Chapter 12
Solutions
1
A solution is composed of two parts: the solute and
the solvent.
Solute
The gas (or solid) in a solution of gases (or solids),
or the component present in the smaller amount.
Solvent
The liquid in the case of a solution of gases or
solids, or the component present in the larger
amount.
12 | 2
Fluids that mix with or
dissolve in each other in
all proportions are said
to be miscible (left).
Fluids that do not
dissolve in each other
are said to be
immiscible (right).
12 | 3
a. The 5 g of Mo is the solute; the 80 g of Cr is the
solvent.
b. MgCl2 is the solute; H2O is the solvent.
c. O2 and N2 are the solutes; Ar is the solvent.
12 | 4
A saturated solution is
in equilibrium with
respect to the amount
of dissolved solute.
The rate at which the
solute leaves the solid
state equals the rate at
which the solute
returns to the solid
state.
12 | 5
The solubility of a solute is the amount that
dissolves in a given quantity of solvent at a given
temperature.
An unsaturated solution is a solution not in
equilibrium with respect to a given dissolved
substance and in which more of the substance can
be dissolved.
12 | 6
12 | 7
A supersaturated solution is a solution that
contains more dissolved substance than a
saturated solution does. This occurs when a
solution is prepared at a higher temperature and is
then slowly cooled. This is a very unstable
situation, so any disturbance causes precipitation.
12 | 8
Solubility can be understood in terms of two
factors:
1. The natural tendency toward disorder
favors dissolving.
2. The relative forces between and within
species must be considered.
Stronger forces within solute species oppose
dissolving.
Stronger forces between species favor dissolving.
12 | 9
For molecular solutions,
this can be summarized
as “Like dissolves like.”
In other words, solutes
dissolve in solvents that
have the same type of
intermolecular forces.
An immiscible solute
and solvent are
illustrated at right.
12 | 10
When considering ionic solutes in water, we need
to examine the hydration energy and the lattice
energy.
12 | 11
The stronger ion-dipole
force between the ion
and the solvent—that is,
hydration energy—
favors dissolving.
A stronger force
between ions—that is,
lattice energy—opposes
dissolving.
12 | 12
The force of attraction
between water and both
a cation and an anion is
illustrated to the left with
lithium fluoride, LiF.
12 | 13
The process of
dissolving occurs at the
surfaces of the solid.
Here we see water
hydrating (dissolving)
ions.
12 | 14
The hydration energy for AB2 must be greater than
the hydration energy for CB2.
12 | 15
In general, solubility depends on temperature.
12 | 16
In most cases, solubility increases with increasing
temperature. However, for a number of
compounds, solubility decreases with increasing
temperature.
The difference is explained by differences in the
heat of solution.
12 | 17
When dissolving absorbs heat (is endothermic),
the temperature of the solution decreases as the
solute dissolves. The solubility will increase as
temperature increases.
When dissolving releases heat (is exothermic), the
temperature of the solution increases as the solute
dissolves. The solubility will decrease as
temperature increases.
12 | 18
Hot packs use an
exothermic solution
process.
Cold packs use an
endothermic solution
process.
12 | 19
Henry’s law describes the effect of pressure on
gas solubility: The solubility of a gas in a liquid is
directly proportional to the partial pressure of the
gas above the solution.
This is expressed mathematically in the equation
S = kHP
where
S = gas solubility
kH = Henry’s law constant for the gas
P = partial pressure of the gas over the solution
12 | 20
In general, pressure has
little or no effect on the
solubility of solids or
liquids in water.
The solubility of a gas
increases as pressure
increases, as illustrated
at right.
12 | 21
?
Helium–oxygen mixtures are sometimes
used as the breathing gas in deep-sea
diving. At sea level (where the pressure is
1.0 atm), the solubility of pure helium in
blood is 0.94 g/mL. What is the solubility of
pure helium at a depth of 1500 feet?
Pressure increases by 1.0 atm for every 33
feet of depth, so at 1500 feet the pressure is
46 atm. (For a helium–oxygen mixture, the
solubility of helium will depend on its initial
partial pressure, which will be less than 1.0
atm.)
12 | 22
P1 = 1.0 atm
S1 = 0.94 g/mL
P2 = 46 atm
S2 = ?
P2
S2  S1 
P1
0.94 g 46 atm
S2 

mL
1.0 atm
43 g
S2 
mL
12 | 23
At high elevations, the partial pressure of oxygen
decreases, decreasing the solubility of oxygen in
water. Fish require a certain minimum level of
dissolved oxygen to survive.
12 | 24
The concentration of a solute can be quantitatively
expressed in several ways:
1.
2.
3.
4.
Molarity
Mass percentage of solute
Molality
Mole fraction
12 | 25
Molarity is the moles of solute per liter of solution.
It is abbreviated as M.
moles of solute
M
liters of solution
12 | 26
Mass percentage of solute is the percentage by
mass of solute in a solution.
Mass percentage of solute 
grams of solute
 100%
grams of solution
12 | 27
?
An experiment calls for 36.0 g of a
5.00% aqueous solution of potassium
bromide. Describe how you would
make up such a solution.
12 | 28
A 5.00% aqueous solution of KBr has 5.00 g KBr
per 100. g solution. The remainder of the 100. g is
water: 95 g.
We can use this ratio to determine the mass of KBr
in 36.0 g solution:
5.00 g KBr
36.0 g solution 
 1.80 g KBr
100.g solution
Since 1.8 g KBr is required for 36.0 g of solution,
the remainder consists of 34.2 g water.
We make the solution by mixing
1.8 g KBr in 34.2 g water.
12 | 29
Molality is the moles of solute per kilogram of
solvent. It is abbreviated as m.
moles of solute
m
kilograms of solvent
12 | 30
?
Iodine dissolves in a variety of organic
solvents. For example, in methylene
chloride, it forms an orange solution.
What is the molality of a solution of
5.00 g iodine, I2, in 30.0 g of methylene
chloride, CH2Cl2?
12 | 31
Mass of solute = 5.00 g I2
Mass of solvent = 30.0 g CH2Cl2
3
5.00 g I2
1 mol I2
10 g
m


30.0 g solvent 253.8 g I2 1 kg
mol
m  0.657
kg
12 | 32
Mole fraction is the moles of component over the
total moles of solution. It is abbreviated C.
moles of solute
Χ
total moles of solution
12 | 33
?
A solution of iodine, I2, in methylene
chloride, CH2Cl2, contains 5.00 g I2 and
56.0 g CH2Cl2. What is the mole
fraction of each component in this
solution?
12 | 34
Mass of solute = 5.00 g I2
Mass of solvent = 56.0 g CH2Cl2
Moles solute 
1 mol I2
5.00 g I2 
 0.01970 mol
253.8 g I2
Moles solvent 
1 mol CH2Cl2
56.0 g I2 
 0.6594 mol
84.93 g CH2Cl2
Total moles  0.01970 mol  0.6594 mol
 0.6791 mol
12 | 35
0.01970 mol I2
Χ I2 
0.6791 mol total
Χ I2  0.0290
Χ CH2Cl2
0.6594 mol I2

0.6791 mol total
Χ CH2Cl2  0.971
12 | 36
?
A bottle of bourbon is labeled 94 proof,
meaning that it is 47% by volume of
alcohol in water. What is the mole
fraction of ethyl alcohol, C2H5OH, in the
bourbon? The density of ethyl alcohol
is 0.80 g/mL.
One liter of bourbon contains 470 mL of alcohol
and 530 mL of water. To solve this problem, we will
convert the volume of ethyl alcohol to mass using
density, and then convert to moles using molar
mass.
12 | 37
0.80 g
1mol
470 mL C2H5OH 

 8.16 mol
1mL 46.08 g
1.00 g 1mol
530 mL H2O 

 29.4 mol
1mL 18.02 g
Total moles  8.16 mol  29.4 mol  37.6 mol
Χ ethanol
8.16 mol C 2H5 OH

37.6 mol total
Χ ethanol  0.22
12 | 38
?
A 3.6 m solution of calcium chloride,
CaCl2, is used in tractor tires to give
them weight. The addition of CaCl2 also
prevents water in the tires from freezing
at temperatures above –20°C. What
are the mole fractions of CaCl2 and
water in such a solution?
The 3.6 m solution contains 3.6 mol CaCl2 in 1.0 kg
of water. To solve this problem, we will convert the
mass of water to moles, and then compute the
mole fractions.
12 | 39
Moles solute = 3.6 mol CaCl2
Moles solvent 
103 g 1 mol
1.0 kg H2O 

 55.5 mol H2O
1 kg 18.02 g
Total moles  3.6 mol  55.5 mol  59.1 mol
Χ CaCl2
3.6 mol

59.1 mol total
Χ CaCl2  0.061
Χ H2O
55.5 mol

59.1 mol total
Χ H2O  0.94
12 | 40
Converting between molality and mole fraction is
relatively simple, because you know the masses or
moles of both the solute and the solvent.
12 | 41
?
A solution contains 8.89 × 10-3 mole
fraction of I2 dissolved in 0.9911 mole
fraction of CH2Cl2 (methylene chloride).
What is the molality of I2 in this
solution?
We will assume we have 1 mole of the solution, so
we begin with 8.89 × 10-3 mol I2 and 0.9911 mol
CH2Cl2. Next, we will convert the moles of solvent
into grams, and then into kilograms. Finally, we will
compute the molality of the solution.
12 | 42
Moles solute = 0.00889 mol I2
84.93 g 1 kg
Kilograms solvent  0.9911 mol 
 3
1 mol
10 g
 0.08417 kg CH2Cl2
0.00889 mol
m
0.08417 kg
mol
m  0.106
kg
12 | 43
Converting between molality and molarity requires
knowing the density of the solution. This enables
you to calculate the mass or volume of the
solution. You can then distinguish the amount of
solute from the amount of solvent, or combine
them to find the volume of solution.
These types of conversions are illustrated in the
following problems.
12 | 44
?
Citric acid, HC6H7O7, is often used in
fruit beverages to add tartness. An
aqueous solution of citric acid is 2.331
m HC6H7O7. What is the molarity of the
solution? The density of the solution is
1.1346 g/mL.
A 2.331 m solution contains 2.331 mol solute in
1.000 kg solvent. We will use this relationship first
to convert the moles of citric acid to grams and
then to find the mass of solution. Using density,
we can then find the volume of solution. Finally,
we will compute the molarity of the solution.
12 | 45
Moles solute = 2.331 mol HC6H7O7
192.14 g
Mass solute  2.331 mol 
 447.88 g
1 mol
Mass of solution  447.88 g  1000.00 g  1447.88 g
1mL
10 3 L
Liters solution  1447.88 g 

1.1346 g mL
 1.2761 L
mol
2.331 mol
 1.827
M
L
1.2761 L
12 | 46
?
An aqueous solution of ethanol is 14.1
M C2H5OH. The density of the solution
is 0.853 g/cm3. What is the molality of
ethanol in the solution?
We will work with 1.00 L of solution. First, we will
convert volume to mass using the density. Then,
we will find the masses of the solute and the
solvent. Finally, we will compute the molality.
12 | 47
1mL 0.853 g
Mass of solution  1.000 L  -3 
10 L
mL
 853 g
46.08 g
Mass of solute  14.1 mol 
 649.7 g
1 mol
Mass of solvent  853 g  649.7 g
1 kg
 203.3 g  3  0.2033 kg
10 g
mol
14.1 mol
 69.4
m
kg
0.2033 kg
12 | 48
Colligative properties of solutions are properties
that depend on the concentration of the solute
molecules or ions in solution but not on the
chemical identity of the solute.
1.
2.
3.
4.
Vapor-pressure lowering
Boiling-point elevation
Freezing-point lowering
Osmotic pressure
12 | 49
The vapor pressure of a solution, P, is less than
the vapor pressure of the pure solvent, P°.
o
Psolution  Psolvent
When the solute is nonvolatile, the vapor pressure
of a solution is the mole fraction of the solvent
times the vapor pressure of pure solvent.
Psolution  X solv entP
o
solv ent
12 | 50
We can explain this relationship by recalling that
evaporation occurs at the surface of the solution.
When a solute is present, less of the surface is
occupied by solvent molecules.
12 | 51
To establish an equilibrium, the gaseous
solvent will condense in the more concentrated
solution until the vapor pressures and
concentrations are equal.
12 | 52
To find the vapor-pressure lowering (a colligative
property), we rearrange the following equation:
o
P  Psolv
ent  Psolution
o
o
P  Psolv

X
P
ent
solv ent solv ent
o
P  Psolv
ent 1- X solv ent 
Since X solute  1- X solv ent 
P  X soluteP
o
solv ent
12 | 53
o
Psolv ent  X solutePsolv
ent
Note that vapor-pressure lowering is directly
proportional to the solute concentration, the
definition of a colligative property.
12 | 54
?
Eugenol, C10H12O2, is the chief
constituent of oil of clove. This pale
yellow liquid dissolves in ethanol,
C2H5OH; it has a boiling point of 255°C.
As a result, we know eugenol’s vapor
pressure is very low; it can be
considered nonvolatile. What is the
vapor-pressure lowering at 20.0°C of a
solution containing 8.56 g of eugenol in
50.0 g of ethanol? The vapor pressure
of ethanol at 20.0°C is 44.6 mmHg.
12 | 55
1mol
Moles solute  8.56 g 
 0.0521 mol
164.22 g
1mol
Moles solvent  50.0 g 
 1.085 mol
46.08 g
Total moles  0.0521 mol  1.085 mol
 1.137 mol
0.0521 mol
Χ solute 
 0.04581
1.137 mol
ΔP  Χ solvent Psolvent
0
ΔP  0.04581  44.6 mmHg
ΔP  2.04 mmHg
12 | 56
When the solute is nonvolatile, it has no
appreciable vapor pressure itself and forms an
ideal solution.
When the solute is volatile, a nonideal solution
results.
12 | 57
The phase diagram on the next slide shows the
changes in freezing and boiling points when a
nonvolatile solute is added to a solvent. The blue
line shows the pure solvent; the purple line shows
the solution.
12 | 58
12 | 59
The boiling point of a solution is higher than the
boiling point of pure solvent. The boiling-point
elevation, Tb, is given by the following equation:
Tb = mKb
12 | 60
The freezing point of a solution is lower than the
freezing point of pure solvent. The freezing-point
depression, Tf, is given by the following equation:
Tf = mKf
12 | 61
?
A solution is made up of eugenol,
C10H12O2, in diethyl ether (“ether”). If
the solution is 0.575 m eugenol in
ether, what are the freezing point and
the boiling point of the solution? The
freezing point and the boiling point of
pure ether are –116.3°C and 34.6°C,
respectively. The freezing-point
depression and boiling-point elevation
constants are 1.79°C/m and 2.02°C/m,
respectively.
12 | 62
m = 0.575 m
Kf = 1.79°C/m
Tf° = –116.3°C
Tf = m × Kf
Kb = 2.02°C/m
Tb° = 34.6°C
Tb = m × Kb
Tf = 0.575 m × 1.79°C/m
Tf = 1.03°C This is the freezing-point depression.
Tf = –116.3 – 1.03 = –117.3°C
Tb = 0.575 m × 2.02°C/m
Tb = 1.16°C This is the boiling-point elevation.
Tb = 34.6 + 1.16 = 35.8°C
12 | 63
?
In a freezing-point depression
experiment, the molality of a solution of
58.1 mg anethole in 5.00 g benzene
was determined to be 0.0784 m. What
is the molar mass of anethole?
First, we will use the freezing-point depression data
to find the molality of the solution. Next, we will use
the molality and mass of solvent to find the moles of
solute. Finally, we will use the moles of solute and
mass of solute to find the molar mass.
12 | 64
Solute mass = 58.1 mg
Solvent mass = 5.00 g
mol solute
m
m = 0.0784 mol/kg
kg solvent
mol solute  m  kg solvent
 0.0784 m  0.00500 kg
 3.92  10- 4 mol
58.1  10-3 g
Molar mass 
 148 g/mol
-4
3.92  10 mol
The molar mass is 148 g/mol.
12 | 65
?
An 11.2-g sample of sulfur was
dissolved in 40.0 g of carbon disulfide.
The boiling-point elevation of carbon
disulfide was found to be 2.63°C. What
is the molar mass of the sulfur in the
solution? What is the formula of
molecular sulfur? The boiling-point
elevation constant, Kb, for carbon
disulfide is 2.40°C/m.
12 | 66
Solute mass = 11.2 g
Solvent mass = 40.0 g
Tb = 2.63°C
Kb = 2.40°C/m
First, we will use the definition of the colligative
property to calculate the molality. Next, we will use
the definition of molality to calculate the moles of
solute. Finally, using the mass and moles of solute,
we will find the molar mass and the molecular
formula of the sulfur.
12 | 67
ΔTb  m  K f
ΔTb
2.63C
m

 1.096 m
C
Kf
2.40
m
mol solute
m
kg solvent
mol solute  m  kg solvent
 1.096 m  0.0400 kg
 0.04383 mol
12 | 68
11.2 g
Molar mass 
 255.5 g/mol
0.04383 mol
The empirical formula for sulfur is S.
The formula mass is 32.065 g/mol.
255.5
n
 8
32.065
The molecular formula S 8 .
12 | 69
Osmosis is the phenomenon of solvent flow
through a semipermeable membrane to equalize
the solute concentration on both sides of the
membrane.
A semipermeable membrane allows solvent
molecules to pass through but not solute
molecules.
12 | 70
Water is the
solvent.
Water will flow
from the left to
the right.
12 | 71
Water is the solvent.
It flows from the beaker into
the thistle tube.
12 | 72
12 | 73
Osmotic pressure, p, is equal to the pressure that,
when applied, just stops osmosis. Osmotic
pressure is a colligative property of a solution.
p = MRT
12 | 74
?
Dextran, a polymer of glucose units, is
produced by bacteria growing in
sucrose solutions. Solutions of dextran
in water have been used as a blood
plasma substitute. At 21°C, what is the
osmotic pressure (in mmHg) of a
solution containing 1.50 g of dextran
dissolved in 100.0 mL of aqueous
solution, if the average molecular mass
of the dextran is 4.0  108 amu?
12 | 75
T = 21°C + 273 = 294 K
Mass solute = 1.50 g
Liters solution = 100.0  10-3 L
Molecular mass = 4.0  108 amu
Molar mass = 4.0  108 g/mol
1mol
Moles solute  1.50 g 
8
4.0  10 g
 3.75  10 9 mol
3.75  10 mol
-8
Molarity 

3.75

10
mol/L
-3
100.0  10 L
-9
12 | 76
π  MRT
mol
L  atm
π  3.75  10
 0.08206
 294 K
L
mol  K
760 mmHg
7
 9.05  10 atm 
1 atm
-8
π  6.9 x 10 mmHg
-4
12 | 77
One can show that 760.0 mmHg is equivalent to
the pressure exerted by a column of water 10.334
m high. Thus each 1.00 mmHg of pressure is
equivalent to the pressure of a 1.36-cm column of
water. If the density of this dextran solution is
equal to that of water, what height of solution
would exert a pressure equal to its osmotic
pressure?
1.36 cm
π  6.9 x 10 mmHg 
1.00 mmHg
-4
π  9.4 x 10 cm
-4
12 | 78
A
B
C
Hypertonic
solution
Isotonic
solution
Hypotonic
solution
Water flows
out of cell.
Water flows
into cell.
Crenation
Hemolysis
12 | 79
Ionic solutes dissolve to form more than one
particle per formula unit. We alter the colligative
property equations to account for this fact by
including i, the number of ions per formula unit:
P = iP°AXB
Tb = iKbm
Tf = iKfm
p = iMRT
12 | 80
?
What is the osmotic pressure at 25.0°C
of an isotonic saline solution (a solution
having an osmotic pressure equal to
that of blood) that contains 0.900 g
NaCl in 100.0 mL of aqueous solution?
Assume that i has the ideal value
(based on the formula).
12 | 81
T = 25°C + 273 = 298 K
Mass solute = 0.900 g NaCl
Volume solution = 100.0 mL = 100.0  10-3 L
Molar mass solute = 58.44 g/mol
i=2
1 mol
Moles solute  0.900 g 
 0.01540 mol
58.44 g
1.540  10-2 mol
Molarity 
 0.1540 mol/L
-3
100.0  10 L
12 | 82
π  iMRT
mol
L  atm
π  2  0.1540
 0.08206
 298 K
L
mol  K
π  7.53 atm
12 | 83
A colloid is a dispersion of particles of one
substance (the dispersed phase) throughout
another substance or solution (the continuous
phase). The dispersed particles range from 1000
pm to 200,000 pm in size—much larger than single
molecules or single ions.
Fog is an example of a colloid. In fog, water
droplets are dispersed through air.
12 | 84
Colloids exhibit the Tyndall effect.
The path of the light is visible through a colloid
because the light is reflected by the relatively
larger-sized particles in the dispersed phase.
12 | 85
Colloid
Solution
12 | 86
Colloids are characterized according to the state
(solid, liquid, or gas) of the dispersed phase and
the state of the continuous phase.
• Fog and smoke are aerosols, which are liquid
droplets or solid particles dispersed throughout
a gas.
• An emulsion consists of liquid droplets
dispersed throughout another liquid (for
example, particles of butterfat dispersed
through homogenized milk).
• A sol consists of solid particles dispersed in a
liquid.
12 | 87
Colloids in which the continuous phase is water
are categorized into two major classes: hydrophilic
colloids and hydrophobic colloids.
Hydrophilic colloid
A colloid in which there is a strong attraction
between the dispersed phase and the continuous
phase (water).
Hydrophobic colloid
A colloid in which there is a lack of attraction
between the dispersed phase and the continuous
phase (water).
12 | 88
Coagulation is the process by which the
dispersed phase of a colloid is made to aggregate
and thereby separate from the continuous phase.
It is analogous to precipitation from a solution.
Curdled milk is an example of coagulation.
12 | 89
When molecules or ions that have both a
hydrophobic end and a hydrophilic end are
dispersed in water, they associate, or aggregate,
to form colloidal-sized particles called micelles.
A colloid in which the dispersed phase consists of
micelles is called an association colloid.
12 | 90
Soap consists of compounds such as sodium
stearate. Sodium stearate is an example of a
molecule with hydrophobic and hydrophilic ends.
Hydrophobic
end
Hydrophilic
end
12 | 91
In water solution, the stearate ions associate to
form micelles in which the hydrocarbon ends point
inward toward one another and away from the
water, and ionic carboxyl groups are on the
outside of the micelle facing the water.
The cleansing action of soap occurs because oil
and grease can be absorbed into the hydrophobic
centers of soap micelles and washed away.
12 | 92
12 | 93
12 | 94
Far left:
Vegetable oil floating
on water (dyed
green).
Left:
When the mixture is
shaken with soap, an
emulsion forms as
the oil droplets are
absorbed into soap
micelles.
12 | 95
Synthetic detergents also form association
colloids. Sodium lauryl sulfate is a synthetic
detergent present in toothpastes and shampoos.
12 | 96
The detergent molecules we have discussed so far
are classified in the trade as anionics, because
they have a negative charge at the hydrophilic
end.
Other detergent molecules are classified as
cationics, because they have a positive charge at
the hydrophilic end.
12 | 97
12 | 98
Many cationic
detergents
also have
germicidal
properties and
are used in
hospital
disinfectants
and in
mouthwashes.
12 | 99