Ch.11
1
Solutions
. . . the components of a mixture
are uniformly intermingled (the
mixture is homogeneous).
2
Solutions
 Solutions are homogeneous mixtures of two or
more pure substances.
 In a solution, the solute is dispersed uniformly
throughout the solvent.
Solution Terminology
 Solution components
 Solute
 Solvent
 Degree of mixing (concentration)
 Solubility
 Saturated solution
 Supersaturated solution
 Miscible, Immiscible (for two liquids)
How Does a Solution Form?
As a solution forms, the solvent pulls solute
particles apart and surrounds, or solvates,
them.
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
If an ionic salt is
soluble in water, it is
because the iondipole interactions
are strong enough to
overcome the lattice
energy of the salt
crystal.
Dissolution of NaCl in Water
8
Energetics of Solution Formation
 Some salts are soluble in water, others are not.
 NaCl is soluble
 AgCl is insoluble
 A substance is soluble if energy is released in the
dissolution process
The Steps in
the Dissolving
Process
10
Steps in Solution Formation
Step 1 - Expanding the solute (endothermic)
Step 2 - Expanding the solvent (endothermic)
Step 3 - Allowing the solute and solvent to interact to
form a solution (exothermic)
Hsoln = Hstep 1 + Hstep 2 + Hstep 3
Hsoln < 0
Exothermic, favors dissolution
Hsoln > 0
Endothermic, dissolution not favored
11
The Heat of Solution
Exothermic – Dissolution is favored!
Endothermic – Dissolution not favored!
12
Mixing and the Trend Toward
Disorder
 For many soluble salts, Hsoln > 0, suggesting they
should not be soluble.
 The tendency toward disorder (entropy)is a
fundamental driving force in all natural processes.
 In some cases this can make an otherwise
unfavorable process occur spontaneously.
 Example: The dissolution of NaCl is endothermic. The
disorder resulting from dissolution favors solubility.
NaCl(s) + heat  Na+(aq) + Cl-(aq)
Hsoln > 0
Tendency Toward Disorder Favors Solution
Ordered State
Disordered State
14
Student, Beware!
Just because a substance disappears when it
comes in contact with a solvent, it doesn’t mean
the substance dissolved.
Student, Beware!
 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.
Ways of Expressing
Concentrations of
Solutions
Solution Composition
moles of solute
1. Molarity (M) =
liters of solution
mass of solute
 100%
2.Mass (weight) percent =
mass of solution
molesA
3.Mole fraction (A) =
total moles in solution
4.
moles of solute
Molality (m) =
kilograms of solvent
18
Mass Percentage
Mass % of A =
mass of A in solution
 100
total mass of solution
Mole Fraction (X)
XA =
moles of A
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.
 Because volume is temperature dependent,
molarity can change with temperature.
Molality (m)
m=
mol of solute
kg of solvent
Because both moles and mass do not change with
temperature, molality (unlike molarity) is not
temperature dependent.
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
Changing Molarity to Molality
If we know the
density of the
solution, we can
calculate the molality
from the molarity,
and vice versa.
Example
Mole Fraction Calculations.
A solution is prepared by dissolving 50.0 g cesium
chloride in 50.0 g water. The volume of the solution is
63.3 mL. For the CsCl solute, calculate
 Mass percent
 Molarity of CsCl, M
 Molality, m
 Mole fraction, XCsCl
A solution is prepared by dissolving 50.0 g cesium chloride in 50.0 g water. The
volume of the solution is 63.3 mL. For the CsCl solute, calculate
Mass Percent
mass of solute
 Mass (weight) percent =
 100%
mass of solution
50.0g/100.0g x 100=50% mass percent
moles
of
solute
•Molarity of CsCl
liters of solution
50.0g x 1mol = .297mol
168.35g
.297mol/.063 L= 4.71 M
26
A solution is prepared by dissolving 50.0 g cesium chloride in
50.0 g water. The volume of the solution is 63.3 mL. For the
CsCl solute, calculate
molesA
 Mole fraction (A) = total moles in solution
(.297 mol CsCl )
.297 mol + (50.0 g / 18.02) =.097
moles of solute
 Molality (m) = kilograms of solvent
.297 mol CsCl / .05 kg water = 5.94 m
27
Try #25 in your book
 density =
10.0 g H3PO4  100. g H 2O
mass

volume
104 mL
 mol H3PO4 = 10.0 g ×
 mol H2O = 100. g ×
= 1.06 g/mL = 1.06 g/cm3
1 mol
97.99 g
1 mol
18.02 g
 mole fraction of H3PO4 =
= 0.102 mol H3PO4
= 5.55 mol H2O
0.102 mol H3PO4
(0.102  5.55) mol
= 0.0180
28
#25 Cont.
  H20 +  H2PO4= 1.0
  H20 = 1.000 – 0.0180 = 0.9820
 molarity =
0.102 mol H3PO4
0.104 L
= 0.981 mol/L
 molality =
0.102 mol H3PO4
0.100 kg
= 1.02 mol/kg
29
Homework #30
 If there are 100.0 mL of wine:
 12.5 mL C2H5OH ×
 87.5 mL H2O ×1.00 g
9.86
87.5  9.86
.86 g C H OH
 molality9=

2
5
0.0875 kg H 2O
= 9.86 g C2H5OH
= 87.5 g H2O
mL
 mass % ethanol =
0.789 g
mL
1 mol
46.07 g
× 100 = 10.1% by mass
= 2.45 mol/kg
30
•
•
•
Nature of Solute and Solvent
Temperature
Pressure
Types of Solutions
 Saturated
 Solvent holds as much
solute as is possible at
that temperature.
 Dissolved solute is in
equilibrium with solid
solute particles.
Types of Solutions
 Unsaturated
 Less than the maximum
amount of solute for
that temperature is
dissolved in the solvent.
Types of Solutions
 Supersaturated
 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.
Like Dissolves Like
 Polar and ionic solutes
Which is more soluble in water?
dissolve in polar solutes
like water. The more ionic
or polar the stronger the
hydration is due to
effective nuclear charge.
 Methanol in water
 NaCl in water
 Nonpolar solutes dissolve
in nonpolar solvents
 Motor oil in kerosene
35
Nature of Solute and Solvent
1.
Predict the relative solubilities in the following
cases:

(a) Br2 in benzene (C6H6) and in water,

(b) KCl in carbon tetrachloride and in liquid
ammonia,

(c) urea (NH2)2CO in carbon disulfide and in
water.
2. Is iodine (I2) more soluble in water or in carbon
disulfide (CS2)?
36
Homework #40
 For ionic compounds, as the charge of the ions increases
and/or the size of the ions decreases, the attraction to water
(hydration) increases.
a. Mg2+; smaller size, higher charge
b. Be2+; smaller
c. Fe3+; smaller size, higher charge
d. F-; smaller
e. Cl-; smaller
f. SO42-; higher charge
37
Solubilities of Some Solids in Water
Effect of
Temperature on
Solubilities of Solids
For a solid dissolved
in a liquid solvent, in
most cases, the
solubility increases
with increasing
temperature.
38
Effect of Temperature
on the Solubilities of
Gases
Solubilities of some
gases in water
For gases in liquids,
solubility
decreases as
temperature
increases.
39
Henry’s Law
Effect of Pressure on Solubility of a Gas
The amount of a gas dissolved in a solution is directly
proportional to the pressure of the gas above the solution.
C = kP
P=
C=
k=
partial pressure of gaseous solute above the
solution
concentration of dissolved gas
a constant
Limitations to Henry’s Law
• Applies to dilute solutions of gases (limiting law)
• Assumes no chemical interaction of gas with the solvent
• O2 dissolves as molecular O2. Henry’s law applies.
• HCl ionizes to H+ and Cl-. Henry’s law does not apply.
40
Demonstration of Henry’s Law
Cgas = kPgas
a) Initial Condition
Increase in Pgas forces more
gas molecules into the solvent.
b) Increased Pressure
c) Final Condition
41
The Effect of Pressure on the
Solubility of Gases
Flash Animation - Click to Continue
42
Expressing Henry’s Law
 Usual form
C = kP
Units of k
L atm/mol
Always include units with Henry’s Law Constant!
43
Using Henry’s Law
 Calculate the molar concentration of O2 in water at 25°C for a
partial pressure of 0.22 atm. The Henry’s law constant for O2 is
3.5 x 10–4 mol/(L·atm).

C = Pk Answer: 7.7 x 10-5 M
 The solubility of CO2 in water is 3.2 x 10–2 M at 25°C and 1 atm
pressure. What is the Henry’s law constant for CO2 in
mol/(L·atm)?

Answer: 3.2 x 10-2 mol/L atm
44
45
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
Boiling Point Elevation and
Freezing Point Depression
Nonvolatile(non
evaporating) solutesolvent interactions
also cause solutions to
have higher boiling
points and lower
freezing points than
the pure solvent.
Boiling Point Elevation
The change in boiling
point is proportional to
the molality of the
solution:
Tb = Kb  m
Tb is added to the normal
boiling point of the solvent.
where Kb is the molal
boiling point elevation
constant, a property of the
solvent.
#57 A solution is prepared by dissolving 27.0 g of
urea in 150.0 g of water. Calculate the boiling
point of the solution. Urea is a nonelectrolyte
ΔTb = Kbm
mol solute 27.0 g N H CO 1000 g 1 mol N H CO
molality = m = kg solvent  150.0 g H O  kg  60.06 g N H CO = 3.00 molal
ΔTb = Kbm = 0.51 C × 3.00 molal = 1.5°C
2
4
2
2
4
2
4
molal
The boiling point is raised from 100.0°C to 101.5°C
(assuming P = 1 atm).
49
Freezing Point Depression
 The change in freezing
point can be found
similarly:
Tf = Kf  m
 Here Kf is the molal
freezing point depression
constant of the solvent.
Tf is subtracted from the normal
freezing point of the solvent.
59. What mass of glycerin (C3H8O3), a nonelectrolyte,
must be dissolved in 200.0 g water to give a solution
with a freezing point of – 1.50 C given the Kf is
1.86C/molal for glycerin?
 ΔTf = Kfm,
 ΔTf = 1.50°C =
1.86 C
×
molal
m, m = 0.806 mol/kg
0.806 mol C3H8O3 92.09 g C3H8O3

=
kg H 2O
mol C3H8O3
 0.200 kg H2O ×
14.8 g C3H8O3
51
Boiling Point Elevation and
Freezing Point Depression
Note that in both
Tb = Kb  m
equations, T does not
depend on what the
solute is, but only on
how many particles are
dissolved.
Tf = Kf  m
Homework # 62
m=
ΔTf
25.0 C

K f 1.86 C kg / mol
= 13.4 mol C2H6O2/kg
 Since the density of water is 1.00 g/cm3, the moles of
C2H6O2 needed are:
1.00 kg H O 13.4 mol C H O
 15.0 L H2O × L H O  kg H O = 201 mol C2H6O2
2
2
2
 ΔTb =
(H20)
Kbm =
2
2
= 11,200 cm3 =
11.2 L
× 13.4 molal = 6.8°C ,
 Volume C2H6O2 = 201 mol C2H6O2 ×
0.51 C
molal
6
62.07 g
1 cm3

mol C 2 H 6 O 2
1.11 g

 Tb = 100.0°C + 6.8°C = 106.8°C
53
54
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
• 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.
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 1 M solution of NaCl does not show
twice the change in freezing point that a 1 M
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 Cl−
reassociate for a short
time, so the true
concentration of
particles is somewhat
less than two times
the concentration of
NaCl.
The van’t Hoff Factor
 Reassociation is more
likely at higher
concentration.
 Therefore, the number
of particles present is
concentration
dependent.
The van’t Hoff Factor
We modify the
previous equations by
multiplying by the
van’t Hoff factor, i
Tf = Kf  m  i
Homework #70
 The solutions of C12H22O11, NaCl and CaCl2 will all have
lower freezing points, higher boiling points and higher
osmotic pressures than pure water.
 The solution with the largest particle concentration will
have the lowest freezing point, the highest boiling point
and the highest osmotic pressure.
 The CaCl2 solution will have the largest effective particle
concentration because it produces three ions per mol of
compound.
 a. pure water
b.
CaCl2 solution
 c. CaCl2 solution
d.
pure water
 e. CaCl2 solution
64
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.
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 = iMRT
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.
Molar Mass from
Colligative Properties
We can use the effects
of a colligative property
such as osmotic
pressure to determine
the molar mass of a
compound.
Colloids:
Suspensions of particles larger than individual
ions or molecules, but too small to be settled
out by gravity.
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
nonpolar, hydrophobic
(water-hating) end.
Colloids in Biological Systems
Sodium stearate
is one example of
such a molecule.
Colloids in Biological Systems
These molecules can
aid in the
emulsification of fats
and oils in aqueous
solutions.
Homework #82
 π = MRT =
0.1mol
0.08206 L atm
mol K
× 298 K = 2.45 atm
 π = 2.45 atm × 760 mm Hg = 1862mm ≈ 2000 mm ≈ 2 m
atm
 The osmotic pressure would support a mercury
column of ≈ 2 m. The height of a fluid column in a tree
will be higher because Hg is more dense than the fluid
in a tree. If we assume the fluid in a tree is mostly H2O,
then the fluid has a density of 1.0 g/cm3. The density
of Hg is 13.6 g/cm3.
L

 Height of fluid ≈ 2 m × 13.6 = 27.2 m ≈ 30 m
76