9
Monoprotic Acid-Base Equilibria
MEASURING pH INSIDE CELLULAR COMPARTMENTS
(a) Mouse macrophage engulfs
two foreign red blood cells as
phagocytosis begins. [From J. P.
Revel, in B. Alberts, D. Bray, J. Lewis,
M. Raff, K. Roberts, and J. D. Watson,
Molecular Biology of the Cell, 2nd ed.
(New York: Garland Publishing, 1989.]
(b) Two macrophages with
ingested 1.6-␮m-diameter
fluorescent beads. (c) Fluorescence
image of panel b. [From K. P.
McNamara, T. Nguyen, G. Dumitrascu,
J. Ji, N. Rosenzweig, and Z. Rosenzweig,
“Synthesis, Characterization, and
Application of Fluorescence Sensing
Lipobeads for Intracellular pH
Measurements,” Anal. Chem. 2001,
73, 3240.]
(a)
(d) Fluorescence spectra of
lipobeads in solutions at pH 5–8.
(e) pH change during
phagocytosis of a single bead by
a macrophage. [From McNamara
et al., ibid.]
(c )
Macrophages are white blood cells that fight infection by ingesting and dissolving foreign
cells—a process called phagocytosis. The compartment containing the ingested foreign cell
merges with compartments called lysosomes, which contain digestive enzymes that are
most active in acid. Low enzyme activity above pH 7 protects the cell from enzymes that
leak into the cell.
One way to measure pH inside the compartment containing the ingested particle and
digestive enzymes is to present macrophages with polystyrene beads coated with a lipid
membrane to which fluorescent (light-emitting) dyes are covalently bound. Panel d shows
that fluorescence intensity from the dye fluorescein depends on pH, but fluorescence from
tetramethylrhodamine does not. The ratio of emission from the dyes is a measure of pH.
Panel e shows the fluorescence intensity ratio changing in 3 s as the bead is ingested and
the pH around the bead drops from 7.3 to 5.7 to allow digestion to commence.
Fluorescein
emission
pH 7.3
Fluorescence intensity ratio
(fluorescein/rhodamine)
Fluorescence intensity
pH 8
(b)
pH 7
Tetramethylrhodamine
emission
pH 6
pH 5
pH 5.7
500
(d )
158
520
540
560
Wavelength (nm)
580
600
0
5
10
Time (s)
15
20
(e)
CHAPTER 9 Monoprotic Acid-Base Equilibria
A
cids and bases are essential to virtually every application of chemistry and for the intelligent use of analytical procedures such as chromatography and electrophoresis. It would be
difficult to have a meaningful discussion of, say, protein purification or the weathering of
rocks without understanding acids and bases. This chapter covers acid-base equilibria and
buffers. Chapter 10 treats polyprotic systems involving two or more acidic protons. Nearly
every biological macromolecule is polyprotic. Chapter 11 describes acid-base titrations. Now
is the time to review fundamentals of acids and bases in Sections 6-5 through 6-7.
9-1 Strong Acids and Bases
What could be easier than calculating the pH of 0.10 M HBr? HBr is a strong acid, so the
reaction
HBr H2O S H3O Br
Table 6-2 gave a list of strong acids and bases
that you must memorize.
goes to completion, and the concentration of H3O is 0.10 M. It will be our custom to write
H instead of H3O, and we can say
Equilibrium constants for the reaction1
pH log[H]
HX H2O T H3O X
log (0.10) 1.00
HCl
HBr
HI
HNO3
Example Activity Coefficient in a Strong-Acid Calculation
Calculate the pH of 0.10 M HBr, using activity coefficients.
Ka 103.9
Ka 105.8
Ka 1010.4
Ka 101.4
HNO3 is discussed in Box 9-1.
Solution
The ionic strength of 0.10 M HBr is 0.10 M, at which the activity coefficient
of H is 0.83 (Table 8-1). Remember that pH is logA H, not log[H]:
pH log[H]H log(0.10)(0.83) 1.08
Box 9-1 Concentrated HNO3 Is Only Slightly Dissociated 2
Strong acids in dilute solution are essentially completely dissociated. As concentration increases, the degree of dissociation
decreases. The figure shows a Raman spectrum of solutions of
nitric acid of increasing concentration. The spectrum measures
scattering of light whose energy corresponds to vibrational energies of molecules. The sharp signal at 1 049 cm1 in the spectrum
of 5.1 M NaNO3 is characteristic of free NO3 anion.
*
A 10.0 M HNO3 solution has a strong signal at 1 049 cm1,
arising from NO3 from dissociated acid. Bands denoted by asterisks arise from undissociated HNO3. As concentration increases,
the 1 049 cm1 signal disappears and signals attributed to undissociated HNO3 increase. The graph shows the fraction of dissociation deduced from spectroscopic measurements. It is instructive to
realize that, in 20 M HNO3, there are fewer H2O molecules than
there are molecules of HNO3. Dissociation decreases because
there is not enough solvent to stabilize the free ions.
98.6 wt % HNO3
*
23.4 M HNO3
*
*
*
1.0
*
19.7 M HNO3
15.7 M HNO3
*
*
*
1
1
2
0.6
0.4
Raman
NMR
0.2
0.0
10.0 M HNO3
H 2O
=3
HNO3
0.8
ractio
F
no
fd
isso
ciatio
n,α
Raman scattering
21.8 M HNO3
48.3 wt % HNO3
0
5
10
15
20
rmal co
o
F
ncentratio
nM
( )
* *
5.1 M NaNO3
Temperature (C)
1 360 cm–1
1 049 cm–1
720 cm–1
Raman spectrum of aqueous HNO3 at 25°C. Signals at 1 360,1 049, and
720 cm1 arise from NO3 anion. Signals denoted by asterisks are from
undissociated HNO3. The wavenumber unit, cm1, is 1/wavelength.
9-1 Strong Acids and Bases
0
25
50
Acid dissociation constant (Ka)
46.8
26.8
14.9
159
Now that we have reminded you of activity coefficients, you can breathe a sigh of relief
because we will neglect activity coefficients unless there is a specific point to be made.
How do we calculate the pH of 0.10 M KOH? KOH is a strong base (completely dissociated), so [OH] 0.10 M. Using Kw [H][OH], we write
[H] From [OH], you can always find [H]:
[H] Kw
1.0 1014
1.0
[OH ]
0.10
Kw
1013 M
pH log[H] 13.00
[OH]
Finding the pH of other concentrations of KOH is pretty trivial:
[OH] (M)
103.00
104.00
105.00
[H] (M)
pH
1011.00
1010.00
109.00
11.00
10.00
9.00
A generally useful relation is
pH pOH pKW 14.00 at 25°C
The temperature dependence of Kw was given
in Table 6-1.
Relation between pH
and pOH:
pH pOH logKw 14.00 at 25°C
(9-1)
The Dilemma
Well, life seems simple enough so far. Now we ask, “What is the pH of 1.0
Applying our usual reasoning, we calculate
Adding base to water cannot lower the pH.
(Lower pH is more acidic.) There must be
something wrong.
[H] Kw /(1.0
108 ) 1.0
108 M KOH?”
106 M 1 pH 6.00
But how can the base KOH produce an acidic solution (pH 6 7) when dissolved in pure
water? It’s impossible.
The Cure
Clearly, there is something wrong with our calculation. In particular, we have not considered
the contribution of OH from the ionization of water. In pure water, [OH] 1.0 107 M,
which is greater than the amount of KOH added to the solution. To handle this problem, we
resort to the systematic treatment of equilibrium.
Kw
Step 1 Pertinent reactions. The only one is H2O ∆ H OH.
Step 2 Charge balance. The species in solution are K, OH, and H. So,
[K] [H] [OH]
If we had been using activities, step 4 is the
only point at which activity coefficients would
have entered.
(9-2)
Step 3 Mass balance. All
comes from the KOH, so
1.0
M.
Step 4 Equilibrium constant expression. Kw [H][OH] 1.0 1014.
Step 5 Count. There are three equations and three unknowns ([H], [OH], [K]) , so we
have enough information to solve the problem.
Step 6 Solve. Because we are seeking the pH, let’s set [H] x. Writing [K] 1.0
108 M in Equation 9-2, we get
K
[K]
[OH] [K] [H] 1.0
108
108 x
Using this expression for [OH] in the Kw equilibrium enables us to solve the problem:
[H][OH] Kw
(x)(1.0
x 2 (1.0
Solution of a quadratic equation:
ax 2 bx c 0
x
b
2b 2 4ac
2a
Retain all digits in your calculator because b 2
is sometimes nearly equal to 4ac. If you round
off before computing b 2 4ac , your answer
may be garbage.
x
1.0
108
9.6
108 x) 1.0
108 )x (1.0
2 (1.0
1014
1014 ) 0
108 ) 2 4(1) (1.0
2(1)
108 M, or 1.1
1014 )
107 M
Rejecting the negative concentration, we conclude that
[H] 9.6
108 M 1 pH log[H] 7.02
This pH is eminently reasonable, because 108 M KOH should be very slightly basic.
160
CHAPTER 9 Monoprotic Acid-Base Equilibria
Figure 9-1 shows the pH calculated for different concentrations of strong base or strong
acid in water. There are three regions:
Region 1 is the only practical case. Unless you were to protect 107 M KOH from the air, the
pH would be overwhelmingly governed by dissolved CO2, not KOH. To obtain a pH near 7,
we use a buffer, not a strong acid or base.
Systematic
treatment
required
here
11
10
9
KOH
8
pH
1. When the concentration is “high” (106 M) , pH is calculated by just considering the
added H or OH. That is, the pH of 105.00 M KOH is 9.00.
2. When the concentration is “low” (108 M) , the pH is 7.00. We have not added
enough acid or base to change the pH of the water itself.
3. At intermediate concentrations of 106 to 108 M, the effects of water ionization and
the added acid or base are comparable. Only in this region is a systematic equilibrium
calculation necessary.
12
7
6
5
HBr
4
3
2
–2 –3 –4 –5 –6 –7 –8 –9 –10
log (concentration)
Water Almost Never Produces 107 M H and 107 M OH
The misconception that dissociation of water always produces 107 M H and 107 M OH
is true only in pure water with no added acid or base. Any acid or base suppresses water ionization, as predicted by Le Châtelier’s principle. In 104 M HBr, for example, the pH is 4. The
concentration of OH is [OH] Kw /[H] 1010 M. But the only source of [OH] is the
dissociation of water. If water produces only 1010 M OH, it must also produce only
1010 M H because it makes one H for every OH. In 104 M HBr solution, water dissociation produces only 1010 M OH and 1010 M H.
Figure 9-1
Calculated pH as a function of
concentration of strong acid or strong base in
water.
Question What concentrations of H and
OH are produced by H2O dissociation in
0.01 M NaOH?
9-2 Weak Acids and Bases
Let’s review the meaning of the acid dissociation constant, Ka, for the acid HA:
Weak-acid
equilibrium:
Ka
HA ∆
H
A
[H][A]
Ka [HA]
(9-3)
A weak acid is one that is not completely dissociated. That is, Reaction 9-3 does not go to
completion. For a base, B, the base hydrolysis constant, Kb, is defined by the reaction
Weak-base
equilibrium:
Kb
B H2O ∆ BH OH
Kb [BH][OH]
[B]
Of course, you know that Ka should really
be expressed in terms of activities, not
concentrations: Ka A HA A /A HA.
Hydrolysis refers to a reaction with water.
(9-4)
A weak base is one for which Reaction 9-4 does not go to completion.
pK is the negative logarithm of an equilibrium constant:
pKw log Kw
pKa log Ka
pKb log Kb
As K increases, pK decreases, and vice versa. Comparing formic and benzoic acids, we see
that formic acid is stronger, with a larger Ka and smaller pKa, than benzoic acid.
O
O
¡
¡
HCOH T H HCO
Formic acid
O
¡
ß COH T
Benzoic acid
Ka 1.80 104
pKa 3.744
Formate
O
¡
ß CO As Ka increases, pKa decreases. The smaller
pKa is, the stronger the acid is.
Ka 6.28 105
pKa 4.202
Benzoate
The acid HA and its corresponding base, A, are said to be a conjugate acid-base pair,
because they are related by the gain or loss of a proton. Similarly, B and BH are a conjugate
pair. The important relation between Ka and Kb for a conjugate acid-base pair, derived in
Equation 6-35, is
Relation between Ka and
Kb for conjugate pair:
9-2
Weak Acids and Bases
K a Kb K w
HA and A are a conjugate acid-base pair.
B and BH also are conjugate.
(9-5)
161
Weak Is Conjugate to Weak
The conjugate base of a weak acid is a weak
base. The conjugate acid of a weak base is a
weak acid. Weak is conjugate to weak.
The conjugate base of a weak acid is a weak base. The conjugate acid of a weak base is a
weak acid. Consider a weak acid, HA, with Ka 104. The conjugate base, A, has
Kb Kw /Ka 1010. That is, if HA is a weak acid, A is a weak base. If Ka were 105, then
Kb would be 109. As HA becomes a weaker acid, A becomes a stronger base (but never a
strong base). Conversely, the greater the acid strength of HA, the less the base strength of
A. However, if either A or HA is weak, so is its conjugate. If HA is strong (such as HCl),
its conjugate base (Cl ) is so weak that it is not a base at all in water.
Using Appendix G
Appendix G lists acid dissociation constants. Each compound is shown in its fully protonated
form. Diethylamine, for example, is shown as (CH3CH2 ) 2NH2, which is really the diethylammonium ion. The value of Ka (1.0 1011 ) given for diethylamine is actually Ka for the
diethylammonium ion. To find Kb for diethylamine, we write Kb Kw /Ka 1.0 1014 /
1.0 1011 1.0 103.
For polyprotic acids and bases, several Ka values are given. Pyridoxal phosphate is given
in its fully protonated form as follows:3
˚
Pyridoxyl phosphate is a derivative of
vitamin B6.
˚
˚
O
O £ CH
¡
˚
OH
HO ® POCH2
˚
HO
N CH3
H
pKa
Ka
1.4 (POH)
3.44 (OH)
6.01 (POH)
8.45 (NH)
3.6
9.8
3.5
0.04
104
107
109
pK1 (1.4) is for dissociation of one of the phosphate protons, and pK2 (3.44) is for the
hydroxyl proton. The third most acidic proton is the other phosphate proton, for which
pK3 6.01, and the NH group is the least acidic (pK4 8.45) .
Species drawn in Appendix G are fully protonated. If a structure in Appendix G has a
charge other than 0, it is not the structure that belongs with the name in the appendix.
Names refer to neutral molecules. The neutral molecule pyridoxal phosphate is not the
species drawn above, which has a 1 charge. The neutral molecule pyridoxal phosphate
is
O
å
å
ß
O ß CH
å
OH
ß POCH2
å
HO
N CH3
H
å
O
We took away a POH proton, not the NH
proton, because POH is the most acidic
group in the molecule (pKa 1.4).
As another example, consider the molecule piperazine:
H2N
NH2
Structure shown for
piperazine in Appendix G
O
• ¢
The acetyl CH3C ß derivative of
°
o-hydroxybenzoic acid is the active
ingredient in aspirin.
ßÍ CO2H
çÇ O™
OCCH3
Acetylsalicylic acid
HN
NH
Actual structure of piperazine,
which must be neutral
9-3 Weak-Acid Equilibria
Let’s compare the ionization of ortho- and para-hydroxybenzoic acids:
ß CO2H
å
OH
o-Hydroxybenzoic acid
(salicylic acid)
pKa 2.97
HO ß
ß CO2H
p-Hydroxybenzoic acid
pKa 4.54
Why is the ortho isomer 30 times more acidic than the para isomer? Any effect that
increases the stability of the product of a reaction drives the reaction forward. In the
ortho isomer, the product of the acid dissociation reaction can form a strong, internal
hydrogen bond.
162
CHAPTER 9 Monoprotic Acid-Base Equilibria
O
䊞
ß CO2H T
ßC
å
OH
åå
O
O ß HN
H
Hydrogen bond
The para isomer cannot form such a bond because the ßOH and ßCO
2 groups are too far
apart. By stabilizing the product, the internal hydrogen bond is thought to make o-hydroxybenzoic acid more acidic than p-hydroxybenzoic acid.
A Typical Weak-Acid Problem
The problem is to find the pH of a solution of the weak acid HA, given the formal concentration of HA and the value of Ka.4 Let’s call the formal concentration F and use the
systematic treatment of equilibrium:
Ka
HA ∆ H A
Reactions:
Kw
H2O ∆ H OH
[H] [A] [OH]
Charge balance:
F
Mass balance:
[A]
(9-6)
[HA]
(9-7)
[H][A]
[HA]
Kw [H][OH]
Ka Equilibrium expressions:
Formal concentration is the total number of
moles of a compound dissolved in a liter. The
formal concentration of a weak acid is the
total amount of HA placed in the solution,
regardless of the fact that some has
changed into A.
(9-8)
There are four equations and four unknowns ([A], [HA], [H], [OH]) , so the problem is
solved if we can just do the algebra.
But it’s not so easy to solve these simultaneous equations. If you combine them, you
will discover that a cubic equation results. At this point, the Good Chemist rides down again
from the mountain on her white stallion to rescue us and cries, “Wait! There is no reason to
solve a cubic equation. We can make an excellent, simplifying approximation. (Besides, I
have trouble solving cubic equations.)”
For any respectable weak acid, [H] from HA will be much greater than [H] from H2O.
When HA dissociates, it produces A. When H2O dissociates, it produces OH. If dissociation
of HA is much greater than H2O dissociation, then [A] W [OH], and Equation 9-6 reduces to
[H] ⬇ [A]
(9-9)
To solve the problem, first set
also is equal
x. Equation 9-9 says that
to x. Equation 9-7 says that [HA] F [A] F x. Putting these expressions into
Equation 9-8 gives
(x)(x)
[H][A]
Ka [HA]
Fx
[H]
[A]
x [H] in weak-acid problems.
Setting F 0.050 0 M and Ka 1.07 103 for o-hydroxybenzoic acid, we can solve
the equation, because it is just a quadratic equation.
x2
1.07
0.050 0 x
x 2 (1.07
x 6.80
[H]
103
103 )x 5.35
105 0
103 (negative root rejected)
[A] x 6.80
103 M
[HA] F x 0.0432 M
pH log x 2.17
Was the approximation [H] ⬇ [A] justified? The calculated pH is 2.17, which means
that [OH] Kw /[H] 1.5 1012 M.
[A] (from HA dissociation) 6.8
For uniformity, we will usually express pH to the
0.01 decimal place, regardless of the number
of places justified by significant figures.
103 M
1 [H] from HA dissociation 6.8
[OH] (from H2O dissociation) 1.5
1 [H] from H2O dissociation 1.5
103 M
1012 M
In a solution of a weak acid, H is derived
almost entirely from HA, not from H2O.
1012 M
The assumption that H is derived mainly from HA is excellent.
9-3 Weak-Acid Equilibria
163
Fraction of Dissociation
1.00
CO2H
0.90
The fraction of dissociation, , is defined as the fraction of the acid in the form A:
OH
Fraction of dissociation (α)
0.80
Fraction of dissociation
of an acid:
pKa = 2.97
0.70
0.50
0.40
0.30
pKa = 4.54
HO
0.10
0
CO2H
–1
–2
–3
–4
–5
log (formal concentration)
x
x
x (F x)
F
(9-10)
For 0.050 0 M o-hydroxybenzoic acid, we find
6.8 103 M
0.14
0.050 0 M
That is, the acid is 14% dissociated at a formal concentration of 0.050 0 M.
The variation of with formal concentration is shown in Figure 9-2. Weak electrolytes
(compounds that are only partially dissociated) dissociate more as they are diluted. o-Hydroxybenzoic acid is more dissociated than p-hydroxybenzoic acid at the same formal concentration because
the ortho isomer is a stronger acid. Box 9-2 and Demonstration 9-1 illustrate weak-acid properties.
0.60
0.20
[A ]
[HA]
[A ]
The Essence of a Weak-Acid Problem
When faced with finding the pH of a weak acid, you should immediately realize that
[H] [A] x and proceed to set up and solve the equation
[H][A]
x2
(9-11)
Equation for weak acids:
Ka
[HA]
Fx
where F is the formal concentration of HA. The approximation [H] [A] would be poor
only if the acid were too dilute or too weak, neither of which constitutes a practical problem.
–6
Figure 9-2 Fraction of dissociation of a
weak electrolyte increases as electrolyte is
diluted. The stronger acid is more dissociated
than the weaker acid at all concentrations.
Box 9-2 Dyeing Fabrics and the Fraction of Dissociation5
Cotton fabrics are largely cellulose, a polymer with repeating
units of the sugar glucose:
These oxygen atoms can bind to dye
CH2OH
O
OH
O LHß O
OH
O
O
HO
CH2OH
O
O
CH2OH
OH
O
HO
OK
O
Structure of cellulose. Hydrogen bonding between glucose units
helps make the structure rigid.
Dyes are colored molecules that can form covalent bonds to fabric. For example, Procion Brilliant Blue M-R is a dye with a blue
chromophore (the colored part) attached to a reactive dichlorotriazine ring:
O
¡
SO3Na
ß
Cl
N
ß
H N
Nß
N
ß
ß
ß
K ⬇1015
a
Cellulose ß CH2OH #¢¢¢
4 cellulose ß CH2O H
To promote dissociation of the cellulose ßCH2OH proton, dyeing
is carried out in sodium carbonate solution with a pH around 10.6.
The fraction of reactive cellulose is given by the fraction of dissociation of the weak acid at pH 10.6:
Cl
Chlorine atoms can be
replaced by oxygen
atoms of cellulose
Procion Brilliant Blue M-R fabric dye
O
¡
Oxygen atoms of the ßCH2OH groups on cellulose can replace Cl
atoms of the dye to form covalent bonds that fix the dye permanently to the fabric:
164
Only about one cellulose ßCH2OH group in 104 is in the reactive
form at pH 10.6.
N
Cl
ßß
ßß
104.4 ⬇ fraction of dissociation
Nß
Nß
Cl
ßß
Dye ß
N
Ka
[RO][H]
[RO]
1015
1
⬇ 10.6
[ROH]
[ROH]
[H ]
10
O ß Cellulose
Nß
ßß
Dye ß
Cl O ß Cellulose
ßß
ßß
Nß
[RO]
[RO]
⬇
[ROH] [RO ]
[ROH]
Because the fraction of dissociation of the very weak acid is so
small, [ROH] W [RO] in the denominator, which is therefore
approximately just [ROH]. The quotient [RO]/[ROH] can be calculated from Ka and the pH:
Ka Blue chromophore
RO ROH
Fraction of dissociation ß SO3Na
HNß
After the fabric has been dyed in cold water, excess dye is
removed with a hot wash. During the hot wash, the second Cl
group of the dye is replaced by a second cellulose or by water
(giving dye ßOH).
The chemically reactive form of cellulose is the conjugate
base:
Chemically reactive
form of cellulose is
deprotonated anion
Cl
CHAPTER 9 Monoprotic Acid-Base Equilibria
Demonstration 9-1 Conductivity of Weak Electrolytes6
The relative conductivity of strong and weak acids is related to their
different degrees of dissociation in aqueous solution. To demonstrate conductivity, we use a Radio Shack piezo alerting buzzer, but
any kind of buzzer or light bulb could easily be substituted. The
voltage required will depend on the buzzer or light chosen.
When a conducting solution is placed in the beaker, the horn
sounds. First show that distilled water and sucrose solution are
nonconductive. Solutions of the strong electrolytes NaCl or HCl
are conductive. Compare strong and weak electrolytes by demonstrating that 1 mM HCl gives a loud sound, whereas 1 mM acetic
acid gives little or no sound. With 10 mM acetic acid, the strength
of the sound varies noticeably as the electrodes are moved away
from each other in the beaker.
When CO2 is absorbed by pure water, the conductivity
increases from dissociation of the H2CO3 (carbonic acid) formed.
Atmospheric CO2 can be measured by conductivity.7
Horn
~20 V
power supply
Cu metal strips
(~1 × 5 cm)
Beaker
Example A Weak-Acid Problem
Find the pH of 0.050 M trimethylammonium chloride.
H
å
N
å
Cl
H3C
H3C
CH3
Trimethylammonium chloride
Solution
We assume that ammonium halide salts are completely dissociated to give
(CH3 ) 3NH and Cl.* We then recognize that trimethylammonium ion is a weak acid,
being the conjugate acid of trimethylamine, (CH3 ) 3N, a weak base. Cl has no basic or
acidic properties and should be ignored. In Appendix G, we find trimethylammonium ion
listed as trimethylamine, but drawn as trimethylammonium ion, with pKa 9.799. So,
Ka 10pKa 109.799 1.59
Cl has no acidic or basic properties because
it is the conjugate base of the strong acid HCl.
If Cl had appreciable basicity, HCl would not
be completely dissociated.
1010
From here, everything is downhill.
Ka
(CH3 ) 3NH ∆ (CH3 ) 3N H
Fx
x
x2
0.050 x
x 2.8
1.59
x
1010
(9-12)
106 M 1 pH 5.55
*R4NX salts are not completely dissociated, because there are some ion pairs, R4NX (aq) (Box 8-1).
Equilibrium constants for R4N X T R4NX (aq) are given below. For 0.050 F solutions, the fraction of
ion pairing is 4% in (CH3)4NBr, 7% in (CH3CH2 ) 4NBr, and 9% in (CH3CH2CH2 ) 4NBr.
R4N
X
Kion pair
R4N
X
Kion pair
Me4N
Bu4N
Me4N
Et4N
Pr4N
Cl
Cl
Br
Br
Br
1.1
2.5
1.4
2.4
3.1
Me4N
Et4N
Pr4N
Bu4N
I
I
I
I
2.0
2.9
4.6
6.0
Me CH3 ß, Et CH3CH2 ß, Pr CH3CH2CH2 ß, Bu CH3CH2CH2CH2 ß
9-3 Weak-Acid Equilibria
165
A handy tip: Equation 9-11 can always be solved with the quadratic formula. However,
an easier method worth trying first is to neglect x in the denominator. If x comes out much
smaller than F, then your approximation was good and you need not use the quadratic formula. For Equation 9-12, the approximation works like this:
x2
x2
⬇
1.59
0.050 x
0.050
1010 1 x 2 (0.050)(1.59
1010 ) 2.8
106 M
The approximate solution (x ⬇ 2.8 106 ) is much smaller than the term 0.050 in the
denominator of Equation 9-12. Therefore, the approximate solution is fine. A reasonable rule
of thumb is to accept the approximation if x comes out to be less than 1% of F.
9-4
Weak-Base Equilibria
The treatment of weak bases is almost the same as that of weak acids.
Kb
B H2O ∆ BH OH
As Kb increases, pKb decreases and the base
becomes stronger.
Kb [BH][OH]
[B]
We suppose that nearly all OH comes from the reaction of B H2O, and little comes from
dissociation of H2O. Setting [OH] x, we must also set [BH] x, because one BH
is produced for each OH. Calling the formal concentration of base F ( [B] [BH]),
we write
[B] F [BH] F x
Plugging these values into the Kb equilibrium expression, we get
A weak-base problem has the same algebra
as a weak-acid problem, except K Kb and
x [OH].
Equation for weak base:
[BH ][OH ]
x2
Kb
[B]
Fx
(9-13)
which looks a lot like a weak-acid problem, except that now x [OH].
A Typical Weak-Base Problem
Consider the commonly occurring weak base cocaine.
N
œ CH3
O
¡
COCH3
H O
H2O
¡
OCC6H5
H
106
Kb 2.6
#£££££4
H
çœ
N
CH3
Cocaine
O
¡
COCH3
H O
OH
¡
OCC6H5
H
If the formal concentration is 0.037 2 M, the problem is formulated as follows:
B H2O T BH OH
0.037 2 x
x
x2
2.6
0.037 2 x
106 1 x 3.1
x
104
Because x [OH], we can write
Question What concentration of OH is
produced by H2O dissociation in this solution?
Was it justified to neglect water dissociation as
a source of OH?
For a base, is the fraction that has reacted
with water.
[H] Kw /[OH] 1.0 1014 /3.1 104 3.2
pH log[H] 10.49
1011
This is a reasonable pH for a weak base.
What fraction of cocaine has reacted with water? We can formulate for a base, called
the fraction of association:
Fraction of association
of a base:
[BH]
x
0.008 3
[BH ] [B]
F
(9-14)
Only 0.83% of the base has reacted.
166
CHAPTER 9 Monoprotic Acid-Base Equilibria
Conjugate Acids and Bases—Revisited
Earlier, we noted that the conjugate base of a weak acid is a weak base, and the conjugate
acid of a weak base is a weak acid. We also derived an exceedingly important relation
between the equilibrium constants for a conjugate acid-base pair: Ka Kb Kw.
In Section 9-3, we considered o- and p-hydroxybenzoic acids, designated HA. Now consider their conjugate bases. For example, the salt sodium o-hydroxybenzoate dissolves to
give Na (which has no acid-base chemistry) and o-hydroxybenzoate, which is a weak base.
The acid-base chemistry is the reaction of o-hydroxybenzoate with water:
ß CO2H OH
ß CO2 H2O T
ß
ß
OH
A (o-hydroxybenzoate)
Fx
Fx
ç
OH
ß CO
2 Na
gives
ç
OH
o-Hydroxybenzoate
HA
x
x2
ß CO2Na
In aqueous solution,
(9-15)
OH
HA and A are a conjugate acid-base pair.
So are BH and B.
x
Kb
From the value of Ka for each isomer, we can calculate Kb for the conjugate base.
Isomer of hydroxybenzoic acid
ortho
para
Ka
Kb( Kw /Ka)
1.07 103
2.9 105
9.3 1012
3.5 1010
Using each value of Kb and letting F 0.050 0 M, we find
pH of 0.050 0 M o-hydroxybenzoate 7.83
pH of 0.050 0 M p-hydroxybenzoate 8.62
These are reasonable pH values for solutions of weak bases. Furthermore, as expected, the
conjugate base of the stronger acid is the weaker base.
Example A Weak-Base Problem
Find the pH of 0.10 M ammonia.
Solution
5
When ammonia is dissolved in water, its reaction is
Kb
NH3 H2O ∆ NH4 OH
4
x
NH4,
In Appendix G, we find ammonium ion,
listed next to ammonia. pKa for
ammonium ion is 9.245. Therefore, Kb for NH3 is
Kb Kw
Ka
1014.00
1.76 105
109.245
To find the pH of 0.10 M NH3, we set up and solve the equation
[NH4][OH]
x2
Kb 1.76 105
[NH3]
0.10 x
x [OH] 1.32 103 M
[H]
9-5
Kw
7.6 1012 M 1 pH log [H] 11.12
[OH]
Buffers
A buffered solution resists changes in pH when acids or bases are added or when dilution
occurs. The buffer is a mixture of an acid and its conjugate base. There must be comparable
amounts of the conjugate acid and base (say, within a factor of 10) to exert significant buffering.
The importance of buffers in all areas of science is immense. At the outset of this chapter, we saw that digestive enzymes in lysosomes operate best in acid, which allows a cell to
protect itself from its own enzymes. If enzymes leak into the buffered, neutral cytoplasm,
they have low reactivity and do less damage to the cell than they would at their optimum pH.
Figure 9-3 shows the pH dependence of an enzyme-catalyzed reaction that is fastest near
9-5
Buffers
Relative rate of reaction
Ammonium
ion
x
Ammonia
Fx
3
2
1
5.0
6.0
7.0
8.0
pH
9.0
10.0
Figure 9-3
pH dependence of the rate of
cleavage of an amide bond by the enzyme
chymotrypsin, which helps digest proteins in
your intestine. [M. L. Bender, G. E. Clement, F. J. Kézdy,
and H. A. Heck, “The Correlation of the pH (pD)
Dependence and the Stepwise Mechanism of
-Chymotrypsin-Catalyzed-Reactions,” J. Am. Chem.
Soc. 1964, 86, 3680.]
O
•
RC ß NHR
Amide bond
167
pH 8.0. For an organism to survive, it must control the pH of each subcellular compartment
so that each reaction proceeds at the proper rate.
Mixing a Weak Acid and Its Conjugate Base
When you mix a weak acid with its conjugate
base, you get what you mix!
If you mix A moles of a weak acid with B moles of its conjugate base, the moles of acid
remain close to A and the moles of base remain close to B. Very little reaction occurs to
change either concentration.
To understand why this should be so, look at the Ka and Kb reactions in terms of
Le Châtelier’s principle. Consider an acid with pKa 4.00 and its conjugate base with
pKb 10.00. Let’s calculate the fraction of acid that dissociates in a 0.10 M solution of HA.
HA
H
T
0.10 x
x
pKa 4.00
A
x
x2
Ka 1 x 3.1
Fx
Fraction of dissociation 103
x
0.031
F
The acid is only 3.1% dissociated under these conditions.
In a solution containing 0.10 mol of A dissolved in 1.00 L, the extent of reaction of A
with water is even smaller.
A H2O T HA OH
0.10 x
x
pKb 10.00
x
x2
Kb 1 x 3.2
Fx
Fraction of association The approximation that the concentrations of
HA and A remain unchanged breaks down
for dilute solutions or at extremes of pH. We will
test the validity of the approximation at the
end of this chapter.
106
x
3.2
F
105
HA dissociates very little, and adding extra A to the solution will make the HA dissociate even less. Similarly, A does not react very much with water, and adding extra HA makes
A react even less. If 0.050 mol of A plus 0.036 mol of HA are added to water, there will
be close to 0.050 mol of A and close to 0.036 mol of HA in the solution at equilibrium.
Henderson-Hasselbalch Equation
The central equation for buffers is the Henderson-Hasselbalch equation, which is merely a
rearranged form of the Ka equilibrium expression.
Ka log xy log x log y
log Ka log
[H][A]
[HA]
[H][A]
[A]
log[H] log
[HA]
[HA]
log[H] log Ka log
L. J. Henderson was a physician who wrote
[H] Ka [acid]/[salt] in a physiology article
in 1908, a year before the word “buffer” and
the concept of pH were invented by the
biochemist S. P. L. Sørensen. Henderson’s
contribution was the approximation of setting
[acid] equal to the concentration of HA
placed in solution and [salt] equal to the
concentration of A placed in solution. In 1916,
K. A. Hasselbalch wrote what we call the
Henderson-Hasselbalch equation in a
biochemical journal.8
˛
Equations 9-16 and 9-17 are sensible only
when the base (A or B) is in the numerator.
When the concentration of base increases, the
log term increases and the pH increases.
168
Henderson-Hasselbalch
equation for an acid:
14243
123
pH
pK a
pH pKa log a
[A]
[HA]
[A ]
b
[HA]
(9-16)
The Henderson-Hasselbalch equation tells us the pH of a solution, provided we know
the ratio of the concentrations of conjugate acid and base, as well as pKa for the acid. If a
solution is prepared from the weak base B and its conjugate acid, the analogous equation is
Henderson-Hasselbalch
equation for a base:
pH pKa log a
[B]
b
[BH ]
pKa applies to
b this acid
(9-17)
where pKa is the acid dissociation constant of the weak acid BH. The important features of
Equations 9-16 and 9-17 are that the base (A or B) appears in the numerator of both equations, and the equilibrium constant is Ka of the acid in the denominator.
CHAPTER 9 Monoprotic Acid-Base Equilibria
Challenge Show that, when activities are included, the Henderson-Hasselbalch equation is
pH pKa log
[A]A
[HA]HA
(9-18)
Properties of the Henderson-Hasselbalch Equation
In Equation 9-16, we see that, if [A] [HA], then pH pKa.
pH pKa log
[A]
pKa log 1 pKa
[HA]
Regardless of how complex a solution may be, whenever pH pKa, [A] must equal [HA].
This relation is true because all equilibria must be satisfied simultaneously in any solution
at equilibrium. If there are 10 different acids and bases in the solution, the 10 forms of Equation 9-16 must all give the same pH, because there can be only one concentration of Hⴙ in
a solution.
Another feature of the Henderson-Hasselbalch equation is that, for every power-of-10
change in the ratio [A]/[HA], the pH changes by one unit (Table 9-1). As the base (A )
increases, the pH goes up. As the acid (HA) increases, the pH goes down. For any conjugate
acid-base pair, you can say, for example, that, if pH pKa 1, there must be 10 times as
much HA as A. Ten-elevenths is in the form HA and one-eleventh is in the form A.
Example Using the Henderson-Hasselbalch Equation
Sodium hypochlorite (NaOCl, the active ingredient of almost all bleaches) was dissolved
in a solution buffered to pH 6.20. Find the ratio [OCl]/[HOCl] in this solution.
When [A] [HA], pH pKa.
Table 9-l Effect of [Aⴚ]/[HA]
on pH
[Aⴚ]/[HA]
pH
100:1
10:1
1:1
1:10
1:100
pKa 2
pKa 1
pKa
pKa 1
pKa 2
In Appendix G, we find that pKa 7.53 for hypochlorous acid, HOCl. The pH
is known, so the ratio [OCl]/[HOCl] can be calculated from the Henderson-Hasselbalch
equation.
Solution
HOCl T H OCl
pH pKa log
[OCl]
[HOCl]
[OCl]
[HOCl]
[OCl]
1.33 log
[HOCl]
6.20 7.53 log
101.33 10log([OCl]/[HOCl]) 0.047 [OCl]
[HOCl]
10log z z
[OCl]
[HOCl]
Finding the ratio [OCl]/[HOCl] requires knowing only the pH and the pKa. We do not
need to know how much NaOCl was added, or the volume.
A Buffer in Action
For illustration, we choose a widely used buffer called “tris,” which is short for tris(hydroxymethyl)aminomethane.
å
BH
pKa 8.072
NH2
å
C
H
HOCH2
CH2OH
HOCH2
å
NH3
å
C
T
CH2OH
HOCH2
HOCH2
B
This form is “tris”
In Appendix G, we find pKa for the conjugate acid of tris listed as 8.072. An example of a salt
containing the BH cation is tris hydrochloride, which is BHCl. When BHCl is dissolved in water, it dissociates to BH and Cl.
9-5 Buffers
169
Example A Buffer Solution
Find the pH of a solution prepared by dissolving 12.43 g of tris (FM 121.135) plus
4.67 g of tris hydrochloride (FM 157.596) in 1.00 L of water.
The concentrations of B and BH added to the solution are
Solution
[B] 12.43 g/L
0.102 6 M
121.135 g/mol
[BH] 4.67 g/L
0.029 6 M
157.596 g/mol
Assuming that what we mixed stays in the same form, we plug these concentrations into
the Henderson-Hasselbalch equation to find the pH:
pH pKa log
[B]
0.102 6
8.072 log
8.61
[BH]
0.029 6
Notice that the volume of solution is irrelevant, because volume cancels in the numerator and denominator of the log term:
The pH of a buffer is nearly independent of
volume.
moles of B/L of solution
moles of BH /L of solution
moles of B
pKa log
moles of BH
pH pKa log
Example Effect of Adding Acid to a Buffer
If we add 12.0 mL of 1.00 M HCl to the solution used in the previous example, what
will be the new pH?
Solution
The key to this problem is to realize that, when a strong acid is added to a
weak base, they react completely to give BH (Box 9-3). We are adding 12.0 mL of
1.00 M HCl, which contains (0.012 0 L)(1.00 mol/L) 0.012 0 mol of H. This much
H will consume 0.012 0 mol of B to create 0.012 0 mol of BH, which is shown
conveniently in a little table:
B
H
S
BH
Initial moles
Final moles
Tris
From HCl
0.102 6
0.090 6
0.012 0
—
644474448
(0.102 6 0.012 0)
0.029 6
0.041 6
644474448
(0.029 6 0.012 0)
Box 9-3 Strong Plus Weak Reacts Completely
A strong acid reacts with a weak base essentially “completely”
because the equilibrium constant is large.
B H T BH
K
1
Ka (for BH )
Weak Strong
base acid
If B is tris(hydroxymethyl)aminomethane, the equilibrium constant for reaction with HCl is
K
1
1
8.072 1.2
Ka
10
108
A strong base reacts “completely” with a weak acid because
the equilibrium constant is, again, very large.
OH HA T A H2O
Strong
base
170
K
1
Kb (for A )
If HA is acetic acid, then the equilibrium constant for reaction
with NaOH is
K
Ka (for HA)
1
1.7
Kb
KW
109
The reaction of a strong acid with a strong base is even more
complete than a strong plus weak reaction:
H OH T H2O
Strong
acid
K
1
1014
KW
Strong
base
If you mix a strong acid, a strong base, a weak acid, and a weak
base, the strong acid and base will neutralize each other until one
is used up. The remaining strong acid or base will then react with
the weak base or weak acid.
Weak
acid
CHAPTER 9 Monoprotic Acid-Base Equilibria
Demonstration 9-2 How Buffers Work
A buffer resists changes in pH because the added acid or base is
consumed by the buffer. As the buffer is used up, it becomes less
resistant to changes in pH.
In this demonstration,9 a mixture containing approximately a
10:1 mole ratio of HSO3 :SO2
3 is prepared. Because pKa for HSO3
is 7.2, the pH should be approximately
pH pKa log
[SO2
1
3 ]
7.2 log
6.2
[HSO3 ]
10
Through 90% completion, the pH should rise by just 1 unit. In the
next 9% of the reaction, the pH will rise by another unit. At the
end of the reaction, the change in pH is very abrupt.
In the formaldehyde clock reaction, formaldehyde is added to
a solution containing HSO3 , SO2
3 , and phenolphthalein indicator.
Phenolphthalein is colorless below a pH of 8 and red above this
pH. The solution remains colorless for more than a minute. Suddenly the pH shoots up and the liquid turns pink. Monitoring the
pH with a glass electrode gave the results in the graph.
When formaldehyde is added, the net reaction is the consumption
of HSO3 , but not of SO2
3 :
Formaldehyde
œ
H2C
ç
B
Bisulfite
H2C £ O SO32
O
SO3H
B
œ
H2C
ç
Sulfite
H2C
œ
ç
œ
ç
OH
(A)
9.0
SO
3
O
8.0
SO
3
(B)
O
SO
3
H2C
B
pH
H2C £ O HSO
3
10.0
HSO3
B
H2C
œ
ç
OH
SO
3
7.0
SO2
3
(In sequence A, bisulfite is consumed directly. In sequence B, the
net reaction is destruction of HSO3 , with no change in the SO2
3
concentration.)
We can prepare a table showing how the pH should change as
the HSO3 reacts.
Percentage of reaction
completed
[SO2
3 ]:[HSO 3 ]
Calculated
pH
0
90
99
99.9
99.99
1:10
1:1
1:0.1
1:0.01
1:0.001
6.2
7.2
8.2
9.2
10.2
6.0
0
30
60
90
Time (s)
120
Graph of pH versus time in the formaldehyde clock reaction.
Procedure: All solutions should be fresh. Prepare a solution
of formaldehyde by diluting 9 mL of 37 wt% formaldehyde to
100 mL. Dissolve 1.5 g of NaHSO310 and 0.18 g of Na2SO3 in
400 mL of water, and add ⬃1 mL of phenolphthalein solution
(Table 11-4). Add 23 mL of formaldehyde solution to the wellstirred buffer solution to initiate the clock reaction. The time of
reaction can be adjusted by changing the temperature, concentrations, or volume.
The information in the table allows us to calculate the pH:
moles of B
moles of BH
0.090 6
8.41
8.072 log
0.041 6
The volume of the solution is irrelevant.
pH pKa log
We see that the pH of a buffer does not change very much when a limited amount of
strong acid or base is added. Addition of 12.0 mL of 1.00 M HCl changed the pH from 8.61
to 8.41. Addition of 12.0 mL of 1.00 M HCl to 1.00 L of unbuffered solution would have lowered the pH to 1.93.
But why does a buffer resist changes in pH? It does so because the strong acid or base is
consumed by B or BH. If you add HCl to tris, B is converted into BH. If you add NaOH,
BH is converted into B. As long as you don’t use up the B or BH by adding too much HCl
or NaOH, the log term of the Henderson-Hasselbalch equation does not change very much
and the pH does not change very much. Demonstration 9-2 illustrates what happens when the
buffer does get used up. The buffer has its maximum capacity to resist changes of pH when
pH pKa. We will return to this point later.
9-5 Buffers
Question Does the pH change in the right
direction when HCl is added?
A buffer resists changes in pH . . .
. . . because the buffer consumes the added
acid or base.
171
Example Calculating How to Prepare a Buffer Solution
How many milliliters of 0.500 M NaOH should be added to 10.0 g of tris hydrochloride to
give a pH of 7.60 in a final volume of 250 mL?
The moles of tris hydrochloride in 10.0 g are (10.0 g)/ (157.596 g/mol) 0.063 5. We can make a table to help solve the problem:
Solution
Reaction with OH:
Initial moles
Final moles
BH
0.063 5
0.063 5 x
OH
x
—
S
B
—
x
The Henderson-Hasselbalch equation allows us to find x, because we know pH and pKa.
mol B
mol BH
x
7.60 8.072 log
0.063 5 x
x
0.472 log
0.063 5 x
x
1 x 0.016 0 mol
100.472 0.063 5 x
pH pKa log
This many moles of NaOH is contained in
0.016 0 mol
0.032 0 L 32.0 mL
0.500 mol/L
Reasons why a calculation would be wrong:
1. You might have ignored activity coefficients.
2. The temperature might not be just right.
3. The approximations that [HA] FHA and
[A] FA could be in error.
4. The pKa reported for tris in your favorite table
is probably not what you would measure in
your lab.
5. You will probably make an arithmetic error
anyway.
0.18
0.16
0.14
Cb (M)
0.12
(a )
0.10
Preparing a Buffer in Real Life!
If you really wanted to prepare a tris buffer of pH 7.60, you would not do it by calculating
what to mix. Suppose that you wish to prepare 1.00 L of buffer containing 0.100 M tris at a
pH of 7.60. You have available solid tris hydrochloride and approximately 1 M NaOH. Here’s
how I would do it:
1. Weigh out 0.100 mol of tris hydrochloride and dissolve it in a beaker containing about
800 mL of water.
2. Place a calibrated pH electrode in the solution and monitor the pH.
3. Add NaOH solution until the pH is exactly 7.60.
4. Transfer the solution to a volumetric flask and wash the beaker a few times. Add the
washings to the volumetric flask.
5. Dilute to the mark and mix.
You do not simply add the calculated quantity of NaOH, because it would not give exactly
the desired pH. The reason for using 800 mL of water in the first step is so that the volume
will be reasonably close to the final volume during pH adjustment. Otherwise, the pH will
change slightly when the sample is diluted to its final volume and the ionic strength changes.
0.08
0.06
0.04
0.02
Buffer capacity (β) (M/pH unit)
0.00
Buffer Capacity11
0.16
Buffer capacity, , is a measure of how well a solution resists changes in pH when strong
acid or base is added. Buffer capacity is defined as
0.14
0.12
0.10
Buffer capacity:
0.08
0.06
(b )
0.04
0.02
0.00
3
Figure 9-4
4
5
6
7
8 9 10 11 12 13
pH
(a) C b versus pH for a solution
containing 0.100 F HA with pKa 5.00.
(b) Buffer capacity versus pH for the same
system reaches a maximum when pH pKa.
The lower curve is the derivative of the
upper curve.
172
dCb
dCa
dpH
dpH
(9-19)
where Ca and Cb are the number of moles of strong acid and strong base per liter needed to
produce a unit change in pH. The greater the buffer capacity, the more resistant the solution
is to pH change.
Figure 9-4a shows Cb versus pH for a solution containing 0.100 F HA with pKa 5.00.
The ordinate (Cb ) is the formal concentration of strong base needed to be mixed with
0.100 F HA to give the indicated pH. For example, a solution containing 0.050 F OH plus
0.100 F HA would have a pH of 5.00 (neglecting activities).
Figure 9-4b, which is the derivative of the upper curve, shows buffer capacity for the
same system. Buffer capacity reaches a maximum when pH pKa. That is, a buffer is most
effective in resisting changes in pH when pH pKa (that is, when [HA] [A]).
CHAPTER 9 Monoprotic Acid-Base Equilibria
In choosing a buffer, seek one whose pKa is as close as possible to the desired pH. The
useful pH range of a buffer is usually considered to be pKa 1 pH unit. Outside this range,
there is not enough of either the weak acid or the weak base to react with added base or acid.
Buffer capacity can be increased by increasing the concentration of the buffer.
Buffer capacity in Figure 9-4b continues upward at high pH (and at low pH, which is not
shown) simply because there is a high concentration of OH at high pH (or H at low pH). Addition of a small amount of acid or base to a large amount of OH (or H) will not have a large
effect on pH. A solution of high pH is buffered by the H2O/OH conjugate acid–conjugate base
pair. A solution of low pH is buffered by the H3O /H2O conjugate acid–conjugate base pair.
Table 9-2 lists pKa values for common buffers that are widely used in biochemistry. The
measurement of pH with glass electrodes, and the buffers used by the U.S. National Institute of
Standards and Technology to define the pH scale, are described in Chapter 15.
Choose a buffer whose pKa is close to the
desired pH.
Buffer pH Depends on Ionic Strength and Temperature
The correct Henderson-Hasselbalch equation, 9-18, includes activity coefficients. Failure to
include activity coefficients is the principal reason why calculated pH values do not agree with
2
measured values. The H2PO
4 /HPO4 buffer has a pKa of 7.20 at 0 ionic strength. At 0.1 M ionic
2
strength, the pH of a 1:1 mole mixture of H2PO
4 and HPO4 is 6.86 (see Problem 9-43). Molecular
biology lab manuals list pKa for phosphoric acid as 6.86, which is representative for ionic strengths
employed in the lab. As another example of ionic strength effects, when a 0.5 M stock solution of
phosphate buffer at pH 6.6 is diluted to 0.05 M, the pH rises to 6.9—a rather significant effect.
Buffer pKa depends on temperature. Tris has an exceptionally large dependence,
0.028 pKa units per degree, near room temperature. A solution of tris with pH 8.07 at 25°C
will have pH ⬇ 8.7 at 4°C and pH ⬇ 7.7 at 37°C.
Changing ionic strength changes pH.
Changing temperature changes pH.
When What You Mix Is Not What You Get
In dilute solution, or at extremes of pH, the concentrations of HA and A are not equal to
their formal concentrations. Suppose we mix FHA moles of HA and FA moles of the salt
NaA. The mass and charge balances are
FHA FA [HA] [A]
Mass balance:
Charge balance:
[Na] [H] [OH] [A]
What you mix is not what you get in dilute
solutions or at extremes of pH.
The substitution [Na] FA and a little algebra lead to the equations
[HA] FHA [H] [OH]
[A] FA [H] [OH]
(9-20)
(9-21)
So far we have assumed that [HA] ⬇ FHA and [A] ⬇ FA, and we used these values in
the Henderson-Hasselbalch equation. A more rigorous procedure is to use Equations 9-20
and 9-21. If FHA or FA is small, or if [H] or [OH] is large, then the approximations
[HA] ⬇ FHA and [A] ⬇ FA are not good. In acidic solutions, [H] W [OH], so [OH]
can be ignored in Equations 9-20 and 9-21. In basic solutions, [H] can be neglected.
Example A Dilute Buffer Prepared from a Moderately Strong Acid
What will be the pH if 0.010 0 mol of HA (with pKa 2.00) and 0.010 0 mol of A are
dissolved in water to make 1.00 L of solution?
Because the solution is acidic (pH ⬇ pKa 2.00), we neglect [OH] in
Equations 9-20 and 9-21. Setting [H] x in Equations 9-20 and 9-21, we use the Ka
equation to find [H]:
T
HA
H
A
Solution
0.010 0 x
Ka [H][A]
[HA]
x
0.010 0 x
(x)(0.010 0 x)
102.00
(0.010 x)
(9-22)
1 x 0.004 14 M 1 pH log[H] 2.38
The concentrations of HA and A are not what we mixed:
[HA] FHA [H] 0.005 86 M
[A] FA [H] 0.014 1 M
HA in this solution is more than 40%
dissociated. The acid is too strong for the
approximation [HA] ⬇ FHA.
In this example, HA is too strong and the concentrations are too low for HA and A to be
equal to their formal concentrations.
9-5 Buffers
173
Table 9-2 Structures and pKa values for common buffersa,b,c,d
Name
N-2-Acetamidoiminodiacetic acid (ADA)
N-Tris(hydroxymethyl)methylglycine
(TRICINE)
Phosphoric acid
ADA
Piperazine-N,N-bis(2-ethanesulfonic acid)
(PIPES)
Citric acid
Glycylglycine
pKae
Structure
O
CH2CO2H
•
œ
H2NCCH2NH
ç
CH2CO2H
Formula
mass
(pKa)/T
(K1)
1.59 (CO2H)
190.15
—
(HOCH2)3CN H2CH2CO 2H
2.02 (CO2H)
179.17
0.003
H3PO4
2.15 (pK1 )
2.48 (CO2H)
98.00
190.15
0.005
—
2.67 (pK1 )
302.37
—
3.13 (pK1 )
192.12
0.002
3.14 (CO2H)
132.12
0.000
3.79 (pK1 )
330.42
—
O3SCH2CH2CH2CH2NH HNCH2CH2CH2CH2SO3 4.29 (pK1 )
358.47
—
(see above)
O3SCH2CH2NH HNCH2CH2SO3
OH
å
HO2CCH2CCH2CO2H
å
CO2H
O
•
H3NCH2CNHCH2CO2H
Piperazine-N,N-bis(3-propanesulfonic
acid) (PIPPS)
Piperazine-N,N-bis(4-butanesulfonic acid)
(PIPBS)
N,N-Diethylpiperazine dihydrochloride
(DEPP2HCl)
Citric acid
Acetic acid
CH3CH2NH HNCH2CH32Cl
4.48 (pK1 )
215.16
—
(see above)
4.76 (pK2 )
4.76
192.12
60.05
0.001
0.000
5.62 (pK1 )
360.49
—
6.27
195.24
0.009
6.40 (pK3 )
192.12
0.002
6.58 (pK1 )
245.23
—
O3SCH2CH2CH2NH HNCH2CH2CH2SO
3
CH3CO2H
N,N-Diethylethylenediamine-N,N-bis
(3-propanesulfonic (DESPEN)
2-(N-Morpholino)ethanesulfonic acid (MES)
O
Citric acid
(see above)
O3SCH2CH2CH2NHCH2CH2HNCH2CH2CH2SO3
å
å
CH3CH2
CH2CH3
NHCH2CH2SO3
N,N,N,N-Tetraethylethylenediamine
dihydrochloride (TEEN2HCl)
(CH3CH2)2NHCH2CH2HN(CH2CH3)22Cl
1,3-Bis[tris(hydroxymethyl)methylamino]
propane hydrochloride (BIS-TRIS
propane2HCl)
(HOCH2)3CNH2(CH2)3NH2C(CH2OH)32Cl
6.65 (pK1 )
355.26
—
ADA
(see above)
6.84 (NH)
190.15
0.007
a. The protonated form of each molecule is shown. Acidic hydrogen atoms are shown in bold type.
b. Many buffers in this table are widely used in biomedical research because of their weak metal binding and physiologic inertness (C. L. Bering, J. Chem. Ed. 1987, 64, 803). In one study,
where MES and MOPS had no discernible affinity for Cu2, a minor impurity in HEPES and HEPPS had a strong affinity for Cu2 and MOPSO bound Cu2 stoichiometrically (H. E. Marsh,
Y.-P. Chin, L. Sigg, R. Hari, and H. Xu, Anal. Chem. 2003, 75, 671). ADA, BICINE, ACES, and TES have some metal-binding ability (R. Nakon and C. R. Krishnamoorthy, Science 1983, 221,
749). Lutidine buffers for the pH range 3 to 8 with limited metal-binding power have been described by U. Bips, H. Elias, M. Hauröder, G. Kleinhans, S. Pfeifer, and K. J. Wannowius, Inorg.
Chem. 1983, 22, 3862.
c. Some data from R. N. Goldberg, N. Kishore, and R. M. Lennen, J. Phys. Chem. Ref. Data 2002, 31, 231. This paper gives the temperature dependence of pKa.
d. Temperature and ionic strength dependence of pKa for buffers: HEPES—D. Feng, W. F. Koch, and Y. C. Wu, Anal. Chem. 1989, 61, 1400; MOPSO—Y. C. Wu, P. A. Berezansky, D. Feng, and
W. F. Koch, Anal. Chem. 1993, 65, 1084; ACES and CHES—R. N. Roy, J. Bice, J. Greer, J. A. Carlsten, J. Smithson, W. S. Good, C. P. Moore, L. N. Roy, and K. M. Kuhler, J. Chem. Eng. Data
1997, 42, 41; TEMN, TEEN, DEPP, DESPEN, PIPES, PIPPS, PIPBS, MES, MOPS, and MOBS—A. Kandegedara and D. B. Rorabacher, Anal. Chem. 1999, 71, 3140. This last set of buffers
was specifically developed for low metal-binding ability (Q. Yu, A. Kandegedara, Y. Xu, and D. B. Rorabacher, Anal. Biochem. 1997, 253, 50).
e. pK a is generally for 25°C and zero ionic strength.
174
CHAPTER 9 Monoprotic Acid-Base Equilibria
Table 9-2 (continued) Structures and pKa values for common buffersa,b,c,d
Name
Structure
3-(N-Morpholino)-2hydroxypropanesulfonic acid (MOPSO)
O
•
H2NCCH2NH2CH2CH2SO3
OH
å
O
NHCH2CHCH2SO
3
Imidazole hydrochloride
HN
N-2-Acetamido-2-aminoethanesulfonic
acid (ACES)
Cl
N
H
PIPES
3-(N-Morpholino)propanesulfonic acid
(MOPS)
Phosphoric acid
(see above)
NHCH2CH2CH2SO
3
O
H3PO4
NHCH2CH2CH2CH2SO 3
4-(N-Morpholino)butanesulfonic acid
(MOBS)
O
N-Tris(hydroxymethyl)methyl-2-aminoethanesulfonic acid (TES)
(HOCH2 ) 3CNH 2CH2CH2SO3
N-2-Hydroxyethylpiperazine-N-2ethanesulfonic acid (HEPES)
HOCH2CH2N
PIPPS
(see above)
N-2-Hydroxyethylpiperazine-N-3propanesulfonic acid (HEPPS)
Glycine amide hydrochloride
NHCH2CH2SO3
NHCH2CH2CH2SO
3
HOCH2CH2N
O
•
H3NCH2CNH2Cl
Formula
mass
(pKa)/T
(K1)
6.85
182.20
0.018
6.90
225.26
0.015
6.99
104.54
0.022
7.14 (pK2 )
302.37
0.007
7.18
209.26
0.012
7.20 (pK2 )
98.00
0.002
7.48
223.29
—
7.55
229.25
0.019
7.56
238.30
0.012
7.97 (pK2 )
330.42
—
7.96
252.33
0.013
pKae
8.04
110.54
—
157.60
0.028
Tris(hydroxymethyl)aminomethane
hydrochloride (TRIS hydrochloride)
(HOCH2)3CNH3Cl
8.07
TRICINE
Glycylglycine
(see above)
(see above)
8.14 (NH)
8.26 (NH)
179.17
132.12
0.018
0.026
N,N-Bis(2-hydroxyethyl)glycine (BICINE)
PIPBS
DEPP2HCl
DESPEN
BIS-TRIS propane2HCl
Ammonia
Boric acid
(HOCH2CH2)2NHCH2CO2
8.33
8.55 (pK2 )
8.58 (pK2 )
9.06 (pK2 )
9.10 (pK2 )
9.24
9.24 (pK1 )
163.17
358.47
207.10
360.49
355.26
17.03
61.83
0.015
—
—
—
—
0.031
0.008
9.39
207.29
0.023
9.88 (pK2 )
245.23
—
10.50
221.32
0.028
Cyclohexylaminoethanesulfonic acid
(CHES)
TEEN2HCl
3-(Cyclohexylamino) propanesulfonic acid
(CAPS)
(see above)
(see above)
(see above)
(see above)
NH4
B(OH) 3
ß NH2CH2CH2SO
3
(see above)
ß NH2CH2CH2CH2SO3
N,N,N,N-Tetraethylmethylenediamine
dihydrochloride (TEMN2HCl)
(CH3CH2)2NHCH2CH2HN(CH2CH3)22Cl
11.01 (pK2 )
231.21
—
Phosphoric acid
Boric acid
H3PO4
B(OH) 3
12.35 (pK3 )
12.74 (pK2 )
98.00
61.83
0.009
—
9-5 Buffers
175
The Henderson-Hasselbalch equation (with activity coefficients) is always true, because
it is just a rearrangement of the Ka equilibrium expression. Approximations that are not
always true are the statements [HA] ⬇ FHA and [A] ⬇ FA.
In summary, a buffer consists of a mixture of a weak acid and its conjugate base. The
buffer is most useful when pH ⬇ pKa. Over a reasonable range of concentration, the pH of a
buffer is nearly independent of concentration. A buffer resists changes in pH because it reacts
with added acids or bases. If too much acid or base is added, the buffer will be consumed and
will no longer resist changes in pH.
Example
Excel’s Goal Seek Tool and Naming of Cells
We saw in Section 6-8 that GOAL SEEK finds solutions to numerical equations. In setting
up Equation 9-22, we made the (superb) approximation [H] W [OH] and neglected
[OH]. With GOAL SEEK, it is easy to use Equations 9-20 and 9-21 without approximations:
Ka [H](FA [H] [OH])
[H][A]
[HA]
FHA [H] [OH]
(9-23)
The spreadsheet illustrates GOAL SEEK and the naming of cells to make formulas
more meaningful. In column A, enter labels for Ka, Kw, FHA, FA, [H], and [OH]. Write
numerical values for Ka, Kw, FHA, and FA in B1:B4. In cell B5, enter a guess for [H].
B
A
1
2
3
4
5
6
7
[H+] =
[OH-] =
C
0.01
1.00E-14
0.01
0.01
Ka =
Kw =
FHA =
FA =
D
E
Reaction quotient
for Ka =
[H+][A-]/[HA]=
0.001222222
1.000E-03 <-Initial guess
1E-11
D4 = H*(FA+H-OH)/(FHA-H+OH)
B6 = Kw/H
Now we want to name cells Bl through B6. Select cell Bl, go to the INSERT menu,
select NAME and then DEFINE. The window will ask if you want to use the name “Ka” that
appears in cell Al. If you like this name, click OK. By this procedure, name the other cells
in column B “Kw”, “FHA”, “FA”, “H”, and “OH”. Now when you write a formula referring to cell B2, you can write Kw instead of B2. Kw is an absolute reference to cell $B$2.
In cell B6 enter the formula “Kw/H” and Excel returns the value 1E-11 for [OH].
The beauty of naming cells is that “Kw/H” is easier to understand than “$B$2/$B$5”.
In cell D4, write “H*(FAH-OH)/(FHA-HOH) ”, which is the quotient in Equation 9-23. Excel returns the value 0.001 222 based on the guess [H] 0.001 in cell B5.
Now use GOAL SEEK to vary [H] in cell B5 until the reaction quotient in cell D4
equals 0.01, which is the value of Ka. In the TOOLS menu, select OPTIONS and go to Calculation. Set Maximum Change to 1e-7 to find an answer within 0.000 000 1 of the desired
answer. In the TOOLS menu, select GOAL SEEK. For Set cell, enter D4. For To value, enter
0.01. For By changing cell, enter H. We just told GOAL SEEK to vary H (cell B5) until cell
D4 equals 0.01 0.000 000 1. Click OK and the answer H 4.142 103 appears in
cell B5. Different starting guesses for H might give negative solutions or might not reach
a solution. Only one positive value of H satisfies Equation 9-23.
Terms to Understand
acid dissociation constant, Ka
base hydrolysis constant, Kb
buffer
buffer capacity
conjugate acid-base pair
fraction of association, (of a base)
fraction of dissociation, (of an acid)
Henderson-Hasselbalch equation
pK
strong acid
strong base
weak acid
weak base
weak electrolyte
Summary
Strong acids or bases. For practical concentrations (106 M) , pH
or pOH can be found by inspection. When the concentration is near
107 M, we use the systematic treatment of equilibrium to calculate
pH. At still lower concentrations, the pH is 7.00, set by autoprotolysis of the solvent.
176
Weak acids. For the reaction HA T H A, we set up and
solve the equation Ka x2 /(F x) , where [H] [A] x, and
[HA] F x. The fraction of dissociation is given by [A]/([HA] [A]) x/F. The term pKa is defined as pKa log Ka.
CHAPTER 9 Monoprotic Acid-Base Equilibria
Weak bases. For the reaction B H2O T BH OH, we set
up and solve the equation Kb x2 /(F x) , where [OH] [BH] x, and [B] F x. The conjugate acid of a weak base is a
weak acid, and the conjugate base of a weak acid is a weak base. For
a conjugate acid-base pair, Ka Kb Kw.
Buffers. A buffer is a mixture of a weak acid and its conjugate
base. It resists changes in pH because it reacts with added acid or
base. The pH is given by the Henderson-Hasselbalch equation:
pH pKa log
[A]
[HA]
prepare the solution. The pH of a buffer is nearly independent of
dilution, but buffer capacity increases as the concentration of buffer
increases. The maximum buffer capacity is at pH pKa, and the
useful range is pH pKa 1.
The conjugate base of a weak acid is a weak base. The weaker
the acid, the stronger the base. However, if one member of a conjugate pair is weak, so is its conjugate. The relation between Ka for an
acid and Kb for its conjugate base in aqueous solution is Ka Kb Kw. When a strong acid (or base) is added to a weak base (or acid),
they react nearly completely.
where pKa applies to the species in the denominator. The concentrations of HA and A are essentially unchanged from those used to
Exercises
9-A. Using activity coefficients correctly, find the pH of 1.0
102 M NaOH.
9-B. Calculate the pH of
(a) 1.0 108 M HBr
(b) 1.0 108 M H2SO4 (H2SO4 dissociates completely to 2H
plus SO2
4 at this low concentration).
9-C. What is the pH of a solution prepared by dissolving 1.23 g of
2-nitrophenol (FM 139.11) in 0.250 L?
9-D. The pH of 0.010 M o-cresol is 6.16. Find pKa for this weak
acid.
CH3
ß
ß
o-Cresol
OH
9-E. Calculate the limiting value of the fraction of dissociation ()
of a weak acid (pKa 5.00) as the concentration of HA approaches
0. Repeat the same calculation for pKa 9.00.
9-F. Find the pH of 0.050 M sodium butanoate (the sodium salt of
butanoic acid, also called butyric acid).
9-G. The pH of 0.10 M ethylamine is 11.82.
(a) Without referring to Appendix G, find Kb for ethylamine.
(b) Using results from part (a), calculate the pH of 0.10 M ethylammonium chloride.
9-H. Which of the following bases would be most suitable for preparing a buffer of pH 9.00: (i) NH3 (ammonia, Kb 1.76 105 );
(ii) C6H5NH2 (aniline, Kb 3.99 1010); (iii) H2NNH2 (hydrazine,
Kb 1.05 106 ) ; (iv) C5H5N (pyridine, Kb 1.58 109 )?
9-I. A solution contains 63 different conjugate acid-base pairs.
Among them is acrylic acid and acrylate ion, with the equilibrium
ratio [acrylate]/[acrylic acid] 0.75. What is the pH of the solution?
H2C¢CHCO2H
9-J. (a) Find the pH of a solution prepared by dissolving 1.00 g of
glycine amide hydrochloride (Table 9-2) plus 1.00 g of glycine
amide in 0.100 L.
O
H2Nß ¡
NH2
Glycine amide
C2H6N2O
FM 74.08
(b) How many grams of glycine amide should be added to 1.00 g
of glycine amide hydrochloride to give 100 mL of solution with pH
8.00?
(c) What would be the pH if the solution in part (a) were mixed
with 5.00 mL of 0.100 M HCl?
(d) What would be the pH if the solution in part (c) were mixed
with 10.00 mL of 0.100 M NaOH?
(e) What would be the pH if the solution in part (a) were mixed
with 90.46 mL of 0.100 M NaOH? (This is exactly the quantity of
NaOH required to neutralize the glycine amide hydrochloride.)
9-K. A solution with an ionic strength of 0.10 M containing
0.010 0 M phenylhydrazine has a pH of 8.13. Using activity coefficients correctly, find pKa for the phenylhydrazinium ion found in
phenylhydrazine hydrochloride. Assume that BH 0.80.
ß NHNH2
Phenylhydrazine
B
ß NHNH3C1
Phenylhydrazine hydrochloride
BHCl
9-L.
Use the GOAL SEEK spreadsheet at the end of the chapter to find the pH of 1.00 L of solution containing 0.030 mol HA
(pKa 2.50) and 0.015 mol NaA. What would the pH be with the
approximations [HA] 0.030 and [A] 0.015?
pKa 4.25
Acrylic acid
Problems
Strong Acids and Bases
9-1. Why doesn’t water dissociate to produce
107 M OH when some HBr is added?
9-2. Calculate the pH of (a) 1.0
102 M KOH.
107
M
H
and
103 M HBr; (b) 1.0
9-3. Calculate the pH of 5.0 108 M HClO4. What fraction of
the total H in this solution is derived from dissociation of
water?
Problems
9-4. (a) The measured pH of 0.100 M HCl at 25°C is 1.092. From
this information, calculate the activity coefficient of H and compare your answer with that in Table 8-1.
(b) The measured pH of 0.010 0 M HCl 0.090 0 M KCl at 25°C
is 2.102. From this information, calculate the activity coefficient of
H in this solution.
(c) The ionic strengths of the solutions in parts (a) and (b) are the
same. What can you conclude about the dependence of activity
coefficients on the particular ions in a solution?
177
Weak-Acid Equilibria
B
A
9-5. Write the chemical reaction whose equilibrium constant is
(a) Ka for benzoic acid, C6H5CO2H
(b) Kb for benzoate ion, C6H5CO
2
(c) Kb for aniline, C6H5NH2
(d) Ka for anilinium ion, C6H5NH3
1
2
3
4
5
6
9-6. Find the pH and fraction of dissociation () of a 0.100 M solution of the weak acid HA with Ka 1.00 105.
9-7. BHClO4 is a salt formed from the base B (Kb 1.00 104 )
and perchloric acid. It dissociates into BH, a weak acid, and ClO4 ,
which is neither an acid nor a base. Find the pH of 0.100 M
BHClO4 .
9-8. Find the pH and concentrations of (CH3 ) 3N and (CH3 ) 3NH in
a 0.060 M solution of trimethylammonium chloride.
9-9. Use the reaction quotient, Q, to explain why the fraction of dissociation of weak acid, HA, increases when the solution is diluted
by a factor of 2.
Using Excel GOAL SEEK
x2/(F-x) =
x=
0.01
1.1111E-03
F=
0.1
Weak-Base Equilibria
9-18. Covalent compounds generally have higher vapor pressure
than ionic compounds. The “fishy” smell of fish arises from amines
in the fish. Explain why squeezing lemon (which is acidic) onto fish
reduces the fishy smell (and taste).
9-19. Find the pH and fraction of association () of a 0.100 M
solution of the weak base B with Kb 1.00 105.
9-20. Find the pH and concentrations of (CH3 ) 3N and (CH3 ) 3NH
in a 0.060 M solution of trimethylamine.
9-10. When is a weak acid weak and when is a weak acid strong?
Show that the weak acid HA will be 92% dissociated when dissolved
in water if the formal concentration is one-tenth of Ka (F Ka /10) .
Show that the fraction of dissociation is 27% when F 10Ka. At
what formal concentration will the acid be 99% dissociated? Compare your answer with the left-hand curve in Figure 9-2.
9-22. Calculate the fraction of association () for 1.00 101,
1.00 102, and 1.00 1012 M sodium acetate. Does increase
or decrease with dilution?
9-11. A 0.045 0 M solution of benzoic acid has a pH of 2.78. Calculate pKa for this acid.
9-24. A 0.10 M solution of a base is 2.0% hydrolyzed ( 0.020).
Find Kb.
9-12. A 0.045 0 M solution of HA is 0.60% dissociated. Calculate
pKa for this acid.
9-13. Barbituric acid dissociates as follows:
O
O
¡
¡
¡
¡
O
O
N: H
5
Ka 9.8 10
NH #£££££4
HN
HN
¡
O
¡
O
Barbituric acid
HA
A
9-21. Find the pH of 0.050 M NaCN.
9-23. A 0.10 M solution of a base has pH 9.28. Find Kb.
9-25. Show that the limiting fraction of association of a base in
water, as the concentration of base approaches 0, is 107 Kb /
(1 107 Kb ). Find the limiting value of for Kb 104 and for
Kb 1010.
Buffers
9-26. Describe how to prepare 100 mL of 0.200 M acetate buffer,
pH 5.00, starting with pure liquid acetic acid and solutions containing ⬃3 M HCl and ⬃3 M NaOH.
9-27. Why is the pH of a buffer nearly independent of concentration?
(a) Calculate the pH and fraction of dissociation of 102.00 M barbituric acid.
(b) Calculate the pH and fraction of dissociation of 1010.00 M
barbituric acid.
9-28. Why does buffer capacity increase as the concentration of
buffer increases?
9-14. Using activity coefficients, find the pH and fraction of dissociation of 50.0 mM hydroxybenzene (phenol) in 0.050 M LiBr.
Take the size of C6H5O to be 600 pm.
9-30. Why is the buffer capacity maximum when pH pKa?
9-15. Cr3 is acidic by virtue of the hydrolysis reaction
Ka1
Cr3 H2O ∆ Cr(OH) 2 H
[Further reactions produce Cr(OH) 2, Cr(OH) 3, and Cr(OH) 4 .] Find
the value of Ka1 in Figure 6-8. Considering only the Ka1 reaction,
find the pH of 0.010 M Cr(ClO4 ) 3. What fraction of chromium is in
the form Cr(OH) 2?
9-16. From the dissociation constant of HNO3 at 25°C in Box 9-1,
find the percent dissociated in 0.100 M HNO3 and in 1.00 M HNO3.
9-17.
Excel GOAL SEEK. Solve the equation x2 /(F x) K
by using GOAL SEEK (Section 6-8). Guess a value of x in cell A4
and evaluate x2 /(F x) in cell B4. Use GOAL SEEK to vary the
value of x until x2 /(F x) is equal to K. Use your spreadsheet to
check your answer to Problem 9-6.
178
9-29. Why does buffer capacity increase as a solution becomes very
acidic (pH ⬇ 1) or very basic (pH ⬇ 13)?
9-31. Explain the following statement: The Henderson-Hasselbalch
equation (with activity coefficients) is always true; what may not be
correct are the values of [A] and [HA] that we choose to use in the
equation.
9-32. Which of the following acids would be most suitable for
preparing a buffer of pH 3.10: (i) hydrogen peroxide; (ii) propanoic
acid; (iii) cyanoacetic acid; (iv) 4-aminobenzenesulfonic acid?
9-33. A buffer was prepared by dissolving 0.100 mol of the weak
acid HA (Ka 1.00 105 ) plus 0.050 mol of its conjugate base
NaA in 1.00 L. Find the pH.
9-34. Write the Henderson-Hasselbalch equation for a solution
of formic acid. Calculate the quotient [HCO2 ]/[HCO2H] at
(a) pH 3.000; (b) pH 3.744; (c) pH 4.000.
9-35. Given that pKb for nitrite ion (NO2 ) is 10.85, find the quotient [HNO2]/[NO2 ] in a solution of sodium nitrite at (a) pH 2.00;
(b) pH 10.00.
CHAPTER 9 Monoprotic Acid-Base Equilibria
9-36. (a) Would you need NaOH or HCl to bring the pH of 0.050 0
M HEPES (Table 9-2) to 7.45?
(b) Describe how to prepare 0.250 L of 0.050 0 M HEPES, pH 7.45.
9-37. How many milliliters of 0.246 M HNO3 should be added to
213 mL of 0.006 66 M 2,2¿ -bipyridine to give a pH of 4.19?
9-38. (a) Write the chemical reactions whose equilibrium constants
are Kb and Ka for imidazole and imidazole hydrochloride, respectively.
(b) Calculate the pH of a solution prepared by mixing 1.00 g of
imidazole with 1.00 g of imidazole hydrochloride and diluting to
100.0 mL.
(c) Calculate the pH of the solution if 2.30 mL of 1.07 M HClO4 are
added.
(d) How many milliliters of 1.07 M HClO4 should be added to 1.00 g
of imidazole to give a pH of 6.993?
9-39. Calculate the pH of a solution prepared by mixing 0.080 0 mol
of chloroacetic acid plus 0.040 0 mol of sodium chloroacetate in
1.00 L of water.
(a) First do the calculation by assuming that the concentrations of
HA and A equal their formal concentrations.
(b) Then do the calculation, using the real values of [HA] and [A]
in the solution.
(c) Using first your head, and then the Henderson-Hasselbalch
equation, find the pH of a solution prepared by dissolving all the
following compounds in one beaker containing a total volume of
1.00 L: 0.180 mol ClCH2CO2H, 0.020 mol ClCH2CO2Na, 0.080
mol HNO3, and 0.080 mol Ca(OH) 2. Assume that Ca(OH) 2 dissociates completely.
9-40. Calculate how many milliliters of 0.626 M KOH should be
added to 5.00 g of MOBS (Table 9-2) to give a pH of 7.40.
9-41. (a) Use Equations 9-20 and 9-21 to find the pH and concentrations of HA and A in a solution prepared by mixing 0.002 00 mol of
acetic acid plus 0.004 00 mol of sodium acetate in 1.00 L of water.
Problems
(b)
After working part (a) by hand, use Excel
find the same answers.
GOAL SEEK
to
9-42. (a) Calculate the pH of a solution prepared by mixing 0.010 0
mol of the base B (Kb 102.00 ) with 0.020 0 mol of BHBr and
diluting to 1.00 L. First calculate the pH by assuming [B] 0.010 0
and [BH] 0.020 0 M. Compare this answer with the pH calculated without making such an assumption.
(b)
After working part (a) by hand, use Excel
find the same answers.
GOAL SEEK
to
9-43. Effect of ionic strength on pKa. Ka for the H2PO4 /HPO2
4
buffer is
Ka [HPO2
4 ][H ]HPO H
2
4
4
[H2PO ]H PO
2
107.20
4
If you mix a 1:1 mole ratio of H2PO4 and HPO2
4 at 0 ionic strength,
the pH is 7.20. Find the pH of a 1:1 mixture of H2PO4 and HPO2
4
at an ionic strength of 0.10. Remember that pH log A H log[H ]H .
9-44. Systematic treatment of equilibrium. The acidity of Al3 is
determined by the following reactions. Write the equations needed
to find the pH of Al(ClO4 ) 3 at a formal concentration F.
1
Al3 H2O ∆ AlOH2 H
2
Al3 2H2O ∆ Al(OH) 2 2H
K22
2Al3 2H2O ∆ Al2 (OH) 4
2 2H
3
Al3 3H2O ∆ Al(OH) 3 (aq) 3H
4
Al3 4H2O ∆ Al(OH) 4 4H
K43
3Al3 4H2O ∆ Al3 (OH) 5
4 4H
179