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Chapter 15

Solutions
LOOKING AHEAD
15.1 Forming Solutions
A. Solubility
• Solubility of Ionic
Substances
• Solubility of Polar
Substances
• Substances Insoluble
in Water
• How Substances Dissolve
B. Solution Composition:
An Introduction
C. Factors Affecting the Rate
of Dissolving
15.2 Describing Solution
Composition
A. Solution Composition:
Mass Percent
B. Solution Composition:
Molarity
C. Dilution
15.3 Properties of Solutions
A. Stoichiometry of
Solution Reactions
B. Neutralization
Reactions
C. Normality
D. Boiling Point and
Freezing Point

518

Bubbles on the
surface of a soap
solution

15
IN

YOUR

LIFE

S

olutions are very important to all of us. In
fact we can walk, talk, laugh, and work
only because millions of chemical
reactions are occurring every second in the
solutions in our bodies. We also encounter
many solutions as we go through a typical
day—the liquid soap we use in our morning
shower, the cranberry juice we drink for
breakfast, the sports drinks we have for lunch,
the gasoline we put in our cars and the
“energy drink” we may have in the middle of
the afternoon. Even the “pure water” from the
water fountain or a water bottle is a solution—
it contains dissolved minerals from the earth,
chlorine compounds to disinfect it, and traces
of many other substances. A solution that we
often take for granted is the one we live in—
the air is a solution of nitrogen (N2), oxygen
(O2), water vapor, carbon dioxide, and traces
of many other gases. Solutions are clearly
essential to our existence.

Brass, a solid solution
of copper and zinc, is
used to make musical
instruments and many
other objects.

WHAT DO YOU KNOW?
Prereading Questions
1. What is meant by the term solution (as in, a solution of salt
and water)?
2. What happens when you mix water and oil?
3. What can you do to get sugar to dissolve more quickly in
water?
4. A certain cookie is made by mixing 150.0 g of dough with
25.0 g of chocolate chips. What is the mass percent of
chocolate chips in the cookie?
5. What is meant by a precipitation reaction?
6. What is the net ionic equation for all reactions of a strong
acid and a strong base?
7. Why is salt scattered on roads during the winter in cold
weather climates?

Solutions • Chapter 15 • 519

SECTION 15.1

Forming Solutions
Key Terms
• Solution
• Solvent
• Solute
• Aqueous solution
• Saturated
• Unsaturated
• Supersaturated
• Concentrated
• Dilute

Solution
A homogeneous mixture

Solvent
The dissolving medium
in a solution

Solute
The substance dissolved
in the solvent to make a
solution

Objectives
• To understand the process of dissolving
• To learn why certain substances dissolve in water
• To learn qualitative terms describing the concentration of a solution
• To understand the factors that affect the rate at which a solid dissolves
A solution is a homogeneous mixture, a mixture in which the components are uniformly intermingled. This means that a sample from one part
is the same as a sample from any other part. For example, the first sip of
coffee is the same as the last sip.
The atmosphere that surrounds us is a gaseous solution containing
O2(g), N2(g), and other gases randomly dispersed. Solutions can also be
solids. For example, brass is a homogeneous mixture—a solution—of copper
and zinc.
These examples illustrate that a solution can be a gas, a liquid, or a
solid (see Table 15.1). The substance present in the largest amount is
called the solvent, and the other substance or substances are called
solutes. For example, when we dissolve a teaspoon of sugar in a glass
of water, the sugar is the solute and the water is the solvent.
Aqueous solutions are solutions with water as the solvent. Because
they are so important, in this chapter we will concentrate on the properties
of aqueous solutions.
Table 15.1

Various Types of Solutions
Example

State of
Solution

Original State
of Solute

State of
Solvent

Aqueous solution

air, natural gas

gas

gas

gas

A solution with water as
a solvent

antifreeze in water

liquid

liquid

liquid

brass

solid

solid

solid

carbonated water (soda)

liquid

gas

liquid

seawater, sugar solution

liquid

solid

liquid

A. Solubility
What happens when you put a teaspoon of sugar in your iced tea and
stir it, or when you add salt to water for cooking vegetables? Why do the
sugar and salt “disappear” into the water? What does it mean when something dissolves—that is, when a solution forms?

520 • Chapter 15 • Solutions

15
Solubility of Ionic Substances
We saw in Chapter 8 that when sodium chloride dissolves in water, the
resulting solution conducts an electric current. This convinces us that the
solution contains ions that can move (this is how the electric current is conducted). The dissolving of solid sodium chloride in water is shown below.

+
NaCl
= Na+

H2O

= Cl–

In the solid state the ions are packed closely together. However, when
the solid dissolves, the ions are separated and dispersed throughout the
solution. The strong ionic forces that hold the sodium chloride crystal
together are overcome by the strong attractions between the ions and the
polar water molecules. This process is represented in Figure 15.1. Notice
that each polar water molecule orients itself in a way to maximize its
attraction with a Cl or Na ion. The negative end of a water molecule is
attracted to a Na ion, while the positive end is attracted to a Cl ion.
The strong forces holding the positive and negative ions in the solid are
replaced by strong water–ion interactions, and the solid dissolves (the
ions disperse).
It is important to remember that when an ionic substance (such as a
salt) dissolves in water, it breaks up into individual cations and anions,
which are dispersed in the water. For instance, when ammonium nitrate,
NH4NO3, dissolves in water, the resulting solution contains NH4 and
NO3 ions, which move around independently. This process can be
represented as

i

nformation

Cations are positive ions.
Anions are negative ions.

H2O(l)

NH4NO3(s) 88n NH4 (aq) ⫹ NO3 (aq)
where (aq) indicates that the ions are surrounded by water molecules.
Anion


+

δ+



+



+



+



+

+

+

+





+

2δ–
δ+



δ+





2δ–

+

+

Figure 15.1

+





+

δ+
Cation

Polar water molecules
interact with the positive
and negative ions of a salt.
These interactions replace
the strong ionic forces
holding the ions together
in the undissolved solid,
thus assisting in the
dissolving process.

15.1 • Forming Solutions • 521

Solubility of Polar Substances
Water also dissolves many nonionic substances. Sugar is one example of a
nonionic solute that is very soluble in water. Another example is ethanol,
C2H5OH. Why is ethanol so soluble in water? The answer lies in the structure of the ethanol molecule.
δ–

H

H

O

H C

C H

H

H

δ+

Polar bond

The molecule contains a polar O—H bond like those in water, which makes
it very compatible with water. Just as hydrogen bonds form among water
molecules in pure water (see p. 490), ethanol molecules can form hydrogen
bonds with water molecules in a solution of the two.
H
H
H

O

H C

C H

H

H

O
H

The sugar molecule is shown below. Common table sugar has the
chemical name sucrose.
H
C
HO

CH2OH

O

C
H

H

OH
C

O

H CH2OH
C
OH

H

C

C
O

H

H

OH C

C

C

OH

H

CH2OH

Notice that this molecule has many polar —OH groups, each of which can
hydrogen-bond to a water molecule. Because of the attractions between
sucrose and water molecules, solid sucrose is quite soluble in water.

Active Reading Questions
1. Does water generally dissolve polar or nonpolar substances? Explain.
2. What is similar about table salt and table sugar that allows them both
to be soluble in water?

522 • Chapter 15 • Solutions

15
Substances Insoluble in Water
Many substances do not dissolve in water. For example, when petroleum
leaks from a damaged tanker, it does not disperse uniformly in the water
(does not dissolve) but rather floats on the surface because its density is
less than that of water. Petroleum is a mixture of molecules like this one.

CH2

CH2

CH3

CH2

CH2
CH2

CH2
CH2

CH2
CH2

CH2
CH2

CH2
CH2

CH2
CH2

CH3
CH2

Since carbon and hydrogen have very similar electronegativities, the
bonding electrons are shared almost equally and the bonds are essentially
nonpolar. The resulting molecule with its nonpolar bonds is not compatible
with the polar water molecules, which prevents it from being soluble in
water. This situation is represented in Figure 15.2.

Figure 15.2
Oil layer

CH2 CH2 CH2 CH2
CH2 CH2 CH2 CH2 CH2
H

O

O

H

H

H

H
H
H

O

O
H
O

H
H

H

O

H

O

An oil layer floating on water. For a substance to
dissolve, the water–water hydrogen bonds must
be broken to make a “hole” for each solute
particle. However, the water–water interactions
will break only if they are replaced by similar
strong interactions with the solute.

H
O

H

Water

H
H

O
H

H

The oil from this tanker spreads out on the water
as fire boats spray water to put out the fire.

15.1 • Forming Solutions • 523

How Substances Dissolve
Notice in Figure 15.2 that the water molecules in liquid water are associated
with each other by hydrogen-bonding interactions. For a solute to dissolve
in water, a “hole” must be made in the water structure for each solute particle. This will occur only if the lost water–water interactions are replaced
by similar water–solute interactions.
In the case of sodium chloride, strong interactions occur between the
polar water molecules and the Na and Cl ions. This allows the sodium
chloride to dissolve.
In the case of ethanol or sucrose, hydrogen-bonding interactions can
occur between the O—H groups on these molecules and water molecules,
making these substances soluble as well.
But oil molecules are not soluble in water, because the many waterwater interactions that would have to be broken to make “holes” for these
large molecules are not replaced by favorable water–solute interactions.

Active Reading Question
Consider two one liter samples of solutions each containing 1.0 mol
solute. One is a solution of table salt and one is a solution of table sugar.
If you could see highly magnified views (molecular level), how would
the solutions compare?
These considerations account for the observed behavior that “like
dissolves like.” In other words, we observe that a given solvent usually
dissolves solutes that have polarities similar to its own. For example,
water dissolves most polar solutes, because the solute–solvent interactions
formed in the solution are similar to the water–water interactions present
in the pure solvent. Likewise, nonpolar solvents dissolve nonpolar solutes.
For example, dry-cleaning solvents used for removing grease stains from
clothes are nonpolar liquids. “Grease” is composed of nonpolar molecules,
so a nonpolar solvent is needed to remove a grease stain.

HANDS-ON CHEMISTRY

• MINI-LAB •

Colors in Motion
Materials
• large clear container
• water
• water soluble markers
• filter paper
Procedure
1. Add water to a large clear container so that it
covers the bottom of the container to a depth
of about _1_ inch. Cover the container.
4
2. Obtain a piece of filter paper and use water
soluble markers to place 4 dots at least _1_ inch
2
apart and _1_ inch from the bottom of the paper.
2

524 • Chapter 15 • Solutions

3. Place the filter paper in the container so that
the bottom of the paper is under water. The
dots should not be submerged.
4. Make observations for the next several minutes.
Results/Analysis
1. Why do some colors separate while others do
not?
2. Why do some colors move farther than others
up the paper?
3. How do the separation of the colors and their
rate of movement relate to solubility and
intermolecular forces?

15
B. Solution Composition: An Introduction
Even for very soluble substances, there is a limit to how much solute
can be dissolved in a given amount of solvent. For example, when you add
sugar to a glass of water, the sugar rapidly disappears at first. However, as
you continue to add more sugar, at some point the solid no longer dissolves
but collects at the bottom of the glass. When a solution contains as much
solute as will dissolve at that temperature, we say it is saturated. If a solid
solute is added to a solution already saturated with that solute, the added
solid does not dissolve. A solution that has not reached the limit of solute
that will dissolve in it is said to be unsaturated. When more solute is
added to an unsaturated solution, it dissolves.
Sometimes when a solid is dissolved to the saturation limit at an elevated temperature and then allowed to cool, all of the solid may remain
dissolved. This type of solution is called a supersaturated solution—it
contains more dissolved solid than a saturated solution will hold at that
temperature. A supersaturated solution is very unstable. Adding a crystal
of the solid will cause immediate precipitation of solid until the solution
reaches the saturation point.
Although a chemical compound always has the same composition, a
solution is a mixture and the amounts of the substances present can vary in
different solutions. For example, coffee can be strong or weak. Strong coffee
has more coffee dissolved in a given amount of water than weak coffee. To
describe a solution completely, we must specify the amounts of solvent and
solute. We sometimes use the qualitative terms concentrated and dilute to
describe a solution. A relatively large amount of solute is dissolved in a
concentrated solution (strong coffee is concentrated). A relatively small
amount of solute is dissolved in a dilute solution (weak coffee is diluted).
As we will see in Section 15.2, the concentration of a solution is a
measure of the amount of solute dissolved in a given volume of solution.
We can use this idea to better understand the terms concentrated and dilute.
For example, compare the two solutions below in which each dot
represents a solute molecule. Which solution is more concentrated?

Solution A
Volume 1.0 L

Saturated
Describes a solution that
contains as much solute
as will dissolve at that
temperature

Unsaturated
Describes a solution in
which more solute can
dissolve than is dissolved
already at that temperature

Supersaturated
Describes a solution that
contains more solute than
a saturated solution will
hold at that temperature

Concentrated
Describes a solution in
which a relatively large
amount of solute is
dissolved in a solution

Dilute
Describes a solution in
which a relatively small
amount of solute is
dissolved in a solution

Solution B
Volume 1.0 L

The volume of each of the solutions is the same. However, the amount
of solute in solution A is less than the amount of solute in solution B.
Because of this, solution B is more concentrated than solution A.
When comparing the concentrations of two solutions you have to
consider both the amount of solute and the volume of solution.
Sodium acetate crystallizing
from a saturated solution

15.1 • Forming Solutions • 525

For example, the following two solutions have the same concentration.

CRITICAL
THINKING
One of the jobs performed
by the U.S. Environmental
Protection Agency (EPA) is
to make sure our drinking
water is safe.
What if the EPA insisted
that all drinking water
had to be 100% pure?
How would it affect our
drinking water supply?

Solution X
Volume 1.0 L

Solution Y
Volume 2.0 L

Although the volume of solution Y is twice the volume of solution X the
amount of solute in solution Y is also twice as much as that in Solution X.

Active Reading Question
Draw pictures of two solutions such that the solution with more solute is
less concentrated than the solution with less solute.
Although these qualitative terms serve a useful purpose, we often need
to know the exact amount of solute present in a given amount of solution.
Now, we will consider various ways to describe the composition of a solution.

C. Factors Affecting the Rate of Dissolving

Tea dissolves much faster in
the hot water on the left
than in the cold water on
the right.

When a solid is being dissolved in a liquid to form a solution, the
dissolving process may occur rapidly or slowly. Three factors affect
the speed of the dissolving process: surface area, stirring, and temperature.
Because the dissolving process occurs at the surface of the solid being
dissolved, the greater the amount of surface area exposed to the solvent, the
faster the dissolving will occur. For example, if we want to dissolve a cube
of sugar in water, how can we speed up the process? The answer is to grind
up the cube into tiny crystals. Because the crystals from the ground-up cube
expose much more surface area to the water than the original cube did, the
sugar dissolves much more quickly.
The dissolving process is also increased by stirring the solution. Stirring
removes newly dissolved particles from the solid surface and continuously
exposes the surface to fresh solvent.
Finally, dissolving occurs more rapidly at higher temperatures. (Sugar
dissolves much more rapidly in hot tea than iced tea.) Higher temperatures
cause the solvent molecules to move more rapidly, thus increasing the rate
of the dissolving process.
In addition to dissolving faster at higher temperatures, most solids are
more soluble at higher temperatures. That is, in most cases more solid will
dissolve in water at 90 °C than in water at 25 °C. The opposite is true for
gases dissolved in water. The solubility of a gas in water typically decreases
as the temperature increases.

L e t ’s R e v i e w
Factors Affecting Dissolving




526 • Chapter 15 • Solutions

Surface area
Stirring
Temperature

15
CHEMISTRY EXPLORERS
Ellen Swallow Richards • 1842–1911

E

llen Swallow Richards was born in Massachusetts in 1842. She was
the first woman in America to be accepted to a scientific school, the
first woman to attend the Massachusetts Institute of Technology (MIT),
and the first woman professional chemist in the nation. She spent her
professional career in research and teaching at MIT, where she founded the
field of home economics.
In the 1880s, Richards worked as water analyst for the Massachusetts
State Board of Health while teaching at MIT. She conducted a two-year survey
analyzing over 100,000 samples of drinking water in Massachusetts. From this
study, she produced the first state water quality standards in the United States,
which in turn led to the first modern sewage treatment plants. Her later studies
in food quality led to state food and drug standards.
The next time you drink a glass of water, make a toast to Ellen Swallow
Richards. It is her pioneering work that led to safe drinking water.

SECTION 15.1
REVIEW QUESTIONS
1 Draw a molecular (microscopic) level
picture to show how salt and sugar look
when dissolved in water.

2 Determine the number of moles of each
type of ion in solution when 1 mole of
each of the following dissolves in water:
a. sodium sulfate
b. ammonium acetate
c. calcium chloride

4 Why does water by itself not dissolve a
grease stain?

5 Can a dilute solution also be saturated?
Explain.

6 Increasing the number of collisions between
solid and solvent particles increases the rate at
which a solid dissolves. Explain how stirring,
increasing surface area, and heating increase
the number of collisions.

d. copper(II) sulfate

3 Chemists often say “like dissolves like.”
What does this mean?

RESEARCH LINKS

15.1 • Forming Solutions • 527

SECTION 15.2

Describing Solution Composition
Key Terms
• Mass percent
• Molarity (M)
• Standard solution
• Dilution

Objectives
• To understand mass percent and how to calculate it
• To understand and use molarity
• To learn to calculate the concentration of a solution made by diluting a
stock solution

A. Solution Composition: Mass Percent
One way of describing a solution’s composition is mass percent,
which expresses the mass of solute present in a given mass of solution.
The definition of mass percent follows:
mass of solute 100%
Mass percent ________________
mass of solution
grams of solute
__________________________________ 100%
grams of solute grams of solvent

i

nformation

The mass of the solution is
the sum of the masses of
the solute and the solvent.

For example, suppose a solution is prepared by dissolving 1.0 g of
sodium chloride in 48 g of water. The solution has a mass of 49 g (48 g
of H2O plus 1.0 g of NaCl), and there is 1.0 g of solute (NaCl) present.
The mass percent of solute, then, is
1.0 g solute
_____________
100% 0.020 100% 2.0% NaCl
49 g solution

HANDS-ON CHEMISTRY

• MINI-LAB •

Rainbow in a Straw
Materials
• five paper cups containing five different colored
solutions of NaCl
• clear, colorless plastic straw
Procedure
1. Obtain five paper cups containing five different
colored solutions. You will also need a clear,
colorless plastic straw.
2. Each of your five solutions has a different
concentration of NaCl. Your task is to order the
solutions from least concentrated to most
concentrated. You may only use the five solutions
and the straw. You may need to experiment a little
and you may make some mistakes along the way.

528 • Chapter 15 • Solutions

Results/Analysis
1. How does concentration of the solution relate to
density of the solution?
2. List the order of solutions from least concentrated
to most concentrated.

15
E XAMPL E 1 5 . 1
Solution Composition: Calculating Mass Percent
A solution is prepared by mixing 1.00 g of ethanol, C2H5OH, with 100.0 g
of water. Calculate the mass percent of ethanol in this solution.
Solution
Where do we want to go?
Mass percent of ethanol ? %
What do we know?
• 1.00 g ethanol (C2H5OH)


100.0 g H2O



mass of solute 100%
mass percent ________________
mass of solution

How do we get there?

(
(

)

grams of C2H5OH
Mass percent C2H5OH _________________
100%
grams of solution
1.00 g C2H5OH
______________________________
100.0 g H2O 1.00 g C2H5OH

)

100%

1.00 g
_______ 100%
101.0 g
0.990% C2H5OH
Practice Problem • Exercise 15.1
A 135-g sample of seawater is evaporated to dryness, leaving 4.73 g of solid
residue (the salts formerly dissolved in the seawater). Calculate the mass
percent of solute present in the original seawater.

E XAMPL E 1 5 . 2
Solution Composition: Determining Mass of Solute
Although milk is not a true solution (it is really a suspension of tiny globules
of fat, protein, and other substrates in water), it does contain a dissolved
sugar called lactose. Cow’s milk typically contains 4.5% by mass of lactose,
C12H22O11. Calculate the mass of lactose present in 175 g of milk.
Solution
Where do we want to go?
Mass lactose in 175 g milk ? g
What do we know?
• 4.5% lactose (C12H22O11) {solute}



175 g milk {solution}
mass of solute 100%
mass percent ________________
mass of solution

15.2 • Describing Solution Composition • 529

How do we get there?
grams of solute
Mass percent _________________ 100%
grams of solution
We now substitute the quantities we know:
Mass of lactose

Mass percent

grams of solute
100% 4.5%
Mass percent
175 g
Mass of milk

We now solve for grams of solute by multiplying both sides by 175 g,
grams of solute
175 g _______________ 100% 4.5% 175 g
175 g
and then dividing both sides by 100%,
4.5% 175 g
100% ______
Grams of solute ______
100%
100%
to give
Grams of solute 0.045 175 g 7.9 g lactose
Practice Problem • Exercise 15.2
What mass of water must be added to 425 g of formaldehyde to prepare a
40.0% (by mass) solution of formaldehyde? This solution, called formalin,
is used to preserve biological specimens.
Hint: Substitute the known quantities into the definition for mass percent,
and then solve for the unknown quantity (mass of solvent).

B. Solution Composition: Molarity
When a solution is described in terms of mass percent, the amount of
solution is given in terms of its mass. However, it is often more convenient
to measure the volume of a solution than to measure its mass. Because of
this, chemists often describe a solution in terms of concentration. We define
the concentration of a solution as the amount of solute in a given volume of
solution. The most commonly used expression of concentration is molarity
(M). Molarity describes the amount of solute in moles and the volume of
the solution in liters.
Molarity is the number of moles of solute per volume of solution in liters.
moles of solute ____
mol
M molarity ________________
L
liters of solution
A solution that is 1.0 molar (written as 1.0 M) contains 1.0 mol of solute
per liter of solution.

Active Reading Question
Adding 1.0 mole of solute to 1.0 L of water does not make a 1.0 M
solution. Why not?

530 • Chapter 15 • Solutions

15
We need to consider both the amount of solute and the volume when
determining the concentration of a solution. For example, consider the
following two solutions where each dot represents 1.0 mol of solute.
Concentration

Concentration

mol solute 2.0 M
______________
2.0
1.0 L solution

mol solute 1.0 M
______________
2.0
2.0 L solution

Solution A
Volume 1.0 L

Solution B
Volume 2.0 L

E XAMPL E 1 5 . 3
Solution Composition: Calculating Molarity, I
Calculate the molarity of a solution prepared by dissolving 11.5 g of solid
NaOH in enough water to make 1.50 L of solution.
Solution
Where do we want to go?
molarity NaOH ? M
What do we know?
• 11.5 g NaOH




V 1.50 L
moles of solute
M ________________
liters of solution
molar mass NaOH 40.0 g/mol

How do we get there?
We have the mass (in grams) of solute, so we need to convert the mass of
solute to moles (using the molar mass of NaOH). Then we can divide the
number of moles by the volume in liters.
Mass of
solute

Moles of
solute
Use molar
mass

Molarity
Moles
Liters

We compute the number of moles of solute, using the molar mass of NaOH
(40.0 g).
mol NaOH 0.288 mol NaOH
_____________
11.5 g NaOH 1
40.0 g NaOH
Then we divide by the volume of the solution in liters.
moles of solute 0.288
mol NaOH 0.192 M NaOH
________________
Molarity ________________
1.50 L solution
liters of solution

15.2 • Describing Solution Composition • 531

EX A MPL E 15. 4
Solution Composition: Calculating Molarity, II
Calculate the molarity of a solution prepared by dissolving 1.56 g of
gaseous HCl into enough water to make 26.8 mL of solution.
Solution
Where do we want to go?
molarity HCl ? M
What do we know?
• 1.56 g HCl (g)




V 26.8 mL
moles of solute
M ________________
liters of solution
molar mass HCl 36.5 g/mol

How do we get there?
First we calculate the number of moles of HCl (molar mass 36.5 g).
mol HCl 0.0427 mol HCl
___________
1.56 g HCl 1
36.5 g HCl
4.27 10 2 mol HCl
Next we change the volume of the solution from milliliters to liters, using
the equivalence statement 1 L 1000 mL, which gives the appropriate
conversion factor.
1L
26.8 mL _________
0.0268 L
1000 mL
2.68 10 2 L
Finally, we divide the moles of solute by the liters of solution.
4.27 10 2 mol HCl 1.59 M HCl
Molarity ______________________
2.68 10 2 L solution
Practice Problem • Exercise 15.4
Calculate the molarity of a solution prepared by dissolving 1.00 g of
ethanol, C2H5OH, in enough water to give a final volume of 101 mL.

It is important to realize that the description of a solution’s composition
may not accurately reflect the true chemical nature of the solute as it is
present in the dissolved state. Solute concentration is always written in
terms of the form of the solute before it dissolves. For example, describing a
solution as 1.0 M NaCl means that the solution was prepared by dissolving
1.0 mol of solid NaCl in enough water to make 1.0 L of solution; it does
not mean that the solution contains 1.0 mol of NaCl units. Actually the
solution contains 1.0 mole of Na ions and 1.0 mole of Cl ions. That is,
it contains 1.0 M Na and 1.0 M Cl .

532 • Chapter 15 • Solutions

15
E XAMPL E 1 5 . 5
Solution Composition: Calculating Ion Concentration from Molarity
Give the concentrations of all the ions in each of the following solutions:
a. 0.50 M Co(NO3)2
b. 1 M FeCl3

Image not available

Solution

for electronic use.

Co(NO3)2 Please refer to the

a. When solid Co(NO3)2 dissolves, it produces ions as follows:

image in the texbook.

H2O(l)

Co(NO3)2(s) 88n Co2 (aq) ⫹ 2NO3 (aq)

21

Co
NO32 NO32

which we can represent as
H2O(l)

1 mol Co(NO3)2(s) 88n 1 mol Co2 (aq) ⫹ 2 mol NO3 (aq)

A solution of
cobalt(II) nitrate

Therefore, a solution that is 0.50 M Co(NO3)2 contains
0.50 M Co2 and (2 ⫻ 0.50) M NO3 , or 1.0 M NO3 .
b. When solid FeCl3 dissolves, it produces ions as follows:
H2O(l)

FeCl3(s) 88n Fe3 (aq) ⫹ 3Cl (aq)
FeCl3

or
H2O(l)

1 mol FeCl3(s) 88n 1 mol Fe3 (aq) ⫹ 3 mol Cl (aq)
A solution that is 1 M FeCl3 contains 1 M Fe3 ions and
3 M Cl ions.

Cl

Fe3
Cl Cl

Practice Problem • Exercise 15.5
Give the concentrations of the ions in each of the following solutions:
a. 0.10 M Na2CO3
b. 0.010 M Al2(SO4)3
Often we need to determine the number of moles of solute present in
a given volume of a solution of known molarity. To do this, we use the
definition of molarity. When we multiply the molarity of a solution by
the volume (in liters), we get the moles of solute present in that sample:
moles of solute
Liters of solution ⫻ molarity ⫽ liters of solution ⫻ ________________
liters of solution
⫽ moles of solute

15.2 • Describing Solution Composition • 533

EX A MPL E 15. 6
Solution Composition: Calculating Number of Moles from Molarity
How many moles of Ag ions are present in 25 mL of a 0.75 M AgNO3
solution?
Solution
Where do we want to go?
Moles Ag in sample solution ? mol
What do we know?
• Molarity of solution 0.75 M

MATH
M

moles of solute
liters of solution

Liters M



V 25 mL



Moles M V

How do we get there?
A 0.75 M AgNO3 solution contains 0.75 M Ag ions and 0.75 M NO3 ions.
Next we must convert the volume given from mL to L.

Moles of solute

1L
0.025 L 2.5 10 2 L
25 mL _________
1000 mL
Now we multiply the volume times the molarity.
0.75 mol Ag
2.5 10 2 L solution _____________ 1.9 10 2 mol Ag
L solution
Practice Problem • Exercise 15.6
Calculate the number of moles of Cl ions in 1.75 L of 1.0 10 3 M AlCl3.

Standard solution
A solution in which the
concentration is
accurately known

A standard solution is a solution whose concentration is accurately
known. When the appropriate solute is available in pure form, a standard
solution can be prepared by weighing out a sample of solute, transferring it
completely to a volumetric flask (a flask of accurately known volume), and
adding enough solvent to bring the volume up to the mark on the neck of
the flask. This procedure is illustrated in Figure 15.3.

Figure 15.3
Steps involved in the
preparation of a standard
aqueous solution. (a) Put a
weighed amount of a substance
(the solute) into the volumetric
flask, and add a small quantity
of water. (b) Dissolve the solid
in the water by gently swirling
the flask (with the stopper in
place). (c) Add more water
(with gentle swirling) until the
level of the solution just reaches
the mark etched on the neck of
the flask. Then mix the solution
thoroughly by inverting the
flask several times.

Wash bottle

Volume marker
(calibration mark)

Weighed
amount
of solute

534 • Chapter 15 • Solutions

(a)

(b)

(c)

15
E XAMPL E 1 5 . 7

To analyze the alcohol content of a certain substance, a chemist needs
1.00 L of an aqueous 0.200 M K2Cr2O7 (potassium dichromate) solution.
How much solid K2Cr2O7 (molar mass 294.2 g/mol) must be weighed
out to make this solution?
Solution
Where do we want to go?
Mass of K2Cr2O7 needed to make solution ? g
What do we know?
• Molarity of solution 0.200 M


V 1.00 L



Molar mass K2Cr2O7 294.2 g/mol



Moles M V

How do we get there?
We need to calculate the number of grams of solute (K2Cr2O7) present (and thus
the mass needed to make the solution). First we determine the number of moles
of K2Cr2O7 present by multiplying the volume (in liters) by the molarity.
0.200 mol K2Cr2O7
0.200 mol K2Cr2O7
1.00 L solution __________________
L solution

MATH
Liters M

Moles of solute

Then we convert the moles of K2Cr2O7 to grams, using the molar mass of
K2Cr2O7 (294.2 g).
294.2 g K2Cr2O7
0.200 mol K2Cr2O7 ________________
58.8 g K2Cr2O7
mol K2Cr2O7
Therefore, to make 1.00 L of 0.200 M K2Cr2O7, the chemist must weigh
out 58.8 g of K2Cr2O7 and dissolve it in enough water to make 1.00 L of
solution. This is most easily done by using a 1.00-L volumetric flask.
Practice Problem • Exercise 15.7
Formalin is an aqueous solution of formaldehyde, HCHO, used as a
preservative for biological specimens. How many grams of formaldehyde
must be used to prepare 2.5 L of 12.3 M formalin?

C. Dilution
To save time and space in the laboratory, solutions that are routinely
used are often purchased or prepared in concentrated form (called stock
solutions). Water (or another solvent) is then added to achieve the molarity
desired for a particular solution. The process of adding more solvent to a
solution is called dilution. For example, the common laboratory acids are
purchased as concentrated solutions and diluted with water as they are
needed. A typical dilution calculation involves determining how much
water must be added to an amount of stock solution to achieve a solution
of the desired concentration. The key to doing these calculations is to
remember that only water is added in the dilution. The amount of solute in
the final, more dilute, solution is the same as the amount of solute in the
original concentrated stock solution.

Dilution
The process of adding
solvent to a solution to
lower the concentration
of solute

15.2 • Describing Solution Composition • 535

For example, consider the following solutions in which each dot represents
1.0 mol of solute particles.

mol solute 8.0 M
______________
Concentration 8.0
1.0 L solution
Solution 1
Volume 1.0 L

i

nformation

If we double the volume of the solution by adding water, we do not change
the moles of solute, but the concentration becomes half of its original value.

Dilution with water
doesn’t alter the number
of moles of solute present.

mol solute 4.0 M
______________
Concentration 8.0
2.0 L solution
Solution 2
Volume 2.0 L

If the volume is doubled again, the concentration is again cut by 2,
becoming one-fourth of the original concentration.

mol solute 2.0 M
______________
Concentration 8.0
4.0 L solution
Solution 3
Volume 4.0 L

i

nformation

The molarities of stock
solutions of the common
concentrated acids are:
Sulfuric (H2SO4)
Nitric (HNO3)
Hydrochloric (HCl)

18 M
16 M
12 M

Notice in all three solutions,
Moles of solute after dilution moles of solute before dilution
The number of moles of solute stays the same but more water is added,
increasing the volume, so the molarity decreases.
Remains constant

M
Decreases

moles of solute
volume (L)
Increases
(water added)

For example, suppose we want to prepare 500. mL of 1.00 M acetic acid,
HC2H3O2, from a 17.5 M stock solution of acetic acid. What volume of the
stock solution is required?
Where do we want to go?
Volume of stock solution needed ? L

536 • Chapter 15 • Solutions

15
What do we know?
• Molarity of stock solution 17.5 M


Top Ten Elements
in Seawater

Solution required

Element

Amount
(kg/mi3)

Oxygen

3.9 1012

Hydrogen

5.0 1011

Chlorine

9.2 1010

How do we get there?
Find the moles of acetic acid needed in the dilute solution.

Sodium

5.3 1010



Sulfur

4.3 109

Calcium

1.9 109

Potassium

1.9 109

Bromine

3.1 108

Carbon

1.3 108

Vdilute 500.0 mL
Mdilute 1.00 M


Moles M V

Convert the volume of the dilute solution to L.
1 L solution
500. mL solution _________________
0.500 L solution
1000 mL solution
Vdilute solution

Convert mL to L

(in mL)



Magnesium 6.1 109

Moles acetic acid needed M V
1.00 mol HC2H3O2
0.500 L solution __________________
0.500 mol HC2H3O2
L solution
Mdilute solution

Find the volume of the stock (concentrated) solution that contains
0.500 mol acetic acid.
Because volume molarity moles, we have
17.5 mol HC2H3O2
V (in liters) __________________
0.500 mol HC2H3O2
L solution
17.5 mol gives
Solving for V by dividing both sides by __________
L solution
0.500 mol HC2H3O2
V ____________________
0.0286 L, or 28.6 mL, of solution
17.5
mol HC2H3O2
__________________
L solution

(

HANDS-ON CHEMISTRY

MATH
Liters M

Moles of solute

)

• MINI-LAB •

Can We Add Concentrations?
Materials
• water
• food coloring
• graduated cylinder
Procedure
1. Place 10.0 mL of water and add 10 drops of food
coloring. Label the concentration of this solution
1.0 D (where D has units of drops/mL). This is
solution A.
2. Add 10.0 mL of water to another cup and add
5 drops of food coloring to the water. This is
solution B.

Results/Analysis
1. What is the concentration of solution B in units
of D?
2. How does this solution C look compared to
solution A and solution B?
3. What is the concentration of solution C in units
of D?
4. Can you add the concentrations of solution A
and solution B to get the concentration of
solution C? Why or why not? How does your
answer to question 2 help you answer this
question?

3. Add half of solution A and half of solution B to a
third cup to make solution C.

15.2 • Describing Solution Composition • 537

Therefore, to make 500. mL of a 1.00 M acetic acid solution, we
take 28.6 mL of 17.5 M acetic acid and dilute it to a total volume of
500. mL. This process is illustrated in Figure 15.4. Because the moles
of solute remain the same before and after dilution, we can write
Initial conditions

M1



Molarity
before
dilution

V1

Final conditions



moles of solute

Volume
before
dilution

M2



Molarity
after
dilution

V2

Volume
after
dilution

Does it make sense?
We can check our calculations on acetic acid by showing that
M1 V1 M2 V2. In the above example, M1 17.5 M,
V1 0.0286 L, V2 0.500 L, and M2 1.00 M, so
mol 0.0286 L 0.500 mol
M1 V1 17.5 ____
L
mol 0.500 L 0.500 mol
M2 V2 1.00 ____
L
and therefore
M1 V1 M2 V2
This shows that the volume (V2) we calculated is correct.

Active Reading Question
You add 95.0 mL of water to 5.0 mL of 5.0 M acetic acid. Compare the
concentration and number of moles of acetic acid in the concentrated
acid to the concentration and number of moles of acetic acid in the
dilute solution.

Figure 15.4
(a) 28.6 mL of 17.5 M acetic acid solution is
transferred to a volumetric flask that already
contains some water.
(b) Water is added to the flask (with swirling)
to bring the volume to the calibration mark,
and the solution is mixed by inverting the flask
several times.
(c) The resulting solution is 1.00 M acetic acid.

538 • Chapter 15 • Solutions

500 mL

(a)

(b)

(c)

15
E XAMPL E 1 5 . 8
Calculating Concentrations of Diluted Solutions
What volume of 16 M sulfuric acid must be used to prepare 1.5 L of a
0.10 M H2SO4 solution?
Solution
Where do we want to go?
Volume of H2SO4 needed to make solution ⫽ ? mL

Image not available
for electronic use.
Please refer to the
image in the textbook.

What do we know?
• Moles solute ⫽ M1 ⫻ V1 ⫽ M2 ⫻ V2
Initial Conditions
(concentrated)

Final Conditions
(dilute)

M1 ⫽ 16 M

M2 ⫽ 0.10 M

V1 ⫽ ?

V2 ⫽ 1.5 L

Approximate dilutions can
be carried out using a
calibrated beaker. Here
concentrated sulfuric acid
is being added to water to
make a dilute solution.

How do we get there?
We can solve the equation
M1 ⫻ V1 ⫽ M2 ⫻ V2
for V1 by dividing both sides by M1
M
M2 ⫻ V2
1 ⫻ V1
________
⫽ ________
M1
M1
to give
M2 ⫻ V2
V1 ⫽ ________
M1
Now we substitute the known values of M2, V2, and M1.
V1

mol (1.5 L)
0.10 ____
(
L )
_________________

⫽ 9.4 ⫻ 10

ⴚ3



mol
16 ____
L

DID YOU KNOW

L

mL ⫽ 9.4 mL
_________
9.4 ⫻ 10ⴚ3 L ⫻ 1000
1L
Therefore, V1 ⫽ 9.4 ⫻ 10ⴚ3 L, or 9.4 mL. To make 1.5 L of 0.10 M H2SO4
using 16 M H2SO4, we must take 9.4 mL of the concentrated acid and
dilute it with water to a final volume of 1.5 L. The correct way to do this
is to add the 9.4 mL of acid to about 1 L of water and then dilute to 1.5 L
by adding more water.

It is always best to add
concentrated acid to
water, not water to
the acid. That way, if
any splashing occurs
accidentally, it is dilute
acid that splashes.

Practice Problem • Exercise 15.8
What volume of 12 M HCl must be taken to prepare 0.75 L of 0.25 M HCl?

15.2 • Describing Solution Composition • 539

HANDS-ON CHEMISTRY

• MINI-LAB •

Good to the Last Drop!
Materials


water



food coloring



graduated cylinder

Procedure
1. Place 10.0 mL, of water and add 20 drops of
food coloring. Label the concentration of this
solution 2.0 D (where D has units of drops/mL).
This is solution A.

Results/Analysis
1. What is the concentration of solution B in
units of D?
2. What is the concentration of solution C in
units of D?
3. What is the concentration of the solution in step
4 in units of D?

2. Take 1.0 mL of solution A and place it in an
empty cup. Add 9.0 mL of water. This is
solution B.
3. Take 1.0 mL of solution B and place it in an empty
cup. Add 9.0 mL of water. This is solution C.
4. Continue this successive dilution until you can no
longer see any color.

SECTION 15.2
REVIEW QUESTIONS
1 Vinegar is made by adding 33 g of acetic acid
to 625 g of water. What is the percent by
mass of acetic acid in this solution of vinegar?

2 Assuming the density of vinegar is 1.0 g/mL,
what is the molarity of vinegar? (Use the
percent by mass of acetic acid in vinegar
from question 1.) The molar mass of acetic
acid is 60.05 g/mol.

3 A solution is labeled “0.450 M magnesium

5 If 4.25 g of CaBr2 is dissolved in enough
water to make 125 mL of solution, what
is the molarity of the solution?

6 What is the key idea to remember when
working with the dilution of a solution?

7 Calculate the new concentration when
50.0 mL of water is added to 725.0 mL
of 1.25 M NaCl.

nitrate.” Calculate the concentration of each
ion present in solution.

4 Calculate the number of moles of KOH in
150.0 mL of a 0.500 M KOH solution.

540 • Chapter 15 • Solutions

RESEARCH LINKS

SECTION 15.3

Properties of Solutions
Objectives
• To learn to solve stoichiometric problems involving solution reactions
• To do calculations involving acid–base reactions
• To learn about normality and equivalent weight
• To use normality in stoichiometric calculations
• To understand the effect of a solute on solution properties

Key Terms
• Neutralization reaction
• Equivalent of an acid
• Equivalent of a base
• Equivalent weight
• Colligative property

A. Stoichiometry of Solution Reactions
Because so many important reactions occur in solution, it is important to
be able to do stoichiometric calculations for solution reactions. The principles
needed to perform these calculations are similar to those developed in
Chapter 9. It is helpful to think in terms of the following steps:

Steps for Solving Stoichiometric Problems
Involving Solutions
Step 1 Write the balanced equation for the reaction. For reactions
involving ions, it is best to write the net ionic equation.

CRITICAL
THINKING

Step 2 Calculate the moles of reactants.
Step 3 Determine which reactant is limiting.
Step 4 Calculate the moles of other reactants or products, as required.
Step 5 Convert to grams or other units, if required.

What if all ionic solids
were soluble in water?
How would it affect
reactions in aqueous
solution?

Active Reading Question
When solving stoichiometry problems for solution reactions, what type
of chemical equation is most convenient to use?

E XAMPL E 1 5 . 9
Solution Stoichiometry: Calculating Mass of Reactants and Products
Calculate the mass of solid NaCl that must be added to 1.50 L of a
0.100 M AgNO3 solution to precipitate all of the Ag ions in the form
of AgCl. Calculate the mass of AgCl formed.
Solution
Where do we want to go?
Mass of AgCl formed ? g
What do we know?
• V 1.50 L


M 0.100 M AgNO3



NaCl(s) AgNO3(aq) n AgCl(s) NaNO3(aq)

15.3 • Properties of Solutions • 541

How do we get there?
Step 1 Write the balanced equation for the reaction.
When added to the AgNO3 solution (which contains Ag and NO3
ions), the solid NaCl dissolves to yield Na and Cl ions. Solid AgCl
forms according to the following balanced net ionic reaction:
Ag (aq) Cl (aq) n AgCl(s)
Step 2 Calculate the moles of reactants.
In this case we must add just enough Cl ions to react with all the
Ag ions present, so we must calculate the moles of Ag ions
present in 1.50 L of a 0.100 M AgNO3 solution. (Remember that a
0.100 M AgNO3 solution contains 0.100 M Ag ions and 0.100 M
NO3 ions.)
0.100 mol Ag
1.50 L ______________ 0.150 mol Ag
L


MATH
Liters M

Moles of Ag present
in 1.5 L of 0.100 M AgNO3

Moles of solute

Step 3 Determine which reactant is limiting.
In this situation we want to add just enough Cl to react with
the Ag present. That is, we want to precipitate all the Ag in
the solution. Thus the Ag present determines the amount of Cl
needed.
Step 4 Calculate the moles of Cl required.
We have 0.150 mol of Ag ions and, because one Ag ion reacts
with one Cl ion, we need 0.150 mol of Cl ,
mol Cl 0.150 mol Cl
__________
0.150 mol Ag 1
1 mol Ag
so 0.150 mol of AgCl will be formed.
0.150 mol Ag 0.150 mol Cl n 0.150 mol AgCl
Step 5 Convert to grams of NaCl required.
To produce 0.150 mol Cl , we need 0.150 mol NaCl. We calculate
the mass of NaCl required as follows:
58.4 g NaCl
0.150 mol NaCl ____________ 8.76 g NaCl
mol NaCl
Moles

Mass
Times molar mass

The mass of AgCl formed is
143.3 g AgCl
0.150 mol AgCl _____________ 21.5 g AgCl
mol AgCl

When aqueous sodium chloride is added to
a solution of silver nitrate, a white silver
chloride precipitate forms.

542 • Chapter 15 • Solutions

15
E XAMPL E 1 5 . 1 0
Solution Stoichiometry: Determining Limiting Reactants and Calculating Mass of Products
When Ba(NO3)2 and K2CrO4 react in aqueous solution, the yellow
solid BaCrO4 is formed. Calculate the mass of BaCrO4 that forms when
3.50 ⫻ 10 3 mol of solid Ba(NO3)2 is dissolved in 265 mL of 0.0100 M
K2CrO4 solution.
Solution
Where do we want to go?
Mass of BaCrO4 formed in the reaction ⫽ ? g
What do we know?
• Amount Ba(NO3)2(s) ⫽ 3.50 ⫻ 10 3 mol




K2CrO4


0.100 M



V ⫽ 265 mL

Ba(NO3)2(s) ⫹ K2CrO4(aq) n BaCrO4(s) ⫹ 2KNO3(aq)

Barium chromate
precipitating

How do we get there?
Step 1 The original K2CrO4 solution contains the ions K and CrO42 .
When the Ba(NO3)2 is dissolved in this solution, Ba2 and NO3
ions are added. The Ba2 and CrO42 ions react to form solid
BaCrO4. The balanced net ionic equation is
Ba2 (aq) ⫹ CrO42 (aq) n BaCrO4(s)
Step 2 Next we determine the moles of reactants. We are told that
3.50 ⫻ 10 3 mol of Ba(NO3)2 is added to the K2CrO4 solution. Each
formula unit of Ba(NO3)2 contains one Ba2 ion, so 3.50 ⫻ 10 3 mol
of Ba(NO3)2 gives 3.50 ⫻ 10 3 mol of Ba2 ions in solution.
3.50 ⫻ 10 3
mol Ba(NO3)2

dissolves to give

3.50 ⫻ 10 3
mol Ba2

Because V ⫻ M ⫽ moles of solute, we can compute the moles of K2CrO4
in the solution from the volume and molarity of the original solution.
First we must convert the volume of the solution (265 mL) to liters.
1L
265 mL ⫻ _________
⫽ 0.265 L
1000 mL
Next we determine the number of moles of K2CrO4, using the
molarity of the K2CrO4 solution (0.0100 M).
0.0100 mol K2CrO4
0.265 L ⫻ ___________________
⫽ 2.65 ⫻ 10 3 mol K2CrO4
L
We know that
2.65 ⫻ 10 3
mol K2CrO4

dissolves to give

2.65 ⫻ 10 3
mol CrO42

so the solution contains 2.65 ⫻ 10 3 mol of CrO42 ions.

15.3 • Properties of Solutions • 543

Step 3 The balanced equation tells us that one Ba2 ion reacts with one
CrO42 . Because the number of moles of CrO42 ions (2.65 10 3)
is smaller than the number of moles of Ba2 ions (3.50 10 3),
the CrO42 will run out first.
Ba2 (aq)



3.50 10 3 mol

CrO42 (aq)

n BaCrO4(s)

2.65 10 3 mol
Smaller
(runs out first)

Therefore, the CrO42 is limiting.
Moles of
CrO42

limits

Moles of
BaCrO4

Step 4 The 2.65 10 3 mol of CrO42 ions will react with 2.65 10 3 mol
of Ba2 ions to form 2.65 10 3 mol of BaCrO4.
2.65 10 3 mol
Ba2



2.65 10 3 mol
CrO42

2.65 10 3 mol
BaCrO4(s)

Step 5 The mass of BaCrO4 formed is obtained from its molar mass
(253.3 g) as follows:
253.3 g BaCrO
2.65 10 3 mol BaCrO4 _______________4 0.671 g BaCrO4
mol BaCrO4
Practice Problem • Exercise 15.10
When aqueous solutions of Na2SO4 and Pb(NO3)2 are mixed, PbSO4
precipitates. Calculate the mass of PbSO4 formed when 1.25 L of 0.0500 M
Pb(NO3)2 and 2.00 L of 0.0250 M Na2SO4 are mixed.
Hint: Calculate the moles of Pb2 and SO42 in the mixed solution, decide
which ion is limiting, and calculate the moles of PbSO4 formed.

B. Neutralization Reactions
So far we have considered the stoichiometry of reactions in solution
that result in the formation of a precipitate. Another common type of
solution reaction occurs between an acid and a base. We introduced these
reactions in Chapter 8. Recall from that discussion that an acid is a substance that furnishes H ions. A strong acid, such as hydrochloric acid,
HCl, dissociates (ionizes) completely in water.
HCl(aq) n H (aq) Cl (aq)
Strong bases are water-soluble metal hydroxides, which are completely
dissociated in water. An example is NaOH, which dissolves in water to
give Na and OH ions.
H2O(l)

NaOH(s) 88n Na (aq) OH (aq)

544 • Chapter 15 • Solutions

15
When a strong acid and strong base react, the net ionic reaction is
H (aq) OH (aq) n H2O(l)
An acid–base reaction is often called a neutralization reaction. When
just enough strong base is added to react exactly with the strong acid in a
solution, we say the acid has been neutralized. One product of this reaction
is always water. The steps in dealing with the stoichiometry of any neutralization reaction are the same as those we followed previously.

Neutralization
reaction
An acid–base reaction

Active Reading Question
What does it mean to say an acid has been neutralized?

E XAMPL E 1 5 . 1 1
Solution Stoichiometry: Calculating Volume in Neutralization Reactions
What volume of a 0.100 M HCl solution is needed to neutralize 25.0 mL
of a 0.350 M NaOH solution?
Solution
Where do we want to go?
Volume of HCl required for neutralization ? L
What do we know?
• HCl molarity 0.100 M




NaOH sample


0.350 M



V 25.0 mL

HCl(aq) NaOH(aq) n H2O(l) NaCl(aq)

How do we get there?
Step 1 Write the balanced equation for the reaction.
Hydrochloric acid is a strong acid, so all the HCl molecules dissociate
to produce H and Cl ions. Also, when the strong base NaOH dissolves, the solution contains Na and OH ions. When these two
solutions are mixed, the H ions from the hydrochloric acid react
with the OH ions from the sodium hydroxide solution to form
water. The balanced net ionic equation for the reaction is
H (aq) OH (aq) n H2O(l)
Step 2 Calculate the moles of reactants.
In this problem we are given a volume (25.0 mL) of 0.350 M NaOH,
and we want to add just enough 0.100 M HCl to provide just
enough H ions to react with all the OH . Therefore, we must
calculate the number of moles of OH ions in the 25.0-mL sample
of 0.350 M NaOH. To do this, we first change the volume to liters
and multiply by the molarity.
mol OH 8.75 10 3 mol OH
1L
_______________
25.0 mL NaOH _________
0.350
1000 mL
L NaOH

Moles of OH present
in 25.0 mL of
0.350 M NaOH

15.3 • Properties of Solutions • 545

Step 3 Determine which reactant is limiting.
This problem requires the addition of just enough H ions to react
exactly with the OH ions present, so the number of moles of OH
ions present determines the number of moles of H that must be
added. The OH ions are limiting.
Step 4 Calculate the moles of H1 required.
The balanced equation tells us that the H and OH ions react in a
1:1 ratio, so 8.75 10 3 mol of H ions is required to neutralize
(exactly react with) the 8.75 10 3 mol of OH ions present.
Step 5 Calculate the volume of 0.100 M HCl required.
Next we must find the volume (V) of 0.100 M HCl required to
furnish this amount of H ions. Because the volume (in liters)
times the molarity gives the number of moles, we have
V
Unknown
volume
(in liters)

0.100 mol H
8.75 10 3 mol H
L
Moles of
H needed

Now we must solve for V by dividing both sides of the equation
mole H .
______________
by 0.100
L
0.100
mol H
_____________
8.75 10 3 mol H
L
____________________
V ______________

0.100 mol H
0.100 mol H
______________
______________
L
L
V 8.75 10 2 L
Changing liters to milliliters, we have
mL 87.5 mL
_________
V 8.75 10 2 L 1000
L
Therefore, 87.5 mL of 0.100 M HCl is required to neutralize 25.0 mL of
0.350 M NaOH.
Practice Problem • Exercise 15.11
Calculate the volume of 0.10 M HNO3 needed to neutralize 125 mL of
0.050 M KOH.
Equivalent of an acid
The amount of acid that
can furnish one mole of
hydrogen ions (H )

Equivalent of a base
The amount of base that
can furnish one mole of
hydroxide ions (OH )

Equivalent weight
The mass (in grams) of
one equivalent of an acid
or a base

C. Normality
Normality is another unit of concentration that is sometimes used, especially when dealing with acids and bases. The use of normality focuses
mainly on the H and OH available in an acid–base reaction. Before we
discuss normality, however, we need to define some terms. One equivalent
of an acid is the amount of that acid that can furnish 1 mol of H ions.
Similarly, one equivalent of a base is defined as the amount of that base
that can furnish 1 mol of OH ions. The equivalent weight of an acid or a
base is the mass in grams of 1 equivalent (equiv) of that acid or base.

546 • Chapter 15 • Solutions

15
The common strong acids are HCl, HNO3, and H2SO4. For HCl and
HNO3 each molecule of acid furnishes one H ion, so 1 mol of HCl can
furnish 1 mol of H ions. This means that
Furnishes 1 mol of H

1 mol HCl 1 equiv HCl
Molar mass (HCl) equivalent weight (HCl)

i

Likewise, for HNO3,
1 mol HNO3 1 equiv HNO3

1 mol HCl

Molar mass (HNO3) equivalent weight (HNO3)
However, H2SO4 can furnish two H ions per molecule, so 1 mol of
H2SO4 can furnish two mol of H . This means that
1 mol
H2SO4

furnishes

2 mol
H

furnishes

1 mol
H

1
2

mol
H2SO4
1
2

mol
H2SO4



1 equiv
H2SO4

nformation

1 mol HCl

produces


1 mol H
1 equiv HCl

1 mol
H

Because each mole of H2SO4 can furnish 2 mol of H , we need to take
only _1_ mol of H2SO4 to get 1 equiv of H2SO4. Therefore,
2
_1_ mol H SO 1 equiv H SO
4
4
2
2
2
and
Equivalent weight (H2SO4) _1_ molar mass (H2SO4)
2
_
1_ (98 g) 49 g
2
The equivalent weight of H2SO4 is 49 g.
The common strong bases are NaOH and KOH. For
NaOH and KOH, each formula unit furnishes one OH ion,
so we can say

Table 15.2

The Molar Masses and Equivalent
Weights of the Common Strong
Acids and Bases
Molar Mass (g)

Equivalent
Weight (g)

1 mol NaOH 1 equiv NaOH

Acid

Molar mass (NaOH) equivalent weight (NaOH)

HCl

36.5

36.5

1 mol KOH 1 equiv KOH

HNO3

63.0

63.0

Molar mass (KOH) equivalent weight (KOH)

H2SO4

98.0

98.0
49.0 _____
2

NaOH

40.0

40.0

KOH

56.1

56.1

These ideas are summarized in Table 15.2.

Base

15.3 • Properties of Solutions • 547

EX A MPL E 15 . 1 2
Solution Stoichiometry: Calculating Equivalent Weight
Phosphoric acid, H3PO4, can furnish three H ions per molecule. Calculate
the equivalent weight of H3PO4.
Solution
Where do we want to go?
Equivalent weight for H3PO4 ? g/equivalent
What do we know?
• Molar mass 98.0 g/mol
How do we get there?
The key point here involves how many protons (H ions) each molecule
of H3PO4 can furnish.
? H

H3PO4
furnishes

Because each H3PO4 can furnish three H ions, 1 mol of H3PO4 can
furnish 3 mol of H ions:
1 mol
H3PO4

furnishes

3 mol
H

So 1 equiv of H3PO4 (the amount that can furnish 1 mol of H ) is one-third
of a mole.
1
3

mol
H3PO4

furnishes

1 mol
H

This means the equivalent weight of H3PO4 is one-third its molar mass.
Equivalent
weight



Molar mass
3

molar mass (H3PO4)
98.0 g
Equivalent weight (H3PO4) ___________________
______ 32.7 g
3
3

Normality (N) is defined as the number of equivalents of
solute per liter of solution.
equivalents
number of equivalents
equiv
Normality N ______________________ ___________ ______
L
liter
liter of solution
This means that a 1 N solution contains 1 equivalent of solute per liter of
solution. Notice that when we multiply the volume of a solution in liters by
the normality, we get the number of equivalents.
equiv
N V ______ L equiv
L

548 • Chapter 15 • Solutions

15
Active Reading Question
What is the normality of a 4.0 M H2SO4 solution?

E XAMPL E 1 5 . 1 3
Solution Stoichiometry: Calculating Normality
A solution of sulfuric acid contains 86 g of H2SO4 per liter of solution.
Calculate the normality of this solution.

i

Solution
Where do we want to go?

To calculate the
concentration of a
solution, first write the
appropriate definition.
Then decide how to
calculate the quantities
shown in the definition.

Normality ? equivalent/L
What do we know?
• Molar mass 98.0 g/mol


H2SO4


nformation

1.00 L




86 g
equivalents
N ___________
L

How do we get there?
To find the number of equivalents present, we must calculate the number
of equivalents represented by 86 g of H2SO4. To do this calculation, we
focus on the definition of the equivalent: it is the amount of acid that furnishes 1 mol of H . Because H2SO4 can furnish two H ions per molecule,
1 equiv of H2SO4 is _1_ mol of H2SO4, so
2
molar mass (H2SO4)
98.0 g
Equivalent weight (H2SO4) ___________________
______ 49.0 g
2
2
We have 86 g of H2SO4
1 equiv H2SO4
86 g H2SO4 ______________
1.8 equiv H2SO4
49.0 g H2SO4
1.8 equiv H2SO4
equiv
N ______ ________________
1.8 N H2SO4
L
1.0 L
Does it make sense?
We know that 86 g is more than 1 equivalent of H2SO4 (49 g) so this
answer makes sense.
Practice Problem • Exercise 15.13
Calculate the normality of a solution containing 23.6 g of KOH in 755 mL
of solution.

15.3 • Properties of Solutions • 549

The main advantage of using equivalents is that 1 equiv of acid contains
the same number of available H ions as the number of OH ions present
in 1 equiv of base. That is,
0.75 equiv (base) will react exactly with 0.75 equiv (acid).
0.23 equiv (base) will react exactly with 0.23 equiv (acid).
And so on.
In each of these cases, the number of H ions furnished by the sample of
acid is the same as the number of OH ions furnished by the sample of base.
n equivalents of any acid will exactly neutralize n equivalents of any base.
n equiv
acid

reacts exactly with

n equiv
base

Because we know that equal equivalents of acid and base are required for
neutralization, we can say that
equiv (acid) equiv (base)
That is,
Nacid Vacid equiv (acid) equiv (base) Nbase Vbase
For any neutralization reaction,
Nacid Vacid Nbase Vbase

CELEBRITY CHEMICAL

Mercury (Hg)

Mercury was isolated from its principal ore, cinnabar (HgS), as early as 500 B.C. Cinnabar was
widely used in the ancient world as a red pigment
(vermilion). Mercury, whose mobility inspired its
naming after the messenger of the gods in Roman
mythology, was of special interest to alchemists.
Alchemists hoped to find a way to change cheap
metals into gold, and it was generally thought that
mercury, which easily forms alloys with many
metals, was the key to this process. Mercury has a
common name of quicksilver.
Mercury is a very dense (13.6 times denser than
water), shiny liquid that has a surprisingly high vapor
pressure for a heavy metal. Because mercury vapor is
quite toxic, it must be stored in stoppered containers
and handled in well-ventilated areas. Many instances
of suspected mercury poisoning occurred in the days
before its hazards were understood.
For example, it is very possible that Sir Isaac
Newton, the renowned physicist, suffered from
mercury poisoning at one time in his career and

550 • Chapter 15 • Solutions

had to “retire” to the country away from his
laboratory to regain his strength. Some have
speculated that Mozart, the famous composer,
may have died of mercury poisoning due to the
mercury salts that were used to treat an illness.
Also, the “mad hatters” of Alice in Wonderland
fame have a historical connection to mercury.
Apparently, the mercury salts used by workers
in London to treat felt used in hats
caused the hatters to become
“mad” over the years.
It turns out that a
change in personality
is a characteristic of
heavy metal
poisoning.
The Mad Hatter’s Tea
Party, an illustration by
Sir John Tenniel, from
Alice’s Adventures in
Wonderland

15
E XAMPL E 1 5 . 1 4
Solution Stoichiometry: Using Normality in Calculations
What volume of a 0.075 N KOH solution is required to react exactly with
0.135 L of 0.45 N H3PO4?
Solution
Where do we want to go?
Volume KOH required to react ? L
What do we know?
• 0.075 N KOH


H3PO4


0.135 L



0.45 N



In neutralization n equivalentsacid equivalentsbase



Nacid Vacid Nbase Vbase

How do we get there?
We know that for neutralization, equiv (acid) equiv (base), or
Nacid Vacid Nbase Vbase
We want to calculate for the volume of base, Vbase, so we solve for Vbase by
dividing both sides by Nbase.
Nbase Vbase
N
acid Vacid
____________
____________
Vbase
Nbase
Nbase
Now we can substitute the given values Nacid 0.45 N, Vacid 0.135 L,
and Nbase 0.075 N into the equation.
Vbase

equiv
0.45 ______ (0.135 L)
N
L
acid Vacid
____________
____________________


0.81 L
equiv
Nbase
______
0.075
L

(

)

This gives Vbase 0.81 L, so 0.81 L of 0.075 N KOH is required to react
exactly with 0.135 L of 0.45 N H3PO4.
Practice Problem • Exercise 15.14
What volume of 0.50 N H2SO4 is required to react exactly with 0.250 L
of 0.80 N KOH?

15.3 • Properties of Solutions • 551

D. Boiling Point and Freezing Point
We saw in Chapter 14 that at an atmospheric pressure of 1 atm water
freezes at 0 °C and boils at 100 °C. If we dissolve a solute such as NaCl in
water, does the presence of the solute affect the freezing point of the water?
Yes. A 1.0 M NaCl solution freezes at about ⫺1 °C and boils at about
104 °C. The presence of solute “particles” in the water extends the liquid
range of water—that is, water containing a solute exists as a liquid over a
wider temperature range than does pure water.
Why does this happen? For example, what effect do the particles have
that causes the boiling point of water to increase? To answer this question
we need to reconsider the boiling process. Recall that for water to boil, bubbles must be able to form in the interior of the liquid. That is, energetic
water molecules in the bubble must be able to exert an internal pressure
large enough to push back the atmospheric pressure.
Now consider the situation in a solution that contains a solute.
Water molecule

Bubble
Solute particle

The forming bubble is now surrounded by solute particles as well as water
molecules. As a result, solute particles may block the pathway of some of
the water molecules trying to enter the bubble. Because fewer H2O molecules can enter the bubble, the molecules that do get inside it must have
relatively high energies to produce the internal pressure necessary to
maintain the bubble.
Compare the situation with and without a solute shown below.
Water molecule

Solute particle
Bubble

Pure water

Solution (contains solute)

To produce the same pressure, the smaller number of molecules found in
the bubble in the solution must have higher energies (speeds) than those
in the bubble in pure water.
Thus, to cause the water in the solution to boil, we must heat the water
to a temperature higher than 100 °C to produce the higher-energy water
molecules needed. In other words, the boiling point of the solution occurs
at a higher temperature than does the boiling point of pure water. The
presence of the solute increases the boiling point of water. The more
solute present, the higher the boiling point, because the smaller number
of H2O molecules able to enter the bubble must have increasing energies.

552 • Chapter 15 • Solutions

15
Note that the raising of the boiling point depends on the number and
not the specific identity of the solute particles. It’s the number that matters—
the more particles present, the more those particles block water molecules
from entering the bubble. A solution property that depends on the number of
solute particles present is called a colligative property. Raising of the boiling
point by a solute is one of the colligative properties of solutions.
Another colligative property of solutions is the lowering of the freezing
point. (As mentioned before, a 1 M NaCl solution freezes at approximately
1 °C.) This property is particularly important in cold areas of the world
where salt is applied to icy roads. When the applied salt dissolves in the
thin layer of water on the surface of ice, forming a very concentrated solution, it causes the ice to melt (it lowers the freezing point of water). The
same property is also used to protect automobile engines. The coolant used

(

in engines is a solution containing ethylene glycol H

H

H

C

C

OH OH

Colligative property
A property that is
dependent only on the
number of solute particles
present in solution

)

H dissolved

in water. The solute (ethylene glycol) raises the boiling point and lowers the
freezing point of the water, thereby protecting the engine from overheating
and freezing.

Active Reading Question
When salt is added to water, what is true about the freezing point of the
solution compared to the freezing point of pure water? What is true about
the boiling point of the solution compared to the boiling point of water?

SECTION 15.3
REVIEW QUESTIONS
1 The key to solving stoichiometry problems
is the mole. How do we find the number of
moles when solutions are mixed to produce
a reaction?

2 How many milliliters of 0.10 M Pb(NO3)2

solution are required to precipitate all the
lead, as PbI2, from 125.0 mL of 0.10 M
NaI solution?

5 What volume of 0.25 M NaOH is required to
neutralize 125.0 mL of 0.15 M HCl?

6 How is normality different from molarity?
7 Is a 1 M H2SO4 solution the same as a
1 N H2SO4 solution? Explain.

8 Other than for taste, why is salt added to
water when cooking pasta?

3 Determine the mass of Ag2CrO4 produced

when 100.0 mL of 0.500 M silver nitrate is
added to 100.0 mL of 0.500 M potassium
chromate.

4 Once you have determined moles of H or

OH in a neutralization reaction how can
you find the volume of the substance? What
is the critical information you need to determine volume?

RESEARCH LINKS

15.3 • Properties of Solutions • 553

Chapter 15 Review
Key Terms

Key Ideas

15.1 Solution

15.1 Forming Solutions

Solvent



A solution is a homogeneous mixture of a solute dissolved in a
solvent.



Substances with similar polarities tend to dissolve with each other
to form a solution.



Water is a very polar substance and tends to dissolve ionic solids or
other polar substances.



Various terms are used to describe solutions:

Solute
Aqueous solution
Saturated
Unsaturated
Supersaturated
Concentrated



Saturated—contains the maximum possible dissolved solid

Dilute



Unsaturated—not saturated



Supersaturated—contains more dissolved solid than should
dissolve at a given temperature



Concentrated—contains a relatively large amount of solute



Dilute—contains a relatively small amount of solute



15.2 Mass percent

Molarity (M)

The rate of dissolution is affected by


Surface area of solute



Stirring



Temperature

15.2 Describing Solution Composition


Standard solution
Dilution

Descriptions of solution composition:
mass of solute 100%
• Mass percent of solute ________________
mass of solution
moles of solute
Molarity ________________
liters of solution
equivalents of solute
• Normality ____________________
liters of solution
Dilution of a solution occurs when additional solvent is added to
lower the concentration of a solution.






No solute is added so
mol solute (before dilution) mol solute (after dilution)



15.3 Neutralization reaction

Equivalent of an acid

15.3 Properties of Solutions


Equivalent of a base

RESEARCH LINKS

554 • Chapter 15 • Solutions

Moles of a dissolved reactant or product can be calculated from the
known concentration and volume of the substance
Mol (concentration)(volume)

Equivalent weight
Colligative property

A standard solution can be diluted to produce solutions with
appropriate concentrations for various laboratory procedures.



The properties of a solvent are affected by dissolving a solute.



The boiling point of a solvent increases as the amount of dissolved
solute increases.



The melting point of a solvent decreases as the amount of dissolved
solute increases.

Chapter 15 Assessment

RESEARCH LINKS

All exercises with blue numbers have answers in the
back of this book.

15.2 Describing Solution Composition

15.1 Forming Solutions

12. Calculate the mass percent of calcium chloride
in each of the following solutions.

A. Solubility
1. Explain why a solution is a homogeneous
mixture. Give two examples of homogeneous
mixtures.
2. In a solution, the substance present in the larg, whereas the
est amount is called the
other substances present are called the
.
3. Discuss how an ionic solute dissolves in water.
How are the strong interionic forces in the solid
overcome to permit the solid to dissolve? How
are the dissolved positive and negative ions
shielded from one another, preventing them
from recombining to form the solid?
4. Why are some molecular solids (such as sugar
or ethyl alcohol) soluble in water, while other
molecular solids (such as petroleum) are insoluble in water? What structural features(s) of some
molecular solids may tend to make them soluble
in water?
B. Solution Composition: An Introduction
5. A solution that contains as much solute as will
dissolve at a given temperature is said to be
.
6. A solution that has not reached its limit of
dissolved solute is said to be
.
7. A solution is a homogeneous mixture and,
unlike a compound, has
composition.
8. The label “concentrated H2SO4” on a bottle
means that there is a relatively
amount of H2SO4 present in the solution.
C. Factors Affecting the Rate of Dissolving
9. What does it mean to increase the surface area
of a solid? Explain why this change causes an
increase in the rate of dissolving.
10. Use a molecular explanation to explain why
increasing the temperature speeds up the rate
of dissolving a solid in a liquid.
11. Explain why the solubility of a gas generally
decreases with an increase in temperature.

A. Solution Composition: Mass Percent

a. 5.00 g of calcium chloride in 95.0 g
of water
b. 1.00 g of calcium chloride in 19.0 g
of water
c. 15.0 g of calcium chloride in 285 g
of water
d. 2.00 mg of calcium chloride in 0.0380 g
of water
13. Calculate the mass, in grams, of NaCl present in
each of the following solutions.
a. 11.5 g of 6.25% NaCl solution
b. 6.25 g of 11.5% NaCl solution
c. 54.3 g of 0.91% NaCl solution
d. 452 g of 12.3% NaCl solution
14. A laboratory assistant prepared a potassium
chloride solution for her class by dissolving
5.34 g of KCl in 152 g of water. What is the
mass percent of the solution she prepared?
15. If 67.1 g of CaCl2 is added to 275 g of water,
calculate the mass percent of CaCl2 in the
solution.
16. What mass of each solute is present in 285 g of
a solution that contains 5.00% by mass NaCl
and 7.50% by mass Na2CO3?
17. A hexane solution contains as impurities 5.2%
(by mass) heptane and 2.9% (by mass) pentane.
Calculate the mass of each component present
in 93 g of the solution.
B. Solution Composition: Molarity
18. A solution that is labeled “0.105 M NaOH”
would contain
mol of NaOH per liter
of solution.
19. How many moles of each ion are present, per
liter, in a solution that is labeled “0.221 M
CaCl2”?
20. If you were to prepare exactly 1.00 L of a 5 M
NaCl solution, you would not need exactly
1.00 L of water. Explain.

Chapter 15 • Assessment • 555

21. For each of the following solutions, the number
of moles of solute is given, followed by the total
volume of solution prepared. Calculate the
molarity.
a. 0.50 mol KBr; 250 mL

d. 0.50 mol KBr; 1.0 L
22. For each of the following solutions, the mass of
the solute is given, followed by the total volume
of the solution prepared. Calculate the molarity.
a. 4.25 g CuCl2; 125 mL

30. Standard silver nitrate solutions are used in the
analysis of samples containing chloride ion.
How many grams of silver nitrate are needed to
prepare 250. mL of a 0.100 M AgNO3 solution?
C. Dilution
31. Calculate the new molarity that results when
250. mL of water is added to each of the
following solutions.

b. 0.101 g NaHCO3; 11.3 mL
c. 52.9 g Na2CO3; 1.15 L
d. 0.14 mg KOH; 1.5 mL

a. 125 mL of 0.251 M HCl

23. If a 45.3-g sample of potassium nitrate is
dissolved in enough water to make 225 mL of
solution, what will be the molarity?
24. An alcoholic iodine solution (“tincture” of
iodine) is prepared by dissolving 5.15 g of
iodine crystals in enough alcohol to make a
volume of 225 mL. Calculate the molarity of
iodine in the solution.
25. Suppose 1.01 g of FeCl3 is placed in a 10.0-mL
volumetric flask, water is added, the mixture is
shaken to dissolve the solid, and then water is
added to the calibration mark of the flask.
Calculate the molarity of each ion present in
the solution.
26. If 495 g of NaOH is dissolved to a final total
volume of 20.0 L, what is the molarity of the
solution?
27. Calculate the number of moles and the number
of grams of the indicated solutes present in each
of the following solution samples.
mol

g

a. 127 mL of 0.105 M HNO3
b. 155 mL of 15.1 M NH3
M KSCN

d. 12.2 mL of 2.45 M HCl
28. What mass (in grams) of NH4Cl is needed to
prepare 450. mL of 0.251 M NH4Cl solution?

556 • Chapter 15 • Solutions

b. 5.51 L of 0.103 M Na3PO4 solution
d. 25.2 mL of 0.00157 M Ca(OH)2 solution

c. 0.50 mol KBr; 750 mL

c. 2.51 L of 2.01 10

a. 10.2 mL of 0.451 M AlCl3 solution
c. 1.75 mL of 1.25 M CuCl2 solution

b. 0.50 mol KBr; 500. mL

3

29. Calculate the number of moles of each ion
present in each of the following solutions.

b. 445 mL of 0.499 M H2SO4
c. 5.25 L of 0.101 M HNO3
d. 11.2 mL of 14.5 M HC2H3O2
32. Many laboratories keep bottles of 3.0 M solutions
of the common acids on hand. Given the following molarities of the concentrated acids, determine
how many milliliters of each concentrated acid
would be required to prepare 225 mL of a 3.0 M
solution of the acid.
Acid

Molarity of Concentrated Reagent

HCl

12.1 M

HNO3

15.9 M

H2SO4

18.0 M

HC2H3O2

17.5 M

H3PO4

14.9 M

33. A chemistry student needs 125 mL of 0.150 M
NaOH solution for an experiment, but the only
solution available in the laboratory is 3.02 M.
Describe how the student could prepare the
solution he needs.
34. How much water must be added to 500. mL of
0.200 M HCl to produce a 0.150 M solution?
(Assume that the volumes are additive.)

15.3 Properties of Solutions

C. Normality

A. Stoichiometry of Solution Reactions

42. Explain why the equivalent weight of H2SO4
is half the molar mass of this substance. How
many hydrogen ions does each H2SO4 molecule
produce when reacting with an excess of OH
ions?

35. One way to determine the amount of chloride
ion in a water sample is to react the sample
with standard AgNO3 solution to produce solid
AgCl.

Ag (aq) Cl (aq) n AgCl(s)
If a 25.0-mL water sample requires 27.2 mL of
0.104 M AgNO3 in such a reaction, what is the
concentration of Cl in the sample?
36. What volume (in mL) of 0.25 M Na2SO4 solution is needed to precipitate all the barium, as
BaSO4(s), from 12.5 mL of 0.15 M Ba(NO3)2
solution?

Ba(NO3)2(aq) Na2SO4(aq) n
BaSO4(s) 2NaNO3(aq)
37. If 36.2 mL of 0.158 M CaCl2 solution is added
to 37.5 mL of 0.149 M Na2CO3, what mass of
calcium carbonate, CaCO3, will be precipitated?
The reaction is

CaCl2(aq) Na2CO3(aq) n
CaCO3(s) 2NaCl(aq)
38. When aqueous solutions of lead(II) ion are
treated with potassium chromate solution, a
bright yellow precipitate of lead(II) chromate,
PbCrO4, forms. How many grams of lead chromate form when a 1.00-g sample of Pb(NO3)2 is
added to 25.0 mL of 1.00 M K2CrO4 solution?
B. Neutralization Reactions
39. What volume of 0.200 M HCl solution is needed
to neutralize 25.0 mL of 0.150 M NaOH
solution?
40. The concentration of a sodium hydroxide solution is to be determined. A 50.0-mL sample of
0.104 M HCl solution requires 48.7 mL of the
sodium hydroxide solution to reach the point
of neutralization. Calculate the molarity of the
NaOH solution.

43. How many equivalents of hydroxide ion are
needed to react with 1.53 equivalents of hydrogen ion? How can you tell when no balanced
chemical equation is given for the reaction?
44. For each of the following solutions, the mass of
solute taken is indicated, along with the total
volume of solution prepared. Calculate the
normality of each solution.
a. 0.113 g NaOH;

10.2 mL

b. 12.5 mg Ca(OH)2; 100. mL
c. 12.4 g H2SO4;

155 mL

45. Calculate the normality of each of the following
solutions.
a. 0.134 M NaOH
b. 0.00521 M Ca(OH)2
c. 4.42 M H3PO4
46. A solution of phosphoric acid, H3PO4, is found
to contain 35.2 g of H3PO4 per liter of solution.
Calculate the molarity and normality of the
solution.
47. What volume of 0.172 N H2SO4 is required to
neutralize 56.2 mL of 0.145 M NaOH?
48. What volume of 0.151 N NaOH is required to
neutralize 24.2 mL of 0.125 N H2SO4? What
volume of 0.151 N NaOH is required to neutralize 24.1 mL of 0.125 M H2SO4?
D. Boiling Point and Freezing Point
49. What is meant by the term “colligative
property”?
50. Explain on a molecular level why the increase in
boiling point is a colligative property.
51. Antifreeze that you use in your car could also be
called “antiboil.” Explain why.

41. What volume of 1.00 M NaOH is required to
neutralize each of the following solutions?
a. 25.0 mL of 0.154 M acetic acid, HC2H3O2
b. 35.0 mL of 0.102 M hydrofluoric acid, HF
c. 10.0 mL of 0.143 M phosphoric acid,
H3PO4
d. 35.0 mL of 0.220 M sulfuric acid, H2SO4

Chapter 15 • Assessment • 557

Critical Thinking
52. Suppose 50.0 mL of 0.250 M CoCl2 solution is
added to 25.0 mL of 0.350 M NiCl2 solution.
Calculate the concentration, in moles per liter,
of each of the ions present after mixing. Assume
that the volumes are additive.
53. Calculate the mass of AgCl formed, and the
concentration of silver ion remaining in solution, when 10.0 g of solid AgNO3 is added to
50. mL of 1.0 10 2 M NaCl solution. Assume
there is no volume change upon addition of
the solid.
54. What mass of BaSO4 will be precipitated from a
large container of concentrated Ba(NO3)2 solution if 37.5 mL of 0.221 M H2SO4 is added?
55. Strictly speaking, the solvent is the component
of a solution that is present in the largest
amount on a mole basis. For solutions involving
water, water is almost always the solvent
because there tend to be many more water
molecules present than molecules of any
conceivable solute. To see why this is so, calculate the number of moles of water present in
1.0 L of water. Recall that the density of water
is very nearly 1.0 g/mL under most conditions.
56. A 14.2-g sample of CaCl2 is added to a 50.0-mL
volumetric flask. After dissolving the salt, water
is added to the calibration mark of the flask.
Calculate the molarity of the solution.
57. Calculate the new molarity when 150. mL
of water is added to each of the following
solutions.
a. 125 mL of 0.200 M HBr
b. 155 mL of 0.250 M Ca(C2H3O2)2
c. 0.500 L of 0.250 M H3PO4
d. 15 mL of 18.0 M H2SO4

558 • Chapter 15 • Solutions

58. Calculate the normality of each of the following
solutions.
a. 0.50 M acetic acid, HC2H3O2
b. 0.00250 M sulfuric acid, H2SO4
c. 0.10 M potassium hydroxide, KOH
59. If 27.5 mL of 3.5 10 2 N Ca(OH)2 solution
is needed to neutralize 10.0 mL of nitric acid
solution of unknown concentration, what is
the normality of the nitric acid?
60. The figures below are molecular-level representations of four aqueous solutions of the same
solute. Arrange the solutions from most to least
concentrated.

Solution A
Volume 1.0 L

Solution B
Volume 4.0 L

Solution C
Volume 2.0 L

Solution D
Volume 2.0 L

Standardized Test Practice
1 Which of the following aqueous solutions has

the greatest number of ions in solution?
A

2.0-L of 1.50 M sodium phosphate

B

3.0-L of 1.50 M sodium chloride

C

2.0-L of 2.00 M potassium fluoride

D 1.0-L of 3.00 M sodium sulfate
2 When a solvent contains as much of the solute

as it can hold, the solution is said to be
A

unsaturated

B

supersaturated

C

saturated

D dilute

Chapter 15
7 You have equal masses of different solutes

dissolved in equal volumes of solution. Which
of the solutes would make the solution with the
highest concentration measured in molarity?
A

NaOH

B

KCl

C

KOH

D LiOH
8 You react 250.0 mL of 0.10 M barium

nitrate with 200.0 mL of 0.10 M potassium
phosphate. What ions are left in solution
after the reaction is complete?
A

potassium ion, nitrate ion, barium ion

3 Which of the following actions does not

B

potassium ion, nitrate ion, phosphate ion

generally increase the rate of a solute
dissolving in a solvent?

C

potassium ion, nitrate ion

A

Increasing the surface area of the solute

B

Stirring the solution

C

Raising the temperature of the solution

D Adding more solute
4 Determine the concentration of a solution

made by dissolving 10.0 g of sodium chloride
in 750.0 mL of solution.
A

0.133 M

B

0.171 M

C

0.228 M

D 0.476 M
5 You have two solutions of sodium chloride.

One is a 2.00 M solution and the other is a
4.00 M solution. You have much more of the
4.00 M solution and you add the solutions
together. Which of the following could be
the concentration of the final solution?
A

3.00 M

B

3.80 M

C

6.00 M

D 8.00 M
6 What is the minimum volume of a 2.00 M

NaOH (aq) solution needed to make 150.0 mL
of a 0.800 M NaOH (aq) solution?
A

60.0 mL

B

90.0 mL

C

120. mL

D barium ion, nitrate ion, phosphate ion
9 What volume of 0.500 M NaOH is needed

to neutralize 45.0 mL of 0.400 M HCl?
A

36.0 mL

B

45.0 mL

C

56.3 mL

D 71.3 mL
10 Which of the following has the largest

equivalent weight?
A

HCl

B

H2SO4

C

H3PO4

D HNO3
11 When a solute is dissolved in a solvent,

the boiling point of the solution is higher
than the boiling point of the pure solvent.
Answer the following questions about this
phenomenon.
a. Explain why adding a solute to a solvent
raises the boiling point.
b. Explain why adding 1 mole of table salt to
water raises the boiling point more than
adding 1 mole of table sugar to the same
amount of water.

D 150. mL

Chapter 15 • Standardized Test Practice • 559


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