Kelter P., Mosher M., Scott A. Chemistry. The Practical Science

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0 5 10 15 20 25 30

Volume added (mL)

Strong base added

to strong acid

Strong base added

to weak acid

Strong acid added

to weak base

Strong acid added

to strong base

FIGURE 18.12

This plot shows a comparison of

the pH for strong-acid–strong-base,

weak-acid–strong-base, and weak-

base–strong-acid titrations using equal

concentrations of acid and base.

e. Addition of a total of 25.00 mL of HCl solution. We are at the equivalence point,

at which we have added exactly the same number of moles of HCl as there

were moles of ammonia at the start of the titration. We now have a solution

that is weakly acidic as a consequence of the conversion of ammonia to the

ammonium ion.

(aq) + H

(aq) 



(aq) K =1.8

moles initial 0.005000

≈

moles added 0.005000

change −0.005000 −0.005000 +0.005000

moles at equilibrium

≈

0 0.005000

[NH

] =

(0.005000 mol)

(0.07500 L)

= 0.06667 M

1.0 ×10

−14

1.8 ×10

−5

= 5.6

−10

We can now solve for the pH of the weak acid.

][NH

]

[NH

]

5.6

−10

0.06667

x = [NH

] = [H

] = 6.1

−6

pH = 5.21

f. Addition of a total of 40.00 mL of HCl solution.

mol excess H

= 0.01500 L

(0.2000 mol)

(1.000 L)

= 0.003000 mol

] =

(0.003000 mol)

(0.09000 L)

= 0.03333 M

pH = 1.48

PRACTICE 18.9

Calculate the pH values and draw the curve for the titration of 25.00 mL of

0.2500 M acetic acid with 0.2500 M sodium hydroxide using the same volumes of

titrant as in our previous titrations: a. 0 mL; b. 5.00 mL; c. 12.50 mL; d. 24.00 mL; e.

25.00 mL; f. 40.00 mL. K

(acetic acid) = 1.8 × 10

−5

See Problems 35, 36, and 38.

798 Chapter 18 Applications of Aqueous Equilibria

–

Orange-red

(formed in 65–98% H

)

Colorless Pink

Colorless

(formed above pH 11)

<–1

–1 8.5

8.5 11 >11

OHC

O HO

–

FIGURE 18.13

The structure of phenolphthalein

changes as we increase the pH

from very low to very high.

We have seen that a successful monoprotic acid–base titration has an analyte

(what you are titrating) that, when titrated with a strong acid or strong base, has

a large K. The reaction is typically fast and reproducible. Polyprotic acid and base

titrations are common, and the analysis of their titration curves presents inter-

esting challenges, although we will not deal with these here.

Indicators

The pH meter, discussed in Chapter 19, can provide very accurate readings of the

pH. Making successful measurements using a pH meter, however, relies on its

having been properly maintained, calibrated, and operated. For speed and

simplicity in a titration, the chemical technician often relies on a chemical pH

reporter. This class of compounds, known as

acid–base indicators or pH indicators,

visually indicates the change in pH as we approach, reach, and pass the equiva-

lence point. Indicators themselves are conjugate acid–base pairs of organic

molecules that change color as they change between their acid and base forms.

Perhaps the best-known example is phenolphthalein, which changes from color-

less to rose-pink as the pH changes from 8.5 to 9.5. Figure 18.13 shows that

phenolphthalein actually has several structural changes (and therefore color

changes) from very low to very high pH.

Why is phenolphthalein such a popular acid–base indicator? Its color change

occurs in the pH region where many acids titrated with strong bases reach their

18.2 Acid–Base Titrations 799

Video Lesson: Acid–Base

Indicators

Video Lesson: CIA

Demonstration: Natural

Acid–Base Indicators

equivalence point. For example, acetic acid has an equivalence point, at which it

is essentially all changed to acetate ion, at about pH 9, and the pH goes from

about 6 to 11 within a small volume of titrant on either side of the equivalence

point. The titration of a strong acid with base (or vice versa) changes very rapidly

between the region of pH 6 to 11. An indicator that itself changes color anywhere

in this area would, generally speaking, be a suitable reporter of the equivalence

point.

In a given titration, how do we know which acid–base indicator to choose? We

want to choose an indicator with a pK

as close as possible to the pH at the equiv-

alence point. A distinct color change is also useful. For example, the pK

phenolphthalein is about 9.5. This is well within the region of the pH at the

equivalence point of titration of acetic acid by sodium hydroxide (pH ≈ 9, de-

pending on acetic acid concentration). As a rule of thumb, the color changes of pH

indicators are visible to

+/−1 pH unit on either side of the indicator pK

. This

means that phenolphthalein will be completely colorless below a pH of 8.5 and

will then be rose-pink from pH 8.5 to about pH 11. It will turn colorless again

above pH 11. A selection of common pH indicators, their color changes, and their

pH ranges is given in Figure 18.14.

During the chemical technician’s analysis, only a few drops of an indicator so-

lution are added to the analyte solution. Why so little? Because the indicator also

undergoes an acid–base reaction. This means that in addition to adding the re-

quired volume to neutralize the analyte, we add just a bit more to cause the color

change in the indicator. For example, if we titrate a solution of acetic acid

800 Chapter 18 Applications of Aqueous Equilibria

Methyl violet

Thymol blue

Bromphenol blue

Methyl orange

Methyl red

Eriochrome black T

Alizarin

Bromthymol blue

Phenolphthalein

Alizarin yellow R

FIGURE 18.14

A selection of common pH indicators, their color changes, and their pH ranges. Our key criterion in se-

lecting the proper indicator is that its pK

should be as close as possible to the equivalence point of the

titration in which it is used. Note that most indicators exhibit a color change over less than 2 pH units.

containing phenolphthalein indicator, we might need 35.27 mL of a sodium hy-

droxide solution to react with the acetic acid itself. Changing the indicator’s

structure (and therefore its color) might require an additional 0.02 mL of the

strong base. The equivalence point is at 35.27 mL, but the point at which you see

the change in indicator color that tells you the titration is ﬁnished, which is called

the

titration endpoint, is at 35.29 mL. Therefore, we would use 35.29 mL as our

number for calculating the concentration of analyte. This could lead to very large

errors if we had a lot of extra indicator in the solution. These errors are reduced

if the endpoint is as close as possible to the equivalence point.

EXERCISE 18.10 Picking an Indicator

What would be a reasonable indicator for the titration of 0.10 M ammonia with

0.10 M HCl?

Solution

The key question is “What is the pH of the solution at the equivalence point?”When

essentially all of the ammonia is converted to ammonium ion, the pH will be

around 5.2, as we saw in Exercise 18.9, part e. As shown in Figure 18.14, several in-

dicators change color around pH =5.2, and methyl red appears to be a good choice.

PRACTICE 18.10

What would be a reasonable indicator for the titration of 0.10 M NaOH with

0.10 M HCl?

See Problems 39–42.

Some natural and commercially prepared indicators are made up of several

colorful organic molecules, and they change color throughout the pH range. No-

table among the natural indicators are the

anthocyanins, which are responsible for

most of the different colors found in vegetables and ﬂowers. There are over 150

naturally occurring anthocyanins in foods such as the red cabbage that we cook

and serve with dinner. The juice from the red cabbage can therefore be used as an

indicator. Figure 18.15 shows the wide range of colors that can be obtained by ad-

justing the pH of red cabbage juice. Commercially prepared solutions of mixtures

of indicators can mimic these color changes, but far more intensely, so far less is

required. For instance, the commercially prepared “universal indicator” is a mix-

ture of thymol blue, methyl red, bromthymol blue, and phenolphthalein. Because

each indicator gains or loses protons at a different pH, the universal indicator has

color changes over a wide pH range, as shown in Figure 18.16.

18.2 Acid–Base Titrations 801

1234567

8 9 10 11 12 13

FIGURE 18.15

The spectrum of colors in each sample of red cabbage juice is caused by the

changes in structure of the anthocyanins as the pH moves from 1 to 13.

pH = 4

pH = 5

pH = 6

pH = 7

pH = 8

pH = 9

pH = 10

FIGURE 18.16

A typical universal in-

dicator is a mixture of

several common

indicators.

Application

HEMICAL ENCOUNTERS:

Anthocyanins and

Universal Indicators

HERE’S WHAT WE KNOW SO FAR

■

Strong-acid–strong-base titrations show a relatively level pH until near the

equivalence point, where the pH rises dramatically.

■

Titration curves in which one component is weak and the other is strong con-

tain four regions, including the initial pH, the buffer region, the equivalence

point region and the post–equivalence point region.

■

The buffer region contains a point at which one-half of the analyte has been

converted to its conjugate. This is called the titration midpoint, and the pH at

this point is equal to the pK of the analyte.

■

The larger the pK of the analyte, the sharper will be the change in pH at the

equivalence point.

■

We can use an indicator to “see” the equivalence point of a titration.

■

We add only a few drops of an indicator to the titration solution so that the

equivalence point and titration endpoint can be as close together as possible.

18.3 Solubility Equilibria

The Paciﬁc Ocean is an incredibly complex heterogeneous system. The bottom

layers of this and other massive waterways are covered with a variety of soils and

sediments, including

calcareous oozes, calcium-containing detritus from dead

single-celled, calcium-based sea life. One of the important compounds within the

oozes is calcium carbonate (CaCO

), some of which is in contact with ocean

water, dissociating to form calcium and carbonate ions. The equation relating

this dissociation is written so that the solid is a reactant and the dissolved ions are

products:

CaCO

(s) 



(aq) + CO

2−

(aq)

The mass-action expression that can be used to determine the solubility of

CaCO

is called the solubility product when the equation is written as shown

above. The solubility product is equal to the product of the concentrations of

the ions (remember that the CaCO

is a solid and is not written as part of the

mass-action expression).

= [Ca

][CO

2−

]

This equilibrium constant is called the solubility product constant (K

) and has the

same conceptual meaning as any other equilibrium constant along with its mass-

action expression. Because the values of K

tend to be very small, the concentra-

tions of ions are quite low and that activities are not important here.

Table 18.3 lists representative K

values for some of the sparingly soluble

salts.

The difﬁculty with describing solubility using a single mass-action

expression is that there are so many other processes that enter into the

chemistry that our typically simple mass-action expression often just

won’t do. The simple calculations we can perform do not always agree with

what we observe in real systems. Let’s take a look at the calcium carbonate

system in a somewhat nonmathematical approach as we discover the

factors that affect the solubility of solids.

Side Reactions That Affect Our Reaction of Interest

The solubility of many ions, when dissolved in an aqueous system, is

affected by side reactions. The solubility of calcium carbonate in a large

ocean-based system is no exception (Figure 18.17).

802 Chapter 18 Applications of Aqueous Equilibria

Application

FIGURE 18.17

Many processes, including the formation

of bicarbonate ion and the reaction of

hydrogen and hydroxide ions, affect the

solubility of calcium carbonate in the

ocean. These stromatolites are forma-

tions of calcium carbonate.

Video Lesson: The Effects of pH

on Solubility

Video Lesson: The Solubility

Product Constant

1. Hydrolysis of carbonate ion. The carbonate ion formed from the dissolution of

calcium carbonate is a base that reacts with water to form bicarbonate ion

and hydroxide ion.

2−

(aq) + H

O(l) HCO

−

(aq) + OH

−

(aq)

Because the carbonate ion is involved in this reaction, some of it is removed

from the calcium carbonate solubility equilibrium. The net result, in accor-

dance with Le Châtelier’s principle, is that more of the calcium carbonate dis-

solves than we would predict.

2. The interplay between the atmosphere and the ocean water. Dissolved CO

from

the air mixes with ocean water to form carbonic acid and, ultimately, hydro-

gen ion and bicarbonate ion.

(aq) + H

O(l)H

(aq)H

(aq) + HCO

−

(aq)

18.3 Solubility Equilibria 803

Selected K

Values at 25°C

Ionic Solid K

(at 25°C) Ionic Solid K

(at 25°C)

Fluorides Hg

CrO

−9

Co(OH)

2.5

−16

BaF

2.4

−5

BaCrO

8.5

−11

Ni(OH)

1.6

−16

MgF

6.4

−9

CrO

9.0

−12

Zn(OH)

4.5

−17

PbF

−8

PbCrO

−16

Cu(OH)

1.6

−19

SrF

7.9

−10

Hg(OH)

−26

CaF

4.0

−11

Carbonates Sn(OH)

−27

NiCO

1.4

−7

Cr(OH)

6.7

−31

Chlorides CaCO

8.7

−9

Al(OH)

−32

PbCl

1.6

−5

BaCO

1.6

−9

Fe(OH)

−38

AgCl 1.6

−10

SrCO

−10

Co(OH)

2.5

−43

* 1.1

−18

CuCO

2.5

−10

ZnCO

−10

Sulﬁdes

Bromides MnCO

8.8

−11

MnS 2.3

−13

PbBr

4.6

−6

FeCO

2.1

−11

FeS 3.7

−19

AgBr 5.0

−13

8.1

−12

NiS 3

−21

* 1.3

−22

CdCO

5.2

−12

CoS 5

−22

PbCO

1.5

−15

ZnS 2.5

−22

Iodides MgCO

−15

SnS 1

−26

PbI

1.4

−8

* 9.0

−15

CdS 1.0

−28

AgI 1.5

−16

PbS 7

−29

* 4.5

−29

Hydroxides CuS 8.5

−45

Ba(OH)

5.0

−3

S 1.6

−49

Sulfates Sr(OH)

3.2

−4

HgS 1.6

−54

CaSO

6.1

−5

Ca(OH)

1.3

−6

1.2

−5

AgOH 2.0

−8

Phosphates

SrSO

3.2

−7

Mg(OH)

8.9

−12

1.8

−18

PbSO

1.3

−8

Mn(OH)

−13

(PO

)

−31

BaSO

1.5

−9

Cd(OH)

2.5

−14

(PO

)

1.3

−32

Pb(OH)

1.2

−15

(PO

)

−39

Chromates Fe(OH)

1.8

−15

(PO

)

−54

SrCrO

3.6

−5

TABLE 18.3

*Contains Hg

ions. K

[Hg

][X

−

]

for Hg

salts, for example.













This interaction generates the bicarbonate ion, which inﬂuences the equilib-

rium shown at the bottom of page 803. The net result is to reduce the effect of

the interaction of carbonate ions with water. The exact change depends on the

amount of carbon dioxide dissolved in seawater.

3. The formation of water from the reaction of hydrogen and hydroxide ions. These

are produced from the processes described in reactions 1 and 2.

1/K

(aq) + OH

−

(aq)H

O(l)

The effect of this reaction is an increase in the concentration of the bicarbon-

ate ion generated from the carbon dioxide equilibrium.

As you can see, these three equilibria interact with others in the ocean as part

of a remarkably complex system in which temperatures and concentrations

change, making the calculation of calcium carbonate solubility at a given tem-

perature most challenging.

Shifting our focus from the oceans to a more controlled setting, we ﬁnd that

side reactions can still confound apparently simple systems. Consider the solubil-

ity of lead(II) iodide (PbI

) in distilled water. The amount of precipitated

lead(II) iodide is related to the initial concentration of iodide as shown in Fig-

ure 18.18. Not accounting for these changes can lead to massive errors in calculating

the solubility of lead iodide.

Molecular-Level Processes

We can take a simple view of the solubility of a salt such as calcium sulfate by con-

sidering only the dissociation reaction:

CaSO

(s)Ca

(aq) + SO

2−

(aq)

Our mass-action expression is

= [Ca

][ SO

2−

]

However, Meites, Pode, and Thomas wrote as early as 1966 that even in the ab-

sence of side reactions, the concentrations we would calculate would be wrong by

about 60%, largely because of the tendency of the calcium and sulfate ions to stay

together as individual

ion pairs. When the small amount of calcium sulfate

dissolves, most of it does not form individual ions. Rather, the ions associate in-

timately with each other. This is especially important in salts of highly charged

(±2 or ±3) cations and anions. The bottom line is that even in the absence of side

reactions, such as the addition of H

to SO

2−

to make HSO

−

in acidic solution,

there are several factors that affect solubility at the molecular level, making many

such calculations challenging. These factors include

804 Chapter 18 Applications of Aqueous Equilibria





–1

–2

–3

–4

–5

–6

–7

–8

0 0.05 0.1 0.15 0.2 0.25 0.3 0.35 0.4 0.45 0.5

–

]

Log [Pb

]

FIGURE 18.18

Even a system that seems as simple as

the solubility of lead(II) iodide isn’t. Most

of the Pb

ions precipitate upon the ad-

dition of a small amount of iodide. How-

ever, a signiﬁcant concentrate of Pb

remains.





1. Formation of ion pairs, as mentioned in the previous paragraph

2. Ion activities, a measure of the effective concentration of ions in solution

3. Thermodynamic measures, including enthalpy and entropy changes in the

solution process

Some simple univalent (both cation and anion singly charged) systems do

give reasonable answers when we do solubility calculations. In these cases, and

others as well, the total number of moles of solute that dissolve per liter of solu-

tion is often called the

molar solubility. For example, we can calculate the molar

solubility of silver bromide (AgBr) by using its solubility product constant and

mass-action expression.

AgBr(s) 



(aq) + Br

−

(aq) K

= 5.0

−13

= [Ag

][Br

−

]

We can set up our table, as we’ve done before. Although the solid AgBr will not

enter into the mass-action expression, we’ll include it in the K

ICE tables for the

reasons mentioned below. In any case, the number of moles of silver bromide that

will dissolve into solution and the number of moles of silver and bromide ions

produced will all be equal, because there is a 1-to-1-to-1 mole ratio in the reac-

tion. We can designate the molar solubility of AgBr as s, in which case the equi-

librium concentrations, [Ag

] and [Br

−

], will also be s.

AgBr(s) 



(aq) + Br

−

(aq)

initial —

change −s +s +s

equilibrium — ss

= [Ag

][Br

−

]

5.0

−13

= s

s = 7.1

−7

M = [Ag

] = [Br

−

]

s = molar solubility of AgBr =7.1

−7

Qualitatively, what would we expect to happen to the solubility if we added a

little sodium bromide, in which the added bromide is a common ion? According

to Le Châtelier’s principle, addition of an ion common to the product would push

the dissociation reaction back to the left, further decreasing the solubility. Cau-

tion: If we add too much bromide ion, the solubility of silver bromide could actu-

ally increase, as a consequence of the formation of soluble species such as AgBr

−

EXERCISE 18.11 How Much Dissolves?

One method of analyzing groundwater for nitrate requires that the chloride ions in

the water sample be removed ﬁrst.This is typically done by adding a solution of silver

ions (Ag

) in order to precipitate the sparingly soluble silver chloride salt (AgCl).

What silver ion concentration is present when silver chloride is added to water?

AgCl(s) 



(aq) + Cl

−

(aq) K

= 1.6 × 10

−10

Solution

Setting up the table, we get

AgCl(s) 



(aq) + Cl

−

(aq)

initial — 00

change −s +s +s

equilibrium — ss

18.3 Solubility Equilibria

805

Application

Then, as usual, we can ﬁnd our equilibrium concentrations by solving the mass-

action expression:

= [Ag

][Cl

−

]

1.6

−10

= (s)(s) = s

s = 1.3

−5

PRACTICE 18.11

Calculate the molar solubility of barium ﬂuoride (BaF

), K

= 2.4

−5

See Problems 49 and 50.

EXERCISE 18.12 Calculating K

The concentration of calcium ions in a saturated solution of calcium ﬂuoride was

found to be 2.15

−4

M. What is the apparent value for the solubility product

constant, K

First Thoughts

This problem is asking the same question as the previous exercise, but in the reverse

direction. The problem gives us the molar solubility of calcium ions, so we’ll exam-

ine the equilibrium expression to determine how to calculate K

Solution

The equilibrium under consideration is

CaF

(s) 



(aq) + 2F

−

(aq)

The mole ratio of Ca

to F

−

is 1-to-2, so if the equilibrium concentration of

is 2.15 × 10

−4

, that of F

−

must be twice as large, or [F

−

] = 4.30 × 10

−4

.We

can substitute these values into the mass-action expression.

= [Ca

][F

−

]

= (2.15

−4

) (4.30

−4

)

= 3.98

−11

Further Insight

The question asked us to calculate the apparent K

value. This was done because

there may be some side reactions, activity considerations, or other factors that affect

the molar solubility of the calcium ﬂuoride salt. In particular, we have neglected the

fact the ﬂuoride ion (F

–

) is a Brønsted base, which will affect the solubility of the

calcium ﬂuoride. (Can you propose how?) In this case, we’ve calculated a value for

that is similar to the actual value, K

= 4.00

−11

PRACTICE 18.12

Calculate the apparent value of K

for lead bromide (PbBr

) if the concentration of

bromide in a saturated solution is 2.1

−2

See Problems 53 and 54.

Solubility, Precipitation, and Gravimetric Analysis

Chemical technicians can determine the concentration of substances in solution

by causing them to form insoluble salt precipitates and weighing these precipitates

or their related solids in a technique called

gravimetric analysis. The quantitation

of a sample on the basis of its mass is among the most powerful tools at the dis-

posal of chemical technicians because good balances are both highly accurate and

806 Chapter 18 Applications of Aqueous Equilibria

Video Lesson: Solubility and the

Common Ion Effect

Video Lesson: Gravimetric

Analysis

precise, and weighing a sample is fast and inexpensive. For exam-

ple, the amount of chloride in a sample is routinely determined by

combining the chloride ion with silver ion, forming the solid silver

chloride as described in the Exercise 18.11.

−

(aq) +Ag

(aq) 



AgCl(s)

Other substances can be routinely determined by gravimetric

analysis. Aluminum ion concentrations can be found via the for-

mation of aluminum hydroxide (Al(OH)

). Igniting (driving off

water at high temperature) the aluminum hydroxide forms alu-

minum oxide (Al

), which can then be weighed. Aluminum can

also be determined by reaction with 8-hydroxyquinoline

ON) to form Al(C

ON)

without subsequent ignition.

(aq) + 3C

ON(aq) 



Al(C

ON)

(s) + 3H

(aq)

The solid forms good crystals that can be weighed after drying.

Sulfur can be determined by reaction of the sulfate (SO

2−

)

with barium ion to form barium sulfate:

2−

(aq) + Ba

(aq) 



BaSO

(s)

Calcium concentrations can be measured by reaction to form

calcium oxalate (CaC

(aq) + C

2−

(aq) 



CaC

(s)

In each of these analyses there are complicating factors, such as the presence

of other elements that can react with the precipitating agents, as well as the com-

plex nature of the precipitation process, so the procedures are a bit more involved

than the simple reactions suggest. In fact, acid–base and other equilibria are

nearly always a vital part of chemistry. Despite all of these concerns, a host of

elements can be determined via precipitation. Understanding solubility equilib-

ria and how we can affect them makes the analyses all the more meaningful.

Precipitation is also used in

metal recovery, in which dissolved metals are re-

claimed from processing wastes. Many metals in industrial efﬂuents (runoff from

manufacturing) are worth recovering because they are environmental hazards or

they waste ﬁnite metal resources. Recycling these metals saves money and the

environment. Such metals, which include copper, mercury, lead, and zinc, are

typically recovered as their sulﬁde salts, although some metals can be precipitated

as their corresponding carbonate salt.

To Precipitate or Not to Precipitate

Often, the formation of a precipitate is not as obvious as simply mixing two solu-

tions containing ions that form a sparingly soluble salt. Let’s say our chemical

technician is interested in mixing two solutions, one containing silver ions and

one containing chloride ions, so that the ﬁnal concentrations are [Ag

] = 1.0 ×

−4

M and [Cl

−

] = 1.0 × 10

−4

M. Will mixing these solutions cause the forma-

tion of the insoluble AgCl? To answer this question, we can perform a calculation

using our mass-action expression. However, because the concentrations do not

reﬂect equilibrium conditions, we will be calculating the reaction quotient (recall

this from Section 16.5) related to the solubility product. We’ll call this Q

AgCl(s) 



−

(aq) +Ag

(aq) K

= 1.6

−10

= [Ag

]

[Cl

−

]

= (1.0

−4

M) (1.0

−4

M) = 1.0

−8

18.3 Solubility Equilibria 807

The concentration of aluminum in solution, like many

metals, can be determined by gravimetric analysis. Here,

aluminum is reacted to form the 8-hydroxyquinoline salt,

which is made pure by recrystallization and weighed.

Application