1. business and industrial
2. energy
3. oil

SECTION 12.2
The Solution Process
Teacher Notes and Answers
SECTION 2 The Solution Process
1.Grind the sugar cube into a powder, shake the
container, or heat the container.
2.In both equilibriums, the rates at which two
processes occur are equal. In one, vaporization
and condensation are balanced. In the other,
dissolution and crystallization are balanced.
3.unsaturated, saturated, supersaturated
4.increases
5.the force between polar molecules and ions in
the crystal
6.polar; nonpolar
7.The rate of dissolution decreases and the
amount of dissolved gas decreases.
8.The solubility will increase.
9.The graph shows a reaction with a negative
enthalpy of solution because the net energy
change is negative. Energy is released after the
solution forms.
10.The dissolution of NaCl in water will not cause
the temperature of the water to increase very
much.
region of the water molecules, and vice versa.
Carbon tetrachloride is nonpolar and will
not form a strong enough attraction to water
molecules to dissolve.
5.The toluene would work better because oil and
toluene are nonpolar, and will dissolve each
other.
6.It will not keep carbon dioxide from escaping
the solution. Henry’s law states that it is
the partial pressure of the same gas above
the solution that keeps that gas in solution.
Increased pressure of only helium will have no
effect on carbon dioxide’s solubility.
Review
1.An unsaturated solution does not contain the
maximum amount of solute, while a saturated
solution does contain the maximum amount of
solute.
2.The particles in hot tea move faster than in iced
tea. Therefore, there are more collisions between
the tea and sugar molecules at the surface of
the sugar grains, and sugar molecules leave the
surface of the grains faster in hot tea than in
cold tea.
3.Continually add sugar to hot water until
undissolved grains remain in the bottom of the
container. Filter the hot solution to remove any
seed crystals. Set the saturated solution aside
until it cools to room temperature. Once it cools,
it will be supersaturated.
4.Ethanol and water are each polar molecules.
The negatively charged region of the ethanol
molecules is attracted to the positively charged
Solutions
1
SECTION 12.2
The Solution Process
In Section 1 you learned about the nature of solutions.
Now you will learn about the process of dissolution—the
dissolving of a solute in a solvent to make a solution.
Several factors affect dissolving.
A crystalline solid dissolves as its particles leave the surface
of the crystal and mix with solvent molecules. The molecules
or ions of the solid are attracted to the solvent. The rate at
which a substance dissolves depends on many factors.
Increasing the Surface Area of the Solute
One way to speed up dissolution is to increase the surface
area of the solute. Crushing larger crystals into smaller ones
increases the amount of crystal surfaces in contact with the
liquid. Dividing something into smaller parts will increase
its exposed surface area, and make it dissolve faster.
Agitating a Solution
As a solid is dissolving, there are more dissolved particles
close to the undissolved crystal than there are away from
the crystal. Stirring or shaking a solution will help spread
dissolved particles throughout the solvent. This makes it
easier for new particles to dissolve into the solvent.
Heating a Solvent
You may have found it more difficult to mix sugar into iced
tea than hot tea. The more energy a solvent’s particles have,
the more they will collide with the dissolving solid, and the
more energy they can transfer to the solid. So, a solid will
dissolve more quickly in a warm liquid than a cool liquid.
Key Terms
solution equilibrium
saturated solution
unsaturated solution
supersaturated solution
solubility
hydration
immiscible
miscible
Henry’s law
effervescence
enthalpy of solution
solvated
large surface area exposed
to solvent—faster rate
CuSO4 •5H2O powdered
Increased surface area
Small surface area exposed
to solvent—slow rate
Solvent
particle
Solute
READING CHECK
1.
Name three ways to make sugar dissolve in water faster.
CuSO4 •5H2O large crystals
The surface area of a powdered solute
is larger. Therefore, the powder will
dissolve faster.
2
CHAPTER 12
Solubility is a measure of how well one substance
dissolves another.
When solid sugar is added to water, sugar molecules leave
the solid surface and mix with the water molecules. These
molecules move about the water and sugar molecules at
random, as do all particles in a liquid. However, some of these
dissolved molecules collide with the crystal and transfer
energy to the crystal. They no longer have enough energy to
pull away from the crystal again. As a result, these molecules
recrystallize, rejoining the crystal lattice.
When the sugar is first added, there is no
solute in the water. Therefore, there is also no
recrystallization of solute. Over time, the
concentration of solute in the water increases,
and the rate of recrystallization increases.
Eventually, if there is enough sugar, the rate
of dissolution and the rate of recrystallization
are equal. At this point, the solution is in
­equilibrium. Solution equilibrium is the
­physical state in which the opposing processes
of dissolution and crystallization of a solute
occur at equal rates.
Recrystallizing
Dissolving
There are three types of solutions.
• A saturated solution contains the maximum amount
of dissolved solute that is possible under the current
conditions. The dissolution of any new solute immediately
results in crystallization to keep the total amount of
solute constant.
• An unsaturated solution contains less than the maximum
amount of dissolved solute possible.
• A supersaturated solution contains more than the maximum
amount of dissolved solute that is normally possible. This
solution is in a state of equilibrium that is unstable.
Critical Thinking
2.
Synthesize How does the equilibrium between a solute and
a solvent compare with the equilibrium between a liquid
and its vapor?
Solutions
3
Mass in grams of NaCH3COO dissolved
in 100 g water at 20˚C
Mass of Solute Added Vs. Mass of Solute Dissolved
60
The graph shows the maximum mass
of sodium acetate, NaC​H3​ ​COO, that
can be dissolved in water.
A. Unsaturated
If a solution is unsaturated,
more solute can dissolve.
No undissolved solute
remains.
40
B. Saturated
If the amount of solute added
exceeds the solubility, some
solute remains undissolved.
20
Solubility = 46.4 g/100 g
0
0
20
40
60
80
100
Mass in grams of NaCH3COO added to 100 g water at 20˚C
Saturated Versus Unsaturated Solutions
One way to tell the difference between a saturated solution
and an unsaturated solution is to add more of the dissolved
substance. If the substance collects at the bottom of the
solution and does not dissolve, the solution is saturated. In an
unsaturated solution, all of the available solute is dissolved.
The maximum amount that can be dissolved is a ratio of
the solute amount to solvent amount. If more solvent is added
to a saturated solution, the solution becomes unsaturated.
Supersaturated Solutions
When a saturated solution is cooled, usually some of the
solute will come out of the solution. This is because the
amount that a solvent can dissolve depends on its
temperature. But sometimes, if the solution is cooled without
being disturbed, it can become a supersaturated solution.
Once a supersaturated solution is disturbed, and crystals
start to form, the recrystalization process will continue
until the solution is saturated without being supersaturated.
Another way to start recrystallization is by introducing a
“seed” crystal. This seed will rapidly grow as molecules of
solute begin to come out of solution.
READING CHECK
3.
4
For a given substance, order an unsaturated, saturated,
and supersaturated solution from smallest amount of
solute to largest.
CHAPTER 12
CONNECT
One example of a supersaturated
solution in nature is honey. If honey
is unprocessed or left undisturbed
for a long period of time, the sugar
in the honey will crystallize.
Solubility of Solutes as a Function of Temperature (in g solute/100. g H2O)
Temperature (°C)
Substance
AgN​O3​ ​
Ba(OH​)​2​
​C12
​ ​​H​22​​O​11​
Ca(OH​)​2​
0
20
40
60
80
100
122
216
311
440
585
733
101.4
—
362
487
1.67
179
0.189
3.89
204
0.173
C​e2​ ​(S​O4​ ​​)​3​
20.8
10.1
KCl
28.0
34.2
KI
128
144
8.22
238
0.141
—
40.1
162
KN​O​3​
13.9
31.6
61.3
LiCl
69.2
83.5
89.8
L​i2​ ​C​O3​ ​
1.54
1.33
NaCl
35.7
35.9
NaN​O​3​
73
87.6
1.17
36.4
102
20.94
287
0.121
—
3.87
—
45.8
0.07
—
51.3
56.3
176
192
206
106
167
245
112
128
98.4
1.01
0.85
37.1
38.0
122
C​O​2 ​(gas at SP)
0.335
0.169
0.0973
0.058
​O​2​(gas at SP)
0.00694
0.00537
0.00308
0.00227
Solubility Values
The solubility of a substance is the amount of that substance
required to form a saturated solution with a certain amount of
solvent at a given temperature. For example, at 20°C the
solubility of sodium acetate is 46.4 g per 100 g of water.
Solubility is usually given in grams of solute per 100 grams of
solvent or grams of solute per 100 milliliters of solvent.
39.2
148
180
—
—
0.00138
0.72
0.00
TIP
In calculations involving
solubility, it is important to
distinguish grams of solvent from
grams of solute. Otherwise, the unit
“grams” will be canceled incorrectly
and the result will be impossible to
interpret.
Like Dissolves Like
Whether a substance is soluble in another substance is often
hard to predict. Lithium chloride is highly soluble in water,
but gasoline is not. Gasoline is highly soluble in benzene,
​C6​ ​​H​6​, but lithium chloride is not.
A rough way of determining whether a substance will
dissolve into another substance is the phrase “like dissolves
like.” For example, polar substances tend to dissolve in polar
solvents, but not in nonpolar solvents.
READING CHECK
4. As temperature increases, the
solubility of a substance
.
Solutions
5
Dissolving Ionic Compounds in Aqueous Solution
The polarity of water molecules plays an important
role in the formation of solutions of ionic compounds
in water. The positive and negative ends of a water
molecule are attracted to different parts of a crystal.
These attractions are strong enough to pull the
Water molecule
molecules or ions out of the crystal and into solution.
Hydration occurs when the attraction of the charged
ends of water molecules dissolve an ionic compound
or a substance whose molecules are polar.
The diagram at the right shows how the process of
hydration works for a lithium chloride crystal. The positive
end of the water molecule, where the hydrogen atoms are, is
attracted to a chloride anion. The negative end, the end away
from the hydrogen atoms, is attracted to a lithium cation.
When the attraction is strong enough to pull the ion out of the
crystal, the ion eventually becomes surrounded by water
molecules. The lithium cations and chloride ions that are in
solution are said to be hydrated.
When some ionic compounds recrystallize out of solution,
the new crystals include water molecules. These compounds
are called hydrates. The formula unit for the crystal structure
contains a specific number of water molecules in a regular
structure. For example, the copper(II) sulfate crystal shown
at the right has the formula CuS​O4​ ​∙5​H2​ ​O. The water can be
removed from the crystal using heat, leaving the anhydrous,
or water-free, salt. When a hydrate dissolves in water, it
breaks up into ions, and the water molecules become part of
the solvent.
Nonpolar Solvents
Ionic compounds are not generally soluble in nonpolar
solvents, such as carbon tetrachloride, CC​l​4​, or toluene,
​C6​ ​​H​5​C​H3​ ​. The nonpolar solvent molecules do not attract the
ions of the crystal strongly enough to overcome the forces
holding the crystals together.
READING CHECK
5.
6
What force of attraction is responsible for the dissolution of
an ionic crystal in water?
CHAPTER 12
Hydrated Li+
LiCl crystal
Hydrated Cl-
When LiCl dissolves, the ions are
hydrated. The attraction between ions
and water molecules is strong enough
that each ion in solution is surrounded
by water molecules.
–
SO42
H2O
H2O
Cu
2+
H 2O
H2O
H2O
–
SO42
Hydrated copper(II) sulfate has the
formula CuS​O4​ ​∙5​H2​ ​O. Heating
releases the water and the anhydrous
crystal CuS​O4​ ​forms.
Liquid Solutes and Solvents
When you get a bottle of salad dressing out of the refrigerator,
you may notice that the various liquids in the bottle have
formed distinct layers. The oil and water in the bottle are
immiscible, which means that they cannot dissolve in one
another. The hydrogen bonding between water molecules
squeezes out the oil molecules. The water and oil form two
layers, with the denser material forming the bottom layer.
However, water and ethanol are miscible, which means
they can dissolve in one another. Ethanol contains an —OH
group on the end, which is a somewhat polar bond.
H H
H−C−C−OH
H H
This —OH group can form hydrogen bonds with water
molecules. The molecular forces in the mixture of water and
ethanol are so similar to those of a pure liquid that the liquids
mix freely with one another.
Another example of miscible liquids is a mixture of fat,
oil, or grease with toluene or gasoline. All of these substances
have nonpolar molecules. The only intermolecular forces
are the weak London dispersion forces. These forces do not
prevent the liquid molecules from moving freely in solution.
READING CHECK
6. A liquid with polar molecules
and a liquid with
molecules are generally miscible.
A liquid with polar molecules and a
liquid with
molecules are generally immiscible.
Hydrogen bond
δ-
δ-
δ+
δ+
δ-
δ-
δ+
δ+
Toluene
Water
molecule,
H2O
Water
(a)
(b)
Ethanol molecule,
C2H5OH
(a) Toluene and water are immiscible.
(b) Ethanol and water are miscible.
Hydrogen bonding between water and
ethanol molecules enhances the ability
of the ethanol to dissolve in water.
Solutions
7
Effects of Pressure on Solubility
Pressure does not generally affect the solubility of liquids or
solids in liquid solvents. However, an increase in the pressure
of a gas can increase the solubility of a gas in a liquid.
A gas and a solvent are normally in a state of equilibrium
in which gas molecules enter or leave the liquid phase at the
same rate.
gas + solvent ⇆ solution
If the pressure on the gas is increased, more gas molecules
will collide with the liquid solvent and the rate of dissolution
increases. Eventually, a new equilibrium is reached with
a higher solubility. SO, an increase in pressure will increase
solubility and result in more of the gas entering the
liquid phase.
Henry’s Law
Henry’s law states that the solubility of a gas in a
liquid is directly proportional to the partial
pressure of that gas on the surface of the liquid.
The law is named after English chemist William
Henry, and applies to liquid gas solutions at
constant temperature.
In the chapter “Gases,” you learned that the
same amount of gas exerts the same amount of
pressure whether it is in a mixture or occupies
a space alone. For this reason, the amount of
the gas that will dissolve in a liquid also does
not depend on the other gases in a mixture.
A manufacturer of carbonated beverages
forces C​O2​ ​ into a bottle with flavored water.
As stated in Henry’s law, the carbon dioxide
will dissolve in the water. When the bottle is
opened, the gas rapidly escapes from the liquid.
This causes the liquid to bubble and fizz, and
is called effervescence.
READING CHECK
7.
8
When happens when the partial pressure of
a gas on a liquid decreases?
CHAPTER 12
(a)
CONNECT
A scuba diver must be aware of
Henry’s law. Deep under water, as
pressure increases, the amount of
air that can dissolve in blood
increases. The extra nitrogen can
affect the nervous system of the
diver and lead to disorientation.
Therefore, scuba divers often use a
mixture of air with less nitrogen
than normal.
In addition, when a scuba diver
returns to the surface, the amount
of nitrogen that can dissolve in the
blood decreases. If the diver
ascends too quickly, the nitrogen
gas will form bubbles in tissues and
blood vessels.
CO2 under high
pressure above solvent
Soluble CO2
molecules
(b)
Air at atmospheric
pressure
Soluble CO2
molecules
CO2 gas
bubble
(a) The unopened bottle has no gas bubbles because
of the pressure of the C​O2​ ​inside the bottle. (b) When
the bottle is opened, the pressure is reduced and some
of the dissolved C​O2​ ​enters the gas phase through
effervescence.
Effects of Temperature on Solubility
As shown in the graph below on the left, an increase in
temperature usually decreases the solubility of a gas in a
liquid. At higher temperatures, the average kinetic energy of
the solvent and the solute particles increases. The attractive
forces between the solvent and solute decrease, and more
molecules are able to escape from the surface of the liquid.
Usually an increase in temperature increases the solubility
of a solid. However, the effect can be more pronounced for
some substances than for other substances.
For example, look at the graphs of potassium nitrate,
KN​O3​ ​, and sodium chloride, NaCl. The solubility of potassium
nitrate increases from 14 g per 100 g of water at 0°C to 167 g
per 100 g of water at 80°C. However, the solubility of sodium
chloride barely changes under the same temperature increase.
Its solubility increases from 36 g per 100 g of water to 38 g per
100 g of water. In some cases, such as that of lithium sulfate,
L​i2​ ​S​O4​ ​, the solubility decreases with increasing temperature.
READING CHECK
8.
What do you expect to happen to the solubility of nitrogen
gas in water if the temperature decreases?
Solubility Vs. Temperature for Some Solid Solutes
Solubility Vs. Temperature Data for Some Gases
Solubility
(mL gas/mL of H2O)
5
4
H2S volume
3
2
CO2 volume
1
0
O2 volume
0
10
20
30
40
50
60
70
80
90 100
Temperature (˚C)
The solubility of gases in water decreases with increasing
temperature. The solubility of a solid in water generally increases
with temperature, although the effect is more pronounced in
some compounds than in others.
Solubility in grams per 100 g of water
260
KNO3
240
220
200
NaNO3
180
160
RbCl
LiCl
140
120
100
80
NH4Cl
60
KCl
NaCl
Li2SO4
40
20
0
0
10 20 30 40 50 60 70 80 90 100
Temperature (˚C)
Solutions
9
A change in energy accompanies solution formation.
The formation of a solution is accompanied by an energy
change. If you dissolve some potassium iodide, KI, in water,
you will find that the outside of the container feels cold to the
touch. However, if you dissolve some sodium hydroxide,
NaOH, the outside of the container feels hot. The formation
of a solid-liquid solution can absorb energy as heat or release
energy as heat.
The net amount of energy that is absorbed as heat by the
solution when a specific amount of solute enters solution is
called the enthalpy of solution. The formation of a solution
can be pictured as a result of three steps, which are
summarized in the graph below. Each step involves the
absorption or release of energy.
1.
Solute particles are separated from the solid. Energy is
required to break the bonds between the particles and the
rest of the solid.
2.
The solute particles push apart the solvent particles. This
requires energy to act against the attractive forces between
the solvent particles.
3.
The solvent particles are attracted to and surround the
solute particles. Instead of breaking apart or pushing
against attractive forces, this involves the forming of new
intermolecular bonds. Energy is released and stored as
potential energy in the bonds.
Critical Thinking
9. Analyze Does the graph below
show a reaction with a positive
enthalpy of solution or a negative
enthalpy of solution? Explain.
Components
Enthalpy (H)
Step 2
Solute
Solvent
Step 3
Solvent particles being
moved apart to allow
solute particles to enter
liquid. Energy absorbed
Step 1
Solute particles becoming
separated from solid.
Energy absorbed
The graph shows the changes in enthalpy that occur during the formation
of a solution.
10
CHAPTER 12
Solvent particles being
attracted to and
solvating solute particles.
Energy released
∆H solution Exothermic
Enthalpies of Solution (kJ/mol solute at 25°C)
Substance
AgN​O​3​(s)
C​H​3​COOH(l)
Enthalpy of solution
+22.59
–1.51
Substance
Enthalpy of solution
KOH(s)
MgS​O4​ ​(s)
–57.61
+15.9
HCl(g)
–74.84
NaCl(s)
HI(g)
–81.67
naN​O​3​(s)
+20.50
KCl(s)
+17.22
naOH(s)
–44.51
KCl​O​3​(s)
+41.38
n​h​3​(g)
–30.50
Kl(s)
+20.33
n​h​4​Cl(s)
+14.78
KN​O​3​(s)
+34.89
n​h​4​N​O3​ ​(s)
+25.69
+3.88
The end result of the dissolving process is the surrounding
of solute particles by solvent particles. A solute particle that is
surrounded by solvent particles is said to be solvated.
Two of the steps of dissolving a substance involve the
absorption of energy. One step involves the release of energy.
If more energy is absorbed than released, the reaction has a
positive enthalpy of solution. As a result, a container in which
the process has occurred will feel colder than it did before.
As the table shows, potassium chloride is an example of a
substance with a positive enthalpy of solution.
The graph on the previous page represents a process in
which more energy is released than absorbed. This substance
has a negative enthalpy of solution. Such a substance will
make its container warmer as it dissolves. Potassium
hydroxide, KOH, makes its container feel hot as it dissolves,
and as the table shows, potassium hydroxide has a negative
enthalpy of solution.
Gases generally have negative enthalpies of solution. This
is because there are few intermolecular forces within a gas.
Little energy has to be absorbed to separate gas molecules
from each other.
READING CHECK
10. NaCl
has a low enthalpy of solution. What does this
suggest about the dissolution of NaCl in water?
Solutions
11
SECTION 12.2 REVIEW
VOCABULARY
1. What is the difference between a saturated and an unsaturated solution?
REVIEW
2. Why would you expect sugar to dissolve faster in hot tea than in iced tea?
3. Explain how you prepare a saturated solution of sugar in water. Then
explain how you would turn this solution into a supersaturated solution.
4. Explain why ethanol will dissolve in water and carbon tetrachloride will not.
Critical Thinking
5. PREDICTING OUTCOMES You get a small amount of lubricating oil on your
clothing. Which would work better to remove the oil, water or toluene?
Explain your answer.
6. INTERPRETING CONCEPTS A commercial “fizz saver” pumps helium under
pressure into a soda bottle to keep gas from escaping. Will this keep C​O​2​in
the drink? Explain.
12
CHAPTER 12