Transcript Document

Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 16
Acids and Bases
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
16
Acids and Bases
16.1 Brønsted Acids and Bases
16.2 Molecular Structure and Acid Strength
Hydrohalic Acids
Oxoacids
Carboxylic Acids
16.3 The Acid-Base Properties of Water
16.4 The pH Scale
16.5 Strong Acids and Bases
Strong Acids
Strong Bases
16.6 Weak Acids and Acid Ionization Constants
The Ionization Constant, Ka
Calculating pH from Ka
Percent Ionization
Using pH to Determine Ka
16.7 Weak Bases and Base Ionization Constants
The Ionization Constant, Kb
Calculating pH from Kb
Using pH to Determine Kb
16
Acids and Bases
16.8 Conjugate Acid-Base Pairs
The Strength of a Conjugate Acid or Base
The Relationship Between Ka and Kb of a Conjugate Acid-Base Pair
16.9 Diprotic and Polyprotic Acids
16.10 Acid-Base Properties of Salt Solutions
Basic Salt Solutions
Acidic Salt Solutions
Neutral Salt Solutions
Salts in Which Both the Cation and the Anion Hydrolyze
16.11 Acid-Base Properties of Oxides and Hydroxides
Oxides of Metals and Nonmetals
Basic and Amphoteric Hydroxides
16.12 Lewis Acids and Bases
16.1
Brønsted Acids and Bases
When a Brønsted acid donates a proton, what remains of the acid is
known as a conjugate base.
Loses a proton
Gains a proton
HCl(aq)
acid
+
H2O(l)
base
⇌ H3O+(aq)
conjugate
acid
+
Cl–(aq)
conjugate
base
The two species HCl and Cl– are known as a conjugate acid-base
pair or simply a conjugate pair.
Brønsted Acids and Bases
When a Brønsted base accepts a proton, the newly formed
protonated species is known as a conjugate acid.
Gains a proton
Loses a proton
NH3(aq)
base
+
H2O(l)
acid
⇌ NH4+(aq)
conjugate
acid
+
OH–(aq)
conjugate
base
Worked Example 16.1
What is (a) the conjugate base on HNO3, (b) the conjugate acid of O2-, (c) the
conjugate base of HSO4-, and (d) the conjugate acid of HCO3-.
Strategy To find the conjugate base of a species, remove a proton from the
formula. To find the conjugate acid of a species, add a proton to the formula. The
word proton, in this context, refers to H+. Thus, the formula and the charge will
both be affected by the addition or removal of H+.
Solution (a) NO3(b) OH(c) SO42(d) H2CO3
Think About It A species does not need to be what we think of as an acid in
order for it to have a conjugate base. For example, we would not refer to the
hydroxide ion (OH-) as an acid – but it does have a conjugate base, the oxide ion
(O2-). Furthermore, a species that can either lose or gain a proton, such as
HCO3-, has both a conjugate base (CO32-) and a conjugate acid (H2CO3).
Worked Example 16.2
Label each of the species in the following equations as an acid, base, conjugate
base, or conjugate acid:
(a) HF(aq) + NH3(aq) ⇌ F-(aq) + NH4+(aq)
(b) CH3COO-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq)
Think
It In a the
Brønsted
acid-base
reaction,
there
is always
Strategy
In About
each equation,
reactant
that loses
a proton
is the
acid and the
anthat
acidgains
and athe
base,
andiswhether
a substance
behaves
an acid orofa one of
reactant
proton
the base.
Each product
is theasconjugate
base depends
on what
it differ
is combined
Water,
for example,
the reactants.
Two species
that
only bywith.
a proton
constitute
a conjugate pair.
behaves as a base when combined with- HCl but behaves as an acid
Solution
(a)
HF loseswith
a proton
a becomes F ; NH3 gains a proton and becomes
when
combined
NH
.
3
NH4+.
HF(aq) + NH3(aq) ⇌ F-(aq) + NH4+(aq)
acid
base
conjugate base conjugate acid
(b) CH3COO- gains a proton to become CH3COOH; H2O loses a proton to
become OH-.
CH3COO-(aq) + H2O(l) ⇌ CH3COOH(aq) + OH-(aq)
base
acid
conjugate acid conjugate base
16.2
Molecular Structure and Acid Strength
The strength of an acid is measured by its tendency to ionize.
HX
→
H+ + X–
Two factors influence ionization:
1) The strength of the H—X bond
2) The polarity of the H—X bond
δ+
δ–
H—X
Molecular Structure and Acid Strength
Hydrohalic acid strength:
HF << HCl < HBr < HI
Biggest factor is bond strength.
Only HF is a weak acid.
Molecular Structure and Acid Strength
Oxoacids:
An oxoacid contains hydrogen, oxygen, and a central nonmetal atom.
Molecular Structure and Acid Strength
To compare oxoacid strength, it is convenient to divide the oxoacids
into two groups:
1) Oxoacids having different central atoms that are from the same
group of the periodic table and that have the same oxidation
number.
HClO3 > HBrO3
Cl is more electronegative; the O—H bond is more polar.
Molecular Structure and Acid Strength
To compare oxoacid strength, it is convenient to divide the oxoacids
into two groups:
2) Oxoacids having the same central atom, but different numbers of
oxygen atoms.
HClO4 > HClO3 > HClO2 > HClO
Worked Example 16.3
Predict the relative strengths of the oxoacids in each of the following groups:
(a) HClO, HBrO, and HIO; (b) HNO3 and HNO2.
Strategy In each group, compare the electronegativies or oxidation numbers of
the central atoms to determine which O–H bonds are the most polar. The more
polar the O–H bond, the more readily it is broken and the stronger the acid.
Solution
(a)About
In a group
withof
different
central
we mustHNO
compare
Think
It Four
the strong
acidsatoms,
are oxoacids:
3,
electronegativities.
electronegativities of the central atoms in this group
HClO4, HClOThe
3, and H2SO4.
decrease as follows: Cl > Br > I.
Acid strength decreases as follows: HClO > HBrO > HIO
(b) These two acids have the same central atom but differ in the number of
oxygen atoms. In a group such as this, the greater the number of attached oxygen
atoms, the higher the oxidation number and the stronger the acid.
HNO3 is stronger than HNO2.
16.3
The Acid-Base Properties of Water
A species that can behave either as a Brønsted acid or a Brønsted
base is called amphoteric.
The acid-base properties of water produces H3O+ and OH– ions in
equilibrium with water in a reaction known as the autoionization of
water.
The equilibrium expression for the autoionization of water is given by:
Kw = [H3O+][OH–] = 1.0 x 10–14 (at 25°C)
Molecular Structure and Acid Strength
An important group of organic acids is the carboxylic acids:
The strength of the acid depends on the nature of the R group.
Acetic acid (Ka = 1.8 x 10–5)
Chloroacetic acid (Ka = 1.4 x 10–3)
The Acid-Base Properties of Water
Kw = [H3O+][OH–] = 1.0 x 10–14 (at 25°C)
Since the product of the concentrations of H3O+ and OH– is equal
to a constant, the relative amount of each obeys a fixed
relationship.
Depending on which ion concentration is in excess, the solution
will be considered acidic or basic.
When [H3O+] = [OH–], the solution is neutral
When [H3O+] > [OH–], the solution is acidic
When [H3O+] < [OH–], the solution is basic
Worked Example 16.4
The concentration of hydronium ions in stomach acid is 0.10 M. Calculate the
concentration of hydroxide ions in stomach acid at 25°C.
Strategy Use the value of Kw to determine [OH-] when [H3O+] = 0.10 M.
Solution Kw = [H3O+][OH-] = 1.0×10-14 at 25°C. Rearranging to solve for [OH-],
-14
1.0×10
[OH-] =
[H3O+]
[OH-]
1.0×10-14
=
0.10
[OH-] = 1.0×10-13 M
Think About It Remember that the equilibrium constants are temperature
dependent. The value of Kw = 1.0×10-14 only at 25°C.
16.4
The pH Scale
The acidity of an aqueous solution depends on the concentration of
hydronium ions, [H3O+].
The pH of a solution is defined as the negative base-10 logarithm of
the hydronium ion concentration (in mol/L)
pH = –log[H3O+]
[H3O+] = 10–pH
In pure water at 25°C, [H3O+] = log1.0 x 10–7
pH = –log(1.0 x 10–7) = 7.00
pH is a dimensionless quantity.
The pH Scale
The pH Scale
Worked Example 16.5
Determine the pH of a solution at 25°C in which the hydronium ion
concentration is (a) 3.5×10-4 M, (b) 1.7×10-7 M, and (c) 8.8×10-11 M.
Strategy Given [H3O+], use pH = –log[H3O+] to solve for pH.
Solution
(a) pH = –log(3.5×10-4) = 3.46
(b) pH = –log(1.7×10-7) = 6.77
(c) pH = –log(8.8×10-11) = 10.06
Think About It When a hydronium ion concentration falls between two
“benchmark” concentrations in Table 16.4, the pH falls between the two
corresponding pH values. In part (c), for example, the hydronium ion
concentration (8.8×10-11 M) is greater than 1.0×10-11 M but less than
1.0×10-10 M. Therefore, we expect the pH to be between 11.00 and 10.00.
Worked Example 16.6
Calculate the hydronium ion concentration in a solution at 25°C in which the pH
is (a) 4.76, (b) 11.95, and (c) 8.01.
Strategy Given pH, use [H3O+] = 10-pH to calculate [H3O+].
Think About It Think About It If you use the calculated
Solution
hydronium ion concentrations to recalculate pH, you will get numbers
+
-4.76 = 1.7×10-5 M
(a) [Hslightly
3O ] = 10
different
from those given in the problem. In part (a), for
example, −log(1.7×10-5) = 4.77. The small difference between this and
+ = 10-11.95 = 1.1×10-12 M
(b) [H4.76
3O ](the
pH given in the problem) is due to a rounding error.
Remember that a concentration derived from a pH with two digits to
+] = 10-8.01 = 9.8×10-9 M
(c) [Hthe
3O right
of the decimal point can have only two significant figures.
Note also that the benchmarks can be used equally well in this
circumstance. A pH between 4 and 5 corresponds to a hydronium ion
concentration between 1.7×10-4 M and 1.0×10-5 M.
The pH Scale
A pOH scale analogous to the pH scale can be defined as the
negative base-10 logarithm of the hydroxide ion concentration.
pOH = –log[OH–]
[OH–] = 10–pOH
From the definition of pH and pOH:
pH + pOH = 14.00
The pH Scale
Worked Example 16.7
Determine the pOH of a solution at 25°C in which the hydroxide ion
concentration is (a) 3.7×10-5 M, (b) 4.1×10-7 M, and (c) 8.3×10-2 M.
Strategy Given [OH-], use pOH = –log[OH-] to calculate pOH.
Solution
(a) pOH = –log(3.7×10-5) = 4.43
(b) pOH = –log(4.1×10-7) = 6.39
(c) pOH = –log(8.3×10-2) = 1.08
Think About It Remember that the pOH scale is, in essence, the reverse of the
pH scale. On the pOH scale, numbers below 7 indicate a basic solution, whereas
number above 7 indicate an acidic solution. The pOH benchmarks (abbreviated in
Table 16.6) work the same way the pH benchmarks do. In part (a), for example, a
hydroxide ion concentration between 1×10-4 M and 1×10-5 M corresponds
to a pOH between 4 and 5.
Worked Example 16.8
Calculate the hydroxide ion concentration in a solution at 25°C in which the
pOH is (a) 4.91, (b) 9.03, and (c) 10.55.
Strategy Given pOH, use [OH-] = 10-pOH to calculate [OH-].
Solution
(a) [OH-] = 10-4.91 = 1.2×10-5 M
(b) [OH-] = 10-9.03 = 9.3×10-10 M
(c) [OH-] = 10-10.55 = 2.8×10-11 M
Think About It Use the benchmark pOH values to determine whether these
solutions are reasonable. In part (a), for example, the pOH between 4 and 5
corresponds to [OH-] between 1×10-4 M and 1×10-5 M.
16.5
Strong Acids and Bases
Strong acid dissociations are not treated as equilibria, rather as
processes that go to completion.
Hydrochloric acid
HCl(aq) + H2O(l)
H3O+(aq) + Cl–(aq)
Hydrobromic acid
HBr(aq) + H2O(l)
H3O+(aq) + Br–(aq)
Hydroiodic acid
HI(aq) + H2O(l)
H3O+(aq) + I–(aq)
Nitric acid
HNO3(aq) + H2O(l)
H3O+(aq) + NO3–(aq)
Chloric acid
HClO3(aq) + H2O(l)
H3O+(aq) + ClO3–(aq)
Perchloric acid
HClO4(aq) + H2O(l)
H3O+(aq) + ClO4–(aq)
Sulfuric acid
H2SO4(aq) + H2O(l)
H3O+(aq) + HSO4–(aq)
Worked Example 16.9
Calculate the pH of an aqueous solution at 25°C that is (a) 0.035 M in HI, (b)
1.2×10-4 M in HNO3, and (c) 6.7×10-5 M in HClO4.
Strategy HI, HNO3, and HClO4 are all strong acids, so the concentration of
hydronium ions in each solution is the same as the stated concentration of the
+] to calculate pH.
acid. Use
pH
=
–log[H
O
3
Think About It Again, note that when a hydronium ion
concentration falls between two of the benchmark concentrations in
Solution
Table 16.4, the pH falls between the two corresponding pH values.
(a) [H3In
O+part
] = 0.035
Mexample, the hydronium ion concentration of
(b), for
pH 1.2×10
= –log(0.035)
1.46 than 1×10-4 M and less than 1×10-3 M.
-4 M is =
greater
Therefore, we expect the pH to be between 4.00 and 3.00.
(b) [H3O+] = 1.2×10-4 M
pH = –log(1.2×10-4) = 3.92
(c) [H3O+] = 6.7×10-5 M
pH = –log(6.7×10-5) = 4.17
Worked Example 16.10
Calculate the concentration on HCl in a solution at 25°C that has pH (a) 4.95,
(b) 3.45, and (c) 2.78.
Strategy Use [H3O+] = 10-pH to convert from pH to [H3O+]. In a strong acid
solution, [H3O+] is equal to the acid concentration.
Solution
(a) [HCl] = [H3O+] = 10-4.95 = 1.1×10-5 M
(b) [HCl] = [H3O+] = 10-3.45 = 3.5×10-4 M
(c) [HCl] = [H3O+] = 10-2.78 = 1.7×10-3 M
Think About It As pH decreases, acid concentration increases.
Strong Acids and Bases
The list of strong bases consists of the hydroxides of alkali metals
and the heaviest alkaline earth metals.
Group 1A hydroxides
Group 2A hydroxides
LiOH(aq)
Li+(aq) + OH–(aq)
NaOH(aq)
Na+(aq) + OH–(aq)
KOH(aq)
K+(aq) + OH–(aq)
RbOH(aq)
Rb+(aq) + OH–(aq)
CsOH(aq)
Cs+(aq) + OH–(aq)
Ca(OH)2(aq)
Ca2+(aq) + 2OH–(aq)
Sr(OH)2(aq)
Sr2+(aq) + 2OH–(aq)
Ba(OH)2(aq)
Ba2+(aq) + 2OH–(aq)
Worked Example 16.11
Calculate the pOH of the following aqueous solutions at 25°C: (a) 0.013 M
LiOH, (b) 0.013 M Ba(OH)2, and (c) 9.2×10-5 M KOH.
Strategy LiOH, Ba(OH)2, and KOH are all strong bases. Use reaction
stoichiometry to determine the hydroxide ion concentration and pOH = –log[OH-]
to determine pOH.
Think
It Theseion
areconcentration
basic pOH values,
which
is what
Solution
(a)About
The hydroxide
is simply
equal
to thewe
should expect
for the
solutions[OH
described
in the =problem.
-] = [LiOH]
concentration
of the base.
Therefore,
0.013 M.Note that
while the solutions inpOH
parts=(a)
and (b) have
the same base
–log(0.013)
= 1.89
concentration, they do not have the same hydroxide concentration
(b) Theand
hydroxide
is twice
that of the base:
thereforeion
doconcentration
not have the same
pOH.
Ba(OH)2(aq) → Ba2+(aq) + 2OH-(aq)
Therefore, [OH-] = 2[Ba(OH)2] = 2(0.013 M) = 0.026 M.
pOH = –log(0.026) = 1.59
(c) The hydroxide ion concentration is equal to the concentration of the base.
Therefore, [OH-] = [KOH] = 9.2×10-5 M.
pOH = –log(9.2×10-5) = 4.04
Worked Example 16.12
An aqueous solution of a strong base has pH 8.15 at 25°C. Calculate the original
concentration of base in the solution (a) if the base is NaOH and (b) if the base is
Ba(OH)2.
Strategy Use pH + pOH = 14.00 to convert from pH to pOH and [OH-] =
10-pOH to determine the hydroxide ion concentration. Consider the stoichiometry
Think About
Alternatively,
wethe
could
determine of
thethe
hydroxide
of dissociation
in eachItcase
to determine
concentration
base itself.
+
-8.15
-7
ion concentration using [H3O ] = 10
= 7.1×10 M and
Solution
pOH
= 14.00-14
– 8.15 = 5.85-6
1.0×10
[OH
]
=
= 1.4×10
-6 M M
[OH-] =7.1×10
10-5.85 =-9 1.41×10
[OH-] isofknown,
is the same
as shown
previously.
(a) TheOnce
dissociation
1 molethe
of solution
NaOH produces
1 mole
of OH-.
Therefore, the
concentration of base is equal to the concentration of hydroxide ion.
[NaOH] = [OH-] = 1.41×10-6 M
(b) The dissociation of 2 mole of Ba(OH)2 produces 2 moles of OH-. Therefore,
the concentration of base is only one-half the concentration of hydroxide ion.
1
[Ba(OH)2] = 2 [OH-] = 7.1×10-7 M
16.6
Weak Acids and Acid Ionization Constants
The ionization of a weak monoprotic acid HA in water is represented
by:
HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)
H3O+   A  
Ka 
HA 
Ka is called the acid ionization constant.
The larger the value of Ka, the stronger the acid.
Solution (at 25 °C)
Ka
pH
0.10 M HF
7.1 x 10–4
2.09
0.10 M CH3COOH
1.8 x 10–5
2.87
Weak Acids and Acid Ionization Constants
Weak Acids and Acid Ionization Constants
Calculate the pH of a 0.50 M HF solution at 25°C.
HF(aq) + H2O(l)
⇌ H3O+(aq)
+
F–(aq)
H3O+  F 
Ka 
 7.1 10 4
HF
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
Initial concentration (M)
0.50
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.50 – x
x
x
Weak Acids and Acid Ionization Constants
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
0.50
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.50 – x
x
x
Initial concentration (M)
H3O+  F 
Ka 
HF
Ka 
 x  x 
0.50  x
 7.1 104
Use quadratic formula to solve
–or –
Since HF is a weak acid, x could be small
compared to 0.50
Weak Acids and Acid Ionization Constants
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
0.50
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.50 – x
x
x
Initial concentration (M)
Ka 
 x  x 
0.50  x
 7.1 10
4
simplifies
Ka 
 x  x 
x2 = (0.50)(7.1 x 10–4) = 3.55 x 10–4
x = 1.9 x 10–2
0.50
 7.1 104
Weak Acids and Acid Ionization Constants
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
Initial concentration (M)
0.50
Change in concentration (M)
–1.9 x 10–2
Equilibrium concentration (M)
0.48
0
0
+1.9 x 10–2 +1.9 x 10–2
1.9 x 10–2
1.9 x 10–2
pH = –log(0.019) = 1.72
The shortcut is acceptable to use if
the calculate value of x is less than
5% of the initial acid concentration
Ka 
0.019M
 100%  3.8%
0.50M
Worked Example 16.13
The Ka of hypochlorous acid (HClO) is 3.5×10-8. Calculate the pH of a solution
at 25°C that is 0.0075 M in HClO.
Strategy Construct an equilibrium table, and express the equilibrium and
concentration of each species in terms of x. Solve for x using the approximation
shortcut, and evaluate whether or not the approximation is valid. Use
pH = –log[H3O+] to determine pH.
HClO(aq) + H2O(l) ⇌ H3O+(aq) + ClO–(aq)
Initial concentration (M)
Change in concentration (M)
0.0075
0
0
–x
+x
+x
x
x
Equilibrium concentration (M) 0.0075 – x
Worked Example 16.13 (cont.)
Solution These equilibrium concentrations are then substituted into the
equilibrium expression to give
(x)(x)
Ka =
= 3.5×10-8
0.0075 – x
Assuming that 0.0075 – x ≈ 0.0075,
Think About It We learned in Section 16.3 that the concentration of
-7 M, yet we use 0 M as the
x2 ion in pure water
hydronium
at
25°C
is
1.0×10
-8
2
= 3.5×10
x = (3.5×10-8)(0.0075)
0.0075
starting
concentration to solve for the pH of a weak acid. The reason for
this is
the get
actual concentration of hydronium ion in pure water is
Solving
forthat
x, we
-5 the
insignificant compared
the amount
x = to
1.62×10by
M ionization of the
2.625
1010 =produced
weak acid. We could use the actual concentration of hydronium as the
initial concentration,
but doing
change the result because (x
According
to the equilibrium
table,so
x =would
[H3O+not
]. Therefore,
+ 1.0×10-7) M ≈ x M. pH
In solving
problems -5of) =this
type, we neglect the
= –log(1.62×10
4.79
small concentration of H+ due to the autoionization of water.
Weak Acids and Acid Ionization Constants
A quantitative measure of the degree of ionization is percent
ionization.
+
HF(aq) + H2O(l) ⇌ H3O (aq) + F–(aq)
Initial concentration (M)
0.50
Change in concentration (M)
–1.9 x 10–2
Equilibrium concentration (M)
0.48
0
0
+1.9 x 10–2 +1.9 x 10–2
1.9 x 10–2
1.9 x 10–2
H3O 
eq
percent ionization 
 100%
HA0
0.019 M
percent ionization 
 100%  3.8%
0.50 M
Weak Acids and Acid Ionization Constants
Calculate the percent ionization of a 1.0 M HF solution at 25°C.
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
1.00
0
Change in concentration (M)
–2.7 x 10–2
+2.7 x 10–2
+2.7 x 10–2
Equilibrium concentration (M)
0.97
2.7 x 10–2
2.7 x 10–2
Initial concentration (M)
percent ionization 
Solution (at 25 °C)
0.027 M
 100%  2.7%
1.0 M
pH
% ionization
0.5 M HF
1.72
3.8
1.0 M HF
1.57
2.7
0
Weak Acids and Acid Ionization Constants
HF(aq) + H2O(l) ⇌ H3O+(aq) + F–(aq)
Solution (at 25 °C)
pH
% ionization
0.5 M HF
1.72
3.8
1.0 M HF
1.57
2.7
Worked Example 16.14
Determine the pH and percent ionization for acetic acid solutions at 25°C with
concentrations (a) 0.15 M, (b) 0.015 M, and (c) 0.0015 M.
Strategy Using the procedure described in Worked Example 16.13, we
construct an equilibrium table and for each concentration of acetic acid, we solve
for the equilibrium concentration of H+. We use pH = –log[H3O+] to find pH, and
the equation below to find percent ionization. Ka for acetic acid is 1.8×10-5.
H3O 
eq
percent ionization 
 100%
HA0
Worked Example 16.14 (cont.)
CH3COOH(aq) + H2O(l) ⇌ H3O+(aq) + CH3COO–(aq)
Solution (a)
0.15
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.15 – x
x
x
Initial concentration (M)
Solving for x gives [H3O+] = 0.0016 M and pH = –log(0.0016) = 2.78.
percent ionization =
0.0016 M
× 100% = 1.1%
0.15 M
(b) Solving the same way as part (a) gives [H3O+] = 5.2×10-4 M and pH = 3.28.
5.2×10-4 M
× 100% = 3.5%
percent ionization =
0.015 M
Worked Example 16.14 (cont.)
Solution (c) Solving the quadratic equation, or using successive approximation
[Appendix 1] gives [H3O+] = 1.6×10-4 M and pH = 3.78.
1.6×10-4 M
× 100% = 11%
percent ionization =
0.0015 M
Think About It Check your work by using the calculated value of Ka to solve
for the pH of a 0.10-M solution of aspirin.
Weak Acids and Acid Ionization Constants
Determine the Ka of a weak acid that has a concentration of 0.25 M
and a pH of 3.47 at 25°C.
HA(aq) + H2O(l)
H3O+   A  
Ka 
?
HA 
⇌ H3O+(aq)
+
A–(aq)
H3O+ = 10–3.47 = 3.39 x 10–4 M
HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)
Initial concentration (M)
0.25
Change in concentration (M)
–3.39 x 10–4
Equilibrium concentration (M)
0.2497
0
0
+3.39 x 10–4 +3.39 x 10–4
3.39 x 10–4
3.39 x 10–4
Weak Acids and Acid Ionization Constants
Determine the Ka of a weak acid that has a concentration of 0.25 M and a
pH of 3.47 at 25°C.
HA(aq) + H2O(l) ⇌ H3O+(aq) + A–(aq)
Initial concentration (M)
0.25
0
Change in concentration (M)
–3.39 x 10–4
Equilibrium concentration (M)
0.2497
H3O+   A  
Ka 
HA 
+3.39 x 10–4 +3.39 x 10–4
3.39 x 10–4
3.39  10 


4
Ka
0
0.2497
3.39 x 10–4
2
 4.6  10 7
Worked Example 16.15
Aspirin (acetylsalicylic acid, HC9H7O4) is a weak acid. It ionizes in water
according to the equation
HC9H7O4(aq) + H2O(l) ⇌ H3O+(aq) + C9H7O4-(aq)
A 0.10-M aqueous solution of aspirin has a pH of 2.27 at 25°C. Determine the Ka
of aspirin.
Strategy Determine the hydronium ion concentration from the pH. Use the
hydronium ion concentration to determine the equilibrium concentrations of the
other species, and plug the equilibrium concentrations into the equilibrium
expressions to evaluate Ka.
Worked Example 16.15 (cont.)
Solution [H3O+] = 10–2.27 = 5.37×10-3 M
To calculate Ka, though, we also need the equilibrium concentrations of C9H7O4and HC9H7O4. The stoichiometry of the reaction tells us that [C9H7O4-] = [H3O+].
Furthermore, the amount of aspirin that has ionized is equal to the amount of
hydronium ion in solution. Therefore, the equilibrium concentration of aspirin is
(0.10 – 5.37×10-3) M = 0.095 M.
HC9H7O4(aq) + H2O(l) ⇌ H3O+(aq) + C9H7O4-(aq)
Think About(M)
It Check your
work by using the calculated
value of 0
Initial concentration
0.10
0
Ka to solve for the pH of a 0.10-M solution of aspirin.
Change in concentration (M)
–0.005
+5.37×10-3 +5.37×10-3
Equilibrium concentration (M)
0.095
5.37×10-3
[H3O+][C9H7O4-]
(5.37×10-3)2
Ka =
= 3.0×10-4
=
[HC9H7O4]
0.095
The Ka of aspirin is 3.0×10-4.
5.37×10-3
16.7
Weak Bases and Base Ionization Constants
The ionization of a weak base is incomplete and is treated in the
same way as the ionization of a weak acid.
B(aq) + H2O(l)
⇌ HB+(aq)
HB+  OH 
Kb 
B
Kb is called the base ionization constant.
The larger the value of Kb, the stronger the base.
+
OH–(aq)
Worked Example 16.16
What is the pH of a 0.040 M ammonia solution at 25°C.
Strategy Construct an equilibrium table, and express equilibrium concentrations
in terms of the unknown x. Plug these equilibrium concentrations into the
equilibrium expression, and solve for x. From the value of x, determine the pH.
NH3(aq) + H2O(l) ⇌ NH4+(aq) + OH-(aq)
0.040
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.040 – x
x
x
Initial concentration (M)
Worked Example 16.16 (cont.)
Solution The equilibrium concentrations are substituted into the equilibrium
expression to give
[NH4+][OH-]
(x)(x)
Kb =
= 1.8×10-5
=
[NH3]
0.040 – x
Assuming that 0.040 – x ≈ 0.040 and solving for x gives
(x)(x)
Think About It It(x)(x)
is a common
error in Kb problems
to forget that
= 1.8×10-5
≈
–x
0.040
x is hydroxide ion0.040
concentration
rather than the hydronium ion
concentration. Always2 make sure that
the pH you calculate for a
x = (1.8×10-5)(0.040) = 7.2×10-7
solution of base is a basic pH, that is, a pH greater than 7.
x = 7.2 107 = 8.5×10-4 M
According to the equilibrium table, x = [OH-]. Therefore, pOH = – log(x):
–log(8.5×10-4) = 3.07
and pH = 14.00 – pOH = 14.00 – 3.07 – 10.93. The pH of a 0.040-M solution of
NH3 at 25°C is 10.93.
Weak Bases and Base Ionization Constants
Worked Example 16.17
Caffeine, the stimulant in coffee and tea, is a weak base that ionizes in water
according to the equation
C8H10N4O2(aq) + H2O(l) ⇌ HC8H10N4O2+(aq) + OH-(aq)
A 0.15-M solution of caffeine at 25°C has a pH of 8.45. Determine the Kb of
caffeine.
Strategy Use pH to determine pOH, and pOH to determine the hydroxide ion
concentration. From the hydroxide ion concentration, use reaction stoichiometry
to determine the other equilibrium concentrations and plus those concentrations
into the equilibrium expression to evaluate Kb.
Worked Example 16.17 (cont.)
Solution pOH = 14.00 – 8.45 – 5.55; [OH-] = 10-5.55 = 2.82×10-6 M
Based on the reaction stoichiometry, [HC8H10N4O2+] = [OH-], and the amount of
hydroxide ion in solution at equilibrium is equal to the amount of caffeine that has
ionized. At equilibrium, therefore,
[C8H10N4O2] = (0.15 – 2.82×10-6) M ≈ 0.15 M
C8H10N4O2(aq) + H2O(l) ⇌ HC8H10N4O2+(aq) + OH-(aq)
Initial Think
concentration
About (M)
It Check your0.15
answer using the calculated
0 Kb to
0
determine the pH ofa 0.15-M solution.
Change in concentration (M)
–2.82×10-6
+2.82×10-6
+2.82×10-6
Equilibrium concentration (M)
0.15
2.82×10-6
2.82×10-6
[HC8H10N4O2+][OH-]
(2.82×10-6)2
Kb =
=
= 5.3×10-11
[C8H10N4O2]
0.15
16.8
Conjugate Acid-Base Pairs
A strong acid ionizes completely in water:
H+(aq) + Cl–(aq)
HCl(aq)
No affinity for
the H+ ion
The chloride ion is a weak conjugate base.
Cl–(aq) + H2O(l)
X
HCl(aq) + OH–(aq)
Conjugate Acid-Base Pairs
A weak acid ionizes to a limited degree in water:
HF(aq)
⇌
H+(aq) +
F–(aq)
Strong affinity
for the H+ ion
The fluoride ion is a strong conjugate base.
F–(aq) + H2O(l) ⇌
HF(aq) + OH–(aq)
Conjugate Acid-Base Pairs
A strong acid has a weak conjugate base.
A weak acid has a strong conjugate base.
A strong base has a weak conjugate acid.
A weak base has a strong conjugate base.
Conjugate Acid-Base Pairs
A simple relationship between the ionization constant of a weak acid
(Ka) and the ionization constant of a weak base (Kb) can be derived:
⇌
H+(aq)
CH3COO– (aq) + H2O(l)
⇌
CH3COOH(aq)
H2O(l)
⇌
H+(aq)
CH3COOH(aq)
H+  CH3COO 
Ka 
CH3COOH
Kb 
CH3COOH OH 
CH3COO 

+
+
CH3COO–(aq)
+
OH–(aq)
OH–(aq)
H+  CH3COO  CH3COOH OH 








H
OH




CH3COO 
CH3COOH
Ka x Kb = Kw
Worked Example 16.18
Determine (a) Kb of the acetate ion (CH3COO-), (b) Ka of the methylammonium
ion (CH3NH3+), (c) Kb of the fluoride (F-), and (d) Ka of the ammonium ion
(NH4+).
Strategy Each species listed is either a conjugate base or a conjugate acid.
Determine the identity of the acid corresponding to each conjugate base and the
identity of the base corresponding to each conjugate acid; then, consult Table 16.7
and 16.8 for their ionization constants. Use the tabulated ionization constants and
Kw = Ka×Kb to calculate each indicated K value.
K
Ka = Kw
b
and
K
Kb = Kw
a
Solution (a) A Kb value is requested, indicating that the acetate ion is a
conjugate base. To identify the corresponding Brønsted acid, add a proton to the
formula to get CH3COOH (acetic acid). The Ka of acetic acid is 1.8×10-5.
Conjugate base CH3
COO-:
1.0×10-14
Kb =
= 5.6×10-10
-5
1.8×10
Worked Example 16.18 (cont.)
Solution (b) A Ka value is requested, indicating that the methylammonium ion is
a conjugate acid. Determine the identity of the corresponding Brønsted base by
removing a proton from the formula to get CH3NH2 (methylamine). The Kb of
methylamine is 4.4×10-4.
1.0×10-14
Conjugate acid CH3NH3 Ka =
= 2.3×10-11
-4
4.4×10
Think About It Because the conjugates of weak acids and bases
(c) F- ishave
the conjugate
of HF;salts
Ka =containing
7.1×10-4these
.
ionization base
constants,
ions have an effect
on the pH of a solution. In Section
-14 16.10 we will use the ionization
1.0×10
-11 to calculate pH for
Conjugate
F-: Kb = acids and-4conjugates
constantsbase
of conjugate
bases
= 1.4×10
7.1×10
solutions containing dissolved salts.
(d) NH4+ is the conjugate acid of NH3; Kb = 1.8×10-5.
+:
Conjugate acid NH4
+:
1.0×10-14
Ka =
= 5.6×10-10
-5
1.8×10
16.9
Diprotic and Polyprotic Acids
Diprotic and polyprotic acids undergo successive ionizations, losing
one proton at a time, and each has a Ka associate with it.
H2CO3(aq)
HCO3– (aq)
⇌
⇌
H+(aq) + HCO3–(aq)
H+  HCO3 
K a1 
H2CO3 
H+(aq) + CO32– (aq)
H+  CO32 

HCO3 
Ka2
Ka1 > Ka2
For a given acid, the first ionization constant is much larger than the
second, and so on.
Diprotic and Polyprotic Acids
Worked Example 16.19
Oxalic acid (H2C2O4) is a poisonous substance used mainly as a bleaching agent.
Calculate the concentrations of all species present at equilibrium in a 0.10-M
solution at 25°C.
Strategy Follow the same procedure for each ionization as for the determination
of equilibrium concentrations for a monoprotic acid. The conjugate base resulting
from the first ionization is the acid for the second ionizations, and its starting
concentration is the equilibrium concentration from the first ionization.
H2C2O4(aq) ⇌ H+(aq) + HC2O4–(aq)
HC2O4– (aq) ⇌ H+(aq) + C2O42– (aq)
Ka1 = 6.5 x 10–2
Ka2 = 6.1 x 10–5
Construct an equilibrium table for each ionization, using x as the unknown in the
first ionization and y as the unknown in the second ionization.
Worked Example 16.19 (cont.)
Strategy
H2C2O4(aq) ⇌ H+(aq) + HC2O4–(aq)
0.10
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
x
Initial concentration (M)
The equilibrium concentration of the hydrogen oxalate (HC2O4-) after the first
ionization becomes the starting concentration for the second ionization.
Additionally, the equilibrium concentration of H+ is the starting concentration for
the second ionization.
HC O – (aq) ⇌ H+(aq) + C O 2– (aq)
2
4
2
4
x
x
0
Change in concentration (M)
–y
+y
+y
Equilibrium concentration (M)
x–y
x+y
y
Initial concentration (M)
Worked Example 16.19 (cont.)
[H+][HC2O4-]
Ka1 =
[H2C2O4]
x2
-2
6.5×10 =
0.10 – x
Applying the approximation and neglecting x in the denominator of the
expression gives
x2
-2
6.5×10 ≈
0.10
Solution
x2 = 6.5×10-3
x = 8.1×10-2 M
Testing the approximation,
8.1×10-2 M
×100% = 81%
0.10 M
Clearly the approximation is not valid, so we must solve the following quadratic
equation:
x2 + 6.5×10-3x – 6.5×10-3 = 0
Worked Example 16.19 (cont.)
Solution The result x = 0.054 M. Thus, after the first ionization, the
concentrations of species in solution are
[H+] = 0.054 M
[HC2O4-] = 0.054 M
[H2C2O4] = (0.10 – 0.054) M = 0.046 M
Rewriting the equilibrium table for the second ionization, using the calculated
value of x, gives the following:
HC2O4– (aq) ⇌ H+(aq) + C2O42– (aq)
Initial concentration (M)
Change in concentration (M)
Equilibrium concentration (M)
[H+][C2O42-]
Ka2 =
[HC2O4-]
0.054
0.054
0
–y
+y
+y
0.054 – y 0.054 + y
6.1×10-5 =
(0.054 + y)(y)
0.054 – y
y
Worked Example 16.19 (cont.)
Solution Assuming that y is very small and applying the approximations
0.054 + y ≈ 0.054 and 0.054 – y ≈ 0.054 gives
(0.054)(y)
= y = 6.1×10-5
0.054
We must test the approximation as follows to see if it is valid:
Think About It Note that the second ionization did not contribute
-5 M
significantly to the H+
concentration.
Therefore, we could determine
6.1×10
×100%
= 0.11%
the pH of this solution by
considering
only the
first ionization. This
0.054
M
is true in general for polyprotic acids where Ka1 is at least
This time, because the ionization constant is much smaller,
the approximation is
1000×K
.
[Note
that
it
is
necessary
to
consider
the
second
a2
valid. At equilibrium,
the concentrations of all species are
ionization to determine the concentration of oxalate ion (C2O42-).]
[H2C2O4] = 0.046 M
[HC2O4-] = (0.054 – 6.1×10-5) = 0.054 M
[H+] = (0.054 + 6.1×10-5) = 0.054 M
[C2O42-] = 6.1×10-5 M
16.10 Acid-Base Properties of Salt Solutions
Salt hydrolysis occurs when ions produced by the dissociation of a
salt react with water to produce either hydroxide ions or hydronium
ions.
Basic salts (conjugates of weak acids):
F–(aq) + H2O(l) ⇌
HF(aq) + OH–(aq)
Acidic salts (conjugates of weak bases)
NH4+(aq) + H2O(l) ⇌
NH3(aq) + H3O+(aq)
Worked Example 16.20
Calculate the pH of a 0.10-M solution of sodium fluoride (NaF) at 25°C.
Strategy A solution of NaF contains Na+ ions and F- ions. The F- ion is the
conjugate base of the weak acid, HF. Use the Ka value for HF (7.1×10-4) and
Kw = Kb×Ka to determine Kb for F-:
1.0×10-14
Kw
Kb = K =
= 1.4×10-11
-4
7.1×10
a
Then, solve this pH problem like any equilibrium problem, using an equilibrium
table.
[HF][OH-]
F (aq) + H2O(l) ⇌ HF(aq) + OH (aq) Kb =
[F-]
F-(aq) + H2O(l) ⇌ HF(aq) + OH-(aq)
0.10
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
x
Initial concentration (M)
Worked Example 16.20 (cont.)
Solution Substituting the equilibrium concentrations into the equilibrium
expression and using the shortcut to solve x, we get
1.4×10-11
x2
≈
=
0.10 – x
x2
0.10
x = (1.4  10 11 )( 0.10 ) = 1.2×10-6 M
Think About It It’s easy to mix up pH and pOH in this type of
problem. Always make a qualitative prediction regarding the pH of a
According to our equilibrium table, x = [OH-]. In this case, the autoionization of
salt solution first, and then check to make sure that your calculated
water makes a significant contribution to the hydroxide ion concentration so the
pH agrees with your prediction. In this case, we would predict a
total concentration will be the sum of 1.2×10-6- M (from the ionization of F-) and
basic pH because the anion in the salt (F ) is the conjugate base of a
1.0×10-7 M (from the autoionization of water). Therefore, we calculate the pOH
weak acid (HF). The calculated pH, 8.05, is indeed basic.
first as
pOH = –log(1.2×10-6 + 1.0×10-7) = 5.95
and then the pH,
pH = 14.00 – pOH = 14.00 – 5.95 = 8.05
The pH of a 0.10-M solution of NaF at 25°C is 8.05.
Worked Example 16.21
Calculate the pH of a 0.10-M solution of ammonium chloride (NH4Cl) at 25°C.
Strategy A solution of NH4Cl contains NH4+ ions and Cl- ions. The NH4+ ion is
the conjugate acid of the weak base, NH3. Use the Kb value for NH3 (1.8×10-5)
and Kw = Kb×Ka to determine Ka for F-:
1.0×10-14
Kw
Ka = K =
= 5.6×10-10
-5
1.8×10
b
Again, we write the balanced chemical equation and the equilibrium expression:
NH4
+ (aq)
+ H2O(l) ⇌ NH3(aq) + H3
O+(aq)
[NH3][H3O+]
Kb =
[NH4+ ]
NH4+ (aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
0.10
0
0
Change in concentration (M)
–x
+x
+x
Equilibrium concentration (M)
0.10 – x
x
x
Initial concentration (M)
Worked Example 16.21 (cont.)
Solution Substituting the equilibrium concentrations into the equilibrium
expression and using the shortcut to solve x, we get
5.6×10-10
x2
≈
=
0.10 – x
x2
0.10
x = (5.6  10 10 )( 0.10 ) = 7.5×10-6 M
According to our equilibrium table, x = [H3O+]. The pH can be calculated as
follows:
pH = –log(7.5×10-6) = 5.12
The pH of a 0.10-M solution of ammonium chloride (at 25°C) is 5.12.
Think About It In this case, we would predict an acidic pH because the cation
in the salt (NH4+) is the conjugate acid of a weak base (NH3). The calculated pH
is acidic.
Acid-Base Properties of Salt Solutions
Small, highly charged metal ions can react with water to produce an
acidic solution.
Acid-Base Properties of Salt Solutions
The pH of salt solutions can be qualitatively predicted by
determining which ions facilitate hydrolysis.
Examples
A cation that will make a solution acidic is
 The conjugate acid of a weak base
NH4+ , CH3NH3+ , C2H5NH3+
 A small, highly charged metal ion (other than Group
1A or 2A)
Al3+ , Cr3+ , Fe3+ , Bi3+
An anion that will make a solution basic is
 The conjugate base of a weak acid
CN– , NO2– , CH3COO–
A cation that will not affect the pH of a solution is
 A Group 1A or heavy Group 2A cation (except Be2+)
Li+ , Na+ , Ba2+
An anion that will not affect the pH of a solution is
 The conjugate base of a strong acid
Cl– , NO3– , ClO4–
Worked Example 16.22
Predict whether a 0.10-M solution of each of the following salts will be basic,
acidic, or neutral: (a) LiI, (b) NH4NO3, (c) Sr(NO3)2, (d) KNO2, (e) NaCN.
Strategy Identify the ions present in each solution, and determine which, if any,
will impact the pH of the solution.
Solution (a) Ions in solution: Li+ and I-. Li+ is a Group 1A cation; I- is the
conjugate base of the strong acid HI. Therefore, neither ion hydrolyzes to any
significant degree. Solution will be neutral.
(b) Ions in solution: NH4+ and NO3-. NH4+ is the conjugate acid of the weak base
NH3; NO3- is the conjugate base of the strong acid HNO3. In this case, the cation
will hydrolyze, making the pH acidic:
NH4+ (aq) + H2O(l) ⇌ NH3(aq) + H3O+(aq)
Worked Example 16.22 (cont.)
Solution (c) Ions in solution: Sr2+ and NO3-. Sr2+ is a heavy Group 2A cation;
NO3- is the conjugate base of the strong acid, HNO3. Neither ion hydrolyzes to
any significant degree.
(d) Ions in solution: K+ and NO2-. K+ is a Group 1A cation; NO2- is the conjugate
base of the weak acid HNO2. In this case, the anion hydrolyzes, thus making the
pH basic:
NO2- (aq) + H2O(l) ⇌ HNO2(aq) + OH-(aq)
(e) Ions in solution: Na+ and CN-. Na+ is a Group 1A cation; CN- is the conjugate
base of the weak acid HCN. In this case, too, the anion hydrolyzes, thus making
the pH basic:
CN- (aq) + H2O(l) ⇌ HCN(aq) + OH-(aq)
Think About It It’s very important that you be able to identify the ions in
solution correctly. If necessary, review the formulas and charges of the common
polyatomic ions.
Acid-Base Properties of Salt Solutions
The pH of a solution that contains a salt in which both the cation and
the anion hydrolyze depends on the relative strengths of the weak
acid and base.
Qualitative predictions can be made using the Kb (of the salts anion)
and the Ka (of the salts cation).
 When Kb > Ka, the solution is basic
 When Kb < Ka, the solution is acidic
 When Kb ≈ Ka, the solution is neutral or nearly neutral
16.11 Acid-Base Properties of Oxides and Hydroxides
Acid-Base Properties of Oxides and Hydroxides
Basic metallic oxides react with water to form metal hydroxides:
Na2O(s) + H2O(l) → 2NaOH(aq)
BaO(s) + H2O(l) → Ba(OH)2(aq)
Acidic oxides reaction with water as follows:
CO2(g) + H2O(l) ⇌ H2CO3(aq)
SO3(g) + H2O(l) ⇌ H2SO4(aq)
Reactions between acidic oxides and bases and those between basic
oxides and acids resemble normal acid-base reactions that produce a
salt and water.
CO2(g) + 2NaOH(aq) → Na2CO3(aq) + H2O(l)
BaO(s) + 2HNO3(aq) → Ba(NO3)2 (aq) + H2O(l)
Acid-Base Properties of Oxides and Hydroxides
Aluminum oxide (Al2O3) is amphoteric.
It can act as an acid:
Al2O3(s) + 6HCl(aq) → 2AlCl3(aq) + 3H2O(l)
Or it can act as a base:
Al2O3(s) + 2NaOH(aq) + 3H2O(l) → 2NaAl(OH)4(aq)
Acid-Base Properties of Oxides and Hydroxides
All the alkali and alkaline earth metal hydroxides, except Be(OH)2,
are basic.
Be(OH)2
Al(OH)3
Sn(OH)2
Pb(OH)2
Acid:
Be(OH)2(s) + 6H+(aq) → 2Be2+(aq) + 2H2O(l)
amphoteric
Cr(OH)3
Cu(OH)2
Zn(OH)2
Cd(OH)2
Base:
Be(OH)2(s) + 2OH–(aq) → Be(OH)42– (aq)
16.12 Lewis Acids and Bases
A Lewis base is a substance that can donate a pair of electrons.
A Lewis acid is a substance that can accept a pair of electrons.
empty
unhybridized 2pz
orbital
Boron trifluoride
a Lewis acid
Ammonia,
a Lewis base
A coordinate covalent bond
16
Acids and Bases
Brønsted Acids and Bases
Basic Salt Solutions
The Acid-Base Properties of Water
Acidic Salt Solutions
The pH Scale
Neutral Salt Solutions
Strong Acids and Bases
Salts in Which Both the Cation and the
The Ionization Constant, Ka
Anion Hydrolyze
Calculating pH from Ka
Oxides of Metals and Nonmetals
Using pH to Determine Ka
Basic and Amphoteric Hydroxides
The Ionization Constant, Kb
Lewis Acids and Bases
Calculating pH from Kb
Using pH to Determine Kb
The Strength of a Conjugate Acid or Base
The Relationship Between Ka and Kb of a Conjugate Acid-Base Pair
Diprotic and Polyprotic Acids
Hydrohalic Acids
Oxoacids
Carboxylic Acids