Chemistry: A Molecular Approach, 1st Ed.
Nivaldo Tro
Chapter 12
Solutions
Roy Kennedy
Massachusetts Bay Community College
Wellesley Hills, MA
2008, Prentice Hall
Solution
• homogeneous mixtures
 composition may vary from one sample to another
 appears to be one substance, though really contains
multiple materials
• most homogeneous materials we encounter are
actually solutions
 e.g., air and sea water
• nature has a tendency toward spontaneous mixing
 generally, uniform mixing is more energetically
favorable
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Solutions
• solute is the dissolved substance
•
•
•
seems to “disappear”
“takes on the state” of the solvent
solvent is the substance solute
dissolves in
does not appear to change state
when both solute and solvent have
the same state, the solvent is the
component present in the highest
percentage
solutions in which the solvent is
water are called aqueous solutions
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Seawater
• drinking seawater will dehydrate you and give
you diarrhea
• the cell wall acts as a barrier to solute moving
• the only way for the seawater and the cell
solution to have uniform mixing is for water to
flow out of the cells of your intestine and into
your digestive tract
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Common Types of Solution
Solution Phase
gaseous solutions
liquid solutions
solid solutions
Solute
Phase
gas
gas
liquid
solid
solid
Solvent
Phase
gas
liquid
liquid
liquid
solid
Example
air (mostly N2 & O2)
soda (CO2 in H2O)
vodka (C2H5OH in H2O)
seawater (NaCl in H2O)
brass (Zn in Cu)
• solutions that contain Hg and some other metal are
•
called amalgams
solutions that contain metal solutes and a metal solvent
are called alloys
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Brass
Type
Color
% Cu
% Zn Density
g/cm3
MP
°C
Tensile
Strength
psi
Uses
Gilding
redish
95
5
8.86
1066
50K
pre-83 pennies,
munitions, plaques
Commercial
bronze
90
10
8.80
1043
61K
door knobs,
grillwork
Jewelry
bronze
87.5
12.5
8.78
1035
66K
costume jewelry
Red
golden
85
15
8.75
1027
70K
electrical sockets,
fasteners & eyelets
Low
deep
yellow
80
20
8.67
999
74K
musical instruments,
clock dials
Cartridge
yellow
70
30
8.47
954
76K
car radiator cores
Common
yellow
67
33
8.42
940
70K
lamp fixtures,
bead chain
Muntz metal
yellow
60
40
8.39
904
70K
nuts & bolts,
brazing rods 6
Solubility
• when one substance (solute) dissolves in another
•
(solvent) it is said to be soluble
salt is soluble in water
bromine is soluble in methylene chloride
when one substance does not dissolve in another it is
said to be insoluble
oil is insoluble in water
• the solubility of one substance in another
depends on two factors – nature’s tendency
towards mixing, and the types of
intermolecular attractive forces
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Spontaneous Mixing
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Solubility
• there is usually a limit to the solubility of one
substance in another
gases are always soluble in each other
two liquids that are mutually soluble are said to be
miscible
alcohol and water are miscible
oil and water are immiscible
• the maximum amount of solute that can be dissolved
•
in a given amount of solvent is called the solubility
the solubility of one substance in another varies with
temperature and pressure
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Mixing and the Solution Process
Entropy
• formation of a solution does not necessarily
lower the potential energy of the system
 the difference in attractive forces between atoms of
two separate ideal gases vs. two mixed ideal gases
is negligible
 yet the gases mix spontaneously
• the gases mix because the energy of the
•
•
system is lowered through the release of
entropy
entropy is the measure of energy dispersal
throughout the system
energy has a spontaneous drive to spread out
over as large a volume as it is allowed
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Intermolecular Forces and the Solution Process
Enthalpy of Solution
• energy changes in the formation of most solutions also
•
involve differences in attractive forces between
particles
must overcome solute-solute attractive forces
 endothermic
• must overcome some of the solvent-solvent attractive
forces
 endothermic
• at least some of the energy to do this comes from
making new solute-solvent attractions
 exothermic
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Intermolecular Attractions
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Relative Interactions and Solution Formation
Solute-to-Solvent
Solute-to-Solvent
Solute-to-Solvent
Solute-to-Solute +
>
Solution Forms
Solvent-to-Solvent
Solute-to-Solute +
=
Solution Forms
Solvent-to-Solvent
Solute-to-Solute + Solution May or
<
Solvent-to-Solvent May Not Form
• when the solute-to-solvent attractions are weaker than
the sum of the solute-to-solute and solvent-to-solvent
attractions, the solution will only form if the energy
difference is small enough to be overcome by the
entropy
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Solution Interactions
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Will It Dissolve?
• Chemist’s Rule of Thumb –
Like Dissolves Like
• a chemical will dissolve in a solvent if it has a similar
•
structure to the solvent
when the solvent and solute structures are similar, the
solvent molecules will attract the solute particles at
least as well as the solute particles to each other
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Classifying Solvents
Solvent
Class
Structural
Feature
Water, H2O
polar
O-H
Methyl Alcohol, CH3OH
Ethyl Alcohol, C2H5OH
Acetone, C3H6O
Toluene, C7H8
polar
polar
polar
nonpolar
O-H
O-H
C=O
C-C & C-H
Hexane, C6H14
Diethyl Ether, C4H10O
nonpolar
nonpolar
C-C & C-H
C-C, C-H & C-O,
(nonpolar > polar)
Carbon Tetrachloride
nonpolar
C-Cl, but symmetrical
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Example 12.1a  predict whether the following
vitamin is soluble in fat or water
OH
The 4 OH groups make
the molecule highly
polar and it will also
H-bond to water.
Vitamin C is water
soluble
OH
H 2C
C
H
H
C
O
C
C
O
C
HO
OH
Vitamin C
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Example 12.1b  predict whether the following
vitamin is soluble in fat or water
The 2 C=O groups are
polar, but their
geometric symmetry
suggests their pulls will
cancel and the molecule
will be nonpolar.
O
H
C
C
CH3
HC
C
C
HC
C
CH
C
H
C
O
Vitamin K3 is fat
soluble
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Vitamin K3
18
Energetics of Solution Formation
• overcome attractions between the solute particles –
•
•
•
endothermic
overcome some attractions between solvent molecules
– endothermic
for new attractions between solute particles and solvent
molecules – exothermic
the overall DH depends on the relative sizes of the DH
for these 3 processes
DHsol’n = DHsolute DHsolvent + DHmix
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Solution Process
1. add energy in to overcome solute-solute attractions
3. form new solute-solvent attractions, releasing energy
2. add energy in to overcome some solvent-solvent attractions
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Energetics of Solution Formation
if the total energy cost for
breaking attractions between
particles in the pure solute and
pure solvent is greater
less thanthan
the the
energy released in making the
new attractions between the
solute and solvent, the overall
process will be endothermic
exothermic
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Heats of Hydration
• for aqueous ionic solutions, the energy added to
overcome the attractions between water molecules and
the energy released in forming attractions between the
water molecules and ions is combined into a term
called the heat of hydration
 attractive forces in water = H-bonds
 attractive forces between ion and water = ion-dipole
 DHhydration = heat released when 1 mole of gaseous ions
dissolves in water
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Heat of Hydration
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Ion-Dipole Interactions
• when ions dissolve in water they become
hydrated
• each ion is surrounded by water molecules
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Solution Equilibrium
• the dissolution of a solute in a solvent is an equilibrium
•
•
•
process
initially, when there is no dissolved solute, the only
process possible is dissolution
shortly, solute particles can start to recombine to
reform solute molecules – but the rate of dissolution >>
rate of deposition and the solute continues to dissolve
eventually, the rate of dissolution = the rate of
deposition – the solution is saturated with solute and no
more solute will dissolve
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Solution Equilibrium
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Solubility Limit
• a solution that has the maximum amount of solute
dissolved in it is said to be saturated
 depends on the amount of solvent
 depends on the temperature
 and pressure of gases
• a solution that has less solute than saturation is said to
•
be unsaturated
a solution that has more solute than saturation is said to
be supersaturated
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How Can You Make a Solvent Hold
More Solute Than It Is Able To?
• solutions can be made saturated at non-room conditions
•
•
– then allowed to come to room conditions slowly
for some solutes, instead of coming out of solution
when the conditions change, they get stuck in-between
the solvent molecules and the solution becomes
supersaturated
supersaturated solutions are unstable and lose all the
solute above saturation when disturbed
 e.g., shaking a carbonated beverage
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Adding Solute to a Supersaturated
Solution of NaC2H3O2
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Temperature Dependence of Solubility
of Solids in Water
• solubility is generally given in grams of solute that will
•
dissolve in 100 g of water
for most solids, the solubility of the solid increases as
the temperature increases
 when DHsolution is endothermic
• solubility curves can be used to predict whether a
solution with a particular amount of solute dissolved in
water is saturated (on the line), unsaturated (below the
line), or supersaturated (above the line)
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Solubility Curves
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Temp KCl NaCl NH4Cl Li2SO4 Ca(OH)2 Ce2(SO4)3 KNO3
•9H2O
0
10
20
30
40
50
60
70
80
90
100
DH
sol’n
27.6
31
34
37
40
42.6
45.5
48.3
51.1
54
56.7
35.7
35.8
36
36.3
36.6
37
37.3
37.8
38.4
39
39.8
29.4
33.3
37.2
41.1
45.8
50.4
55.2
60.2
65.6
71.3
77.3
35.3
354
34.2
33.5
32.7
32.5
31.9
29.9
0.185
0.176
0.165
0.153
0.141
0.128
0.116
0.106
0.094
0.085
0.077
984
222
844
-1720
-927
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30.7
21.4
9.84
7.24
5.63
13.9
21.2
31.6
45.3
61.3
3.87
106
167
203
245
32
Solubility of Some Salts in Water
120
110
Solubility, g salt in 100 g water
100
90
80
KCl
70
NaCl
60
NH4Cl
50
Li2SO4
40
Ce2(SO4)3•9H2O
30
KNO3
20
10
0
0
10
20
30
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50
60
70
Temperature, °C
80
90
100
110
33
Temperature Dependence of Solubility
of Gases in Water
• solubility is generally given in moles of solute
that will dissolve in 1 Liter of solution
• generally lower solubility than ionic or polar
covalent solids because most are nonpolar
molecules
• for all gases, the solubility of the gas decreases
as the temperature increases
the DHsolution is exothermic because you do not need
to overcome solute-solute attractions
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Temp
0
10
20
30
40
50
60
70
80
90
N2
0.00294
0.00231
0.00190
0.00162
0.00139
0.00122
0.00105
0.00085
0.00066
0.00038
O2
0.00695
0.00537
0.00434
0.00359
0.00308
0.00266
0.00227
0.00186
0.00138
0.00079
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Ar
0.00025
9.95 x 10
-5
CO2 acetylene
0.335
0.2
0.232
0.15
0.169
0.117
0.126
0.094
0.0973
0.0761
0.0576
NH3
89.5
68.4
52.9
41.0
31.6
23.5
16.8
11.1
6.5
3.0
35
Solubility of Gases in Water at Various
Temperatures
0.4
Solubility, g
in 100 g water
0.35
0.3
0.25
CO2
acetylene
0.2
0.15
0.1
0.05
0
0
10
20
30
40
50
60
70
80
90
100 110
Temperature, °C
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Solubility of Gases in Water at Various Temperatures
Solubility, g
in 100 g water
0.008
0.007
0.006
0.005
N2
O2
0.004
0.003
0.002
0.001
0
0
10
20
30
40
50
60
70
80
90
100
110
Temperature, °C
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Pressure Dependence of Solubility of
Gases in Water
• the larger the partial pressure of a gas in contact
with a liquid, the more soluble the gas is in the
liquid
38
Henry’s Law
• the solubility of a gas (Sgas)
is directly proportional to its
partial pressure, (Pgas)
Sgas = kHPgas
• kH is called Henry’s Law
Constant
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Relationship between Partial Pressure
and Solubility of a Gas
Partial Pressure
100
200
300
400
500
600
700
800
900
1000
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O2
CO2
0.000571
0.001142
0.001713
0.002284
0.002855
0.003426
0.003997
0.004568
0.005139
0.005711
0.022237
0.044474
0.066711
0.088947
0.111184
0.133421
0.155658
0.177895
0.200132
0.222368
40
Solubility of Gases in Water at Various
Pressures (temp = 20°C)
Solubility, g in 100 g water
0.25
0.2
0.15
O2
CO2
0.1
0.05
0
0
200
400
600
800
1000
1200
Partial Pressure
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persrst
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Ex 12.2 – What pressure of CO2 is required to
keep the [CO2] = 0.12 M at 25°C?
Given: S = [CO2] = 0.12 M,
Find: P of CO2, atm
Concept Plan:
[CO2]
S
P
kH
P
Relationships: S = kHP, kH = 3.4 x 10-2 M/atm
Solve:
S
0.12 M
P

 3.5 atm
2
-1
kH 3.1410 M  atm
Check: the unit is correct, the pressure higher than 1 atm
meets our expectation from general experience
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Concentrations
• solutions have variable composition
• to describe a solution, need to describe components
•
•
and relative amounts
the terms dilute and concentrated can be used as
qualitative descriptions of the amount of solute in
solution
concentration = amount of solute in a given amount of
solution
 occasionally amount of solvent
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Solution Concentration
Molarity
• moles of solute per 1 liter of solution
• used because it describes how many
molecules of solute in each liter of solution
• if a sugar solution concentration is 2.0 M,
1 liter of solution contains 2.0 moles of
sugar, 2 liters = 4.0 moles sugar, 0.5 liters =
1.0 mole sugar
moles of solute
molarity =
liters of solution
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Molarity and Dissociation
• the molarity of the ionic compound allows you to
•
•
•
•
determine the molarity of the dissolved ions
CaCl2(aq) = Ca+2(aq) + 2 Cl-1(aq)
A 1.0 M CaCl2(aq) solution contains 1.0 moles of
CaCl2 in each liter of solution
1 L = 1.0 moles CaCl2, 2 L = 2.0 moles CaCl2
Because each CaCl2 dissociates to give one Ca+2 =
1.0 M Ca+2
1 L = 1.0 moles Ca+2, 2 L = 2.0 moles Ca+2
Because each CaCl2 dissociates to give 2 Cl-1 =
2.0 M Cl-1
1 L = 2.0 moles Cl-1, 2 L = 4.0 moles Cl-1
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Solution Concentration
Molality, m
• moles of solute per 1 kilogram of solvent
defined in terms of amount of solvent, not solution
like the others
• does not vary with temperature
because based on masses, not volumes
moles of solute
molality, m 
kg of solvent
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Percent
• parts of solute in every 100 parts solution
• mass percent = mass of solute in 100 parts
solution by mass
if a solution is 0.9% by mass, then there are 0.9
grams of solute in every 100 grams of solution
or 0.9 kg solute in every 100 kg solution
Mass of Solute, g
Mass Percent 
100%
Mass of Solution, g
Mass of Solute  Mass of Solvent  Mass of Solution
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Percent Concentration
Part (solute)
Percent 
100%
Whole (solution)
Mass of Solute, g
Mass Percent 
100%
Mass of Solution, g
Mass of Solute  Mass of Solvent  Mass of Solution
Mass of Solute, g
Percent Mass/Volum e 
100%
Volume of Solution, mL
Mass of Solute  Volume of Solvent  Volume of Solution
Volume of Solute, mL
Volume Percent 
100%
Volume of Solution, mL
Volume of Solute  Volume of Solvent  Volume of Solution
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Using Concentrations as
Conversion Factors
• concentrations show the relationship between
the amount of solute and the amount of solvent
 12%(m/m) sugar(aq) means 12 g sugar  100 g solution
 or 12 kg sugar  100 kg solution; or 12 lbs.  100 lbs. solution
 5.5%(m/v) Ag in Hg means 5.5 g Ag  100 mL solution
 22%(v/v) alcohol(aq) means 22 mL EtOH  100 mL solution
• The concentration can then be used to convert the
amount of solute into the amount of solution, or vice
versa
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Preparing a Solution
• need to know amount of solution and
concentration of solution
• calculate the mass of solute needed
start with amount of solution
use concentration as a conversion factor
5% by mass 5 g solute  100 g solution
“Dissolve the grams of solute in enough solvent to
total the total amount of solution.”
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Example - How would you prepare 250.0 g of
5.00% by mass glucose solution (normal glucose)?
Given:
Find:
Equivalence:
Solution Map:
250.0 g solution
g Glucose
5.00 g Glucose  100 g solution
g solution
g Glucose
5.00 g Glucose
100 g solution
Apply Solution Map:
5.00 g glucose
250.0 g solution 
 12.5 g glucose
100 g solution
Answer:
Dissolve 12.5 g of glucose in enough water to total 250.0 g
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Solution Concentration
PPM
• grams of solute per 1,000,000 g of solution
• mg of solute per 1 kg of solution
• 1 liter of water = 1 kg of water
for water solutions we often approximate the kg of
the solution as the kg or L of water
grams solute x 106
grams solution
mg solute
mg solute
kg solution
L solution
53
Ex 12.3 – What volume of 10.5% by mass soda
contains 78.5 g of sugar?
Given: 78.5 g sugar
Find: volume, mL
Concept Plan: g solute
100 g sol' n
10.5g sugar
g sol’n
1 mL sol' n
1.04g sol' n
mL sol’n
Relationships: 100 g sol’n = 10.5 g sugar, 1 mL sol’n = 1.04 g
Solve:
100g
1 mL
78.5 g sugar 

 719 mL
78.5 g sugar 1.04g
Check: the unit is correct, the magnitude seems reasonable as
the mass of sugar  10% the volume of solution
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Solution Concentrations
Mole Fraction, XA
• the mole fraction is the fraction of the moles of one
•
•
•
component in the total moles of all the components
of the solution
total of all the mole fractions in a solution = 1
unitless
the mole percentage is the percentage of the moles
of one component in the total moles of all the
components of the solution
 = mole fraction x 100%
mole fraction of A = XA =
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moles of components A
total moles in the solution
55
Ex 12.4a – What is the molarity of a solution prepared
by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to
make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: M
Concept Plan: g C2H6O2
mol C2H6O2
mol
M
M
L
mL sol’n
L sol’n
Relationships: M = mol/L, 1 mol C2H6O2 = 62.07 g, 1 mL = 0.001 L
C 2 H 6O 2
Solve:
0.27171mol
mol
C H O
17.2 g C2 H 6O2 
M
2
6  20.2771 mol
62.07g C2 H 6O 2
0.515L
0.001 L
mL  M
 0.515 L
M 5150.538
1 mL
Check: the unit is correct, the magnitude is reasonable
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Ex 12.4b – What is the molality of a solution prepared
by mixing 17.2 g of C2H6O2 with 0.500 kg of H2O to
make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: m
Concept Plan: g C2H6O2
mol C2H6O2
mol
m
m
kg
kg H O
2
Relationships: m = mol/kg, 1 mol C2H6O2 = 62.07 g
C2 H
Solve:
0.2717mol
1 mol
C6OH2 O
17.2 g Cm
2 H
6O 2 
2
6  20.2771 mol
62.07g C2 H 6O 2
0.500kg H 2O
m  0.554M
Check: the unit is correct, the magnitude is reasonable
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Ex 12.4c – What is the percent by mass of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of
H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: %(m/m)
Concept Plan:
g C2H6O2 %  g solute100%
g sol'n
g solvent
g sol’n
%
Relationships: 1 kg = 1000 g
Solve:
H 2O
17.2g C1000
H gO
0.500kg H 2O  2 6 2  5.00102 g H 2O
%
1 kg H 2O  100%
517.2 g sol' n
g C2H6O2  500 g H2O  517.2 g sol'n
%17.23.33%
Check: the unit is correct, the magnitude is reasonable
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Ex 12.4d – What is the mole fraction of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of
H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: C
Concept Plan: g C2H6O2
mol C2H6O2   molA
moltotal
g H2O
mol H2O
C
Relationships: C = molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
1 mol
Solve:
0.27
71Cmol
C22H6O2
2 H 6O
17.2 g C2H6O2  62.07g C2H6O2
 0.2771 mol
0.2771 mol
C2 H6O2  27.75 mol H 2O
1000g 1 mol H O
0.500kg H 2O-
3
  9.8910
1 kg

2
18.02g H 2O
 27.75 mol H 2O
Check: the unit is correct, the magnitude is reasonable
Tro, Chemistry: A Molecular Approach
59
Ex 12.4d – What is the mole percent of a solution
prepared by mixing 17.2 g of C2H6O2 with 0.500 kg of
H2O to make 515 mL of solution?
Given: 17.2 g C2H6O2, 0.500 kg H2O, 515 mL sol’n
Find: C
Concept Plan: g C2H6O2
mol C2H6O2  %  molA 100%
moltotal
g H2O
mol H2O
C%
Relationships: C = molA/moltot, 1 mol C2H6O2 = 62.07 g, 1 mol H2O = 18.02 g
1 mol
Solve:
2 H 6CO22H 6O 2
0.27
71Cmol
17
.
2
g
C
H
O

 %  2 6 2 62.07g C H O  0.2771 mol 100%
0.2771 molC2H6O22 6 272 .75 molH 2O
1000g 1 mol H 2O
.500
kg H 2%
O

 27.75 mol H 2O
0%
 0.989
1 kg
18.02g H 2O
Check: the unit is correct, the magnitude is reasonable
Tro, Chemistry: A Molecular Approach
60
Converting Concentration Units
• assume a convenient amount of solution






given %(m/m), assume 100 g solution
given %(m/v), assume 100 mL solution
given ppm, assume 1,000,000 g solution
given M, assume 1 liter of solution
given m, assume 1 kg of solvent
given X, assume you have a total of 1 mole of solutes in the solution
• determine amount of solution in non-given unit(s)
 if assume amount of solution in grams, use density to convert to mL and
then to L
 if assume amount of solution in L or mL, use density to convert to grams
• determine the amount of solute in this amount of solution, in
grams and moles
• determine the amount of solvent in this amount of solution, in
grams and moles
• use definitions to calculate other units
Tro, Chemistry: A Molecular Approach
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62
Vapor Pressure of Solutions
• the vapor pressure of a solvent above a
solution is lower than the vapor pressure of
the pure solvent
the solute particles replace some of the solvent
molecules at the surface
Eventually,
reAddition
of aequilibrium
nonvolatileissolute
The pure solvent
establishes
an
established,
smaller
number
reduces thebut
ratea of
vaporization,
liquidmolecules
 vapor equilibrium
ofdecreasing
vapor
– therefore
the
the amount
of vapor
vapor pressure will be lower
Tro, Chemistry: A Molecular Approach
63
Thirsty Solutions
• a concentrated solution will draw solvent
molecules toward it due to the natural drive for
materials in nature to mix
• similarly, a concentrated solution will draw pure
solvent vapor into it due to this tendency to mix
• the result is reduction in vapor pressure
Tro, Chemistry: A Molecular Approach
64
Thirsty Solutions
When equilibrium
Beakers
with equalis
liquid levelsthe
established,
of pure
liquid
solvent
level
in and
the solution
a solution
are place
beaker
is in
higher
a bellthan
jar.
Solvent
the
solution
molecules
level in the
evaporate
pure
solvent
from
beaker
each–
onethirsty
the
and fillsolution
the bell jar,
establishing
grabs
and holds
an solvent
equilibrium
vapor
more effectively
with the
liquids in the beakers.
Tro, Chemistry: A Molecular Approach
65
Raoult’s Law
• the vapor pressure of a volatile solvent above a
solution is equal to its mole fraction of its
normal vapor pressure, P°
Psolvent in solution = solvent∙P°
since the mole fraction is always less than 1, the
vapor pressure of the solvent in solution will always
be less than the vapor pressure of the pure solvent
Tro, Chemistry: A Molecular Approach
66
Ex 12.5 – Calculate the vapor pressure of water in a
solution prepared by mixing 99.5 g of C12H22O11 with
300.0 mL of H2O
Given:
Find:
99.5 g C12H22O11, 300.0 mL H2O
PH2O
Concept Plan: g C12H22O11
mL H2O
mol C12H22O11
g H2 O
mol H2O

molA
moltotal
P    P
CH2O
PH2O
Relationships: P°H2O = 23.8 torr, 1 mol C12H22O11 = 342.30 g, 1 mol H2O = 18.02 g
Solve:
1 molC12HH
22O
11
2O
99.5g C12H 22O11  16.65 mol
 0.2907 mol
C.98
12 H
22O11

0
2
8
gC
H22
O.11
0.2907 mol342.30
C12H 22
O12
16
65 mol H 2O
11
1.00g 1 molH 2O
 16.65 molH 2O

PH2O   H2O 1 mL
P  H2O18.02
 0.g98
2
8
H 2O 23.8 torr
300.0mL H 2O 
PH2O  23.4 torr
Tro, Chemistry: A Molecular Approach
67
Ionic Solutes and Vapor Pressure
• according to Raoult’s Law, the effect of solute on
•
•
the vapor pressure simply depends on the number
of solute particles
when ionic compounds dissolve in water, they
dissociate – so the number of solute particles is a
multiple of the number of moles of formula units
the effect of ionic compounds on the vapor
pressure of water is magnified by the dissociation
 since NaCl dissociates into 2 ions, Na+ and Cl, one
mole of NaCl lowers the vapor pressure of water twice
as much as 1 mole of C12H22O11 molecules would
Tro, Chemistry: A Molecular Approach
68
Effect of Dissociation
Tro, Chemistry: A Molecular Approach
69
Ex 12.6 – What is the vapor pressure of H2O when 0.102
mol Ca(NO3)2 is mixed with 0.927 mol H2O @ 55°C?
Given: 0.102 mol Ca(NO3)2, 0.927 mol H2O, P° = 118.1 torr
Find: P of H2O, torr
Concept Plan:
PH2O
H2O, P°H2O
P    P
Relationships: P = ∙P°
Solve: Ca(NO3)2  Ca2+ + 2 NO3  3(0.102 mol solute)
927 mol H O
0
.
P



P
H 2O  0.27518118.1torr
H 2O
 H 2O

H 2O
30.102molsolute  0.927 molH2O
PH 2O  88.8 torr
 H2O  0.7518
Check: the unit is correct, the pressure lower than the
normal vapor pressure makes sense
Tro, Chemistry: A Molecular Approach
70
Raoult’s Law for Volatile Solute
• when both the solvent and the solute can evaporate,
•
both molecules will be found in the vapor phase
the total vapor pressure above the solution will be the
sum of the vapor pressures of the solute and solvent
 for an ideal solution
Ptotal = Psolute + Psolvent
• the solvent decreases the solute vapor pressure in the
same way the solute decreased the solvent’s
Psolute = solute∙P°solute and Psolvent = solvent∙P°solvent
Tro, Chemistry: A Molecular Approach
71
Ideal vs. Nonideal Solution
• in ideal solutions, the made solute-solvent
interactions are equal to the sum of the broken
solute-solute and solvent-solvent interactions
ideal solutions follow Raoult’s Law
• effectively, the solute is diluting the solvent
• if the solute-solvent interactions are stronger or
weaker than the broken interactions the solution
is nonideal
Tro, Chemistry: A Molecular Approach
72
Vapor Pressure of a
Nonideal Solution
• when the solute-solvent interactions are stronger than
the solute-solute + solvent-solvent, the total vapor
pressure of the solution will be less than predicted by
Raoult’s Law
 because the vapor pressures of the solute and solvent are
lower than ideal
• when the solute-solvent interactions are weaker than
the solute-solute + solvent-solvent, the total vapor
pressure of the solution will be larger than predicted by
Raoult’s Law
Tro, Chemistry: A Molecular Approach
73
Deviations from Raoult’s Law
Tro, Chemistry: A Molecular Approach
74
Tro, Chemistry: A Molecular Approach
75
Freezing Point Depression
• the freezing point of a solution is lower than the freezing
point of the pure solvent
 for a nonvolatile solute
 therefore the melting point of the solid solution is lower
• the difference between the freezing point of the solution
and freezing point of the pure solvent is directly
proportional to the molal concentration of solute particles
FPsolvent – FPsolution) = DTf = m∙Kf
• the proportionality constant is called the Freezing Point
Depression Constant, Kf
 the value of Kf depends on the solvent
 the units of Kf are °C/m
Tro, Chemistry: A Molecular Approach
76
Kf
Tro, Chemistry: A Molecular Approach
77
Ex 12.8 – What is the freezing point of a 1.7 m aqueous
ethylene glycol solution, C2H6O2?
Given: 1.7 m C2H6O2(aq)
Find: Tf, °C
Concept Plan:
m
DTf
DT f  m  K f
Relationships: DTf = m ∙Kf, Kf for H2O = 1.86 °C/m, FPH2O = 0.00°C
Solve: DT f  m  K f ,H 2O
FPH O  FPsol'n  DT f

 1.7 m  1.86
DT f  3.2 C
C
m

2
0.00C  FPsol'n  3.2C
FPsol'n  3.2C
Check: the unit is correct, the freezing point lower than
the normal freezing point makes sense
Tro, Chemistry: A Molecular Approach
78
Boiling Point Elevation
• the boiling point of a solution is higher than the boiling
point of the pure solvent
 for a nonvolatile solute
• the difference between the boiling point of the solution
•
and boiling point of the pure solvent is directly
proportional to the molal concentration of solute
particles
BPsolution – BPsolvent) = DTb = m∙Kb
the proportionality constant is called the Boiling Point
Elevation Constant, Kb
 the value of Kb depends on the solvent
 the units of Kb are °C/m
Tro, Chemistry: A Molecular Approach
79
Ex 12.9 – How many g of ethylene glycol, C2H6O2, must be
added to 1.0 kg H2O to give a solution that boils at 105°C?
Given: 1.0 kg H2O, Tb = 105°C
Find: mass C2H6O2, g
Concept Plan:
DTb
DTb  m  Kb
kg H2O
mol C2H6O2
molsolute
m
kg solvent
m
62.07 g
1 mol
g C2H6O2
Relationships: DTb = m ∙Kb, Kb H2O = 0.512 °C/m, BPH2O = 100.0°C
MMC2H6O2 = 62.07 g/mol, 1 kg = 1000 g
Solve:
C


105



.
0

100
.
0

C

m
0.512
DT  m  K
m
b
b
m  9.77 m
9.77molC2 H6O2 62.07g C2 H 6O2
1.0 kg H 2O 

 6.1102 g C2 H6O2
1.0 kg H 2O
1 molC2 H 6O2
Tro, Chemistry: A Molecular Approach
80
Osmosis
• osmosis is the flow of solvent through a semipermeable membrane from solution of low
concentration to solution of high concentration
• the amount of pressure needed to keep osmotic
flow from taking place is called the osmotic
pressure
• the osmotic pressure, P, is directly proportional
to the molarity of the solute particles
R = 0.08206 (atm∙L)/(mol∙K)
P = MRT
Tro, Chemistry: A Molecular Approach
81
Tro, Chemistry: A Molecular Approach
82
Ex 12.10 – What is the molar mass of a protein if 5.87 mg
per 10 mL gives an osmotic pressure of 2.45 torr at 25°C?
Given: 5.87 mg/10 mL, P = 2.45 torr, T = 25°C
Find: molar mass, g/mol
Concept Plan:
M
P,T
P  MRT
mL
0.001 L
1 mL
L
mol protein
mol  M  L
Relationships: PMRT, T(K)=T(°C)+273.15, R= 0.08206atm∙L/mol∙K
Solve:
M = mol/L, 1 mL = 0.001 L, MM = g/mol, 1 atm = 760 torr
2.45 torr
atmL
T (K) T (C) 1273.15
L
1.318 mol protein M 0.08206
298K 
6
mol
K protein
10.0 mL 
 760 torr/atm

1
.
3
1
8

10
mol
 25  273.15
 298
0.001
mLK
1 L P  MRT
M  1.318104 M

MM 
5.87103 g
1.318106 mol
Tro, Chemistry: A Molecular Approach

 4.45103 g/mol
83
Colligative Properties
• colligative properties are properties whose value
depends only on the number of solute particles, and not
on what they are
 Vapor Pressure Depression, Freezing Point Depression,
Boiling Point Elevation, Osmotic Pressure
• the van’t Hoff factor, i, is the ratio of moles of solute
•
particles to moles of formula units dissolved
measured van’t Hoff factors are often lower than you
might expect due to ion pairing in solution
Tro, Chemistry: A Molecular Approach
84
Tro, Chemistry: A Molecular Approach
85
AAn
hyperosmotic
hyposmotic
isosmoticsolution
solution
solution
has
has
has
athe
lower
a higher
same
osmotic
osmotic
pressure
pressure
than
as
than
the
the
solution
solution
solution
inside
inside
inside
the
thethe
cell
cell
cell
– –asas
–aas
result
a result
a result
there
there
there
is isa is
net
noa net
flow
netthe
flow
offlow
water
of
water
into
of water
the
outcell,
of
into
the
causing
orcell,
out causing
itoftothe
swell
cell.
it to shrivel
Tro, Chemistry: A Molecular Approach
86
Colloids
• a colloidal suspension is a heterogeneous
mixture in which one substance is dispersed
through another
most colloids are made of finely divided particles
suspended in a medium
• the difference between colloids and regular
suspensions is generally particle size – colloidal
particles are from 1 to 100 nm in size
Tro, Chemistry: A Molecular Approach
87
Properties of Colloids
• the particles in a colloid exhibit Brownian
motion
• colloids exhibit the Tyndall Effect
scattering of light as it passes through a suspension
colloids scatter short wavelength (blue) light more
effectively than long wavelength (red) light
Tro, Chemistry: A Molecular Approach
88
Tro, Chemistry: A Molecular Approach
89
Tro, Chemistry: A Molecular Approach
90
Soap and Micelles
• soap molecules have both a hydrophilic
(polar/ionic) “head” and a hydrophobic
(nonpolar) “tail”
• part of the molecule is attracted to water, but the
majority of it isn’t
• the nonpolar tail tends to coagulate together to
form a spherical structure called a micelle
Tro, Chemistry: A Molecular Approach
91
Tro, Chemistry: A Molecular Approach
92
Soap Action
• most dirt and grease is made of nonpolar
molecules – making it hard for water to remove
it from the surface
• soap molecules form micelles around the small
oil particles with the polar/ionic heads pointing
out
• this allows the micelle to be attracted to water
and stay suspended
Tro, Chemistry: A Molecular Approach
93
The polar heads of the micelles attract them to
the water, and simultaneously repel other
micelles so they will not coalesce and settle out.
Tro, Chemistry: A Molecular Approach
94
Tro, Chemistry: A Molecular Approach
95