood cells kills bacteria. Calculate the percent dissociation of
(a) 0.40 M HClO; (b) 0.035 M HClO (Ka = 2.9 x 10–8).
PLAN: Percent dissociation of HClO =
[HClO]dissoc
x 100
[HClO]init
The value of [HClO]init is given. To calculate [HClO]dissoc, we write the balanced
equation for the dissociation of HClO, set up a reaction table with x = [HClO]dissoc =
[ClO–] = [H3O+], and use the expression for Ka. We assume that HClO dissociates to
a very small extent since Ka is small (2.9 x 10–8).
SOLUTION: (a) Writing the balanced equation and the expression for Ka:
H3O+(aq) + ClO– (aq)
HClO(aq) + H2O(l)
Ka =
18-47
[H3O+][ClO–]
[HClO]
= 2.9 x 10–8
Sample Problem 18.9
Concentration (M) HClO(aq) + H2O(l)
Initial
Change
Equilibrium
H3O+(aq) + ClO–(aq)
0.40
–x
-
0
+x
0
+x
0.40 – x
-
x
x
Since Ka is small, we will assume that x << 0.40 and [HClO] ≈ 0.40 M.
Ka =
[H3O+][ClO–]
[HClO]
= 2.9 x
10–8
=
x2
0.40
x ≈ √ (0.40)(2.9 x 10–8) = 1.1 x 10–4 M = [HClO]dissoc
% Dissociation =
18-48
1.1 x 10–4 M
0.40 M
x 100 = 0.028% (< 5%; assumption is justified.)
Sample Problem 18.9
(b) Performing the same calculations using [HClO]init = 0.035 M:
Ka =
[H3O+][ClO–]
[HClO]
= 2.9 x 10–8 =
x2
0.035
x ≈ √(0.035)(2.9 x 10–8)= 3.2 x 10–5 M = [HClO]dissoc
% Dissociation =
18-49
3.2 x 10–5 M
0.035 M
x 100 = 0.091% (< 5%; assumption is justified.)
Polyprotic Acids
A polyprotic acid is an acid with more than one ionizable
proton. In solution, each dissociation step has a different
value for Ka:
H3PO4(aq) + H2O(l)
H2PO4
HPO4
–(aq)
2–(aq)
+ H2O(l)
+ H2O(l)
H2PO4
–(aq)
HPO4
2–(aq)
PO4
3–(aq)
+ H3
O+(aq)
+ H3
+ H3
O+(aq)
O+(aq)
Ka1 >> Ka2 >> Ka3
Ka1 =
Ka2 =
Ka3 =
[H3O+][H2PO4–]
[H3PO4]
[H3O+][HPO42–]
[H2PO4–]
[H3O+][PO43–]
[HPO4
]
2–
= 7.2x10–3
= 6.3x10–8
= 4.2x10–13
We usually neglect [H3O+] produced after the first dissociation.
18-50
Table 18.5
18-51
Successive Ka values for Some Polyprotic Acids at
25°C
Sample Problem 18.10
Calculating Equilibrium Concentrations
for a Polyprotic Acid
PROBLEM: Ascorbic acid (H2C6H6O6; H2Asc for this problem),
known as vitamin C, is a diprotic acid (Ka1 = 1.0x10–5 and
Ka2 = 5x10–12) found in citrus fruit. Calculate [H2Asc],
[HAsc–], [Asc2–], and the pH of 0.050 M H2Asc.
PLAN: We first write the dissociation equations and the associated
Ka expressions. Since Ka1 >> Ka2, we can assume that the
first dissociation produces almost all the H3O+. Also, since Ka1
is small, the amount of H2ASc that dissociates can be
neglected. We set up a reaction table for the first dissociation,
with x equal to [H2Asc]dissoc, and solve for [H3O+] and [HAsc-].
SOLUTION:
H2Asc(aq) + H2O(l)
HAsc–(aq) + H2O(l)
18-52
Hasc–(aq)
+ H3
O+(aq)
Asc2–(aq) + H3O+(aq)
Ka1 =
Ka2 =
[HAsc–][H3O+]
= 1.0x10-5
[H2Asc]
[Asc2–][H3O+]
[HAsc–]
= 5x10-12
Sample Problem 18.10
Concentration (M) H2Asc(aq) + H2O(l)
Initial
Change
Equilibrium
HAsc–(aq) + H3O+(aq)
0.050
–x
-
0
+x
0
+x
0.050 – x
-
x
x
x2
[HAsc–][H3O+]
–5
≈
Ka1 =
= 1.0x10 =
0.050 – x
[H2Asc]
x2
0.050
x = [H3O+] = [Asc–] = √(0.050)(1.0 x 10–5) = 7.1x10–4 M
pH = –log[H3O+] = –log(7.1x10–4) = 3.15
18-53
Sample Problem 18.10
Checking assumptions:
1. [H3O+]
from
[H3O+]
from
HAsc-
<< [H3O+]
from H2Asc
≈ √ [HAsc–](Ka2)
HAsc-
This is even less than [H3O+]
: For any second dissociation,
= √ (7.1 x 10–4)(5 x 10–12)
from H2O
, so the assumption is justifed.
2. [H2Asc]dissoc << [H2Asc]init:
7.1 x 10–4 M x 100 = 1.4% (< 5%; assumption is justified).
0.050 M
18-54
= 6 x 10–8 M
Weak Bases
A Brønsted-Lowry base is a species that accepts an H+.
For a weak base that dissolves in water:
B(aq) + H2O(l)
BH+(aq) + OH–(aq)
The base-dissociation or base-ionization constant is
given by:
[BH+][OH–]
Kb =
[B]
Note that no base actually dissociates in solution, but ions are
produced when the base reacts with H2O.
18-55
Figure 18.9 Abstraction of a proton from water by the base
methylamine.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Lone pair of N
pair binds H+
18-56
Table 18.6
18-57
Kb Values for Some Molecular (Amine) Bases at
25°C
Sample Problem 18.11
Determining pH from Kb and Initial [B]
PROBLEM: Dimethylamine, (CH3)2NH, a key intermediate in
detergent manufacture, has a Kb of 5.9x10–4. What is the
pH of 1.5 M (CH3)2NH?
PLAN: We start with the balanced equation for the reaction of the
amine with H2O, remembering that it is a weak base. We then
write the expression for Kb, set up a reaction table and solve
for [OH–]. From [OH–] we can calculate [H3O+] and pH.
We make similar assumptions to those made for weak acids.
Since Kb >> Kw, the [OH–] from H2O is negligible. Since Kb is
small, we can assume that the amount of amine reacting is
also small, so [(CH3)2NH] ≈ [(CH3)2NH]init.
SOLUTION: (CH3)2NH(aq) + H2O(l)
[(CH3)2NH2+][OH–]
Kb =
[(CH3)2NH]
18-58
(CH3)2NH2+(aq) + OH–(aq)
Sample Problem 18.11
Concentration (M) (CH3)2NH(aq) + H2O(l)
Initial
Change
Equilibrium
1.5
-x
1.5 - x
-
(CH3)2NH2+(aq) + OH–(aq)
0
+x
x
Since Kb is small, x << 1.5 and 1.5 – x ≈ 1.5
2
[(CH3)2NH2+][OH–]
x
Kb =
= 5.9 x 10–4 ≈
[(CH3)2NH]
1.5
x = [OH–] = 3.0 x 10–2 M
Check assumption:
3.0 x 10–2 M x 100 = 2.0% (< 5%; assumption is justified).
1.5 M
18-59
0
+x
x
Sample Problem 18.11
–14
K
1.0x10
w
–13 M
[H3O+] =
=
=
3.3
x
10
[OH–]
3.0x10–2
pH = –log (3.3 x 10–13) = 12.48
18-60
Anions of Weak Acids as Weak Bases
The anions of weak acids often function as weak bases.
–]
[HA][OH
A-(aq) + H2O(l)
HA(aq) + OH-(aq) Kb =
[A–]
A solution of HA is acidic, while a solution of A- is basic.
HF(aq) + H2O(l)
H3O+(aq) + F–(aq)
HF is a weak acid (much weaker than H3O+), so this
equilibrium lies to the left.
[HF] >> [F–], but [H3O+]from HF >> [OH–] from H O;
the solution is therefore acidic.
2
18-61
If NaF is dissolved in H2O, it dissolves completely, and Fcan act as a weak base:
F–(aq) + H2O(l)
HF(aq) + OH–(aq)
Equilibrium favors the formation of the weaker acid and
base. Since HF (although a weak acid) is a stronger
acid than H2O, this equilibrium also lies to the left.
[F–] >> [HF], but [OH–] from F >> [H3O+ ] from H O;
the solution is therefore basic.
-
18-62
2
Ka and Kb for a Conjugate Acid-Base Pair
HA + H2O
A- + H2O
2H2O
H3O+ + AHA + OH–
H3O+ + OH–
Kc for the overall equation = K1 x K2, so
[H3O+][A–] x [HA][OH–] = [H O+][OH–]
3
–
[HA]
[A ]
Ka
x
Kb
=
Kw
This relationship is true for any conjugate acid-base pair.
18-63
Sample Problem 18.12
Determining the pH of a Solution of A-
PROBLEM: Sodium acetate (CH3COONa, or NaAc for this problem)
has applications in photographic development and textile
dyeing. What is the pH of 0.25 M NaAc at 25ºC? Ka of
acetic acid (HAc) is 1.8 x 10–5.
PLAN: Sodium salts are soluble in water and acetate is the anion of
HAc so it acts as a weak base. We write the base dissociation
equation and the expression for Kb, and solve for [OH-]. We
recall that any soluble ionic salt dissociates completely in
solution, so [Ac–]init = 0.25 M.
SOLUTION:
Ac–(aq) + H2O(l)
HAc(aq) + OH–(aq)
[HAc][OH–]
Kb =
[Ac–]
18-64
Sample Problem 18.12
Concentration (M) Ac–(aq) + H2O(l)
Initial
0.25
-x
Change
0.25 - x
Equilibrium
Kb of
Kb =
Ac-
Kw
=
Ka
5.6x10-10
HAc(aq) + OH–(aq)
0
0
+x
+x
x
x
1.0x10–14
–10 M
=
=
5.6x10
1.8x10–5
x2
[HAc][OH–]
=
≈
–
[Ac ]
0.25
so x = [OH–] = 1.2x10–5 M
Checking the assumption:
1.2x10–5 x 100 = 4.8x10-3% (< 5%; assumption is justified)
0.25
18-65
Sample Problem 18.12
–14
K
1.0x10
w
–10 M
[H3O+] =
=
=
8.3x10
[OH–]
1.2x10–5
pH = – log (8.3x10–10) = 9.08
18-66
Acid Strength of Nonmetal Hydrides
For nonmetal hydrides (E-H), acid strength depends on:
• the electronegativity of the central nonmetal (E), and
• the strength of the E-H bond.
Across a period, acid strength increases.
Electronegativity increases across a period, so the acidity of E-H
increases.
Down a group, acid strength increases.
The length of the E-H bond increases down a group and its bond
strength therefore decreases.
18-67
Figure 18.10
The effect of atomic and molecular properties on
nonmetal hydride acidity.
Copyright  The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Bond strength
decreases, so acidity
increases
6A(16)
18-68
7A(17)
H 2O
HF
H 2S
HCl
H2Se
HBr
H2Te
HI
Electronegativity
increases, so
acidity increases
Acid Strength of Oxoacids
All oxoacids have the acidic H bonded to an O atom.
Acid strength of oxoacids depends on:
• the electronegativity of the central nonmetal (E), and
• the number of O atoms around E.
For oxoacids with the same number of O atoms, acid
strength increases as the electronegativity of E increases.
For oxoacids with different numbers of O atoms, acid
strength increases with the number of O atoms.
18-69
Figure 18.11
18-70
The relative strengths of oxoacids.
Hydrated Metal Ions
Some hydrated metal ions are able to transfer an H+ to
H2O. These metal ions will form acidic solutions.
Consider a metal ion in solution, Mn+:
Mn+(aq) + H2O(l) → M(H2O)xn+(aq)
If Mn+ is small and highly charged, it will withdraw enough
e- density from the O-H bonds of the bound H2O molecules
to release H+:
M(H2O)xn+(aq) + H2O(l)
18-71
M(H2O)x-1OH(n-1)+(aq) + H3O+(aq)
Figure 18.12
18-72
The acidic behavior of the hydrated Al3+ ion.
Salts that Yield Neutral Solutions
A salt that consists of the cation of a strong base and
the anion of a strong acid yields a neutral solution.
NaNO3
Na+ is the cation of
NaOH, a strong base.
NO3– is the anion of
HNO3, a strong acid.
This solution will be neutral, because neither Na+ nor
NO3– will react with H2O to any great extent.
18-73
Salts that Yield Acidic Solutions
A salt that consists of the cation of a weak base and
the anion of a strong acid yields an acidic solution.
NH4Cl
NH4 + is the cation of
NH3, a weak base.
Cl– is the anion of
HCl, a strong acid.
This solution will be acidic, because NH4+ will react with
H2O to produce H3O+:
NH4+(aq) + H2O(l)
18-74
NH3(aq) + H3O+(aq)
Salts that Yield Acidic Solutions
A salt that consists of a small, highly charged metal
cation and the anion of a strong acid yields an acidic
solution.
Fe(NO3)3
Fe3+ is a small, highly
charged metal cation.
NO3 - is the anion of
HNO3, a strong acid.
This solution will be acidic, because the hydrated Fe3+ ion
will react with H2O to produce H3O+:
Fe(H2O)63+(aq) + H2O(l)
18-75
Fe(H2O)5OH2+(aq) + H3O+(aq)
Salts that Yield Basic Solutions
A salt that consists of the anion of a weak acid and the
cation of a strong base yields a basic solution.
CH3COONa
CH3COO– is the anion of
CH3COOH, a weak acid.
Na+ is the cation of
NaOH, a strong base.
This solution will be basic, because CH3COO– will react
with H2O to produce OH–:
CH3COO–(aq) + H2O(l)
18-76
CH3COOH(aq) + OH–(aq)
Sample Problem 18.13
Predicting Relative Acidity of Salt Solutions
from Reactions of the Ions with Water
PROBLEM: Predict whether aqueous solutions of the following are
acidic, basic, or neutral, and write an equation for the
reaction of any ion with water:
(a) Potassium perchlorate, KClO4 (b) Sodium benzoate, C6H5COONa
(c) Chromium(III) nitrate, Cr(NO3)3
PLAN: We identify the cation and anion from the formula for each salt.
Depending on an ion’s ability to react with water, the solution
will be neutral (strong-acid anion with strong-base cation or
small, highly charged metal cation), acidic (weak-base cation
with strong-acid anion), or basic (weak-acid anion and strongbase cation).
SOLUTION:
(a) K+ is the cation of a strong base (KOH) while ClO4– is the anion of
a strong acid (HClO4). This solution will be neutral.
18-77
Sample Problem 18.13
(b) Na+ is the cation of a strong base (NaOH) while the benzoate
anion (C6H5COO–) is the anion of a weak acid (benzoic acid). The
benzoate ion will react with H2O to produce OH– ions:
C6H5COO–(aq) + H2O(l)
C6H5COOH(aq) + OH– (aq)
This solution will be basic.
(c) NO3– is the anion of a strong acid (HNO3) and will not react with
H2O to any great extent. Cr3+ is a small metal cation with a fairly
high charge density. It will become hydrated and the hydrated ion
will react with H2O to form H3O+ ions:
Cr(H2O)63+(aq) + H2O(l)
This solution will be acidic.
18-78
Cr(H2O)5OH2+(aq) + H3O+(aq)
Salts of Weakly Acidic Cations and
Weakly Basic Anions
If a salt that consists of the cation of a weak base and
the anion of a weak acid, the pH of the solution will
depend on the relative acid strength or base strength of
the ions.
NH4CN
NH4+ is the cation of a
weak base, NH3.
18-79
CN– is the anion of a
weak acid, HCN.
NH4+(aq) + H2O(l)
NH3(aq) + H3O+(aq)
CN–(aq) + H2O(l)
HCN(aq) + OH– (aq)
The reaction that proceeds farther to the right determines the
pH of the solution, so we need to compare the Ka of NH4+
with the Kb of CN–.
Ka of NH4
Kb of
+
CN–
–14
K
1.0x10
w
=
=
Kb of NH3
1.76x10–5
= 5.7x10–10
Kw
1.0x10–14
=
=
Ka of HCN
6.2x10–10
= 1.6x10–5
Since Kb of CN– > Ka of NH4+, CN– is a stronger base than
NH4+ is an acid. A solution of NH4CN will be basic.
18-80
Table 18.7
18-81
The Acid-Base Behavior of Salts in Water
Sample Problem 18.14
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, at 25°C is acidic, basic, or neutral.
PLAN: Zn2+ is a small, highly charged metal cation, while HCOO– is
the anion of a weak acid. Both will react with H2O, so to
determine the acidity of the solution we must compare the Ka
of the hydrated Zn2+ ion with the Kb of the HCOO– ion.
SOLUTION:
Zn(H2O)62+(aq) + H2O(l)
Zn(H2O)5OH+(aq) + H3O+(aq)
HCOO–(aq) + H2O(l)
HCOOH(aq) + OH–(aq)
Ka of Zn(H2O)62+ = 1x10–9 (from Appendix C.)
Kw
1.0x10–14
–
Kb of HCOO =
=
= 5.6x10–11
–4
Ka of HCOOH
1.8x10
Ka for Zn(H2O)62+ >> Kb HCOO–, therefore the solution is acidic.
18-82
The Leveling Effect
All strong acids and bases are equally strong in water.
All strong acids dissociate completely to form H3O+, while
all strong bases dissociate completely to form OH–.
In water, the strongest acid possible is H3O+ and the
strongest base possible is OH–.
H2O exerts a leveling effect on any strong acid or base.
18-83
The Lewis Acid-Base Definition
A Lewis base is any species that donates an electron
pair to form a bond.
A Lewis acid is any species that accepts an electron pair
to form a bond.
The Lewis definition views an acid-base reaction as the
donation and acceptance of an electron pair to form
a covalent bond.
18-84
Lewis Acids and Bases
A Lewis base must have a lone pair of electrons to
donate.
Any substance that is a Brønsted-Lowry base is also a Lewis base.
A Lewis acid must have a vacant orbital (or be able to
rearrange its bonds to form one) to accept a lone pair
and form a new bond.
Many substances that are not Brønsted-Lowry acids are Lewis
acids.
The Lewis definition expands the classes of acids.
18-85
Electron-Deficient Molecules as Lewis Acids
B and Al often form electron-deficient molecules, and
these atoms have an unoccupied p orbital that can accept
a pair of electrons:
BF3 accepts an electron pair from ammonia to form a covalent bond.
18-86
Lewis Acids with Polar Multiple Bonds
Molecules that contain a polar multiple bond often function
as Lewis acids:
The O atom of an H2O molecule donates a lone pair to the S of SO2,
forming a new S‒O σ bond and breaking one of the S‒O p bonds.
18-87
Metal Cations as Lewis Acids
A metal cation acts as a Lewis acid when it dissolves in
water to form a hydrated ion:
The O atom of an H2O molecule donates a lone pair to an available
orbital on the metal cation.
18-88
Figure 18.13
18-89
The Mg2+ ion as a Lewis acid in chlorophyll.
Sample Problem 18.15
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: We examine the formulas to see which species accepts the
electron pair (Lewis acid) and which donates it (Lewis base)
in forming the adduct.
SOLUTION:
(a) The H+ ion accepts the electron pair from OH–. H+ is the Lewis
acid and OH– is the Lewis base.
(b) BCl3 accepts an electron pair from Cl–. Cl– is the Lewis base and
BCl3 is the Lewis acid.
(c) An O atom from each H2O molecule donates an electron pair to
K+. H2O is therefore the Lewis base, and K+ is the Lewis acid.
18-90