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

Подождите немного. Документ загружается.

about 160 km south-southwest of Dublin (Figure 4.11), receives periodic ship-

ments of a concentrated aqueous HFSA solution from a chemical manufacturer

in Spain. The concentration of this HFSA solution, shipped in 4000-gallon con-

tainers, is about 22 M HFSA. The solution is then diluted with water to 7.8 M and

shipped to cities and towns all over Ireland for use in the nation’s many water

treatment plants. At these plants, the HFSA solution is further diluted in the

water supply to a ﬁnal ﬂuoride ion concentration of 1 ppm.

How do the workers know how much to dilute? There are really two questions

here. One is a mathematics question, and one is a chemical process question

(“How do I prepare this solution?”). In terms of the chemical process, it has long

been known that unless you are adding two of the exact same liquids to each

other, the volumes are not additive. That is, adding 100 mL of ethanol to 100 mL

of water does not give 200 mL of solution. Rather, it gives a volume that is a little

bit less than 200 mL. The practical outcome of this is that for laboratory-scale

preparation of a solution of known molarity, we must use volumetric glassware

(shown in Figure 4.12), in which the etched mark on the ﬂask indicates the listed

volume to a very high accuracy, typically within +/−0.05% of the stated volume.

We add the reagent to the ﬂask and then dilute with solvent (water, in this case)

to the mark.

We can now deal with the mathematics question. Let’s assume we wish to pre-

pare 500.0 mL of 7.8 M HFSA solution from a sample of 22 M HFSA. To do this,

we need to calculate the volume of 22 M HFSA that can be added to the volu-

metric ﬂask and diluted with water. How many moles of HFSA will be in the ﬁnal

solution? We are given the molarity (7.8 M) and volume (500.0 mL) of the ﬁnal

solution, so we can do the following calculation:

7.8 mol HFSA

L HFSA solution

× 0.5000 L solution = 3.9 mol HFSA

What volume of the concentrated (22 M) HFSA solution must we dispense into

the volumetric ﬂask and dilute with water in order to give us 3.9 moles of HFSA?

3.9 mol HFSA ×

1 L HFSA

22 mol HFSA

= 0.1773 L HFSA solution

= 177.3 mL HFSA solution = 180 mL HFSA solution (to 2 signiﬁcant ﬁgures)

This means that we can measure 180 mL of HFSA solution and dilute to 500.0 mL

in our volumetric ﬂask and, after mixing, have a 7.8 M HFSA solution.

138 Chapter 4 Solution Stoichiometry and Types of Reactions

Dublin

New Ross

tlantic

Ocean

Irish

Sea

IRELAND

50 Miles0

Scale

FIGURE 4.11

New Ross, Ireland, is about 100 miles

(160 km) from Dublin.

FIGURE 4.12

Using a volumetric ﬂask. A solute is added to the volumetric ﬂask and dissolved in water. The resulting solution

is then diluted to the mark on the neck of the ﬂask. The result is a precise and accurate volume of solution.

Visualization: Dilution

Video Lesson: CIA

Demonstration: Dilutions

Tutorial: Dilution

Note that in our calculations, we have compared the number of moles of solute

in the concentrated solution that we are measuring to the number of moles of the

diluted solution that we want to prepare. Mathematically, this can be written:

number of moles measured

number of moles needed

from the concentrated solution in the dilute solution

Because we calculated the number of moles needed in the dilute (“ﬁnal”) solution

by multiplying its concentration (C

ﬁnal

) by its total solution volume (V

ﬁnal

), and

because this is equal to the number of moles of concentrated (“initial”) solution

measured, we can say that

initial

× V

initial

= C

ﬁnal

× V

ﬁnal

Our goal is to ﬁnd the volume of the initial solution of HFSA to dilute, so we can

solve for V

initial

ﬁnal

× V

ﬁnal

initial

Using our data for HFSA, we ﬁnd that

initial

7.8 M × 0.5000 L

22 M

= 0.177 L = 180 mL

The total volume of 22 M HFSA needed to prepare large, intercity shipments of

7.8 M HFSA is far larger, but the problem-solving strategy is the same.

EXERCISE 4.5 Practice with Dilution of Solutions

Hydrochloric acid is typically shipped as a 12 M aqueous solution. If we are to do a

laboratory analysis that requires 250.0 mL of a 1.0 M solution, how many milliliters

of the concentrated solution should be diluted?

First Thoughts

What range of volumes would make sense as an answer? We want to dilute our solu-

tion by a factor of 12, to go from 12 M to 1.0 M. If the total volume of the solution is

to be 250.0 mL, and we are diluting a solution with water by a factor of 12,this would

suggest that we need about 20 mL of the concentrated solution (roughly 1/12 of

250.0 mL). The remainder of the solution is water, added to dilute the concentrated

acid.

Solution

initial

ﬁnal

× V

ﬁnal

initial

1.0 M × 250.0mL

12 M

= 20.8mL= 21 mL

Further Insights

It is quite common for industrially prepared concentrated acids to be labeled for

shipment on the basis of grams of acid per grams of solution,

g acid

g solution

. This is

known as weight-to-weight percent, or (w/w). Hydrochloric acid, for example, is

shipped as a 37% (w/w) solution and has a density of 1.18 grams of acid per milli-

liter of solution. How do we relate w/w to molarity? Here are the ﬂowchart and the

calculation.

g soln

mL soln

1000 mL

mol HCl

gHCl

g soln

−−−−−→

gHCl

mL soln

−−−−−→

gHCl

L soln

−−−−−→

mol HCl

L soln

37 g HCl

100 g solution

1.18 g solution

1 mL solution

1000 mL solution

1 L solution

1 mole HCl

36.5gHCl

= 12 M

4.2 The Concentration of Solutions 139

PRACTICE 4.5

Industrially prepared nitric acid (HNO

) has a concentration of 16 molar. How

many milliliters of nitric acid are required to prepare 2.00 L of 0.15 M HNO

See Problems 29 and 30.

4.3 Stoichiometric Analysis of Solutions

The concepts of concentration and solution stoichiometry are put to important

use every day to examine all manner of solutions. In medicine, they are used to an-

alyze bodily ﬂuids for evidence of illness and to prepare pharmaceuticals in appro-

priate amounts. In industry, they are used to ensure that the correct quantities of

dissolved reactants are mixed together. In forensic chemistry they can determine

the levels of drugs and poisons in a victim or suspect of crime. And in the food

industry they make it possible to measure and modify the nutritional content of

foods.

For example, the amount of vitamin C in a sample of fruit juice can be deter-

mined by adding iodine to the fruit juice via the following reaction:

+ I

→ C

+ 2HI

vitamin C brownish-red colorless colorless

colorless

140 Chapter 4 Solution Stoichiometry and Types of Reactions

The controlled addition of a solution of known concentration to react with the

substance of interest to determine its concentration is called a

titration. In the

titration of fruit juice with iodine shown in Figure 4.13, we slowly add iodine so-

lution from a

buret (a laboratory device used to precisely and accurately add

small known quantities of solution) to a sample of fruit juice. When all the vita-

min C in the sample has reacted, excess iodine will begin to accumulate. The

point at which the reactants have been completely consumed is known as the

equivalence point (very close to the visual end point) of the titration. We can de-

tect when the reaction has reached this stage by including a compound, in this

case starch, that reacts with the excess iodine to produce a color change.Although

the iodine itself is colored, it is often difﬁcult to see this color, especially in fruit

juice. However, the addition of starch results in a very obvious color change.

starch + I

→ I

−starch

colorless brownish-red deep blue

Here is how we can use this reaction to determine the vitamin C content of a

fruit juice in moles per liter and percent by mass. We begin by preparing a solu-

tion of a known concentration of iodine that turns out to be, for example,

0.01052 M I

. This is known as a standard solution, because we use it as a standard

of known concentration against which other solutions can be tested. We then

measure out 50.00 mL (0.05000 L) of fruit juice and add some starch to react with

excess iodine. By adding our standard solution of iodine from a buret, we titrate

the fruit juice until a color change is observed. Typically, this process is repeated

several times to minimize error. In our hypothetical titrations, let’s assume that

Vitamin C

Starch–iodine end point.

Application

we averaged 16.97 mL of iodine solution before the color change was observed.

How much vitamin C is in the fruit juice? We can quantify the vitamin C content

of the fruit juice using the following strategy:

1. Write the balanced chemical equation relating the reaction of iodine with

vitamin C, C

2. Use dimensional analysis to convert the units of 16.97 mL of iodine solution

into grams of vitamin C.

+ I

→ C

+ 2HI

1 mole + 1 mole 1 mole + 2 moles

16.97 mL I

solution ×

1LI

solution

1000 mL I

solution

0.01052 mol I

1LI

solution

1 mol vitamin C

1 mol I

176.1 g vitamin C

1 mol vitamin C

= 0.03144 g vitamin C

That is the number of grams of vitamin C in our 50.00 mL (0.05000 L) sample

of fruit juice. We can determine the amount of vitamin C in 1 L by doing the

following calculation:

0.03144 g vitamin C

50.00 mL fruit juice solution

1000 mL fruit juice solution

1 L fruit juice solution

0.6288 g vitamin C

L fruit juice solution

Our calculations reveal that there are 0.03144 g of vitamin C in our 50.00 mL

sample of fruit juice. Alternatively, this is equal to 0.6288 g of vitamin C per liter

of fruit juice. If we assume that the density of the fruit juice is

1.0g

, what is the

concentration of vitamin C in ppm?

Recall from Table 4.2 that parts per million (ppm) of vitamin C can be

expressed as

mg vitamin C

L fruit juice solution

. We know the vitamin C concentration in

, so

all we need to do is convert from grams to milligrams of vitamin C.

0.6288 g vitamin C

L fruit juice solution

1000 mg vitamin C

1 g vitamin C

628.8 mg vitamin C

L fruit juice solution

= 628.8 ppm vitamin C

4.3 Stoichiometric Analysis of Solutions 141

FIGURE 4.13

The process of titration.

Video Lesson:

Acid–Base

Titrations

Video Lesson: CIA

Demonstration:

Titrations

142 Chapter 4 Solution Stoichiometry and Types of Reactions

Water is such a precious resource that

countries have been known to threaten

one another with war over water rights. In the United

States, individual states quarrel and even sue each other

over access to fresh water. For example, Nebraska and

Kansas have fought for decades over who owns the water

that ﬂows from Colorado through Nebraska and into

Kansas via the Republican River.

We are so used to thinking of water as essential and

beneﬁcial that it is easy to overlook the many chemical

situations in which too much water can be very undesir-

able. What would be the effect of too much water in the

Republican River? On a smaller scale, what would be the

effect of too much water in a chocolate bar or potato

chips? Canned cooking fats and oils certainly do not

beneﬁt from the presence of water; and there are limits

to the amount of water that can be in pharmaceutical

products that are ingested as tablets, gelcaps, and caplets.

Therefore, it is important to be able to determine how

much water is present in samples of foods, medicines,

and other industrial products, even if the quantities of

water involved are very small.

One chemical method

used to quantify the water

content of a wide range of

samples, such as chewing

gum, jelly beans, and peanuts,

is the Karl Fischer titration,

named after the scientist who

devised the basic method in

1935. The Karl Fischer titra-

tion makes use of the reaction

of iodine (I

) with the water in

the sample being analyzed.

The reaction is performed in

the presence of organic

solvents such as pyridine

N) and methanol

(CH

OH). The precise details

of the chemical reaction are

quite complex, to the extent that even now, more than 70

years after the method was ﬁrst developed, the exact

processes involved in the reaction are still the subject of

research. We can summarize it, however, as follows:

(Reactants) +I

+ H

O → (products)

The chemical details vary with the actual method used.

In all cases, however, the crucial quantitative fact that al-

lows the water content to be measured is that the iodine

and water always react in a known mole ratio, which is 1:1

in the case shown above. The end point of the titration,

when we can tell that all the water has been consumed, is

How do we know?

How to test for small amounts of water

The Republican River supplies Harlan County Reservoir with water

used for irrigation and recreation.

Hot oil spatters violently

when water, such as that

on the surface of this

turkey, is present.

FIGURE 4.14

Automated Fischer titration

apparatus.

marked by a distinctive color change, which enables us

to determine the quantity of iodine required to achieve

that color change. The amount of water that must

have reacted with that quantity of iodine can then be

calculated.

Modern laboratory instrumentation has allowed

the titration process to be automated, as shown in

Figure 4.14. Instead of direct observation of a color

change, the automated instruments may rely on mea-

surement of a ﬂow of electrons to excess iodine as soon

as all the water has reacted. This ﬂow of electrons is due

to the process I

+ 2e

−

→ 2I

−

. Another option is to use

the reversal of the process shown above—namely

−

→ I

+ 2e

−

—to generate the iodine needed to react

with the water.

The details vary depending on the machines used,

but automated Karl Fischer titrations allow accurate de-

termination of the water content in tiny samples con-

taining hardly any water at all. What do we mean by

“hardly any” and “tiny”? The method is useful down to

the level of 50 ppm water in a sample that can have a

volume as small as 10 microliters.

EXERCISE 4.6 Practice with Titration

A common type of titration is the reaction of a solution of sodium hydroxide (with

a known concentration) with a solution of an acid (with an unknown concentra-

tion). The result can be used to determine the molarity of the acid solution. What is

the molarity of a solution of 50.00 mL of HCl if 31.98 mL of 0.1253 M NaOH is re-

quired to react with it? The products of this reaction are sodium chloride and water.

First Thoughts

We should begin this problem as we begin all stoichiometry problems in chemistry,

with a balanced chemical equation. From the problem we determine the equation

for the titration of hydrochloric acid with sodium hydroxide:

HCl(aq) + NaOH(aq) →H

O(l) + NaCl(aq)

Our ﬂowchart for the dimensional analysis can be expressed as follows:

mL NaOH → mol NaOH →mol HCl → M HCl

Remember to change the milliliters of NaOH and HCl to liters in order to work with

molarity in the proper units.

Solution

From the equation, 1 mole of HCl reacts with 1 mole of NaOH. Therefore, we can

calculate the molarity of the HCl. (Remember to construct the ﬂowchart ﬁrst.)

31.98 mL NaOH ×

1 L NaOH solution

1000 mL NaOH solution

0.1253 mol NaOH

1 L NaOH solution

1 mol HCl

1 mol NaOH

= 0.004007 mol HCl

Now that we know the quantity of HCl in moles, we can determine the molarity by

dividing by the volume of the HCl.

0.004007 mol HCl

50.00 mL HCl solution

1000 mL HCl solution

1 L HCl solution

0.08014 mol HCl solution

1 L HCl solution

= 0.08014 M HCl

Since the titration used more of the HCl solution than of the NaOH solution, we

would expect the HCl solution to be more dilute. Our answer makes sense.

Further Insights

Titration is just one (albeit one very important) technique for ﬁnding out how

much of something you have. We will learn about many other “quantitative analy-

sis” procedures in this textbook, based on physical properties such as color, electri-

cal conductivity, and the formation of a precipitate.

PRACTICE 4.6

The equation for the reaction between a sodium hydroxide solution and dilute sul-

furic acid is

2NaOH(aq) + H

(aq) → Na

(aq) + 2H

O(l)

What is the molarity of a solution of 25.00 mL of H

if 22.25 mL of 0.100 M

NaOH is required to reach equivalence?

See Problems 47–54.

4.3 Stoichiometric Analysis of Solutions 143

Video Lesson: Solving Titration

Problems

144 Chapter 4 Solution Stoichiometry and Types of Reactions

4.4 Types of Chemical Reactions

The vast number of aqueous reactions that have been identiﬁed since the begin-

ning of chemistry would stun the ancient alchemists. Fortunately, most of these

reactions fall within certain general types or categories because of key similarities

among them. Three of the more important types of chemical reactions are

■

Precipitation reactions

■

Acid–base reactions

■

Oxidation–reduction reactions

In order to more fully understand the three types of processes, we will take this

opportunity to build on our ability to write chemical equations in aqueous solu-

tions. Our ﬁrst step is determining what happens to the individual components

when they are added to water. We’ll use our knowledge of electrolytes to assist us

in this process.

Molecular and Ionic Equations

The reaction that was the focus of Exercise 4.6 includes three strong electrolytes:

hydrochloric acid (HCl), sodium hydroxide (NaOH), and sodium chloride

(NaCl). The reaction equation, containing the individual components written as

compounds, is known as the

molecular equation. The molecular equation illus-

trates that the individual substances exist as hydrated compounds in the aqueous

solution. The molecular equation representing the reaction of hydrochloric acid

and sodium hydroxide can be written

HCl(aq) + NaOH(aq) →H

O(l) + NaCl(aq)

molecular equation

However, as we have already discussed, strong electrolytes dissolve in water, dis-

sociate into their component ions, and become relatively independent within the

solution. For this reason, all the individual ions can be shown separately, in what

is known as a

complete ionic equation.

(aq) + Cl

−

(aq) + Na

(aq) + OH

−

(aq) → Na

(aq) + Cl

−

(aq) + H

O(l)

complete ionic equation

Note that only the strong electrolytes have been written as individual ions.Unlike

the electrolytes, molecules such as water do not dissociate into ions to any appre-

ciable extent. They are not written as individual ions.

Complete ionic equations list all substances as they exist in the solution,

whether or not they participate in the reaction. For example, in the complete

ionic equation above, the sodium (Na

) and chloride (Cl

−

) ions are present at

both the start and the end of the reaction, yet they do not participate at all during

the reaction; they “sit on the sidelines,” like spectators at a sporting event. For this

reason, they are known as

spectator ions. The real action during this reaction is

the combination of hydrogen ions (H

) and hydroxide ions (OH

−

) to form water

O). If we remove the spectator ions from the equation, we can simplify the

equation to better show the combination of hydrogen ions and hydroxide ions:

(aq) + Cl

−

(aq) + Na

(aq) + OH

−

(aq) → Na

(aq) + Cl

−

(aq) + H

O(l)

The result represents the overall, or “net,” chemical change that occurs in this reac-

tion and is known as the

net ionic equation:

(aq) + OH

−

(aq) → H

O(l)

net ionic equation

4.5 Precipitation Reactions 145

HERE’S WHAT WE KNOW SO FAR

■

Molecular equations show the formulas of all reactants and products but do

not indicate whether any of the compounds really exist as ions in solution.

■

Complete ionic equations show all of the dissolved ions present in an equation

individually, along with all other reactants and products.

■

Net ionic equations show only those dissolved ions, and other reactants and

products, that actually participate in or result from the reaction concerned.

EXERCISE 4.7 The Three Equations

Aqueous hydrochloric acid and aqueous sodium carbonate react to form aqueous

sodium chloride, water, and gaseous carbon dioxide. Write the molecular, complete

ionic, and net ionic equations for this reaction.

Solution

The molecular equation is

2HCl(aq) + Na

(aq) → 2NaCl(aq) +H

O(l) + CO

(g)

In the complete ionic equation, all ionic compounds are written as separated ions in

solution:

(aq) + 2Cl

−

(aq) + 2Na

(aq) + CO

2−

(aq)

→ 2Na

(aq) + 2Cl

−

(aq) + H

O(l) + CO

(g)

The net ionic equation is completed by removing the spectator ions:

(aq) + CO

2−

(aq) → H

O(l) + CO

(g)

PRACTICE 4.7

Identify the molecular, complete ionic, and net ionic equations that describe the re-

action of potassium oxalate (K

) and nitric acid (HNO

) to form potassium

nitrate (KNO

) and oxalic acid (H

). Assume that oxalic acid is the only non-

electrolyte in the reaction.

See Problems 43–46, 59–62, and 91.

4.5 Precipitation Reactions

An important challenge for chemists trying to clean up industrial wastewater is to

remove the ions of “heavy metals” (often very dense metals, 5 g/mL or greater)

such as lead, mercury, and cadmium, which are toxic to humans and other life.

This is particularly important in mining operations, where aqueous waste

streams from the mine can contain abnormally high levels of heavy metals. One

way to clean up the wastewater is to treat it with sulﬁde ions (S

2−

), because most

nonalkali metal ions will combine with sulﬁde ions to form a solid “precipitate.”

The word

precipitate is both a noun and a verb. It is used as a noun to refer to any

solid material that forms within a solution, and as a verb to describe that process

in action. We can say that metal sulﬁdes form a precipitate (noun). Alternatively,

we can say that metal sulﬁdes will precipitate (verb) out of solution as soon as

they form. We say that a precipitate is

insoluble because not very much of it can

dissolve into the solvent. On the other hand, a relatively large amount of a

soluble

substance can dissolve into the solvent.About 2 kg of sucrose, table sugar, can dis-

solve in a liter of water at 25°C. That is highly soluble compared to silver chloride,

which has a solubility of about 2 mg per liter of water at 25°C.

Water drainage from a mine is often tainted with

high levels of sulﬁde, sulfate, and other ions.

Application

HEMICAL ENCOUNTERS:

Focus on Lead Sulﬁde

Video Lesson: Precipitation

Reactions

Tutorial: Precipitation Reactions

Visualization: Precipitation

Reactions

Industrial wastewater can be treated with sulfate-reducing bacteria. These

bacteria convert sulfate ions into sulﬁde ions as part of their natural biological ac-

tivity (Figure 4.15). In the presence of heavy metals, the sulﬁde generated by the

bacteria can help to reduce the amount of soluble heavy metals by making insol-

uble precipitates. For example, the net ionic equation for the reaction of lead(II)

ions with sulﬁde ions is

(aq) + S

2−

(aq) → PbS(s)

The ions are initially in the aqueous phase and so are dissolved in water. When

they meet and combine, they form the insoluble solid precipitate lead(II) sulﬁde.

This is an example of a

precipitation reaction, a reaction in which an insoluble pre-

cipitate is formed from soluble reactants. We can precipitate lead(II) sulﬁde in the

laboratory by mixing a solution of a soluble lead(II) compound with a solution

of a soluble sulﬁde. This is illustrated in Figure 4.16, which shows a solution of

ammonium sulﬁde being added to a lead(II) nitrate solution. The visible precip-

itate forms via the following molecular equation:

(NH

)

S(aq) + Pb(NO

)

(aq) → 2NH

(aq) + PbS(s)

The lead(II) sulﬁde is a solid that essentially doesn’t dissolve or ionize, so we note

this in all written forms of the equation. For example, the complete ionic equa-

tion is

2NH

(aq)+S

2−

(aq)+Pb

(aq)+2NO

−

(aq)→2NH

(aq)+2NO

−

(aq)+PbS(s)

The net ionic equation more clearly shows this precipitation reaction. The am-

monium and nitrate ions are spectator ions, so the net ionic equation reduces to

(aq) + S

2−

(aq) → PbS(s)

which is simply the precipitation of lead(II) sulﬁde.

How can we predict when a precipitation reaction will occur? For most com-

mon ions, the answer has been determined experimentally and can be summa-

rizedbyasetofsolubility rules. Table 4.3 illustrates these rules in detail. Com-

pounds generally follow these rules with a few exceptions, some of which are also

noted in the table. We can use this information to write and balance the molecu-

lar equation for a reaction, determine whether any of the compounds form insol-

uble precipitates, and write the complete ionic and net ionic equations.

146 Chapter 4 Solution Stoichiometry and Types of Reactions

FIGURE 4.15

Sulfate-reducing bacteria. Spring water

in the ﬂoodplain of the Prairie Dog Town

Fork of the Red River that ﬂows through

the panhandle of Texas has exposed

black mud. The black sediment is formed

from the interaction of dissolved metal

ions and sulﬁde produced from sulfate-

reducing bacteria.

Solubility Rules for Ionic Compounds

Soluble Insoluble

Group IA and ammonium Most carbonates are insoluble except

compounds are soluble. Group IA carbonates and (NH

)

Acetate, chlorate, perchlorate, and Most phosphates are insoluble except

nitrate compounds are soluble. Group IA phosphates and (NH

)

Most chlorides, bromides, and iodides Most sulﬁdes are insoluble except

are soluble except those of Ag

,Hg

, Group IA sulﬁdes and (NH

)

and Pb

Most sulfates are soluble except those Most hydroxides are insoluble except

of Ca

,Sr

,Ba

,Ag

,Hg

, Group IA hydroxides and Ca(OH)

and Pb

. Sr(OH)

, and Ba(OH)

TABLE 4.3

FIGURE 4.16

Lead sulﬁde precipitates from the

combination of its soluble ions.

Visualization: Solubility Rules

EXERCISE 4.8 Identifying Precipitation Reactions

Which combinations of the following aqueous solutions will produce precipitates:

aluminum bromide, barium hydroxide, magnesium sulfate, and nickel(II) iodide?

Use the solubility rules to guide you in your decision.

First Thoughts

You need to consider all the possible combinations of positive and negative ions in

order to identify those that will produce an insoluble product. The four solutions in

this question contain Al

,Br

−

,Ba

,OH

−

,Mg

,SO

2−

,Ni

, and I

−

Solution

Table 4.3 reveals the following solubilities for all the possible combinations of ions:

aluminum hydroxide—insoluble aluminum sulfate—soluble

aluminum iodide—soluble barium bromide—soluble

barium sulfate—insoluble barium iodide—soluble

magnesium bromide—soluble magnesium hydroxide—insoluble

magnesium iodide—soluble nickel(II) bromide—soluble

nickel(II) hydroxide—insoluble nickel(II) sulfate—soluble

The solubility rules indicate that we could make four insoluble precipitates, in the

following precipitation reactions:

❚ Add aluminum bromide solution to barium hydroxide solution, to form a

precipitate of aluminum hydroxide.

(aq) + 3OH

−

(aq) →Al(OH)

(s)

❚ Add barium hydroxide solution to magnesium sulfate solution, to form a

precipitate of barium sulfate, mixed with a precipitate of magnesium hydroxide.

(aq) +2OH

−

(aq) →Mg(OH)

(s) and Ba

(aq) +SO

2−

(aq) →BaSO

(s)

❚ Add nickel(II) iodide solution to barium hydroxide solution, to form a precipitate

of nickel(II) hydroxide.

(aq) + 2OH

−

(aq) → Ni(OH)

(s)

Further Insights

The result when we add a barium hydroxide solution to a magnesium sulfate solu-

tion reveals a complication. Precipitation reactions may produce a mixture of pre-

cipitates if the cations of each dissolved compound form insoluble products with

the anions of the other compound with which they are mixed. When we prepare a

compound by precipitation, we must ensure that the only precipitate that forms is

the desired one. We will discuss ways of doing this in Chapter 18.

PRACTICE 4.8

Which combination of the following solutions will produce a precipitate:

AgNO

(aq), NaCl(aq), Na

S(aq), ZnSO

(aq)?

See Problems 57–66, 89, and 90.

EXERCISE 4.9 The Stoichiometry of a Precipitation Reaction

Calculate the mass of the precipitate that is formed when 1.30 L of 0.0200 M AlBr

solution is added to 3.00 L of 0.0350 M NaOH solution.

First Thoughts

First we map out our plan of attack. We will do the following steps:

1. Write the balanced molecular equation (we will not need to write the net ionic

equation).

4.5 Precipitation Reactions 147