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4: Reactions That Form Water: Acids and Bases

# 4: Reactions That Form Water: Acids and Bases

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16.4 The pH Scale

529

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

531

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

532

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