Chapter 13
Properties of Solutions
13.1 The Solution
Process
Solutions


Solutions are homogeneous mixtures of two or more pure
substances.
In a solution, the solute is dispersed uniformly throughout the
solvent (present in greatest amount).
Solutions
The intermolecular forces
between solute and
solvent particles must be
strong enough to compete
with those between solute
particles and those
between solvent particles.
How Does a Solution Form?
As a solution forms, the solvent pulls solute particles
apart and surrounds, or solvates, them.
How Does a Solution Form
If an ionic salt is soluble
in water, it is because
the ion-dipole
interactions are strong
enough to overcome
the lattice energy of the
salt crystal.
Energy Changes in Solution

Simply put, three
processes affect the
energetics of solution:



separation of solute
particles,
separation of solvent
particles,
new interactions between
solute and solvent.
Energy Changes in Solution
The enthalpy change
of the overall process
depends on H for
each of these steps.
Why Do Endothermic
Processes Occur?
Things do not tend to
occur spontaneously
(i.e., without outside
intervention) unless the
energy of the system is
lowered.
Why Do Endothermic
Processes Occur?
Yet we know the in
some processes, like
the dissolution of
NH4NO3 in water, heat
is absorbed, not
released.
Enthalpy Is Only Part of the Picture
The reason is that
increasing the disorder or
randomness (known as
entropy) of a system
tends to lower the energy
of the system.
Enthalpy Is Only Part of the Picture
So even though enthalpy
may increase, the overall
energy of the system can
still decrease if the
system becomes more
disordered.
Student, Beware!
Just because a substance disappears when it comes in contact
with a solvent, it doesn’t mean the substance dissolved.
 Dissolution is a physical change — you can get back the
original solute by evaporating the solvent.
 If you can’t, the substance didn’t dissolve, it reacted.
13.1 GIST


Why doesn’t NaCl dissolve in nonpolar
solvents such as hexane, C6H14?
Label the following processes as exothermic
or endothermic:



Forming solvent-solute interactions
Breaking solvent-solvent interactions
Silver chloride, AgCl, is essentially insoluble
in water. Would you expect a significant
change in entropy of the system when 10 g of
AgCl is added to 500 mL of water?
Sample Exercise 13.1
1) Water vapor reacts with excess solid
sodium sulfate to form a hydrate:

a)
b)
Na2SO4(s) + 10 H2O(g) → Na2SO4●10H2O(s)
Does the system become more or less ordered
in the process?
Does the entropy of the system increase or
decrease?
Practice
2) Does the entropy increase or
decrease when the stopcock is
opened allowing mixing of the two
gases?
13.2 Saturated Solutions
and Solubility
Types of Solutions

Saturated


In a saturated solution,
the solvent holds as
much solute as is
possible at that
temperature.
Dissolved solute is in
dynamic equilibrium with
solid solute particles.
Types of Solutions

Unsaturated

If a solution is
unsaturated, less solute
than can dissolve in the
solvent at that
temperature is dissolved
in the solvent.
Types of Solutions

Supersaturated


In supersaturated solutions, the solvent holds more
solute than is normally possible at that temperature.
These solutions are unstable; crystallization can usually
be stimulated by adding a “seed crystal” or scratching
the side of the flask.
p.535 GIST: Is a supersaturated solution of sodium acetate
a stable equilibrium solution?
13.3 Factors Affecting
Solubility
Factors Affecting Solubility

Chemists use the axiom “like dissolves like."


Polar substances tend to dissolve in polar solvents.
Nonpolar substances tend to dissolve in nonpolar solvents.
Factors Affecting Solubility
The more similar the
intermolecular
attractions, the more
likely one substance is
to be soluble in
another.
Factors Affecting Solubility
Glucose (which has
hydrogen bonding) is very
soluble in water, while
cyclohexane (which only
has dispersion forces) is
not.
p.538 GIST: Suppose the
hydrogens on the OH groups
in glucose were replaced with
methyl groups, CH3. Would
you expect the solubility of
the resulting molecule to be
higher than, lower than, or
about the same as the
solubility of glucose?
Factors Affecting Solubility


Vitamin A is soluble in nonpolar compounds (like
fats).
Vitamin C is soluble in water.
Sample Exercise 13.2 Predicting Solubility Patterns
Predict whether each of the following substances is
more likely to dissolve in the nonpolar solvent carbon
tetrachloride (CCl4) or in water: C7H16, Na2SO4, HCl,
and I2.
Sample Exercise 13.2 Predicting Solubility Patterns
Practice Exercise
Arrange the following substances in order of increasing
solubility in water:
Gases in Solution


In general, the solubility
of gases in water
increases with
increasing mass.
Larger molecules have
stronger dispersion
forces.
Gases in Solution


The solubility of liquids
and solids does not
change appreciably
with pressure.
The solubility of a gas
in a liquid is directly
proportional to its
pressure.
Henry’s Law
Sg = kPg
where
 Sg is the solubility of the
gas,
 k is the Henry’s Law
constant for that gas in
that solvent, and
 Pg is the partial pressure
of the gas above the
liquid.
Exercise 13.13


1) Calculate the concentration of CO2 in a
soft drink that is bottled with a partial
pressure of CO2 of 4.0 atm over the liquid at
25ºC. The Henry’s law constant for CO2 in
water at this temperature is 3.1 x 10-2 mol/Latm.
2) Calculate the concentration of CO2 in a
soft drink after the bottle is opened and
equilibrates at 25ºC under a CO2 partial
pressure of 3.0 x 10-4 atm.
Temperature
Generally, the solubility
of solid solutes in liquid
solvents increases with
increasing temperature.
Temperature

The opposite is true of
gases.


Carbonated soft drinks
are more “bubbly” if
stored in the refrigerator.
Warm lakes have less O2
dissolved in them than
cool lakes.
13.4 Ways of Expressing
Concentration
Mass Percentage
Mass % of A =
mass of A in solution
 100
total mass of solution
Parts per Million and
Parts per Billion
Parts per Million (ppm)
ppm =
mass of A in solution
 106
total mass of solution
Parts per Billion (ppb)
ppb =
mass of A in solution
 109
total mass of solution
Sample Exercise 13.4 Calculation of Mass-Related Concentrations
(a) A solution is made by dissolving 13.5 g of glucose
(C6H12O6) in 0.100 kg of water. What is the mass
percentage of solute in this solution? (b) A 2.5-g
sample of groundwater was found to contain 5.4 µg of
Zn2+. What is the concentration of Zn2+ in parts per
million?
Practice Exercise
a) Calculate the mass percentage of NaCl in a solution
containing 1.50 g of NaCl in 50.0 g of water. (b) A
commercial bleaching solution contains 3.62 mass %
sodium hypochlorite, NaOCl. What is the mass of
NaOCl in a bottle containing 2.50 kg of bleaching
solution?
(
Mole Fraction (X)
moles of A
XA =
total moles in solution

In some applications, one needs the mole
fraction of solvent, not solute — make sure
you find the quantity you need!
Molarity (M)
M=


mol of solute
L of solution
You will recall this concentration measure
from Chapter 4.
Since volume is temperature-dependent,
molarity can change with temperature.
Molality (m)
m=
mol of solute
kg of solvent
Since both moles and mass do not change
with temperature, molality (unlike molarity) is
not temperature-dependent.
Changing Molarity to Molality
If we know the density
of the solution, we
can calculate the
molality from the
molarity and vice
versa.
Sample Exercise 13.5 Calculation of Molality
A solution is made by dissolving 4.35 g glucose
(C6H12O6) in 25.0 mL of water at 25 °C.
Calculate the molality of glucose in the
solution. Water has a density of 1.00 g/mL.
Practice
What is the molality of a solution made by
dissolving 36.5 g of naphthalene (C10H8) in
425 g of toluene (C7H8)?
Sample Exercise 13.6 Calculation of Mole Fraction and Molality
An aqueous solution of hydrochloric acid
contains 36% HCl by mass. (a) Calculate the
mole fraction of HCl in the solution. (b)
Calculate the molality of HCl in the solution.
Practice

1) A commercial bleach contains 3.62 mass
% NaOCl in water. Calculate:


A) the molality
B) the mole fraction of NaOCl
Sample Exercise 13.7 Calculation of Molality Using the Density of a Solution
A solution with a density of 0.876 g/mL
contains 5.0 g of toluene (C7H8) and 225 g of
benzene. Calculate the molarity of the
solution.
Practice

1) A solution containing equal masses of
glycerol (C3H8O3) and water has a density of
1.10 g/mL. Calculate:



A) The molality of glycerol
B) The mole fraction of glycerol
C) The molarity of glycerol in the solution
13.5 Colligative
Properties
Colligative Properties


Changes in colligative properties depend
only on the number of solute particles
present, not on the identity of the solute
particles.
Among colligative properties are




Vapor pressure lowering
Boiling point elevation
Melting point depression
Osmotic pressure
Vapor Pressure
Because of solutesolvent intermolecular
attraction, higher
concentrations of
nonvolatile solutes
make it harder for
solvent to escape to the
vapor phase.
Vapor Pressure
Therefore, the vapor
pressure of a solution is
lower than that of the
pure solvent.
Raoult’s Law
PA = XAPA
where


XA is the mole fraction of compound A, and
PA is the normal vapor pressure of A at that
temperature.
NOTE: This is one of those times when you want
to make sure you have the vapor pressure of
the solvent.
Sample Exercise 13.8 Calculation of Vapor-Pressure Lowering
Glycerin (C3H8O3) is a nonvolatile
nonelectrolyte with a density of 1.26 g/mL at
25 °C. Calculate the vapor pressure at 25
°C of a solution made by adding 50.0 mL of
glycerin to 500.0 mL of water. The vapor
pressure of pure water at 25 °C is 23.8 torr
(Appendix B), and its density is 1.00 g/mL.
Practice


1) The vapor pressure of pure water at 110ºC
is 1070 torr. A solution of ethylene glycol and
water has a vapor pressure of 1.00 atm at
110ºC. Assuming that Raoult’s law is
obeyed, what is the mole fraction of ethylene
glycol in the solution?
P.547 GIST: Adding 1 mol of NaCl to 1 kg of
water reduces the vapor pressure of water
more than adding 1 mol of C6H12O6. Explain.
Ideal Solutions



Ideal solutions obey Raoult’s Law
Real liquid-liquid solutions best approximate ideal
behavior when the solute concentration is low
and when the solute and solvent have similar
molecular sizes and similar types of
intermolecular attractions.
If a solution is not ideal, the observed vapor
pressure will either by higher or lower than
predicted using Raoult’s Law
Deviation from Raoult’s Law

negative deviation



ex. acetone and water
when solvent has special attraction to solute
observed vapor pressure will be lower than
predicted
Deviation from Raoult’s Law

positive deviation



ex. ethanol and hexane
when solvent has especially weak interactions
with solute
observed vapor pressure will be higher than
predicted
Volatile Solutions

when both components are
volatile
Ptotal  PA  PB   A  P   B  P

A

where


PA and PB are partial pressures
PA° and PB° are vapor pressure of
pure A and B

B
A solution is prepared by mixing 5.81 g of
acetone (C3H6O, molar mass = 58.1 g/mol)
and 11.9 g of chloroform (HCCl3, molar
mass = 119.4 g/mol). At 35oC, this
solution has a total pressure of 260. torr.
Is this an ideal solution? The vapor
pressure of pure acetone and pure
chloroform at 35oC are 345 torr and 293
torr respectively.
Boiling Point Elevation




a nonvolatile solute lowers the vapor pressure
a higher T is required to reach the 1 atm of
pressure which defines boiling point
a nonvolatile solute elevates the boiling point
of the solvent
the amount of the elevation depends on
concentration of the solute
T  K b  msolute
p.549 GIST: An unknown solute dissolved in water
causes the boiling point to increase by 0.51°C. Does
this mean that the concentration of the solute is 1.0m?
Freezing Point Depression




vapor pressure of solid and liquid are equal at
freezing point
nonvolatile solute lowers the vapor pressure so a
lower T is needed to decrease the vapor pressure
to that of the solid
a nonvolatile solute depresses the freezing
point of the solvent
the amount of the depression depends on
concentration of the solute
T  K f  msolute
Boiling Point Elevation and
Freezing Point Depression
Note that in both
equations, T does not
depend on what the
solute is, but only on
how many particles are
dissolved.
Tb = Kb  m
Tf = Kf  m
Example 1


18.00 g of glucose are added to 150.0 g of
water. The boiling point constant is 0.51
C*kg/mol.
What is the boiling point of the solution?
Example 2



What mass of C2H6O2
(M=62.1 g/mol) needs to be
added to 10.0 L H2O to make
a solution that freezes at 23.3°C?
density is 1.00 g/mL
boiling point constant is
1.86°C*kg/mol
Exercise 13.9


1) Automotive antifreeze consists of ethylene
glycol (C2H6O2), a nonvolative nonelectrolyte.
Calculate the boiling point and freezing point
of a 25.0 mass % solution of ethylene glycol
in water.
2) Calculate the freezing point of a solution
containing 0.600 kg of CHCl3 and 42.0 g of
eucalyptol (C10H18O), a fragrant substance
found in the leaves of eucalyptus trees.
Colligative Properties of
Electrolytes
Since these properties depend on the number of particles
dissolved, solutions of electrolytes (which dissociate in
solution) should show greater changes than those of
nonelectrolytes.
Colligative Properties of
Electrolytes
However, a 1M solution of NaCl does not show
twice the change in freezing point that a 1M
solution of methanol does.
van’t Hoff Factor
One mole of NaCl in
water does not really
give rise to two moles
of ions.
van’t Hoff Factor
Some Na+ and Clreassociate for a short
time, so the true
concentration of particles
is somewhat less than
two times the
concentration of NaCl.
van’t Hoff Factor


Reassociation is more
likely at higher
concentration.
Therefore, the number
of particles present is
concentrationdependent.
van’t Hoff Factor

We modify the previous
equations by
multiplying by the van’t
Hoff factor, i.
Tf = Kf  m  i
Exercise 13.10


1) List the following aqueous solutions in
order of their expected freezing point: 0.050m
CaCl2, 0.15m NaCl, 0.10m HCl, 0.050m
HC2H3O2, and 0.10m C12H22O11.
2) Which of the following solutes will produce
the largest increase in boiling point upon
addition ot 1 kg of water: 1 mol Co(NO3)2, 2
mol KCl, or 3 mol of ethylene glycol
(C2H6O2)?
Osmosis


Some substances form semipermeable
membranes, allowing some smaller
particles to pass through, but blocking
other larger particles.
In biological systems, most
semipermeable membranes allow water to
pass through, but solutes are not free to
do so.
Semi-permeable Membrane


a partition that allows solvent particles to pass
through but not solute particles
separates a solution and a pure solvent
Osmosis


the flow of solvent through a semipermeable membrane
when the system has reached
equilibrium, the water levels are
different since there is greater
hydrostatic pressure in a solution
Osmosis
In osmosis, there is net movement of solvent from
the area of higher solvent concentration (lower
solute concentration) to the are of lower solvent
concentration (higher solute concentration).
Osmotic Pressure
The pressure required to stop osmosis,
known as osmotic pressure, , is
=(
n
)
RT = MRT
V
where M is the molarity of the solution.
If the osmotic pressure is the same on both sides
of a membrane (i.e., the concentrations are the
same), the solutions are isotonic.
Osmosis in Blood Cells

If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.

Water will flow out of the
cell, and crenation
results.
Osmosis in Cells

If the solute
concentration outside
the cell is less than that
inside the cell, the
solution is hypotonic.

Water will flow into the
cell, and hemolysis
results.
Sample Exercise 13.11 Calculations Involving Osmotic Pressure
The average osmotic pressure of blood is 7.7
atm at 25 °C. What molarity of glucose
(C6H12O6) will be isotonic with blood?
Practice
What is the osmotic pressure at 20 °C of a
0.0020 M sucrose (C12H22O11) solution?
 P.533 GIST: Of two KBr solutions, one is
0.5m and the other is 0.20m, which is
hypotonic with respect to the other?
Example 3


When 1.00x10-3 g of a protein is mixed with
water to make 1.00 mL of solution, the
osmotic pressure is 1.12 torr at 25.0°C.
Find the molar mass of this protein.
Sample Exercise 13.12 Molar Mass for Freezing-Point Depression
A solution of an unknown nonvolatile
nonelectrolyte was prepared by
dissolving 0.250 g of the substance in
40.0 g of CCl4. The boiling point of the
resultant solution was 0.357 °C higher
than that of the pure solvent. Calculate
the molar mass of the solute.
Sample Exercise 13.12 Molar Mass for Freezing-Point Depression
Practice Exercise
Camphor (C10H16O) melts at 179.8 °C,
and it has a particularly large freezingpoint-depression constant,
Kf = 40.0 °C/m. When 0.186 g of an
organic substance of unknown molar mass
is dissolved in 22.01 g of liquid camphor,
the freezing point of the mixture is found
to be 176.7 °C. What is the molar mass
of the solute?
Sample Exercise 13.13 Molar Mass from Osmotic Pressure
The osmotic pressure of an aqueous
solution of a certain protein was measured
to determine the protein’s molar mass. The
solution contained 3.50 mg of protein
dissolved in sufficient water to form 5.00
mL of solution. The osmotic pressure of
the solution at 25 °C was found to be
1.54 torr. Treating the protein as a
nonelectrolyte, calculate its molar mass.
Practice

A sample of 2.05 g of polystyrene of uniform
polymer chain length was dissolved in
enough toluene to form 0.100 L of solution.
The osmotic pressure of this solution was
found to be 1.21 kPa at 25ºC. Calculate the
molar mass of polystyrene.
13.6 Colloids
Colloids
Suspensions of particles larger than individual ions
or molecules, but too small to be settled out by
gravity are called colloids.
Tyndall Effect


Colloidal suspensions can
scatter rays of light.
This phenomenon is
known as the Tyndall
effect.
Colloids in Biological Systems
Some molecules have a
polar, hydrophilic
(water-loving) end and a
non-polar, hydrophobic
(water-hating) end.
Colloids in Biological Systems
These molecules can
aid in the emulsification
of fats and oils in
aqueous solutions.
Sample Integrative Exercise Putting Concepts Together
A 0.100-L solution is made by dissolving 0.441 g of CaCl2(s) in
water. (a) Calculate the osmotic pressure of this solution at 27
°C, assuming that it is completely dissociated into its
component ions. (b) The measured osmotic pressure of this
solution is 2.56 atm at 27 °C. Explain why it is less than the
value calculated in (a), and calculate the van’t Hoff factor, i, for
the solute in this solution. (See the “A Closer Look” box on
Colligative Properties of Electrolyte Solutions in Section 13.5.)
(c) The enthalpy of solution for CaCl2 is
ΔH = –81.3 kJ/mol. If the final temperature of the solution is 27
°C, what was its initial temperature? (Assume that the density
of the solution is 1.00 g/mL, that its specific heat is 4.18 J/g-K,
and that the solution loses no heat to its surroundings.)