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

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surrounded by their individual hydration spheres. We can represent the hydra-

tion sphere in a chemical equation by writing the symbol (aq) after the hydrated

molecules, ions, or atoms to specify their aqueous phase:

NaCl(s) → Na

(aq) + Cl

−

(aq)

There are many substances that are not made up of cations and anions. Some

of them dissolve in water, such as the nutrient molecule glucose (C

) and

the molecule ethanol (C

O) found in alcoholic beverages. Why do they dis-

solve in water? These molecules possess a feature similar to water: The electrons

are polarized toward speciﬁc regions of the molecule. This leaves a slightly posi-

tive charge and a slightly negative charge on different regions of the molecules.

These partial charges can interact with the partial charges on water molecules.

The strong interaction allows the molecules to mix easily with water and dissolve

as shown in Figure 4.5. And, as is true with ions, a cage of water molecules sur-

rounds the dissolved solute molecules.

(s) → C

(aq)

Acids and bases are common substances, and many of them dissolve readily in

water. For example, nitric acid, HNO

, dissolves in water to produce the nitrate

ion,NO

−

(aq),and the hydrogen ion,H

(aq).According to one deﬁnition of these

common substances, acids produce a hydrogen ion when dissolved in water. On

128 Chapter 4 Solution Stoichiometry and Types of Reactions

–

Polar

bond

Polar

bond

Polar

bond

Polar

bond

Polar

bond

(a) (b)

Polar

bond

–

FIGURE 4.5

The partial charges on polar covalent

bonds, such as the O—H bonds in

ethanol and glucose, can interact with

the partial charges on water molecules

and allow molecules to dissolve in water.

Ethanol

An electrostatic potential map of glucose and

ethanol. The electrostatic potential is plotted

on the surface of a computer-generated

model of the molecules. Note how the electron

density is distributed in each molecule, and

compare these models to that of water

(Figure 4.3).

D-glucose

the other hand, sodium hydroxide (NaOH)

is a common base that dissolves in water, re-

sulting in Na

(aq) and OH

−

(aq). Accord-

ing to the same deﬁnition, bases produce

hydroxide ion in solution.

Electrolytes

You may have encountered sports drinks

sold with the claim that they help maintain

a healthy electrolyte balance by supplying

ions, sugar (to supply calories), and water

in an appropriate combination. The key in-

gredients quoted on the label of a leading

brand of sports drink can be found in Fig-

ure 4.6. The sodium and potassium in the

drink are present as hydrated Na

and K

ions. They are accompanied by hydrated

anions such as Cl

−

(chloride ions) so that

the entire solution has no net electrical

charge. Ionic compounds, such as sodium chloride and potassium chloride,

which dissociate in water to release free ions, are known as

electrolytes. The name

comes from the ability of solutions containing electrolytes to conduct electricity

because of the presence of mobile ions that can carry the electric current.

Electrolytes are of great medical importance, because our blood and cells

must contain an appropriate mixture of electrolytes to maintain good health. The

presence of electrolytes in blood and cells allows the body to absorb the right

amount of water from the gut and to excrete water via the kidneys. Maintaining

the optimal balance of electrolytes is especially important after strenuous exer-

cise, in which sweat containing water, sodium ions, and potassium ions leaves the

body as part of a vital cooling mechanism. The ions of electrolytes also play a cen-

tral role in creating the nerve impulses that allow our sense organs to work, our

muscles to move, and our brains to think.

Substances such as sodium chloride, which dissociate completely into ions

when they dissolve, are called

strong electrolytes. In fact, most ionic compounds

that dissolve in water fall into this category of electrolyte. A few molecules, such

as hydrogen chloride (HCl), are also considered strong electrolytes. The equation

below shows that gaseous HCl ﬁrst dissolves, and then dissociates in the water, to

form hydrated ions.

HCl(g) → HCl(aq) → H

(aq) + Cl

−

(aq)

Other electrolytes, such as acetic acid (vinegar, CH

COOH), are called weak

electrolytes

, because even though they may dissolve completely in water, the for-

mation of ions occurs to a much lesser extent, as shown in Figure 4.7. Note the dif-

ferent type of arrow,





, that deﬁnes a reaction that does not proceed completely

to products.

COOH(l) → CH

COOH(aq)





(aq) + CH

COO

−

(aq)

Those substances, such as glucose (C

), that dissolve in water without

forming ions are called

nonelectrolytes. The nonelectrolytes dissolve and associate

with water to form a hydration sphere, but they do not form ions. The equation

that represents a nonelectrolyte’s addition to water does not show the dissocia-

tion step found in the previous two equations.

(s) → C

(aq)

4.1 Water—A Most Versatile Solvent 129

HCl

Electrostatic potential map of HCl.

Note the very low electron density

at the hydrogen end of the

molecule.

Application

HEMICAL

NCOUNTERS:

Sports Drinks and

Electrolyte Balance

FIGURE 4.6

The key ingredients of a sports drink in-

clude sodium and potassium cations.

Visualization: Electriﬁed Pickle

Video Lesson: CIA

Demonstration: The Electric

Pickle

130 Chapter 4 Solution Stoichiometry and Types of Reactions

COOH(aq)

Acetic acid

Acetate

Water

COOH(aq) + H

O(l) CH

COO

–

(aq) + H

(aq)

Acetic

acid

FIGURE 4.7

Weak electrolytes (such as acetic acid)

do not dissociate completely in water.

Deﬁning Electrolytes

How do we know whether a compound is a strong electrolyte, a weak electrolyte, or

a nonelectrolyte?

Often the type of compound, assuming it dissolves in water, can

tell us whether it will dissociate. For example, most ionic compounds that dis-

solve in water are strong electrolytes, as are the strong acids and bases. Weak acids

and bases do not dissociate completely in aqueous solution and are therefore

weak electrolytes. Nonelectrolytes include the water-soluble but nonionic com-

pounds. Although there are many exceptions to these rules, the basic trends, sum-

marized in Table 4.1, are valid.

Experimentation can provide evidence as to whether a particular solution

contains a strong electrolyte, a weak electrolyte, or a nonelectrolyte. One such

experiment is shown in Figure 4.8. Using a solution as the connector in a light

bulb circuit, we can measure the ability of the solution to conduct electricity. The

FIGURE 4.8

The effect that each type of electrolyte

has on the conductivity of a solution as

measured by the brightness of a light

bulb. From left to right are solutions of

sodium chloride, acetic acid, and glucose.

(a) (b) (c)

Visualization: Electrolyte

Behavior

Visualization: Electrolytes

4.2 The Concentration of Solutions 131

Electrolytes and Types of Compounds

Ionic Strong Weak Molecular

Compounds Acids/Bases Acids/Bases Compounds

Strong Electrolyte Yes Yes No Sometimes

Weak Electrolyte Sometimes No Yes Sometimes

Nonelectrolyte No No No Sometimes

TABLE 4.1

Strong and weak acids and bases are described in Section 4.6.

brightness of a light bulb demonstrates the type of electrolyte in solution. What’s

happening in Figure 4.8? Because ions can “carry” an electrical charge, the solu-

tion containing the greatest number of ions will have the brightest light bulb.

4.2 The Concentration of Solutions

Hyponatremia is a serious condition characterized by a low sodium level in the

blood. The organization USA Track and Field recently released new guidelines

calling for long-distance runners to avoid drinking excessive amounts of water

during long runs, because doing so could lead to excessively diluted blood and a

severely reduced sodium level. The result, hyponatremia, gives rise to a high fever,

nausea, and, ultimately, heat stroke. Hyponatremia occurs when the sodium con-

centration is less than 130 mmol (mmol = 10

−3

mol) of sodium ions per liter of

blood.

The

concentration, the amount of solute per volume of solution, is often

quoted in moles of solute per liter of solution, in grams of solute per liter of

solution, or in various other ways. For example, long-distance runners should

consume sports drinks instead of water. A typical sports drink has 110 mg Na

per 8-ounce serving, which won’t reduce the sodium level in your blood. An

8-ounce serving of this drink contains 110 mg sodium ions. A larger quantity of

the drink—say, 1 quart (32 ounces)—contains 440 mg Na

, but the concentration

remains the same: 110 mg Na

per 8-ounce serving. Concentration is an intensive

property (see Section 1.4) because it doesn’t change with the volume of solution.

A useful concentration unit is known as

molarity. This unit speciﬁcally indi-

cates the moles of solute per liter of solution, as illustrated in Figure 4.9.

Molarity =

moles of solute

liter of solution

= M

Application

Moles of

solute

Liters of

solvent

Solution with

known concentration

in molarity

Molarity =

moles solute

liters solution

Compound

with a

known mass

Solvent

with a

known volume

FIGURE 4.9

The basic concepts of determining concen-

tration, measured in moles per liter.

Video Lesson: Concentrations

of Solutions

For example, 600.0 mg of aspirin in a soluble aspirin tablet might be dissolved in

orange juice to form 150.0 mL (150.0 cm

) of solution. We can calculate the con-

centration of the aspirin solution once we have calculated the number of moles

of aspirin and converted the volume of the juice into liters. We can use dimen-

sional analysis to convert the units of milligrams of aspirin per milliliter of solu-

tion into moles of aspirin per liter of solution:

600.0 mg aspirin

150.0 mL solution

1 g aspirin

1000 mg aspirin

1 mol aspirin

180.2 g aspirin

1000 mL

0.02220 mol aspirin

L solution

The molarity is 0.02220 mole of aspirin per liter, or 0.02220 molar (written

0.02220 M).

The relationship

Molarity =

moles of solute

liter of solution

can be used to calculate the

number of moles of a solute if you know the molarity and the volume. In fact, this

relationship can be used in many ways as a dimensional analysis factor. The exer-

cises and practices that follow will give you an opportunity to use this new factor

to answer questions.

EXERCISE 4.1 Calculating Molarity

Sodium hydroxide (NaOH) is one of the most important substances used in the

chemical industry, with over 9.5 billion kg manufactured in 2004, ranking it among

the top ten chemicals produced in the United States. It is used in chemical manu-

facturing processes, such as in making soaps and detergents. It is also used to break

down the lignin that holds cellulose together in wood so that the cellulose can be

made into paper. Dilute solutions of NaOH are often used in industry to verify the

quality of medicines produced.

a. What is the molarity of NaOH in a solution that contains 3.48 g of sodium

hydroxide in 500.0 mL of solution?

b. How many grams of NaOH are required to prepare 617 mL of 1.200 M NaOH

solution?

Solution

a. Molarity is expressed in units of

moles of solute

liter of solution

. We can solve for the

molarity by converting the units of

g NaOH

mL soln

into

moles of NaOH

liter of solution

using

dimensional analysis. First, we construct a ﬂowchart such as we introduced

in Chapter 1:

1000 mL

mol NaOH

g NaOH

mL soln

−−−−−→

g NaOH

L soln

−−−−−−−→

moles of NaOH

liter of solution

Then we complete the calculation:

3.48 g NaOH

500.0 mL solution

1000 mL

1 mol NaOH

40.00 g NaOH

0.174 mol NaOH

= 0.174 M NaOH

132 Chapter 4 Solution Stoichiometry and Types of Reactions

b. Given the molarity and the volume, we can solve for moles of NaOH and then

change to grams of NaOH using the molar mass of the compound. Again, we’ll

do it all in one step by converting units.

L soln

mL soln

mol NaOH

L soln

g NaOH

mol NaOH

mL soln −−−−−−−→L soln −−−−−−−→mol NaOH −−−−−−−→g NaOH

Then we complete the calculation:

617 mL solution×

1000 mL

1.200 mol NaOH

1 L solution

40.00 g NaOH

mol NaOH

= 29.6gNaOH

PRACTICE 4.1

a. What is the molarity of K

in the typical muscle cell, which contains 6.5 g K

per liter of solution?

b. What is the molarity of Na

in a sports drink that contains 110 mg Na

per

8-ounce serving? (1 ounce = 29.6 mL)

c. How many grams of ethanol (C

O) do we need to prepare 600.0 mL of a

1.200 molar ethanol solution?

See Problems 15–17, 19–20, 29, 31, 32, and 34.

EXERCISE 4.2 Calculating Moles

a. How many moles of calcium hydroxide, Ca(OH)

, are found in 250 mL of a

0.800 molar solution of this compound?

b. If we assume that Ca(OH)

is a strong electrolyte, how many moles of Ca

ions will be present in 250 mL of 0.800 M Ca(OH)

? How many moles of

−

ions will be present?

First Thoughts

This problem is different from the previous one because it asks for the number of

moles in a solution that is already prepared, rather than asking for the concentration

of a solution (part a of Exercise 4.1) or the mass of a reagent required to prepare

a solution (part b). Still, the problem-solving approach is similar. We note what

we start with, what we want, and how we get there. You can use a ﬂowchart to help

you stay on track.

Solution

a. Again, we use dimensional analysis to solve the problem. Note that we must

convert the milliliters of solution into liters so that we can use the molarity

term. Our ﬂowchart for this calculation is constructed ﬁrst:

1000 mL

mol Ca(OH)

L soln

mL soln −−−−−−−→L soln −−−−−−−→mol Ca(OH)

Then

250 mL solution ×

1000 mL

0.800 mol Ca(OH)

1 L solution

= 0.20 mol Ca(OH)

b. We start by writing the balanced equation that describes what happens to

Ca(OH)

when it is added to water.

Ca(OH)

(s) → Ca

(aq) + 2OH

−

(aq)

4.2 The Concentration of Solutions 133

Then we can calculate the answer by performing a dimensional analysis. Our

calculation can be accomplished using the mole ratio relating the number of

moles of Ca(OH)

to the number of moles of Ca

. Our ﬂowchart is written

ﬁrst:

1000 mL

mol Ca(OH)

L soln

mol Ca

mol Ca(OH)

mL soln −−−−−→L soln −−−−−−−→mol Ca(OH)

−−−−−−−−→mol Ca

Then

250 mL solution ×

1000 mL

0.800 mol Ca(OH)

1 L solution

1 mol Ca

1 mol Ca(OH)

= 0.20 mol Ca

Again, we need to use the mole ratio to indicate the relationship between

Ca(OH)

and OH

−

. We’ll use the same ﬂowchart for the Ca

determination,

but we’ll modify the last step to show the mole ratio between Ca(OH)

and

−

. The calculation is

250 mL solution ×

1000 mL

0.800 mol Ca(OH)

1 L solution

2 mol OH

−

1 mol Ca(OH)

= 0.40 mol OH

−

Further Insights

Although it might seem like “busy work” to include the mole ratio in a calculation

that can be determined by counting the number of ions in the formula, it is very

important to include mole ratios along with the equation that shows the dissolution

process. If we avoid this step now, we will ﬁnd it harder to solve more complex prob-

lems in later chapters.

PRACTICE 4.2

How many moles of the strong electrolyte sodium phosphate (Na

) will be in

5.00 L of a 0.77 M solution? How many moles of hydrated sodium ions will be in the

solution?

See Problems 13, 14, 37, and 38.

EXERCISE 4.3 Calculating Volumes

What volume of a 0.150 M solution of ethanol (C

O) will contain 12.5 moles of

ethanol?

First Thoughts

This is another unit conversion problem that we can solve using dimensional analy-

sis. However, in this case, we’re using the molarity in a different way: to determine

the volume rather than moles, grams, or molarity, as in previous problems.

Solution

First, we write our ﬂowchart:

L soln

mol ethanol

mol ethanol −−−−−−−→L ethanol

Then

12.5 mol ethanol ×

L solution

0.150 mol ethanol

= 83.3 L solution

134 Chapter 4 Solution Stoichiometry and Types of Reactions

Further Insights

In questions like this, the identity of the solute is irrelevant provided we are dealing

with moles, rather than masses. For example, the answer we obtained for the volume

of 0.150 M ethanol containing 12.5 moles of ethanol would be equally valid for the

volume of a 0.150 M solution of glucose containing 12.5 moles of glucose. Both re-

quire a volume of 83.3 L. In a similar way, if we were asked to calculate the number

of moles of a compound in a given volume of solution, we would not need to know

the identity of the compound if we were given the molarity of the solution. How-

ever, if we were asked about speciﬁc numbers of moles of ions of a strong electrolyte, we

would need the formula of the compound to determine the number of times the

ions or atoms appear in the formula.

PRACTICE 4.3

a. What volume of a 3.40 M solution of copper sulfate will contain 4.76 moles of

copper sulfate?

b. What volume of a 2.25 M solution of Ca(NO

)

will contain 5.5 moles of

calcium nitrate? What volume will contain 5.5 moles of nitrate ions?

See Problems 18, 21, and 22–24.

Parts per Million, Parts per Billion, and so on

Drinking water standards, which indicate the maximum permissible level of

harmful contaminants in the water that we consume, are dictated in the United

States by the Environmental Protection Agency (EPA). Other countries have their

own agencies responsible for water standards, such as La Secretaría de Medio

Ambiente y Recursos Naturales (SEMERNAT, The Federal Agency of the Envi-

ronment and Natural Resources) in Mexico. If water in the United States has

more than these maximum levels, the EPA declares it unsafe to drink. For exam-

ple, as of January 23, 2006, the maximum allowed level of arsenic in drinking

water is 1.3 × 10

−7

M. We can use molarity to discuss the concentrations of pol-

lutants such as arsenic, but the resulting numbers are, in many cases, very small.

When the concentrations get this small, we often ﬁnd it easier to use an alterna-

tive measure of concentration. Speciﬁcally, we can talk about the concentration

of arsenic in terms of

parts per million (ppm),or parts per billion (ppb),or even

parts per trillion (ppt).

We are already familiar with the related, but larger, unit “percent.” What does

percent mean? In a compound that is “1 percent nitrogen by mass,” the nitrogen

contributes 1 gram out of every 100 grams to the total mass. In the

same way, a level of one part per million (1 ppm), means the chem-

ical contributes 1 gram out of every million grams of the total mass.

Similarly, one part per billion (1 ppb) corresponds to a level of

1 gram out of every billion grams of the total. One part per trillion

(ppt) corresponds to 1 gram out of every trillion grams.

These concentration units correspond to very small levels in-

deed. We can get an idea of exactly how small from the following

approximate comparisons, as illustrated in Figure 4.10:

■

One part per million is roughly equivalent to a drop of ink in a

12-gallon bucket of water.

■

One part per billion is roughly equivalent to a drop of ink in a

large tanker truck full of water.

■

One part per trillion is roughly equivalent to a drop of ink in a

12-million-gallon reservoir of water.

4.2 The Concentration of Solutions 135

Application

HEMICAL

ENCOUNTERS:

Maximum Levels

of Chemicals in

Drinking Water

An environmental chemist sampling the

water in a river.

Let’s look at these concentration units in a slightly different way.As an example,

the EPA maximum level for hexachlorobenzene (C

), a pesticide used on wheat

in the United States until 1965, is 1 ppb. This means that the maximum allowable

level in drinking water should be 1 gram of C

per billion grams of solution.

1ppbC

1gC

g solution

If we use

1 g water

1 mL water

as the density of water and make the key assumption that the

solution is so dilute that its density is about equal to that of water, then

Density of very dilute aqueous solutions

1 g solution

1 mL solution

This means that we can express the concentration of hexachlorobenzene as the

number of grams of C

per liter of solution:

1gC

g solution

1 g solution

1 mL solution

mL solution

1 L solution

−6

L solution

1 µgC

L solution

↑↑ ↑

(one ppb C

) (density of the solution) (convert mL to L)

The bottom line is that in dilute aqueous solutions, we can express parts per

billion as

ppb solute =

µg solute

L solution

This relationship, the number of grams of solute per liter of solution, is more

conceptually understandable than the actual deﬁnition of parts per billion. For

our hexachlorobenzene example, 1 liter of solution that contains 1.0 µg hexa-

chlorobenzene is said to have a concentration of 1.0 ppb hexachlorobenzene. We

can even use this deﬁnition to convert from other concentrations to ppb. What is

the maximum level of arsenic, in ppb, if the molarity of a solution at this maxi-

mum level is 1.3 × 10

−7

M arsenic?

Our ﬂowchart looks like this:

gAs

mol As

−6

mol As

L solution

−−−−−→

gAs

L solution

−−−−−→

µgAs

L solution

= ppb As

Then

1.3 ×10

−7

mol As

1 L solution

74.92 g As

1 mol As

1 µgAs

−6

gAs

9.7 µgAs

L solution

= 9.7ppb

Note the relationships among parts per million, parts per billion, and parts per

trillion in Table 4.2.

136 Chapter 4 Solution Stoichiometry and Types of Reactions

Hexachlorobenzene

(a) (b) (c)

FIGURE 4.10

To visualize the idea of parts per million,

per billion, and per trillion, consider that

one drop of ink could be placed in these

quantities of water. (a) Placing that drop

of ink in a 12-gallon bucket results in

1 ppm. (b) Placing it in a tanker truck

results in 1 ppb. (c) Placing it in a

12-million-gallon reservoir results in

1 ppt.

4.2 The Concentration of Solutions 137

Although we more often hear about how small concentrations of compounds

in water can be harmful, very small levels of some chemicals in drinking water

can actually be helpful to human health. For example, the presence of tiny

amounts of ﬂuoride (F

−

) in drinking water is recognized as beneﬁcial in the pre-

vention of tooth decay. The U.S. Public Health Service recommends a level of be-

tween 0.7 and 1.2 ppm F

−

. The effects of such small levels in preventing tooth

decay have been reported as “striking” by the American Dental Association.

EXERCISE 4.4 Converting ppm to Molarity

An acceptable midrange value for ﬂuoride ion (F

−

) in drinking water is 1.0 ppm. To

what concentration of ﬂuoride, in M, does this correspond?

First Thoughts

Using the mass-per-volume relationship we established in Table 4.2 enables us to

solve the problem with

1.0 ppm =

1.0mgF

−

L solution

Solution

Remember to construct the ﬂowchart before you start the calculation.

1.0mgF

−

L solution

−3

−

1mgF

−

1 mol F

−

19.00 g F

−

= 5.3 ×10

−5

M F

−

Further Insights

High concentrations of ﬂuoride ion can be very harmful to your health. In fact,

some would argue that ﬂuoride at any level is harmful. The debate surrounding

ﬂuoridation of drinking water continues.

PRACTICE 4.4

Cyanide ion (CN

−

) concentrations of 0.200 ppm in drinking water are considered

the upper limit for human consumption. What is this concentration in M?

See Problems 25–28, 32, 35, and 36.

Dilution

How do municipal water treatment plant workers prepare water so that it contains

a ﬂuoride ion concentration of about 1 part per million?

They use a concentrated

source of ﬂuorine, hydroﬂuorosilicic acid, H

SiF

(“HFSA”), which they then di-

lute in the drinking water. HFSA reacts with water in a fairly complex way that re-

leases ﬂuoride ion into the water. In Ireland, a company in the town of New Ross,

Common Units Used in Expressing “Parts per” Concentration of “X”

Mass-to-Mass Mass-to-Volume

Unit Relationship Relationship

parts per million (ppm)

g solution

mg X

L solution

parts per billion (ppb)

g solution

µgX

L solution

parts per trillion (ppt)

g solution

ng X

L solution

Mass-to-volume relationship assumes a density of 1g/mL.

TABLE 4.2

Application