1
Acid-Base Equilibria
1
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
Contents and Concepts
2
Learning Objectives
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
2
1. Acid-Ionization Equilibria (cont)
a. State the assumption that allows for using approximations when solving problems.
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.
b. Calculate concentrations of species in a solution of
a diprotic acid. 4
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
4. Acid–Base Properties of Salt Solutions
– a. Write the hydrolysis reaction of an ion
to form an acidic solution.
– b. Write the hydrolysis reaction of an ion
to form a basic solution.
– c. Predict whether a salt solution is acidic, basic, or neutral.
– d. Obtain Ka from Kb or Kb from Ka.
– e. Calculating concentrations of species in a salt solution.
6
3
Solutions of a Weak Acid or Base with
Another Solute
5. Common-Ion Effect
– a. Explain the common-ion effect.
– b. Calculate the common-ion effect on acid ionization (effect of a strong acid).
– c. Calculate the common-ion effect on
acid ionization (effect of a conjugate base).
7
6. Buffers
– a. Define buffer and buffer capacity.
– b. Describe the pH change of a buffer
solution with the addition of acid or base.
– c. Calculate the pH of a buffer from given
volumes of solution.
– d. Calculate the pH of a buffer when a strong
acid or a strong base is added.
– e. Learn the Henderson–Hasselbalch equation.
– f. State when the Henderson–Hasselbalch equation can be applied
8
7. Acid–Base Titration Curves
– a. Define equivalence point.
– b. Describe the curve for the titration of a
strong acid by a strong base.
– c. Calculate the pH of a solution of a
strong acid and a strong base.
– d. Describe the curve for the titration of a
weak acid by a strong base.
– e. Calculate the pH at the equivalence
point in the titration of a weak acid by a strong base.
9
4
7. Acid–Base Titration Curves (cont)
– f. Describe the curve for the titration of a weak base by a strong acid.
– g. Calculate the pH of a solution at several points of a titration of a weak base by a strong acid.
10
Solutions of a Weak Acid or Base
• 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
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
5
• Acetic acid is a weak acid. It reacts with
water as follows:
• HC2H3O2(aq) + H2O(l) H3O+(aq) + C2H3O2
-(aq)
13
• The equilibrium constant for the reaction of
a weak acid with water is called the acid-ionization constant (or acid-dissociation
constant), Ka.
[ ][ ][ ]HA
A OH3a
−+
=K
• Liquid water is not included in the
equilibrium constant expression.
WHY? 14
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.
– Because acetic acid is a weak electrolyte, it ionizes to a small extent in water.
– When acetic acid is added to water it reacts as follows.
(aq)OHC(aq)OH 2323
−−−−++++++++)l(OH)aq(OHHC 2232
++++
15
6
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).
– Consider the generic monoprotic acid, HA.
)aq(A)aq(OH)l(OH)aq(HA 32
−−−−++++ ++++++++
– The corresponding equilibrium expression is:
]OH][HA[
]A][OH[K
2
3c
−−−−++++
====
16
– Since the concentration of water remains relatively constant, we rearrange the equation to
get:
]HA[
]A][OH[K]OH[K 3
c2a
−−−−++++
========
– Thus, Ka , the acid-ionization constant, equals the constant [H2O]Kc.
]HA[
]A][OH[K 3
a
−−−−++++
====
– Table 17.1 lists acid-ionization constants for various weak acids.
17
18
7
HCN
Highest pH
Lowest pH
HF
HNO2
HC2H3O2
HCN
HF
HNO2
HC2H3O2
1.7 × 10-5
4.5 × 10-4
Ka
6.8 × 10-4
4.9 × 10-10
In order of decreasing Ka:
19
B-8
Experimental Determination of Ka
• 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.
– With weak acids, the pH can be used to determine the equilibrium composition of ions in the solution.
20
Calculations with Ka
• Given the value of Ka and the initial concentration of HA, you can calculate the equilibrium concentration of all species.
• Given the value of Ka and the initial
concentration of HA, you can calculate the degree of ionization and the percent ionization.
• Given the pH of the final solution and the initial
concentration of HA, you can find the value
of Ka and the percent ionization.21
8
We can be given pH, percent or degree ionization, initial concentration, and Ka.
From pH, we can find [H3O+].
From percent or degree ionization, we can find Ka.
Using what is given, we can find the other quantities.
22
A Problem To Consider
• 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 acid-ionization constant for this acid at 25°C and the percent 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.
– We will abbreviate the formula for nicotinic acid as HNic.
23
– Let x be the moles per liter of product formed.
)aq(Nic)aq(OH)l(OH)aq(HNic 32
−−−−++++++++++++
Starting 0.012 0 0Change -x +x +x
Equilibrium 0.012-x x x
– The equilibrium-constant expression is:
]HNic[
]Nic][OH[K 3
a
−−−−++++
====
– Substituting the expressions for the equilibrium concentrations, we get
)x012.0(
xK
2
a −−−−====
24
9
– We can obtain the value of x from the given pH.
)pHlog(anti]OH[x 3−−−−========
++++
)39.3log(antix −−−−====
00041.0101.4x4 ====××××====
−−−−
– Substitute this value of x in our equilibrium expression.
– Note first, however, that
012.001159.0)00041.0012.0()x012.0( ≅≅≅≅====−−−−====−−−−
the concentration of unionized acid remains virtually unchanged.
25
– Substitute this value of x in our equilibrium expression.
522
a 104.1)012.0(
)00041.0(
)x012.0(
xK
−−−−××××====≅≅≅≅−−−−
====
– To obtain the degree of dissociation:
034.00.012
0.00041ondissociatiofDegree ========
– The percent ionization is obtained by multiplying by 100, which gives 3.4%.
26
• 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.
• a. What is the acid-ionization constant, Ka, for phenol at 25°C?
• b. What is the degree of ionization?
Another Problem
27
10
HC6H5O(aq) + H2O(l) H3O+(aq) + C6H5O
-(aq)
Initial 0.10 0 0
Change –x +x +x
Equilibrium 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.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
[[[[ ]]]][[[[ ]]]][[[[ ]]]]OHHC
OHC OH
56
563a
−−−−++++
====K
0.10
)10 (3.7 26
a
−×
=K
-10
a 10 1.4 ×=K 29
5
6
10 3.7ionization of Degree
0.10
10 3.7
0.100 ionization of Degree
−
−
×=
×==
x
• The degree of ionization is the ratio of
ionized concentration to original concentration:
• Percent ionization is the degree of ionization
× 100%:
• Percent ionization = 3.7 x 10-3% or 0.0037%
30
11
•Simplifying Assumption for Acid and Base Ionizations
•The equilibrium concentration of the acid is most often ([HA]0 – x).
•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
•This is a valid assumption when the ratio of [HA]0to Ka is > 103. If it is not valid, you must use the quadratic equation to solve the problem.
31
•Para-hydroxybenzoic acid is used to make certain dyes. What are the concentrations of
this acid, of hydronium ion, and of para-hydroxybenzoate 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.
We will use the generic formula HA for para-hydroxybenzoic acid and the following equilibrium:
HA + H2O H3O+ + A-
32
HA(aq) + H2O(l) H3O+(aq) + A-(aq)
Initial 0.200 0 0
Change –x +x +x
Equilibrium 0.200 – x x x
[ ][ ][ ]
.HA
A OH3a
−+
=K
x
x
-0.20010 2.6
25
=×−
0.200 - 0.200 so 1000, 10 2.6
0.2005
≈>×
−x
33
12
)10 (2.3 pH ]O[H log pH -33 -- ×==
+
2610 5.2 x=×−
0.200
2
a
xK =
]O[H 10 2.3 3
3 +−=×= Mx
2.64 pH =
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.
35
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.
• The degree of ionization of a weak acid depends on both the Ka and the concentration of the acid solution (see Figure 16.3).
36
13
Figure 16.3: Variation of
Percent Ionization of a Weak Acid with Concentration
37
Calculations With Ka
• How do you know when you can use this simplifying assumption?
– then this simplifying assumption of ignoring the subtracted x gives an acceptable error of less than 5%.
– It can be shown that if the acid concentration, Ca, divided by the Ka exceeds 100, that is,
100K
Cif
a
a >>>>
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
14
– Note that
11103.3
0036.0K
C4
a
a ====××××
==== −−−−
which is less than 100, so we must solve the equilibrium equation exactly.
Abbreviate the formula for acetylsalicylic acid asHAcs 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
– These data are summarized below.
)aq(Acs)aq(OH)l(OH)aq(HAcs 32
−−−−++++++++++++
Starting 0.0036 0 0Change -x +x +x
Equilibrium 0.0036-x x x
– The equilibrium constant expression is
a3 K
]HAcs[
]Acs][OH[====
−−−−++++
– If we substitute the equilibrium concentrations and the Ka into the equilibrium constant expression, we get 41
a3 K
]HAcs[
]Acs][OH[====
−−−−++++
42
103.3)x0036.0(
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
0)102.1(x)103.3(x642 ====××××−−−−××××++++
−−−−−−−−
42
15
– Now substitute into the quadratic formula.
a2
ac4bbx
2 −−−−±±±±−−−−====
– Now substitute into the quadratic formula.
2
)102.1(4)103.3()103.3(x
6244 −−−−−−−−−−−− ××××−−−−××××±±±±××××−−−−====
– The lower sign in ± gives a negative root which we can ignore
– Taking the upper sign, we get
43 104.9]OH[x
−−−−++++××××========
43
– Now we can calculate the pH.
03.3)104.9log(pH4 ====××××−−−−====
−−−−
Check your answer by substituting the hydrogenIon concentration into the equilibrium expression andCalculate Ka
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, . . .
• The next example illustrates how to do
calculations for a polyprotic acid.
45
16
Polyprotic Acids
• Some acids have two or more protons (hydrogen ions) to donate in aqueous
solution. These are referred to as polyprotic acids.
– The first proton is lost completely followed by a weak ionization of the hydrogen sulfate ion, HSO4
-.
– Sulfuric acid, for example, can lose two protons in aqueous solution.
)aq(HSO)aq(OH)l(OH)aq(SOH 43242
−−−−++++ ++++→→→→++++
)aq(SO)aq(OH)l(OH)aq(HSO2
4324
−−−−++++−−−−++++++++
46
– For a weak diprotic acid like carbonic acid, H2CO3, two simultaneous equilibria must be considered.
)aq(HCO)aq(OH)l(OH)aq(COH 33232
−−−−++++ ++++++++
)aq(CO)aq(OH)l(OH)aq(HCO2
3323
−−−−++++−−−−++++++++
– Each equilibrium has an associated acid-ionization constant.
– For the loss of the first proton
7
32
331a 103.4
]COH[
]HCO][OH[K
−−−−
−−−−++++
××××========
47
– Each equilibrium has an associated acid-ionization constant.
– For the loss of the second proton
11
3
2
332a 108.4
]HCO[
]CO][OH[K
−−−−
−−−−
−−−−++++
××××========
– In general, the second ionization constant, Ka2, for a polyprotic acid is smaller than the first ionization constant, Ka1.
– In the case of a triprotic acid, such as H3PO4, the third ionization constant, Ka3, is smaller than the second one, Ka2.
48
17
– When several equilibria occur at once, it might appear complicated to calculate equilibrium compositions.
– However, reasonable assumptions can be made that simplify these calculations as we show in the next example.
49
– 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.
A Problem To Consider
• 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.
– The pH can be determined by simply solving the equilibrium problem posed by the first ionization.
50
– If we abbreviate the formula for ascorbic acid as H2Asc, then the first ionization is:
(aq)AscH(aq)OH)l(OH)aq(AscH 322
−−−−++++ ++++++++
Starting 0.10 0 0Change -x +x +x
Equilibrium 0.10-x x x
(aq)AscH(aq)OH)l(OH)aq(AscH 322
−−−−++++ ++++++++
– The equilibrium constant expression is
1a
2
3 K]AscH[
]HAsc][OH[====
−−−−++++
51
18
– Substituting into the equilibrium expression
52
109.7)x10.0(
x −−−−××××====−−−−
– Assuming that x is much smaller than 0.10, you get
0028.0108.2x
)10.0()109.7(x
3
52
====××××≅≅≅≅
××××××××≅≅≅≅
−−−−
−−−−
– The hydronium ion concentration is 0.0028 M, so
55.2)0028.0log(pH ====−−−−====
52
– The ascorbate ion, Asc2-, which we will call y, is produced only in the second ionization of H2Asc.
(aq)Asc(aq)OH)l(OH)aq(HAsc2
32
−−−−++++−−−− ++++++++
– Assume the starting concentrations for HAsc- and H3O
+ to be those from the first equilibrium.
(aq)Asc(aq)OH)l(OH)aq(HAsc2
32
−−−−++++−−−− ++++++++
(aq)Asc(aq)OH)l(OH)aq(HAsc2
32
−−−−++++−−−− ++++++++
Starting 0.0028 0.0028 0Change -y +y +y
Equilibrium 0.0028-y 0.0028+y y
53
– The equilibrium constant expression is
2a
23 K
]HAsc[
]Asc][OH[====
−−−−
−−−−
++++
– Substituting into the equilibrium expression
12106.1
)y0028.0(
)y)(y0028.0( −−−−××××====−−−−
++++
– Assuming y is much smaller than 0.0028, the equation simplifies to
12106.1
)0028.0(
)y)(0028.0( −−−−××××====
54
19
– Hence,
122106.1]Asc[y
−−−−−−−− ××××====≅≅≅≅
– The concentration
– of
– the ascorbate ion equals Ka2.
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 C4H4O6
2− 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.
56
Next, we will use the second acid-ionization equilibrium to find [C4H4O6
2-].
H2A(aq) +
H2O(l) H3O+(aq) + HA-
(aq)
Initial 0.10 0 0
Change –x +x +x
Equilibrium 0.10 – x x x
[ ][ ][ ]AH
HA OH
2
31a
−+
=Kx
x
-0.1010 9.2
24
=×−
.assumption gsimplifyin the use cannot We
100 10 9.2
0.104
<×
−
57
20
-5-4 10 9.2- 10 9.21 ×=×== cba
a
acbbx
2
42−±−
=
10 9.210 9.2 245 xx =×−×−−
24 )(0.1010 9.2 xx =−×−
0 10 9.2 - 10 9.2 -5-42=××+ xx
2(1)
)10 4(1)(-9.2)10 (9.2)10 (9.2 -5244×−×±×−
=
−−
x
3424
10 9.6 10 4.62
)10 (1.92)10 (9.2 −−
−−
×±×−=×±×−
=x
58
2-3 10 1.0and10 9.1 −×−=×= xx
ion.concentrat negative a have to impossible physically
is it because value negative the eliminate We
M 10 9.1]O[H 3
3
−+×=
M 10 9.1][HA 3−−×=
m.equilibriu ionization
-acid second the for these use Now we
M 10 9.0A][H 4
2
−×=
are ionsconcentrat
the m,equilibriu ionization acid first the of end the At
59
HA-(aq) + H2O(l) H3O+(aq) + A2-
(aq)
Initial 0.0091 0.0091 0
Change –x +x +x
Equilibrium
0.0091 – x 0.0091 + x x
[ ][ ][ ]
.HA
A OH-
23
2a
−+
=K)- (0.0091
) (0.009110 4.3 5
x
xx+=×
−
0.0091 - 0.0091 and 0.0091 0.0091
that assume can We
≈≈+ xx60
21
0.0091
0.009110 4.3 5 x
=×−
valid. asw assumption urO
Mx 10 4.3 5−×=
M 10 4.3 ]OH[C 5-2
644−
×=
M 10 9.1 ]OH[HC 3-
644−
×=
M 10 9.1 ]O[H 33
−+×=
M 10 9.0 ]OHC[H 46442
−×=
)10 (9.1 logpH 3−×−= 2.04pH =
61
Base-Ionization Equilibria
• Equilibria involving weak bases are treated
similarly to those for weak acids.
– Ammonia, for example, ionizes in water as follows.
)aq(OH)aq(NH)l(OH)aq(NH 423
−−−−++++++++++++
– The corresponding equilibrium constant is:
]OH][NH[
]OH][NH[K
23
4c
−−−−++++
====
62
– The concentration of water is nearly constant.
]NH[
]OH][NH[K]OH[K
3
4c2b
−−−−++++
========
– In general, a weak base B with the base ionization
)aq(OH)aq(HB)l(OH)aq(B 2
−−−−++++ ++++++++
has a base ionization constant equal to
]B[
]OH][HB[Kb
−−−−++++
====Table 17.2 lists
ionization constants for
some weak bases.63
22
64
• 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.
• Ammonium becomes ammonium ion:
NH3+ H2O � NH4+ + OH-
• Ethylamine becomes ethyl ammonium ion:
C2H5NH2 + H2O � C2H5NH3+ + OH-
65
• Dimethylamine becomes
dimethylammonium ion:
(CH3)2NH2 + H2O � (CH3)2NH3+ + OH-
• Pyridine becomes pyridinium ion:
C5H5N+ H2O � C5H5NH+ + OH-
• Hydrazine becomes hydrazinium ion:
N2H4+ H2O � N2H5+ + OH-
66
23
We can be given pH, initial concentration,
and Kb.
From the pH, we can find first [H3O+] and
then [OH-]. Using what is given, we can
find the other quantities.
We can also use a simplifying assumption: When [B]0 / Kb > 103, the expression
([B]0 – x) = [B]0.
67
68
Go To 11C Slides
Slide 41
Figure 16.12: Curve for the titration of a
strong acid by a strong base.
69
24
• 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.
– However, the pH changes rapidly from a pH of about 3 to a pH of about 11.
Titration of a Strong Acid by a Strong Base
– To detect the equivalence point, you need an acid-base indicator that changes color within the pH range 3-11.
– Phenolphthalein can be used because it changes color in the pH range 8.2-10. (see Figure 15.10)
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.
– Because the reactants are a strong acid and a strong base, the reaction is essentially complete.
)l(OH)l(OH)aq(OH)aq(OH 223++++→→→→++++
−−−−++++
– We get the amounts of reactants by multiplying the volume of each (in liters) by their respective molarities.
71
mol00250.0100.0L0250.0OHMol Lmol
3====××××====
++++
mol00100.0100.0L0100.0OHMol Lmol ====××××====
−−−−
– All of the OH- reacts, leaving an excess of H3O+
l0.00100)mo00250.0(OHExcess 3−−−−====
++++
++++==== OHmol0.00150 3
– 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)
M0429.0L0.0350
mol00150.0]OH[ 3
========++++
72
25
• 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.
– Hence,
368.1)0429.0log(pH ====−−−−====
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• The titration of a weak acid by a strong base gives
a somewhat different curve.
– The pH range of these titrations is shorter.
– The equivalence point will be on the basic sidesince the salt produced contains the anion of a weak acid.
– Figure 17.13 shows the curve for the titration of nicotinic acid with NaOH.
Titration of a Weak Acid by a Strong Base
74
Figure 17.13: Curve for the titration of a weak
acid by a strong base.
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26
A Problem To Consider
• 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.
– 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.
– 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.
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– The molar amount of acetate ion formed equals the initial molar amount of acetic acid.
solnL1
e ionmol acetat0.10solnL1025
3 ××××××××−−−− e ionmol acetat102.5
3-××××====
– The total volume of the solution is 50.0 mL. Hence,
M050.0L1050
mol102.5ionconcentratmolar
3
3-
====××××
××××====
−−−−
– The hydrolysis of the acetate ion follows the method given in an earlier section of this chapter.
– 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. The pH is 8.73 77
• 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.
Titration of a Strong Acid by a Weak Base
78
27
What would the titration of a weak acid with a
weak base appear?
What would the titration of a weak base with a
weak acid appear?
79
Figure 16.14: Curve for the titration of a weak
base by a strong acid.
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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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28
Operational Skills
• 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
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