4: Reactions That Form Water: Acids and Bases
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16.4 The pH Scale
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Now we compute the pH.
pH ϭ Ϫlog[Hϩ] ϭ Ϫlog(1.0 ϫ 10Ϫ11) ϭ 11.00
b. In this case we are given the [Hϩ] and we can compute the pH.
pH ϭ Ϫlog[Hϩ] ϭ Ϫlog(1.0) ϭ 0
We next solve the Kw expression for [OHϪ].
[OHϪ] ϭ
Kw
[Hϩ]
ϭ
1.0 ϫ 10Ϫ14
ϭ 1.0 ϫ 10Ϫ14 M
1.0
Now we compute the pOH.
pOH ϭ Ϫlog[OHϪ] ϭ Ϫlog(1.0 ϫ 10Ϫ14) ϭ 14.00 ■
We can obtain a convenient relationship between pH and pOH by starting with the Kw expression [Hϩ][OHϪ] ϭ 1.0 ϫ 10Ϫ14 and taking the negative
log of both sides.
Andrew Syred/Science Photo Library/Photo Researchers, Inc.
Ϫlog([Hϩ][OHϪ]) ϭ Ϫlog(1.0 ϫ 10Ϫ14)
Because the log of a product equals the sum of the logs of the terms—that is,
log(A ϫ B) ϭ log A ϩ log B—we have
Ϫlog[Hϩ] Ϫlog[OHϪ] ϭ Ϫlog(1.0 ϫ 10Ϫ14) ϭ 14.00
pH
pOH
which gives the equation
pH ϩ pOH ϭ 14.00
Red blood cells can exist only
over a narrow range of pH.
This means that once we know either the pH or the pOH for a solution, we
can calculate the other. For example, if a solution has a pH of 6.00, the pOH
is calculated as follows:
pH ϩ pOH ϭ 14.00
pOH ϭ 14.00 Ϫ pH
pOH ϭ 14.00 Ϫ 6.00 ϭ 8.00
EXAMPLE 16.7
Calculating pOH from pH
The pH of blood is about 7.4. What is the pOH of blood?
SOLUTION
pH ϩ pOH ϭ 14.00
pOH ϭ 14.00 Ϫ pH
ϭ 14.00 Ϫ 7.4
ϭ 6.6
The pOH of blood is 6.6.
Self-Check EXERCISE 16.4 A sample of rain in an area with severe air pollution has a pH of 3.5. What is
the pOH of this rainwater?
See Problems 16.45 and 16.46. ■
530 Chapter 16 Acids and Bases
It is also possible to find the [Hϩ] or [OHϪ] from the pH or pOH. To find the
[Hϩ] from the pH, we must go back to the definition of pH:
pH ϭ Ϫlog[Hϩ]
or
ϪpH ϭ log[Hϩ]
To arrive at [Hϩ] on the right-hand side of this equation we must “undo” the
log operation. This is called taking the antilog or the inverse log.
Inverse log (ϪpH) ϭ inverse log (log[Hϩ])
Inverse log (ϪpH) ϭ [Hϩ]
MATH SKILL BUILDER
This operation may involve a 10x
key on some calculators.
There are different methods for carrying out the inverse log operation on
various calculators. One common method is the two-key inv log sequence.
(Consult the user’s manual for your calculator to find out how to do the
antilog or inverse log operation.) The steps in going from pH to [Hϩ] are as
follows:
Steps for Calculating [H؉] from pH
Step 1 Take the inverse log (antilog) of ϪpH to give [Hϩ] by using the
inv log keys in that order. (Your calculator may require different
keys for this operation.)
Step 2 Press the minus [Ϫ] key.
Step 3 Enter the pH.
For practice, we will convert pH ϭ 7.0 to [Hϩ].
Measuring the pH of the water
in a river.
David Woodfall/Stone/Getty Images
pH ϭ 7.0
ϪpH ϭ Ϫ7.0
16.4 The pH Scale
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The inverse log of Ϫ7.0 gives 1 ϫ 10Ϫ7.
[Hϩ] ϭ 1 ϫ 10Ϫ7 M
This process is illustrated further in Example 16.8.
EXAMPLE 16.8
Calculating [H؉] from pH
The pH of a human blood sample was measured to be 7.41. What is the [Hϩ]
in this blood?
SOLUTION
pH ϭ 7.41
ϪpH ϭ Ϫ7.41
[Hϩ] ϭ inverse log of Ϫ7.41 ϭ 3.9 ϫ 10Ϫ8
[Hϩ] ϭ 3.9 ϫ 10Ϫ8 M
Notice that because the pH has two decimal places, we need two significant figures for [Hϩ].
Self-Check EXERCISE 16.5 The pH of rainwater in a polluted area was found to be 3.50. What is the [Hϩ]
for this rainwater?
See Problems 16.49 and 16.50. ■
A similar procedure is used to change from pOH to [OHϪ], as shown in
Example 16.9.
EXAMPLE 16.9
Calculating [OH؊] from pOH
The pOH of the water in a fish tank is found to be 6.59. What is the [OHϪ]
for this water?
SOLUTION
We use the same steps as for converting pH to [Hϩ], except that we use the
pOH to calculate the [OHϪ].
pOH ϭ 6.59
ϪpOH ϭ Ϫ6.59
[OHϪ] ϭ inverse log of Ϫ6.59 ϭ 2.6 ϫ 10Ϫ7
[OHϪ] ϭ 2.6 ϫ 10Ϫ7 M
Note that two significant figures are required.
Self-Check EXERCISE 16.6 The pOH of a liquid drain cleaner was found to be 10.50. What is the [OHϪ]
for this cleaner?
See Problems 16.51 and 16.52. ■
C H E M I S T R Y I N F OCUS
Garden-Variety Acid–Base Indicators
When placed in an acidic solution, most of
the basic form of the indicator is converted to
the acidic form by the reaction
What
can flowers tell us about acids and
bases? Actually, some flowers can tell us whether
the soil they are growing in is acidic or basic. For
example, in acidic soil, bigleaf hydrangea blossoms will be blue; in basic (alkaline) soil, the
flowers will be red. What is the secret? The pigment in the flower is an acid–base indicator.
Generally, acid–base indicators are dyes that
are weak acids. Because indicators are usually
complex molecules, we often symbolize them as
HIn. The reaction of the indicator with water can
be written as
HIn(aq) ϩ H2O(l) 3
4 H3Oϩ(aq) ϩ InϪ(aq)
To work as an acid–base indicator, the conjugate acid–base forms of these dyes must have different colors. The acidity level of the solution will
determine whether the indicator is present mainly
in its acidic form (HIn) or its basic form (InϪ).
InϪ(aq) ϩ Hϩ(aq) n HIn(aq)
When placed in a basic solution, most of the
acidic form of the indicator is converted to the
basic form by the reaction
HIn(aq) ϩ OHϪ(aq) n InϪ(aq) ϩ H2O(l)
It turns out that many fruits, vegetables,
and flowers can act as acid–base indicators. Red,
blue, and purple plants often contain a class of
chemicals called anthocyanins, which change
color based on the acidity level of the surroundings. Perhaps the most famous of these plants is
red cabbage. Red cabbage contains a mixture of
anthocyanins and other pigments that allow it
to be used as a “universal indicator.” Red cabbage juice appears deep red at a pH of 1–2, purple at a pH of 4, blue at a pH of 8, and green at
a pH of 11.
16.5 Calculating the pH of Strong
Acid Solutions
OBJECTIVE:
To learn to calculate the pH of solutions of strong acids.
In this section we will learn to calculate the pH for a solution containing a
strong acid of known concentration. For example, if we know a solution contains 1.0 M HCl, how can we find the pH of the solution? To answer this
question we must know that when HCl dissolves in water, each molecule dissociates (ionizes) into Hϩ and ClϪ ions. That is, we must know that HCl is a
strong acid. Thus, although the label on the bottle says 1.0 M HCl, the solution contains virtually no HCl molecules. A 1.0 M HCl solution contains Hϩ
and ClϪ ions rather than HCl molecules. Typically, container labels indicate
the substance(s) used to make up the solution but do not necessarily describe
the solution components after dissolution. In this case,
1.0 M HCl S 1.0 M Hϩ and 1.0 M ClϪ
Therefore, the [Hϩ] in the solution is 1.0 M. The pH is then
pH ϭ Ϫlog[Hϩ] ϭ Ϫlog(1.0) ϭ 0
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