1. Acid-Ionization Equilibria (cont)
a. State the assumption that allows for using
approximations when solving problems.
Acid-Base
Equilibria
b. Calculate concentrations of species in a weak acid
solution using Ka (quadratic formula).
2. Polyprotic Acids
a. State the general trend in the ionization constants
of a polyprotic acid.
1
b. Calculate concentrations of species in a solution of
4
a diprotic acid.
Contents and Concepts
Solutions of a Weak Acid or Base
1. Acid-Ionization Equilibria
2. Polyprotic Acids
3. Base-Ionization Equilibria
4. Acid–Base Properties of Salt Solutions
Solutions of a Weak Acid or Base with Another
Solute
5. Common-Ion Effect
6. Buffers
7. Acid–Base Titration Curves
2
3. Base-Ionization Equilibria
– a.
Write the chemical equation for a
weak base undergoing ionization
in aqueous solution.
– b.
Define base-ionization constant.
– c.
Calculate concentrations of
species in a weak base solution
using Kb.
5
Learning Objectives
4. Acid–Base Properties of Salt Solutions
Solutions of a Weak Acid or Base
• Acid-Ionization Equilibria
– a.
Write the chemical equation for a weak
acid undergoing acid ionization in
aqueous solution.
– b.
Define acid-ionization constant and
degree of ionization.
– c.
Determine Ka from the solution pH.
– d.
Calculate concentrations of species in
a weak acid solution using Ka
(approximation method).
3
– a.
– b.
– c.
– d.
– e.
Write the hydrolysis reaction of an ion
to form an acidic solution.
Write the hydrolysis reaction of an ion
to form a basic solution.
Predict whether a salt solution is acidic,
basic, or neutral.
Obtain Ka from Kb or Kb from Ka.
Calculating concentrations of species
in a salt solution.
6
1
Solutions of a Weak Acid or Base with
Another Solute
5. Common-Ion Effect
– a.
– b.
7. Acid–Base Titration Curves (cont)
Explain the common-ion effect.
Calculate the common-ion effect on
acid ionization (effect of a strong acid).
Calculate the common-ion effect on
acid ionization (effect of a conjugate
base).
– c.
– f.
– g.
Describe the curve for the titration of a
weak base by a strong acid.
Calculate the pH of a solution at
several points of a titration of a weak
base by a strong acid.
10
7
Solutions of a Weak Acid or Base
6. Buffers
– a.
– b.
– c.
– d.
– e.
– f.
Define buffer and buffer capacity.
Describe the pH change of a buffer
solution with the addition of acid or base.
Calculate the pH of a buffer from given
volumes of solution.
Calculate the pH of a buffer when a strong
acid or a strong base is added.
Learn the Henderson–Hasselbalch equation.
State when the Henderson–Hasselbalch
equation can be applied
• The simplest acid-base equilibria are those in
which a single acid or base solute reacts with
water.
– In this chapter, we will first look at solutions
of weak acids and bases.
– We must also consider solutions of salts,
which can have acidic or basic properties
as a result of the reactions of their ions
with water.
11
8
7. Acid–Base Titration Curves
– a.
– b.
– c.
– d.
– e.
Define equivalence point.
Describe the curve for the titration of a
strong acid by a strong base.
Calculate the pH of a solution of a
strong acid and a strong base.
Describe the curve for the titration of a
weak acid by a strong base.
Calculate the pH at the equivalence
point in the titration of a weak acid by a
strong base.
9
The simplest acid–base equilibria are those
in which a weak acid or a weak base
reacts with water.
• We can write an acid equilibrium reaction
for the generic acid, HA.
• HA(aq) + H2O(l)
H3O+(aq) + A-(aq)
12
2
Acid-Ionization Equilibria
• For a weak acid, the equilibrium
concentrations of ions in solution are
determined by the acid-ionization constant
(also called the acid-dissociation constant).
• Acetic acid is a weak acid. It reacts with
water as follows:
– Consider the generic monoprotic acid, HA.
+
HA (aq ) + H 2O(l )
H3O+(aq) + C2H3O2-(aq)
• HC2H3O2(aq) + H2O(l)
−
H 3O (aq ) + A (aq )
– The corresponding equilibrium expression is:
+
13
• The equilibrium constant for the reaction of
a weak acid with water is called the acidionization constant (or acid-dissociation
constant), Ka.
[H O ][A ]
+
Ka =
−
[ H O ][ A ]
Kc = 3
[ HA ][ H 2O ]
16
– Since the concentration of water remains
relatively constant, we rearrange the equation to
get:
+
K a = [ H 2O ]K c =
−
−
[ H 3O ][ A ]
[ HA ]
– Thus, Ka , the acid-ionization constant, equals the
constant [H2O]Kc.
3
[HA ]
+
Ka =
• Liquid water is not included in the
equilibrium constant expression.
WHY?
14
−
[ H 3O ][ A ]
[HA ]
– Table 17.1 lists acid-ionization constants for
various weak acids.
17
Acid-Ionization Equilibria
• Acid ionization (or acid dissociation) is the
reaction of an acid with water to produce
hydronium ion (hydrogen ion) and the
conjugate base anion.
– When acetic acid is added to water it reacts as
follows.
HC2 H 3O 2 (aq ) + H 2O(l )
+
−
H 3O (aq) + C 2 H 3O 2 (aq)
– Because acetic acid is a weak electrolyte, it
ionizes to a small extent in water.
15
18
3
We can be given pH, percent or degree ionization,
initial concentration, and Ka.
B-8
In order of decreasing Ka:
HF
Ka
6.8 × 10-4
From pH, we can find [H3O+].
Highest pH
From percent or degree ionization, we can find Ka.
HCN
HNO2
HC2H3O2
4.5 × 10-4
HC2H3O2
1.7 × 10-5
HNO2
HCN
4.9 × 10-10
Using what is given, we can find the other
quantities.
HF
Lowest pH
22
19
A Problem To Consider
Experimental Determination of Ka
• Nicotinic acid is a weak monoprotic acid with the
formula HC6H4NO2. A 0.012 M solution of nicotinic
acid has a pH of 3.39 at 25°C. Calculate the acidionization constant for this acid at 25°C and the
percent ionization
• The degree of ionization of a weak
electrolyte is the fraction of molecules that
react with water to give ions.
– Electrical conductivity or some other colligative
property can be measured to determine the degree
of ionization.
– It is important to realize that the solution was made
0.012 M in nicotinic acid, however, some molecules
ionize making the equilibrium concentration of
nicotinic acid less than 0.012 M.
– With weak acids, the pH can be used to determine
the equilibrium composition of ions in the solution.
– We will abbreviate the formula for nicotinic acid as
HNic.
23
20
Calculations with Ka
• Given the value of Ka and the initial
concentration of HA, you can calculate the
equilibrium concentration of all species.
– Let x be the moles per liter of product formed.
HNic(aq ) + H 2O(l )
Starting
Change
Equilibrium
• Given the value of Ka and the initial
concentration of HA, you can calculate the
degree of ionization and the percent ionization.
−
0
+x
x
0
+x
x
– The equilibrium-constant expression is:
+
Ka =
• Given the pH of the final solution and the initial
concentration of HA, you can find the value
of Ka and the percent ionization.
0.012
-x
0.012-x
+
H 3O (aq ) + Nic (aq )
−
[H 3O ][ Nic ]
[HNic ]
Ka =
x2
( 0.012 − x )
– Substituting the expressions for the equilibrium
concentrations, we get
21
24
4
– We can obtain the value of x from the given
pH.
H3O+(aq) + C6H5O-(aq)
HC6H5O(aq) + H2O(l)
+
x = [ H 3O ] = anti log( −pH )
x = anti log(−3.39)
Initial
Change
−4
x = 4.1 × 10 = 0.00041
Equilibrium
– Substitute this value of x in our equilibrium expression.
– Note first, however, that
(0.012 − x) = (0.012 − 0.00041) = 0.01159 ≅ 0.012
the concentration of unionized acid remains
virtually unchanged.
0.10
0
0
–x
+x
+x
0.10 – x
x
x
• We were told that pH = 5.43. That allows us to
find [H3O+] = 3.7 × 10-6 M = x = [C6H5O-].
• Now we find [HC6H5O] = 0.10 – x = 0.10 M.
25
– Substitute this value of x in our equilibrium expression.
2
Ka =
2
x
≅ ( 0.00041) = 1.4 × 10 −5
( 0.012 − x )
( 0.012)
– To obtain the degree of dissociation:
Degree of dissociati on =
0.00041
= 0.034
0.012
– The percent ionization is obtained by multiplying by 100,
which gives 3.4%.
26
28
• Finally, we write the expression for Ka and
substitute the concentrations we now
know.
• HC6H5O + H2O
H3O+ + C6H5O• [H3O+] = [C6H5O-] = 3.7 × 10-6 M
• [HC6H5O] = 0.10 M
[H O ] [C H O ]
+
Ka =
Ka =
3
−
6
5
[HC 6H5 O]
(3.7 × 10 −6 )2
0.10
K a = 1.4 × 10 -10
29
• The degree of ionization is the ratio of
ionized concentration to original
concentration:
−6
Another Problem
• Sore-throat medications sometimes contain the
weak acid phenol, HC6H5O. A 0.10 M solution
of phenol has a pH of 5.43 at 25°C.
x
3.7 × 10
=
0.100
0.10
Degree of ionization = 3.7 × 10 −5
Degree of ionization =
• Percent ionization is the degree of ionization
× 100%:
• a. What is the acid-ionization
constant, Ka, for phenol at 25°C?
• Percent ionization = 3.7 x 10-3% or 0.0037%
• b. What is the degree of ionization?
27
30
5
•Simplifying Assumption for Acid and Base
Ionizations
•The equilibrium concentration of the acid is most
often ([HA]0 – x).
Ka =
x2
0.200
5.2 × 10 −6 = x 2
•If x is much, much less than [HA]0, we can
assume that subtracting x makes no difference to
[HA]:
•([HA]0 – x) = [HA]0
x = 2.3 × 10 −3 M = [H3O + ]
pH = - log [H3O + ] = - pH (2.3 × 10 -3 )
•This is a valid assumption when the ratio of [HA]0
to Ka is > 103. If it is not valid, you must use the
quadratic equation to solve the problem.
pH = 2.64
31
•Para-hydroxybenzoic acid is used to make
certain dyes. What are the concentrations of
this acid, of hydronium ion, and of parahydroxybenzoate ion in a 0.200 M aqueous
solution at 25°C? What is the pH of the
solution and the degree of ionization of the
acid? The Ka of this acid is 2.6 × 10-5.
34
Calculations With Ka
• Once you know the value of Ka, you can
calculate the equilibrium
concentrations of species HA, A-,
and H3O+ for solutions of different
molarities.
We will use the generic formula HA for parahydroxybenzoic acid and the following equilibrium:
H3O+ + A-
HA + H2O
32
H3O+(aq) + A-(aq)
0
0
HA(aq) + H2O(l)
0.200
Initial
Change
–x
Equilibrium 0.200 – x
2.6 × 10
+x
x
x
Calculations With Ka
• Note that in our previous example, the degree of
dissociation was so small that “x” was negligible
compared to the concentration of nicotinic acid.
• It is the small value of the degree of ionization that
allowed us to ignore the subtracted x in the
denominator of our equilibrium expression.
[H O ][A ].
+
Ka =
+x
35
−
3
[HA ]
−5
• The degree of ionization of a weak acid depends on
both the Ka and the concentration of the acid solution
(see Figure 16.3).
x2
=
0.200 - x
0.200
> 1000, so 0.200 - x ≈ 0.200
2.6 × 10 −5
33
36
6
– Note that
Ca
Ka
Figure 16.3:
Variation of
Percent
Ionization of a
Weak Acid with
Concentration
= 0.0036
3.3 × 10
−4
= 11
which is less than 100, so we must solve the
equilibrium equation exactly.
Abbreviate the formula for acetylsalicylic acid as
HAcs and let x be the amount of H3O+ formed per
liter.
– The amount of acetylsalicylate ion is also x mol; the
amount of nonionized acetylsalicylic acid is (0.0036-x)
mol.
40
37
Calculations With Ka
– These data are summarized below.
• How do you know when you can use this
simplifying assumption?
Ca
Ka
−
H 3O (aq ) + Acs (aq)
Starting
0.0036
Change
-x
Equilibrium 0.0036-x
– It can be shown that if the acid concentration, Ca,
divided by the Ka exceeds 100, that is,
if
+
HAcs(aq ) + H 2O(l )
0
+x
x
0
+x
x
– The equilibrium constant expression is
+
−
[ H 3O ][ Acs ]
= Ka
> 100
[ HAcs]
– then this simplifying assumption of ignoring the
subtracted x gives an acceptable error of less
than 5%.
38
– If the simplifying assumption is not valid,
you can solve the equilibrium equation
exactly by using the quadratic equation.
– The next example illustrates this with a solution
of aspirin (acetylsalicylic acid), HC9H7O4, a
common headache remedy.
– The molar mass of HC9H7O4 is 180.2 g.
From this we find that the sample contained
0.00180 mol of the acid.
Hence, the concentration of the acetylsalicylic acid
is 0.00180 mol/0.500 L = 0.0036 M (Retain two
significant figures, the same number of significant
figures in Ka).
39
– If we substitute the equilibrium concentrations and
the Ka into the equilibrium constant expression, we
get
41
+
−
[ H 3O ][ Acs ]
= Ka
[ HAcs]
x2
= 3.3 × 10 −4
−
( 0.0036 x )
– You can solve this equation exactly by using the
quadratic formula.
Rearranging the preceding equation to put it in the
form ax2 + bx + c = 0, we get
−
−
x 2 + ( 3.3 × 10 4 )x − (1.2 × 10 6 ) = 0
42
7
Polyprotic Acids
– Now substitute into the quadratic formula.
2
x=
− b ± b − 4ac
2a
• Some acids have two or more protons
(hydrogen ions) to donate in aqueous
solution. These are referred to as polyprotic
acids.
– Now substitute into the quadratic formula.
x=
− ( 3.3 × 10−4 ) ± ( 3.3 × 10−4 )2 − 4(1.2 × 10−6 )
2
– Sulfuric acid, for example, can lose two protons in
aqueous solution.
– The lower sign in ± gives a negative root which we
can ignore
−
+
H 2SO 4 (aq ) + H 2O( l ) → H 3O (aq ) + HSO 4 (aq )
−
– Taking the upper sign, we get
HSO 4 (aq ) + H 2O( l )
+
−
x = [H 3O ] = 9.4 × 10 4
43
– Now we can calculate the pH.
+
2−
H 3O (aq ) + SO 4 (aq )
– The first proton is lost completely followed by a weak
46
ionization of the hydrogen sulfate ion, HSO4-.
– For a weak diprotic acid like carbonic acid, H2CO3,
two simultaneous equilibria must be considered.
−
pH = − log(9.4 × 10 4 ) = 3.03
H 2CO3 (aq) + H 2O( l )
Check your answer by substituting the hydrogen
Ion concentration into the equilibrium expression and
Calculate Ka
−
HCO 3 (aq ) + H 2O( l )
−
+
H 3O (aq) + HCO 3 (aq)
+
2−
H 3O (aq ) + CO 3 (aq )
– Each equilibrium has an associated acid-ionization
constant.
– For the loss of the first proton
+
K a1 =
−
[ H 3O ][ HCO 3 ]
= 4.3 × 10 −7
[ H 2CO 3 ]
44
• Polyprotic Acids
• A polyprotic acid has more than one acidic
proton—for example, H2SO4, H2SO3,
H2CO3, H3PO4.
• These acids have successive ionization
reactions with Ka1, Ka2, . . .
47
– Each equilibrium has an associated acid-ionization
constant.
– For the loss of the second proton
+
K a2 =
2−
[ H 3O ][CO 3 ]
= 4.8 × 10 −11
−
[ HCO 3 ]
– In general, the second ionization constant, Ka2, for
a polyprotic acid is smaller than the first ionization
constant, Ka1.
• The next example illustrates how to do
calculations for a polyprotic acid.
– In the case of a triprotic acid, such as H3PO4, the
third ionization constant, Ka3, is smaller than the
second one, Ka2.
45
48
8
– Substituting into the equilibrium expression
x2
= 7.9 × 10−5
(0.10 − x)
– When several equilibria occur at once, it might
appear complicated to calculate equilibrium
compositions.
– Assuming that x is much smaller than 0.10, you get
−
x 2 ≅ (7.9 × 10 5 ) × (0.10)
– However, reasonable assumptions can be made
that simplify these calculations as we show in the
next example.
x ≅ 2.8 × 10
−3
= 0.0028
– The hydronium ion concentration is 0.0028 M, so
pH = − log(0.0028) = 2.55
52
49
A Problem To Consider
– The ascorbate ion, Asc2-, which we will call y, is
produced only in the second ionization of H2Asc.
• Ascorbic acid (vitamin C) is a diprotic acid,
H2C6H6O6. What is the pH of a 0.10 M solution?
What is the concentration of the ascorbate ion,
C6H6O62- ?
The acid ionization constants are Ka1 = 7.9 x 10-5
and Ka2 = 1.6 x 10-12.
−
+
−
H 3O (aq) + Asc 2 (aq)
HAsc (aq ) + H 2O(l )
– Assume the starting concentrations for HAsc- and
H3O+ to be those from the first equilibrium.
−
– For diprotic acids, Ka2 is so much smaller than Ka1
that the smaller amount of hydronium ion produced in
the second reaction can be neglected.
– The pH can be determined by simply solving the
equilibrium problem posed by the first ionization.
+
−
HAsc (aq ) + H 2O( l )
Starting
Change
Equilibrium
−
H 3O (aq) + Asc2 (aq)
HAsc (aq ) + H 2O(l )
0.0028
-y
0.0028-y
+
0.0028
+y
0.0028+y
50
– If we abbreviate the formula for ascorbic acid as
H2Asc, then the first ionization is:
+
Starting
Change
Equilibrium
−
H 3O (aq) + HAsc (aq)
0.10
-x
0.10-x
0
+x
x
+
[H 3O + ][Asc2 ] =
K a2
−
[HAsc ]
– Substituting into the equilibrium expression
(0.0028 + y )( y ) =
−
1.6 × 10 12
(0.0028 − y )
0
+x
x
– Assuming y is much smaller than 0.0028, the
equation simplifies to
– The equilibrium constant expression is
−
[H 3O ][HAsc ] =
K a1
[H 2 Asc]
53
−
−
H 3O (aq) + HAsc (aq)
H 2 Asc(aq ) + H 2O(l )
0
+y
y
– The equilibrium constant expression is
+
H 2 Asc(aq ) + H 2O(l )
−
H 3O (aq) + Asc 2 (aq)
51
(0.0028)(y )
= 1.6 × 10−12
(0.0028)
54
9
9.2 × 10 −4 (0.10 − x ) = x 2
– Hence,
−
y ≅ [ Asc2 ] = 1.6 × 10
9.2 × 10 −5 − 9.2 × 10 −4 x = x 2
−12
x 2 + 9.2 × 10 -4 x - 9.2 × 10 -5 = 0
– The concentration
– of
– the ascorbate ion equals Ka2.
b = 9.2 × 10 -4
a =1
x=
x=
x=
c = - 9.2 × 10 -5
− b ± b 2 − 4ac
2a
− (9.2 × 10 −4 ) ± (9.2 × 10 −4 )2 − 4(1)(-9.2 × 10 -5 )
2(1)
− (9.2 × 10 −4 ) ± (1.92 × 10 −2 )
= −4.6 × 10 − 4 ± 9.6 × 10 −3
2
55
• Tartaric acid, H2C4H4O6, is a diprotic acid
used in food products. What is the pH of a
0.10 M solution? What is the concentration
of the C4H4O62− ion in the same solution?
• Ka1 = 9.2 × 10−4; Ka2 = 4.3 × 10−5.
First, we will use the first acid-ionization equilibrium
to find [H+] and [HC4H4O6-]. In these calculations,
we will use the generic formula H2A for the acid.
Next, we will use the second acid-ionization
equilibrium to find [C4H4O62-].
H2A(aq)
+
0.10
Initial
Change
–x
H2O(l)
[H O ][HA ]
+
K a1 =
3
+x
+x
x
x
[H2 A ]
−
9.2 × 10 − 4 =
x2
0.10 - x
0.10
< 100
9.2 × 10 − 4
We cannot use the simplifyin g assumption .
x = 9.1× 10 -3
and x = −1.0 × 10 −2
We eliminate the negative value because it is
physically impossible to have a negative concentration.
At the end of the first acid ionization equilibrium, the
concentrations are
[H3O + ] = 9.1× 10 −3 M
[H2 A] = 9.0 × 10 −4 M
[HA − ] = 9.1 × 10 −3 M
Now we use these for the second acid ionization equilibrium.
56
H3O+(aq) + HA(aq)
0
0
Equilibrium 0.10 – x
58
59
HA-(aq) +
Initial
0.0091
Change
Equilibriu
m
H3O+(aq) + A2(aq)
0.0091
0
–x
+x
+x
0.0091 – x
0.0091 + x
x
[H O ][A ].
[HA ]
2−
+
Ka2 =
H2O(l)
3
-
4.3 × 10 −5 =
(0.0091 + x ) x
(0.0091 - x )
We can assume that
57
0.0091 + x ≈ 0.0091 and 0.0091 - x ≈ 0.0091
60
10
4.3 × 10 −5 =
0.0091 x
0.0091
x = 4.3 × 10 −5 M
Our assumption was valid.
2-
[C 4H4O6 ] = 4.3 × 10 −5 M
-
[HC 4H4O6 ] = 9.1 × 10 −3 M
[H3 O + ] = 9.1 × 10 −3 M
[H2C4H4O 6 ] = 9.0 × 10 −4 M
pH = −log (9.1× 10 −3 )
pH = 2.04
61
Base-Ionization Equilibria
• Writing Kb Reactions
• The bases in Table 16.2 are nitrogen
bases; that is, the proton they accept adds
to a nitrogen atom. Next we’ll practice
writing the Kb reactions.
• Equilibria involving weak bases are treated
similarly to those for weak acids.
– Ammonia, for example, ionizes in water as
follows.
+
NH3 (aq) + H2O(l)
−
NH4 (aq) + OH (aq)
– The corresponding equilibrium constant is:
+
64
−
[NH 4 ][OH ]
Kc =
[ NH 3 ][H 2O]
• Ammonium becomes ammonium ion:
NH3+ H2O NH4+ + OH• Ethylamine becomes ethyl ammonium ion:
C2H5NH2 + H2O C2H5NH3+ + OH-
62
65
– The concentration of water is nearly constant.
+
• Dimethylamine becomes
dimethylammonium ion:
(CH3)2NH2 + H2O (CH3)2NH3+ + OH-
−
[NH 4 ][OH ]
K b = [ H 2O]K c =
[NH 3 ]
– In general, a weak base B with the base ionization
B(aq ) + H 2O(l )
+
−
HB (aq ) + OH (aq )
• Pyridine becomes pyridinium ion:
C5H5N+ H2O C5H5NH+ + OH-
has a base ionization constant equal to
+
−
[HB ][OH ]
Kb=
[B ]
Table 17.2 lists
ionization
constants for
some weak bases.
63
• Hydrazine becomes hydrazinium ion:
N2H4+ H2O N2H5+ + OH66
11
Titration of a Strong Acid by a Strong Base
We can be given pH, initial concentration,
and Kb.
• Figure 16.12 shows a curve for the titration of HCl
with NaOH.
– At the equivalence point, the pH of the solution is
7.0 because it contains a salt, NaCl, that does
not hydrolyze.
From the pH, we can find first [H3O+] and
then [OH-]. Using what is given, we can
find the other quantities.
– However, the pH changes rapidly from a pH of
about 3 to a pH of about 11.
– To detect the equivalence point, you need an
acid-base indicator that changes color within the
pH range 3-11.
We can also use a simplifying assumption:
When [B]0 / Kb > 103, the expression
([B]0 – x) = [B]0.
– Phenolphthalein can be used because it changes
color in the pH range 8.2-10. (see Figure 15.10)
67
70
A Problem To Consider
• Calculate the pH of a solution in which 10.0 mL of
0.100 M NaOH is added to 25.0 mL of 0.100 M
HCl.
Go To 11C Slides
– Because the reactants are a strong acid and a
strong base, the reaction is essentially complete.
Slide 41
+
−
H 3O ( aq ) + OH ( aq ) → H 2 O ( l ) + H 2 O ( l )
– We get the amounts of reactants by multiplying the
volume of each (in liters) by their respective
molarities.
68
Figure 16.12: Curve for the titration of a
strong acid by a strong base.
71
+
Mol H 3O = 0.0250L × 0.100 mol L = 0.00250 mol
−
Mol OH = 0.0100L × 0.100 mol L = 0.00100 mol
– All of the OH- reacts, leaving an excess of H3O+
+
Excess H 3O = (0.00250 − 0.00100)mol
= 0.00150 mol H 3O +
– You obtain the H3O+ concentration by dividing the
mol H3O+ by the total volume of solution (=0.0250 L
+ 0.0100 L=0.0350 L)
+
[ H 3O ] =
69
0.00150 mol
= 0.0429 M
0.0350 L
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A Problem To Consider
• Calculate the pH of a solution in which 10.0
mL of 0.100 M NaOH is added to 25.0 mL
of 0.100 M HCl.
• Calculate the pH of the solution at the
equivalence point when 25.0 mL of 0.10 M
acetic acid is titrated with 0.10 M sodium
hydroxide. The Ka for acetic acid is 1.7 x 10-5.
– Hence,
pH = − log( 0.0429) = 1.368
– At the equivalence point, equal molar
amounts of acetic acid and sodium
hydroxide react to give sodium acetate.
– First, calculate the concentration of the acetate
ion.
73
– In this case, 25.0 mL of 0.10 M NaOH is needed to
react with 25.0 mL of 0.10 M acetic acid.
76
– The molar amount of acetate ion formed equals
the initial molar amount of acetic acid.
Titration of a Weak Acid by a Strong Base
25 × 10
• The titration of a weak acid by a strong base gives
a somewhat different curve.
−3
L soln ×
0.10 mol acetate ion = 2.5 × 10 -3 mol acetat e ion
1 L soln
– The total volume of the solution is 50.0 mL.
Hence,
– The pH range of these titrations is shorter.
– The equivalence point will be on the basic side
since the salt produced contains the anion of a
weak acid.
– Figure 17.13 shows the curve for the titration of
nicotinic acid with NaOH.
molar concentration =
2.5 × 10-3 mol =
0.050 M
−
50 × 10 3 L
– The hydrolysis of the acetate ion follows the method
given in an earlier section of this chapter.
74
Figure 17.13: Curve for the titration of a weak
acid by a strong base.
– You find the Kb for the acetate ion to be 5.9 x 10-10 and
that the concentration of the hydroxide ion is 5.4 x 10-6.
77
The pH is 8.73
Titration of a Strong Acid by a Weak Base
• The titration of a weak base with a strong
acid is a reflection of our previous example.
– Figure 17.14 shows the titration of NH3 with HCl.
– In this case, the pH declines slowly at first, then
falls abruptly from about pH 7 to pH 3.
– Methyl red, which changes color from yellow at pH
6 to red at pH 4.8, is a possible indicator.
75
78
13
Operational Skills
What would the titration of a weak acid with a
weak base appear?
• Calculating the common-ion effect on acid
ionization
• Calculating the pH of a buffer from given volumes
of solution
• Calculating the pH of a solution of a strong acid
and a strong base
• Calculating the pH at the equivalence point in the
titration of a weak cid with a strong base
What would the titration of a weak base with a
weak acid appear?
79
82
Figure 16.14: Curve for the titration of a weak
base by a strong acid.
80
Operational Skills
• Determining Ka (or Kb) from the solution pH
• Calculating the concentration of a species in a
weak acid solution using Ka
• Calculating the concentration of a species in a
weak base solution using Kb
• Predicting whether a salt solution is acidic, basic,
or neutral
• Obtaining Ka from Kb or Kb from Ka
• Calculating concentrations of species in a salt
solution
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14