Chemistry: Atoms First
Julia Burdge & Jason Overby
Chapter 13
Physical Properties of
Solutions
Kent L. McCorkle
Cosumnes River College
Sacramento, CA
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
13
Physical Properties of Solutions
13.1 Types of Solutions
13.2 A Molecular View of the Solution Process
The Importance of Intermolecular Forces
Energy and Entropy in Solution Formation
13.3 Concentration Units
Molality
Percent by Mass
Comparison of Concentration Units
13.4 Factors that Affect Solubility
Temperature
Pressure
13.5 Colligative Properties
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
Electrolyte Solutions
13.6 Calculations Using Colligative Properties
13.7 Colloids
13.1
Types of Solutions
A solution is a homogeneous mixture of two or more substances.
A solution consists of a solvent and one or more solutes.
Types of Solutions
Solutions can be classified by the amount of solute dissolved.
An unsaturated solution is one that contains less solute than the
solvent has the capacity to dissolve at a specific temperature.
Types of Solutions
Solutions can be classified by the amount of solute dissolved.
A saturated solution is one that contains the maximum amount of
solute that will dissolve in a solvent at a specific temperature.
Types of Solutions
Supersaturated solutions are generally unstable.
13.2
A Molecular View of the Solution Process
Solvation occurs when solute molecules are separated from one
another and surrounded by solvent molecules.
Solvation depends on three types of interactions:
1) Solute-solute interactions
2) Solvent-solvent interactions
3) Solute-solvent interactions
A Molecular View of the Solution Process
ΔHsoln = ΔH1 + ΔH2 + ΔH3
Separated solute Separated solvent
Energy
Step 2
ΔH2 > 0
Separated solute
Step 3
ΔH3 < 0
Solvent
Solution
Step 1
ΔH1 > 0
Solute
Solvent
ΔHsoln > 0
The Importance of Intermolecular Forces
“Like dissolves like”
Two substances with similar type and magnitude of intermolecular
forces are likely to be soluble in each other.
Toluene, C7H8
Octane, C8H18
Both non-polar liquids,
solution forms when mixed
Two liquids are said to be miscible if they are completely soluble in
each other in all proportions.
The Importance of Intermolecular Forces
“Like dissolves like”
Two substances with similar type and magnitude of intermolecular
forces are likely to be soluble in each other.
Water, H2O
Octane, C8H18
Polar and non-polar liquids,
solution does not form when mixed
The Importance of Intermolecular Forces
“Like dissolves like”
Two substances with similar type and magnitude of intermolecular
forces are likely to be soluble in each other.
Water, H2O
Ethanol, C2H6O
Both polar liquids,
solution forms when mixed
Worked Example 13.1
Determine for each solute whether the solubility will be greater in water, which is
polar, or in benzene (C6H6), which is nonpolar: (a) Br2, (b) sodium iodide (NaI),
(c) carbon tetrachloride, and (d) formaldehyde (CH2O).
Strategy Consider the structure of each solute to determine whether or not it is
polar. For molecular solutes, start with a Lewis structure and apply the VSEPR
theory. We expect polar solutes, including ionic compounds, to be more soluble in
water. Nonpolar solutes will be more soluble in benzene.
Solution (a) Bromine is a homonuclear diatomic molecule and is nonpolar.
Bromine is more soluble in benzene.
(b) Sodium iodide is ionic and more soluble in water.
Worked Example 13.1 (cont.)
Solution (c) Carbon tetrachloride has the following Lewis structure:
With four electron domains around the central atom, we expect a tetrahedral
Think About It Remember that molecular formula alone is not
arrangement. A symmetrical arrangement of identical bonds results in a nonpolar
sufficient to determine the shape or polarity of a polyatomic
molecule. Carbon tetrachloride is more soluble in benzene.
molecule. It must be determined by starting with a correct Lewis
structure and applying VSEPR theory.
(d) Formaldehyde has the following Lewis structure:
Crossed arrows represent individual bond dipoles. This molecule is polar and
can form hydrogen bonds in water. Formaldehyde is more soluble in water.
13.3
Concentration Units
The amount of solute relative to the volume of a solution or to the
amount of solvent in a solution is called concentration.
Molarity:
molarity = M =
moles of solute
liters of solution
Mole fraction:
mole fraction of component A =  A =
moles of A
sum of moles of all components
Concentration Units
Molality (m) is the number of moles of solute dissolved in 1 kg
(1000 g) solvent:
molality = m =
moles of solute
mass of solvent in kg 
Percent by Mass:
percent by mass =
mass of solute
100 %
mass of solute + mass of solvent
Worked Example 13.2
A solution is made by dissolving 170.1 g of glucose (C6H12O6) in enough water to
make a liter of solution. The density of the solution is 1.062 g/mL. Express the
concentration in (a) molality, (b) percent by mass, and (c) parts per million.
Strategy Use the molar mass of glucose to determine the number of moles of
glucose in a liter of solution. Use the density (in g/L) to calculate the mass of a
liter of solution. Subtract the mass of glucose from the mass of solution to
determine the mass of water. The molar mass of glucose is 180.2 g/mol.
170.1 g
= 0.9440 mol glucose per liter of solution
180.2 g/mol
1062 g
1 liter of solution ×
= 1062 g
L
Solution (a)
1062 g – 170.1 g = 892 g water = 0.892 kg water
0.9440 mol glucose
= 1.06 m
0.892 kg water
Worked Example 13.2 (cont.)
Solution
170.1 g
(b)
× 100% = 16.02% glucose by mass
1062 g solution
(c)
170.1 g
× 1,000,000 = 1.602×105 ppm glucose
1062 g solution
Think About It Pay careful attention to units in problems such as this. Most
require conversions between grams and kilograms and/or liters and milliliters.
Worked Example 13.3
“Rubbing alcohol” is a mixture of isopropyl alcohol (C3H7OH) and water that is
70 percent isopropyl alcohol by mass (density = 0.79 g/mL at 20°C). Express
the concentration of rubbing alcohol in (a) molarity and (b) molality.
Strategy (a) Use density to determine the total mass of a liter of solution, and
use percent by mass to determine the mass of isopropyl alcohol in a liter of
solution. Convert the mass of isopropyl alcohol to moles, and divide moles by
liters of solution to get molarity.
(b) Subtract the mass of C3H7OH from the mass of solution to get the mass of
water. Divide moles of C3H7OH by the mass of water (in kg) to get molality.
The mass of a liter of rubbing alcohol is 790 g, and the molar mass of isopropyl
alcohol is 60.09 g/mol.
Worked Example 13.3 (cont.)
Solution
70 g C3H7OH
553 g C3H7OH
(a) 790 g solution
=
×
L solution
100 g solution
L solution
553 g C3H7OH
1 mol
9.20 mol C3H7OH
×
=
= 9.2 M
L solution
60.09 g C3H7OH
L solution
(b) 790 g solution – 553 g C3H7OH = 237 g water = 0.237 kg water
9.20 mol C3H7OH
= 39 m
0.237 kg water
Rubbing alcohol is 9.2 M and 39 m in isopropyl alcohol.
Think About It Note the large difference between molarity and molality in this
case. Molarity and molality are the same (or similar) only for very dilute
aqueous solutions.
13.4
Factors That Affect Solubility
Temperature affects the solubility of most substances.
Factors That Affect Solubility
Pressure greatly influences the solubility of a gas.
Henry’s law states that the solubility of a gas in a liquid is
proportional to the pressure of the gas over the solution.
c = kP
c
molar concentration
(mol/L)
P pressure (atm)
k
proportionality
constant called
Henry’s law constant
Worked Example 13.4
Calculate the concentration of carbon dioxide in a soft drink that was bottled
under a partial pressure of 5.0 atm CO2 at 25°C (a) before the bottle is opened
and (b) after the soda has gone “flat” at 25°C . The Henry’s law constant for CO2
in water at this temperature is 3.1×10-2 mol/L∙atm. Assume that the partial
pressure of CO2 in air is 0.0003 atm and that Henry’s law constant for the soft
drink is the same as that for water.
Strategy Use c = kP and the given Henry’s law constant to solve for the molar
concentration (mol/L) of CO2 at 25°C and the two pressures given.
Solution (a) c = (3.1×10-2 mol/L∙atm)(5.0 atm) = 1.6×10-1 mol/L
(b) c = (3.1×10-2 mol/L∙atm)(0.0003 atm) = 9×10-6 mol/L
Think About It With a pressure approximately 15,000 smaller in part (b) than
in part (a), we expect the concentration of CO2 to be approximately 15,000
times smaller–and it is.
13.5
Colligative Properties
Colligative properties are properties that depend on the number of
solute particles in solution.
Colligative properties do not depend on the nature of the solute
particles.
The colligative properties are:
 vapor-pressure lowering
 boiling-point elevation
 freezing-point depression
 osmotic pressure
Colligative Properties
Raoult’s law states that the partial pressure of a solvent over a
solution is given by the vapor pressure of the pure solvent times the
mole fraction of the solvent in the solution.
P1  1P1
P1 partial pressure of
solvent over solution
P° vapor pressure of
pure solvent
χ1 mole fraction of solvent
ΔP vapor pressure lowering
χ2 mole fraction of solute
P   2P1
Worked Example 13.5
Calculate the vapor pressure of water over a solution made by dissolving 225 g of
glucose in 575 g of water at 35°C. (At 35°C,
H OP° = 42.2 mmHg.)
2
Strategy Convert the masses of glucose and water to moles, determine the mole
fraction of water, and use P1 = χ1P°1 to find the vapor pressure over the solution.
The molar masses of glucose and water are 180.2 and 18.02 g/mol, respectively.
Solution
Think About It This problem can also be solved using Equation
to calculate the vapor-pressure lowering,
22513.5
g glucose
575 gΔP.
water
= 1.25 mol glucose and
= 31.9 mol water
180.2 g/mol
18.02 g/mol
χwater =
31.9 mol water
= 0.962
1.25 mol glucose + 31.9 mol water
PH O= χwaterP° H=O 0.962 × 42.4 mmHg = 40.6 mmHg
2
2
The vapor pressure of water over the solution is 40.6 mmHg.
Colligative Properties
If both components of a solution
are volatile, the vapor pressure of
the solution is the sum of the
individual partial pressures.
PA   A PAo
PB   B PBo
o
A A
o
B B
PT   P   P
Benzene
Toluene
Colligative Properties
Benzene
Toluene
PT   A PAo   B PBo
An ideal solution obeys Raoult’s
law.
Colligative Properties
Solutions boil at a higher
temperature than the pure
solvent.
Tb  Tb  Tbo
Tb  Kb m
ΔTb boiling point elevation
Kb boiling point elevation
constant (°C/m)
m molality
Colligative Properties
Solutions freeze at a lower
temperature than the pure
solvent.
Tf  Tfo  Tf
Tf  K f m
ΔTf freezing point
depression
Kf freezing point
depression constant
(°C/m)
m
molality
Colligative Properties
Worked Example 13.6
Ethylene glycol [CH2(OH)CH2(OH)] is a common automobile antifreeze. It is
water soluble and fairly nonvolatile (b.p. 197°C). Calculate (a) the freezing
point and (b) the boiling point of a solution containing 685 g of ethylene glycol in
2075 g of water.
Strategy Convert grams of ethylene glycol to moles, and divide by the mass of
water in kilograms to get molal concentration. Use molal concentrations and ΔTb
= Kbm Think
and ΔTAbout
The lowers
molar mass
of ethylene
Because it both
the freezing
pointglycol
and (C2H6O2)
f = Kfm,Itrespectively.
is 62.07
g/mol.
are 1.86°C/m
0.52°C/m,
respectively.
raises
theKboiling
antifreeze
is usefuland
at both
temperature
f and Kpoint,
b for water
extremes.
Solution
685 g C2H6O2
11.04 mol C2H6O2
= 11.04 mol C2H6O2 and
= 5.32 m C2H6O2
62.07 g/mol
2.075 kg water
(a) ΔTf = Kfm = (1.86°C/m)(5.32 m) = 9.89°C
The freezing point of the solution is (0 – 9.89)°C = – 9.89°C
(b) ΔTb = Kbm = (0.52°C/m)(5.32 m) = 2.8°C
The boiling point of the solution is (100.0 + 2.8)°C = 102.8°C
Colligative Properties
Osmosis is the selective passage of solvent
molecules through a porous membrane from a
more dilute solution to a more concentrated one.
Colligative Properties
Osmotic pressure () of a solution is the pressure required to stop
osmosis.
  MRT
 Osmotic pressure (atm)
M molarity (moles/L)
R gas constant (0.08206 L∙atm/mol∙K)
T absolute temperature (kelvins)
Colligative Properties
Electrolytes undergo dissociation when dissolved in water.
The van’t Hoff factor (i) accounts for this effect.
actual number of particles in solution after dissociation
i
number of formulas units initially dissolved in solution
Tf  iKf m
Tb  iKb m
  iMRT
Colligative Properties
The van’t Hoff factor (i) is 1 for all nonelectrolytes:
C12H22O11(s)
H2O
C12H22O11(aq)
1 particle dissolved, i = 1
For strong electrolytes i should be equal to the number of ions:
NaCl(s)
H2O
Na+(aq) + Cl–(aq)
2 particles dissolved, i = 2
Na2SO4(s)
H2O
2Na+(aq) + SO42–(aq)
3 particles dissolved, i = 3
Colligative Properties
The van’t Hoff factor (i) is usually smaller than predicted due to the
formation of ion pairs.
An ion pair is made up of one or more cations and one or more
anions held together by electrostatic forces.
ion pair
Colligative Properties
The van’t Hoff factor (i) is usually smaller than predicted due to the
formation of ion pairs.
An ion pair is made up of one or more cations and one or more
anions held together by electrostatic forces.
Colligative Properties
Concentration has an effect on experimentally measured van’t Hoff
factors (i).
Worked Example 13.7
The osmotic pressure of a 0.0100 M potassium iodide (KI) solution at 25°C is
0.465 atm. Determine the experiment van’t Hoff factor for KI at this
concentration.
Strategy Use osmotic pressure to calculate the molar concentration of KI, and
divide by the nominal concentration of 0.01000 M. R = 0.08206 L∙atm/K∙mol,
and T = 298 K.
Think About It The calculated van’t Hoff factor for KI is 2. The
experimental
Hofffor
factor
Solution
Solving πvan’t
= MRT
M, must be less than or equal to the
calculated value.
π
0.465 atm
M=
=
= 0.0190 M
RT
(0.08206 L∙atm/K∙mol)(298 K)
i=
0.0190 M
= 1.90
0.0100 M
The experimental van’t Hoff factor for KI at this concentration is 1.90.
Worked Example 13.8
Quinine was the first drug widely used to treat malaria, and it remains the
treatment of choice for severe cases. A solution prepared by dissolving 10.0 g of
quinine in 50.0 mL of ethanol has a freezing point 1.55°C below that of pure
ethanol. Determine the molar mass of quinine. (The density of ethanol is 0.789
g/mL.) Assume that quinine is a nonelectrolyte.
Strategy Use ΔTf = Kfm to determine the molal concentration of the solution.
Use the density of ethanol to determine the mass of the solvent. The molal
concentration of quinine multiplied by the mass of ethanol (in kg) gives moles of
quinine. The mass of quinine (in grams) divided by moles of quinine gives the
molar mass. Kf for ethanol is 1.99°C/m.
Solution mass of ethanol = 50.0 mL × 0.789 g/mL = 39.5 g or 3.95×10-2 kg
Solving ΔTf = Kfm for molal concentration
m=
ΔTf
1.55°C
=
= 0.779 m
Kf
1.99°C/
m
Worked Example 13.8 (cont.)
Solution The solution is 0.779 m in quinine (i.e., 0.779 mol of quinine/kg
ethanol solvent.)
0.779 mol quinine
(3.95×10-2 kg ethanol) = 0.00308 mol quinine
kg ethanol
molar mass of quinine =
10.0 g quinine
= 325 g/mol
0.00308 mol quinine
Think About It Check the result using the molecular formula of quinine:
C20H24N2O2 (324.4 g/mol). Multistep problems such as this one require careful
tracking of units at each step.
Worked Example 13.9
A solution is prepared by dissolving 50.0 g of hemoglobin (Hb) in enough water
to make 1.00 L of solution. The osmotic pressure of the solution is measured and
found to be 14.3 mmHg at 25°C. Calculate the molar mass of hemoglobin.
(Assume that there is no change in volume when the hemoglobin is added to
water.)
Strategy Use π = MRT to calculate the molarity of the solution. Because the
solution volume is 1 L, the molarity is equal to the number of moles of
hemoglobin. Dividing the given mass of hemoglobin by the number of moles
gives the molar mass. R = 0.08206 L∙atm/K∙mol, T = 298 K, and
π = 14.3 mmHg/(760 mmHg/atm) = 1.88×10-2 atm.
Worked Example 13.9 (cont.)
Solution Rearranging π = MRT to solve for molarity we get,
π
1.88×10-2 atm
M=
=
= 7.69×10-4 M
RT
(0.08206 L∙atm/K∙mol)(298 K)
Thus, the solution contains 7.69×10-4 moles of hemoglobin.
molar mass of hemoglobin =
50.0 g
= 6.50×104 g/mol
-4
7.69×10 mol
Think About It Biological molecules can have very high molar masses.
13.6
Calculations Using Colligative Properties
Percent dissociation is the percentage of dissolved molecules (or
formula units, in the case of an ionic compound) that separate into
ions in a solution.
Strong electrolytes should have complete, or 100%, dissociation,
however, experimentally determined van’t Hoff factors indicate that
this is not the case.
Percent dissociation of a strong electrolyte is more complete at lower
concentration.
Percent ionization of weak electrolytes is also dependent on
concentration.
Worked Example 13.10
A solution that is 0.100 M in hydrofluoric acid (HF) has an osmotic pressure of
2.64 atm at 25°C. Calculate the percent ionization of HF at this concentration.
Strategy Use the osmotic pressure and π = MRT to determine the molar
concentration of the particles in solution. Compare the concentration of particles
to the nominal concentration (0.100 M) to determine what percentage of the
original HF molecules are ionized. R = 0.08206 L∙atm/K∙mol, and T = 298 K.
Solution Rearranging π = MRT to solve for molarity,
M=
π
2.64 atm
=
= 0.108 M
RT
(0.08206 L∙atm/K∙mol)(298 K)
Worked Example 13.10 (cont.)
Solution The concentration of dissolved particles is 0.108 M. Consider the
ionization of HF:
HF(aq) ⇌ H+(aq) + F-(aq)
According to this equation, if x HF molecules ionize, we get x H+ ions and x Fions. Thus, the total concentration of particles in solution will be the original
concentration
of HF minus
which
givesthe
thelower
concentration
of intact HF
Think About
It Forx,weak
acids,
the concentration,
the
+ and F-):
molecules,
plus
whichionization.
is the concentration
ions (Hof
greater
the2x,
percent
A 0.010 Mofsolution
HF has an
osmotic pressure of 0.30 atm, corresponding to 23 percent
– x) +of2x
0.100
x
ionization. A 0.0010 (0.100
M solution
HF= has
an +osmotic
pressure of
3.8×10-2 atm, corresponding to 56 percent ionization.
Therefore, 0.108 = 0.100 + x and x = 0.008. Because we earlier defined x as the
amount of HF ionized, the percent ionization is given by
percent ionization =
0.008 M
×100% =8%
0.100 M
At this concentration HF is 8 percent ionized.
13.7
Colloids
A colloid is a dispersion of particles of one substance throughout
another substance. Colloid particles are much larger than the normal
solute molecules.
Categories of colloids:
 aerosols
 foams
 emulsions
 sols
 gels
Colloids
Examples of colloids
Colloids
Colloids with water as the dispersing medium can be categorized as
hydrophilic (water loving) or hydrophobic (water fearing).
Hydrophilic groups on the
surface of a large molecule
stabilize the molecule in
water.
Colloids
Colloids with water as the dispersing medium can be categorized as
hydrophilic (water loving) or hydrophobic (water fearing).
Negative ions are adsorbed onto the surface of hydrophobic colloids.
The repulsion between like charges prevents aggregation of the
articles.
Colloids
Hydrophobic colloids can be stabilized by the presence of
hydrophilic groups on their surface.
Colloids
Emulsification is the process of stabilizing a colloid that would
otherwise not stay dispersed.
13
Key Concepts
Types of Solutions
A Molecular View of the Solution Process
The Importance of Intermolecular Forces
Energy and Entropy in Solution Formation
Concentration Units
Molality
Percent by Mass
Comparison of Concentration Units
Factors that Affect Solubility
Temperature
Pressure
Colligative Properties
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
Electrolyte Solutions
Calculations Using Colligative Properties
Colloids