1 Water, pH and “p” notation
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17.1  Water, pH, and “p” Notation
775
Often, for convenience, we omit the water molecule that carries the hydrogen ion and
write H+ in place of H3O+. The equilibrium equation for the autoionization of water then
simplifies to
���
H2O �
� H+ + OHThe equation for Kw based on this is likewise simplified.
[H+] [OH] = Kw
(17.2)
In pure water, the concentrations of H+ and OH produced by the
autoionization are equal because the ions are formed in equal numbers. It’s
been found that the concentrations have the following values at 25 °C.
Temperature (°C)
Kw
0
1.5 × 1015
10
3.0 × 1015
20
6.8 × 1015
25
1.0 × 1014
30
1.5 × 1014
37a
2.5 × 1014
40
3.0 × 1014
50
5.5 × 1014
60
9.5 × 1014
[H+] = [OH] = 1.0 × 107 mol L1
Therefore, at 25 °C,
Kw = (1.0 × 107) × (1.0 × 107)
Kw = 1.0 × 1014
(17.3)
As with other equilibrium constants, the value of Kw varies with temperature (see Table 17.1). Unless stated otherwise we will deal with systems
at 25 °C.
+
Kw at Various Temperatures
Table 17.1

Effect of Solutes on [H ] and [OH ]
a
Water’s autoionization takes place in any aqueous solution, and because of
Normal body temperature
the effects of other solutes, the molar concentrations of H+ and OH may
not be equal. Nevertheless, their product, Kw, is the same. Thus, although Equations
17.1–17.3 were derived for pure water, they also apply to dilute aqueous solutions. The
significance of this must be emphasized. In any aqueous solution, the product of [H+] and
[OH] equals Kw, although the two molar concentrations may not actually equal each other.
Criteria for Acidic, Basic, and Neutral Solutions
One of the consequences of the autoionization of water is that in any aqueous solution,
there are always both H3O+ and OH ions, regardless of what solutes are present. This means
that in a solution of the acid HCl there is some OH, and in a solution of the base NaOH,
there is some H3O+. We call a solution acidic or basic depending on which ion has the
largest concentration.
A neutral solution is one in which the molar concentrations of H3O+ and OH are
equal. An acidic solution is one in which some solute has made the molar concentration of
H3O+ greater than that of OH. On the other hand, a basic solution exists when the molar
concentration of OH exceeds that of H3O+.
Neutral solution [H3O+] = [OH]
Acidic solution
[H3O+] > [OH]
Basic solution
[H3O+] < [OH]
n Because of the autoionization of
water, even the most acidic solution
has some OH, and even the most
basic solution has some H3O+.
If we consider the autoionization constant of water, we can look at some quantitative
relationships of acids and bases. For instance, we can take a solution that has an easily
measurable 0.0020 M concentration of H3O+ and calculate an extremely small OH concentration that exists in the same solution. Rearranging Equation 17.3, we get
[OH − ] =
Kw
1.0 × 10 −14
=
= 5.0 × 10 −12 M
[H3O+ ]
2.0 × 10 −3
A similar calculation allows us to determine the hydronium ion concentration if we know
the hydroxide ion concentration.
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776 Chapter 17  Acid–Base Equilibria in Aqueous Solutions
Practice Exercises
17.1  Commercial, concentrated, hydrochloric acid has a concentration of 12 moles per
liter. If the molarity of hydronium ions is assumed to also be 12 M, what is the concentration of hydroxide ions? (Hint: Use the relationships just discussed.)
17.2  An aqueous solution of sodium bicarbonate, NaHCO3, has a molar concentration
of hydroxide ion of 7.8 × 106 M. What is the molar concentration of hydrogen ion? Is
the solution acidic, basic, or neutral?
The pH Concept
In most solutions of weak acids and bases, the molar concentrations of H+ and OH are
very small, like those we just discussed. Writing and comparing two exponential values is not
always easy. A Danish chemist, S. P. L. Sørenson (1868–1939), suggested an easier approach.
To make comparisons of small values of [H+] easier, Sørenson defined a quantity that
he called the pH of the solution as follows.
pH = log [H+]
Definition of pH
The properties of logarithms let us rearrange Equation 17.4 as follows.
n Calculating pH
pH = log [H+]
= log (3.6 × 104)
= (3.44)
= 3.44
Calculating [H+]
log [H+] = pH
log [H+] = 12.58
log [H+] = 12.58
[H+] = antilog (12.58)
[H+] = 1012.58
[H+] = 2.6 × 1013 M
n If you still have questions
about these calculations, see
the mathematics review in
Appendix A.
(17.4)1
[H+] = 10pH
(17.5)
Equation 17.4 can be used to calculate the pH of a solution if its molar concentration of
H+ is known. On the other hand, if the pH is known and we wish to calculate the molar
concentration of hydrogen ions, we apply Equation 17.5. For example, if the hydrogen
ion concentration, [H+], is 3.6 × 104, we take the logarithm of this value (which is
3.44) and change its sign so that the result is a pH of 3.44. With digital calculators, it’s
as simple as pressing the log key and then changing the sign. Determining the hydrogen
ion concentration using Equation 17.5 requires that we change the sign of the pH and
then take the antilogarithm. For instance, if we are given a pH of 12.58 we first change it
to 12.58. We then take the antilogarithm according to the instructions that came with
your calculator. In this case the result is 2.6 × 1013 = [H+].
Acidic, Basic, and Neutral Solutions
One meaning attached to pH is that it is a measure of the acidity of a solution. Hence, we
may define acidic, basic, and neutral in terms of pH values. At 25 °C, in pure water, or in
any solution that is neutral,
[H+] = [OH] = 1.0 × 107 M
n Neutral pH is 7.00 only for 25 °C.
At other temperatures a neutral
solution still has [H+] = [OH], but
because Kw changes with temperature,
so do the values of [H+] and [OH].
This causes the pH of the neutral
solution to differ from 7.00. For
example, at normal body temperature
(37 °C), a neutral solution has [H+] =
[OH] = 1.6 × 107, so the
pH = 6.80.
Therefore, by Equation 17.4, the pH of a neutral solution at 25 °C is 7.00.2
An acidic solution is one in which [H+] is larger than 107 M and so has a pH less than
7.00. Thus, as a solution’s acidity increases, its pH decreases.
A basic solution is one in which the value of [H+] is less than 107 M and so has a pH
that is greater than 7.00. As a solution’s acidity decreases, its pH increases. These simple
relationships may be summarized as follows. At 25 °C:
pH = 7.00
Neutral solution
pH < 7.00
Acidic solution
pH > 7.00
Basic solution
The pH values for some common substances are given on a pH scale in Figure 17.1.
1
In this equation and similar ones, like Equation 17.6, the logarithm of only the numerical part of the
bracketed term is taken. The physical units must be mol L1, but they are set aside for the calculation.
2
Recall that the rule for significant figures in logarithms is that the number of decimal places in the logarithm of
a number equals the number of significant figures in the number. For example, 3.2 × 105 has just two
significant figures. The logarithm of this number, displayed on a pocket calculator, is 4.494850022. We
write the logarithm, when correctly rounded, as 4.49.
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17.1  Water, pH, and “p” Notation
Most
acidic
1
0.1 M HCI
2
0.3 M citric acid
Lemon juice
0.1 M acetic acid
Pickles, sour
Sauerkraut
Orange juice
Carbonic acid
(saturated)
Beer
0.03 M boric acid
3
4
5
6
pH 7
“p” Notation
8
The logarithmic definition of pH, Equation 17.4, has proved to be so useful that it has
been adapted to quantities other than [H+]. Thus, for any quantity X,
we may define a term, pX, in the following way.
9
pX =  log X
Milk
Neutral
Defining the
pfunction, pX
pOH =  log [OH]
Pure water
Blood
0.1 M NaHCO3
0.3 M borax
10
Milk of magnesia
11
(17.6)
For example, to express small concentrations of hydroxide ion,
we can define the pOH of a solution as
Soft drinks
One of the deceptive features of pH is how much the hydrogen ion concentration
changes with a relatively small change in pH. Thus, because of the logarithmic scale, a
change of one pH unit corresponds to a tenfold change in the hydrogen ion concentration. For example, at pH 6 the hydrogen ion concentration is 1 × 106 mol L1. At
pH 5, it’s 1 × 105 mol L1, or ten times greater. That’s a major difference.
Solutions having hydrogen ion concentrations greater than 1 M can be prepared and
they should have negative pH values. However, at such high concentrations it is usually
easier to simply write the molarity of the hydrogen ions. For example, in a solution where
the value of [H+] is 2 M, the “2” is a simple enough number; we can’t simplify it further
by using its corresponding pH value. When [H+] is 2.00 M, the pH, calculated by
Equation 17.4, would be log (2.00) or 0.301. There is nothing wrong with negative
pH values, but they offer no advantage over just writing the actual value of [H+].
777
0.1 M ammonia
0.05 M Na2CO3
0.03 M Na3PO4
Limewater
12
13
0.1 M KOH
14
1.0 M NaOH
Most
basic
Similarly, for Kw we can define pKw as follows.
pKw =  log Kw
The numerical value of pKw at 25 °C equals log (1.0 × 1014) or (14.00). Thus,
pKw = 14.00 (at 25 °C)
Figure 17.1  The pH scale.
Common substances and their
approximate pH values.
A useful relationship among pH, pOH, and pKw can be derived by taking the
logarithm of all terms in Equation 17.1 or 17.2 and changing the sign to obtain
Equation 17.7:
pH + pOH = pKw = 14.00
(at 25 °C)
(17.7)
This tells us that in an aqueous solution of any solute at 25 °C, the sum of pH and pOH
is 14.00.
Relating pH and pOH to pKw
pH Calculations
Let’s now look at some pH calculations using Equations 17.4 and 17.7. These are calculations you will be performing often in the rest of this chapter.
If we know the [H+] of a solution—let’s use 6.3 × 105 M as an example—we can
calculate the pH by taking the logarithm and changing the sign. The pH we calculate
should be 4.20. We now have two ways to determine the pOH. Using Equation 17.2
allows us to determine [OH], and then we can take the logarithm and change the sign
to calculate the pOH. An easier method is to use Equation 17.7 and simply subtract
the pH from 14.00. This subtraction tells us that the pOH must be 9.80. Equations
17.4 and 17.7 tell us that if we know one of the four terms, [H+], [OH], pH, or
pOH, we can calculate the other three. Another way to look at this helps us with solving a problem. If we are given any one of these four values, it is the same as knowing
all of them. For instance, if you know the pOH but a problem requires the [H+], we
know how to easily calculate the needed value. The following practice exercises make
use of these calculations.
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n Calculating pH and pOH
pH = log (6.3 × 105)
= (4.20)
= 4.20
pKw = 14.00 = 4.20 + pOH
pOH = 14.00  4.20
= 9.80
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778 Chapter 17  Acid–Base Equilibria in Aqueous Solutions
Practice Exercises
17.3  Water draining from old coal and mineral mines often has pH values of 4.0 or
lower. The cause is thought to stem from reactions of ground water and oxygen with iron
pyrite, FeS. What are the pOH, [H+], and [OH] in an acid mine drainage sample that
has a pH of 4.25? (Hint: Recall that pH, pOH, [H+], and [OH] are all interrelated using
the Kw and pKw.)
17.4  Because rain washes pollutants out of the air, the lakes in many parts of the world
have undergone pH changes. In a New England state, the water in one lake was found to
have a hydrogen ion concentration of 3.2 × 105 mol L1. What are the calculated pH
and pOH values of the lake’s water? Is the water acidic or basic?
17.5  Find the values of [H+] and [OH] that correspond to each of the following values
of pH. State whether each solution is acidic or basic.
(a) 2.90 (the approximate pH of lemon juice)
(b) 3.85 (the approximate pH of sauerkraut)
(c) 10.81 (the pH of milk of magnesia, an antacid)
(d) 4.11 (the pH of orange juice, on the average)
(e) 11.61 (the pH of dilute, household ammonia)
pH Meters
One of the remarkable things about pH is that it can be easily measured using
an instrument called a pH meter (Figure 17.2). An electrode system sensitive to
the hydrogen ion concentration in a solution is first dipped into a standard solution of known pH to calibrate the instrument. Once calibrated, the apparatus
can then be used to measure the pH of any other solution simply by immersing
the electrodes into it. The accuracy of pH meter measurements is usually ± 0.02
pH units.
17.2  pH of Strong Acid
Figure 17.2  A pH meter. After the
instrument has been calibrated with a solution
of a known pH, the combination electrode is
dipped into the solution to be tested and the
pH is read from the meter. (Richard Megna/
Fundamental Photographs.)
and Base Solutions
Many solutes affect the pH of an aqueous solution. In this section we examine
how strong acids and bases behave and how to calculate the pH of their solutions. Weak acids and bases have similar effects, which we will discuss in the next
section.
Strong Acids and Bases
n If necessary, review the list of
strong acids on page 170.
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In Section 16.3 we saw that strong acids and bases are considered to be 100% dissociated
in an aqueous solution. This makes calculating the concentrations of H+ and OH in
their solutions a relatively simple task.
When the solute is a strong monoprotic acid, such as HCl or HNO3, the molarity of H+
will be the same as the molarity of the strong acid. Thus, a 0.010 M solution of HCl contains 0.010 mol L1 of H+ and a 0.0020 M solution of HNO3 contains 0.0020 mol L1
of H+.
To calculate the pH of a solution of a strong monoprotic acid, we use the molarity of
the H+ obtained from the stated molar concentration of the acid. Thus, the 0.010 M HCl
solution mentioned above has a pH of 2.00.
For strong bases, calculating the OH concentration is similarly straightforward. A
0.050 M solution of NaOH contains 0.050 mol L1 of OH because the base is fully dissociated and each mole of NaOH releases one mole of OH when it dissociates. For bases
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17.2  pH of Strong Acid and Base Solutions
779
such as Ba(OH)2 we have to recognize that two moles of OH are released by each mole
of the base.
Ba(OH)2(s) → Ba2+(aq) + 2OH(aq)
Therefore, if a solution contained 0.010 mol Ba(OH)2 per liter, the concentration of
OH would be 0.020 M. Of course, once we know the OH concentration we can
calculate pOH, from which we can calculate the pH. Working through the following
practice exercises will help to solidify these methods.
17.6  Calculate the [H+], pH, and pOH in 0.0050 M HNO3. (Hint: HNO3 is a strong
acid.)
17.7  Calculate the pOH, [H+], and pH in a solution made by weighing 1.20 g of KOH
and dissolving it in sufficient water to make 250.0 mL of solution.
17.8  Rhododendrons are shrubs that produce beautiful flowers in the springtime. They
only grow well in soil that has a pH that is 5.5 or slightly lower. What is the hydrogen ion
concentration in the soil moisture if the pH is 5.5?
Practice Exercises
Effect of Solute on the Ionization of Water
In the preceding calculations, we have made a critical and correct assumption—namely,
that the autoionization of water contributes negligibly to the total [H+] in a solution of an
acid and to the total [OH] in a solution of a base.3 Let’s take a closer look at this.
In a solution of an acid, there are actually two sources of H+. One is from the ionization
of the acid itself and the other is from the autoionization of water. Thus,
[H+]total = [H+]from solute + [H+]from H2O
+
Except in very dilute solutions of acids, the amount of H contributed by the water
([H+]from H2O) is small compared to the amount of H+ contributed by the acid ([H+]from solute).
For instance, in 0.020 M HCl the molarity of OH is 5.0 × 1013 M. The only source of
OH in this acidic solution is from the autoionization of water, and the amounts of OHand H+ formed by the autoionization of water must be equal. Therefore, [H+]from H2O
also equals 5.0 × 1013 M. If we now look at the total [H+] for this solution, we have
n A similar equation applies to the
equilibrium OH concentration in a
solution of a base.
[OH]equilib =
[OH]from solute + [OH]from H2O
[H+]total = 0.020 M + 5.0 × 1013 M
(from HCl)
(from H2O)
= 0.020 M (rounded correctly)
In any solution of an acid, the autoionization of water is suppressed by the H+ furnished by the solute. It’s simply an example of Le Châtelier’s principle. If we look at
the autoionization reaction, we can see that if some H+ is provided by an external source
(an acidic solute, for example), the position of equilibrium will be shifted to the left.
H2O
H+ + OH−
Adding H+ from a solute
causes the position of
equilibrium to shift to the left.
As the results of our calculation have shown, the concentrations of H+ and OH from the
autoionization reaction are reduced well below their values in a neutral solution (1.0 ×
107 M). Therefore, except for very dilute solutions (106 M or less), we will assume that
all of the H+ in the solution of an acid comes from the solute. Similarly, we’ll assume that
in a solution of a base, all the OH comes from the dissociation of the solute.
3
The autoionization of water cannot be neglected when working with very dilute solutions of acids and bases.
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780 Chapter 17  Acid–Base Equilibria in Aqueous Solutions
17.3  Ionization Constants, Ka and Kb
As you’ve learned in Chapter 16, weak acids and bases are incompletely ionized in water
and exist in solution as molecules that are in equilibria with the ions formed by their
reactions with water. To deal quantitatively with these equilibria it is essential that you
be able to write correct chemical equations for the equilibrium reactions, from which
you can then obtain the corresponding correct equilibrium laws. Fortunately, there is a
pattern that applies to the way these substances react. Once you’ve learned how to write
the correct equation for one acid or base, you can write the correct equation for any other.
Reaction of a Weak Acid with Water
In aqueous solutions, all weak acids behave the same way. They are Brønsted acids and,
therefore, proton donors. Some examples are HC2H3O2, HSO4, and NH4+. In water,
these participate in the following equilibria.
���
HC2H3O2 + H2O �
� H3O+ + C2H3O2���
HSO4 + H2O �
� H3O+ + SO42���
NH4+ + H2O �
� H3O+ + NH3
Notice that in each case, the acid reacts with water to give H3O+ and the corresponding
conjugate base. We can represent these reactions in a general way using HA to represent
the formula of the weak acid and A to represent the anion.
General equation for the
ionization of a weak acid
n Some call Ka the acid dissociation
constant.
���
HA + H2O �
� H3O+ + A
(17.8)
As you can see from the equations above, HA does not have to be electrically neutral; it
can be a molecule such as HC2H3O2, a negative ion such as HSO4 or a positive ion such
as NH4+. (Of course, the actual charge on the conjugate base will then depend on the
charge on the parent acid.)
Following the procedure developed in Chapter 15, we can also write a general equation for the equilibrium law, omitting the concentration of liquid water as usual, for
Equation 17.8,
[H3O+ ][[ A− ]
= Ka
[HA ]
(17.9)
The new constant Ka is called an acid ionization constant. Abbreviating H3O+ as H+, the
equation for the ionization of the acid can be simplified as
���
HA �
� H+ + Afrom which the expression for Ka is obtained directly.
[H+ ][[ A− ]
[HA ]
Ka =
(17.10)
You should learn how to write the chemical equation for the ionization of a weak acid
and be able to write the equilibrium law corresponding to its Ka. For example, if we have
a weak acid such as HNO2 we can write the ionization reaction as
���
HNO2 �
� H+ + NO2and the corresponding equilibrium law is
The meat products shown here
contain nitrite ion as a preservative. (Andy Washnik)
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Ka =
[H+ ][[NO2− ]
[HNO2 ]
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17.3  Ionization Constants, Ka and Kb 781
17.9  For each of the following acids, write the equation for its ionization in water and
the appropriate expression for Ka: (a) HC2H3O2, (b) (CH3)3NH+, (c) H3PO4. (Hint:
Determine the conjugate base for each of these acids.)
Practice Exercises
17.10  For each of the following acids, write the equation for its ionization in water and
the appropriate expression for Ka: (a) HCHO2, (b) (CH3)2NH2+, (c) H2PO4.
For weak acids, values of Ka are usually quite small and can be conveniently represented
in a logarithmic form similar to pH. Thus, we can define the pKa of an acid as
pKa = log Ka
n Also, Ka = antilog (pKa) or
The strength of a weak acid is determined by its value of Ka; the larger the Ka, the stronger
and more fully ionized the acid. Because of the negative sign in the defining equation for
pKa, the stronger the acid, the smaller is its value of pKa. The values of Ka and pKa for some
typical weak acids are given in Table 17.2. A more complete list is located in Appendix C.8.
17.11  Use Table 17.2 to find all the acids that are stronger than acetic acid and weaker
than formic acid. (Hint: It may be easier to focus on the Ka values since they are directly
related to acid strength.)
Ka =10pKa
n The values of Ka for strong acids are
very large and are not tabulated. For
many strong acids, Ka values have not
been measured.
Practice Exercises
17.12  Two acids, HA and HB, have pKa values of 3.16 and 4.14, respectively. Which is
the stronger acid? What are the Ka values for these acids?
Table 17.2
Ka and pKa Values for Weak Monoprotic Acids at 25 °C
Name of Acid
Iodic acid
Chloroacetic acid
Nitrous acid
Hydrofluoric acid
Cyanic acid
Formic acid
Barbituric acid
Hydrazoic acid
Acetic acid
Butanoic acid
Propanoic acid
Hypochlorous acid
Hydrocyanic acid
Phenol
Hydrogen peroxide
Formula
HIO3
HC2H2O2Cl
HNO2
HF
HOCN
HCHO2
HC4H3N2O3
HN3
HC2H3O2
HC4H7O2
HC3H5O2
HOCl
HCN
HC6H5O
H2O2
Ka
pKa
1
0.77
3
2.85
4
3.34
4
3.46
4
3.7
4
3.74
5
4.01
5
4.60
5
4.74
5
4.82
5
4.89
8
7.52
10
9.31
1.3 × 10
10
9.89
2.4 × 10
12
11.62
1.7 × 10
1.4 × 10
4.6 × 10
3.5 × 10
2 × 10
1.8 × 10
9.8 × 10
2.5 × 10
1.8 × 10
1.5 × 10
1.3 × 10
3.0 × 10
4.9 × 10
Reaction of a Weak Base with Water
As with weak acids, all weak bases behave in a similar manner in water. They are weak
Brønsted bases and are therefore proton acceptors. Examples are ammonia, NH3, and
acetate ion, C2H3O2. Their reactions with water are
���
NH3 + H2O �
� NH4+ + OH���
C2H3O2 + H2O �
� HC2H3O2 + OH
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