Chapter 12: Solutions and Their
Physical Properties
Dr. Chris Kozak
Memorial University of Newfoundland, Canada
Which of the following do you think is the least soluble in water? A. B. C. 1
Mole Fraction and Mole Percent
χ=
Amount of component i (in moles)
Total amount of all components (in moles)
χ1 + χ2 + χ3 + …χn = 1
Mole % i = χi × 100%
Slide 3
Molarity and Molality
Molarity (M) =
Amount of solute (in moles)
Volume of solution (in litres)
Note: this uses the total volume of the resulting solution.
Eg. 10 ml ethanol in 100 ml water gives 110 ml solution.
Molality (m) =
Amount of solute (in moles)
Mass of solvent (in kilograms)
Note: this uses only the mass of the solvent, not the solution.
Eg. 0.20 mol ethanol in 100 g water ignores the mass of ethanol.
Slide 4
2
Intermolecular Forces and Solutions
Compare to Fig 12.6 in Tro
ΔHb
ΔHc
ΔHa
Non-ideal
Ideal
Non-ideal
ΔHsoln = ΔHa + ΔHb + ΔHc
Slide 5
Intermolecular Forces in Mixtures
•  Magnitude of ΔHa, ΔHb, and
ΔHc depend on
intermolecular forces.
•  Ideal solution
–  Forces are similar between all
combinations of components.
ΔHsoln = 0
Slide 6
3
Ideal Solutions
Substances with similar structures may give an ideal solution
Benzene
Toluene
In general, “like dissolves like”
Slide 7
Non-ideal Solutions – Type 1
•  Adhesive forces greater than
cohesive forces.
Cl3CH
OC(CH3)2
•  Formation of these solutions is
exothermic
•  Eg. Chloroform and acetone:
“Hydrogen-bond-like” interaction
stronger than in the pure liquids
•  It’s really a type of dipole-dipole
interaction that is stronger in the
mixture than in the pure solvents
dipole-dipole interactions
ΔHsoln < 0
Slide 8
4
Non-ideal Solutions –Type 2
•  Adhesive forces are less than
cohesive forces.
•  At the limit these solutions are
heterogeneous.
•  Eg. mixtures of polar and nonpolar liquids.
•  Although endothermic, they
can still occur due to:
–  Raising temperature
–  Entropy, ΔS (Chemistry 1051)
(CH3)2CO
CS2
dipole-induced dipole interactions
ΔHsoln > 0
Slide 9
Immiscible mixtures
Slide 10
5
Ionic Solutions
•  Aqueous solutions of ionic
compounds are due to iondipole interactions (like
dissolves like)
•  If strong enough to overcome interionic forces in
crystal, dissolving occurs
•  “Hydration of ions”
•  Hydration is exothermic and
increases with increasing
intermolecular force
•  Ionic solids are typically
insoluble in non-polar
solvents
For ions, ΔHhyd increases with
increasing charge and
decreases with ionic size
Slide 11
Hydration Energy (Enthalpy)
Sample ΔHhyd (kJ/mol): Na+ = -406, K+ = -322, Mg2+ = -1921
Steps involved in hydration:
NaCl(s) → Na+(g) + Cl-(g)
ΔHlattice
>0
Na+(g) + xs H2O(l) → Na+(aq) ΔHhydration < 0
Cl-(g) + xs H2O(l) → Cl-(aq)
ΔHhydration < 0
ΔHsoln = ΔHlattice + ΔHHydNa + ΔHHydCl ≈ + 5 kJ/mol
Why does salt dissolve if it’s endothermic?
Entropy, ΔS, and Free Energy, ΔG (Chemistry 1051)
Slide 12
6
+
+
electrostatic energy ∝
2+
charge A × charge B
distance
7
Solubility Curves
•  Area below curve is
unsaturated
Supersaturated
Unsaturated
•  Area above curve is
supersaturated
•  Solubility is a function
of temperature
•  Can obtain a
supersaturated
solution by cooling a
saturated one
Slide 15
Solubility of Gases
•  Most gases are less
soluble in water as
temperature increases.
•  In organic solvents the
reverse is often true.
•  Solubility curves of Noble
gases can be complex
Slide 16
8
Henry’s Law
•  Solubility of a gas increases
with increasing pressure.
C = k Pgas
C is solubility, k is a constant (varies with gas)
C
=
Pgas
k=
23.54 mL N2/L
1.00 atm
= 23.54 ml N2/L/atm
To increase solubility to 100 mL N2/L, we need a Pgas of:
Pgas =
C
k
=
100 mL N2/L
23.54 ml N2/L/atm
= 4.25 atm
Slide 17
Henry’s Law
Decreasing pressure
causes carbonated water
to fizz
(solubility is decreased)
Increasing pressure
causes more gas to
enter solvent
Slide 18
9
Henry’s Law state that as the pressure of a gas above a solvent is increased the solubility of the gas increases according to, Determine the solubility of oxygen in water assuming the pressure above the soluKon is 0.21 bar. Henry’s law constants are given in the table. Rank Responses 1 2 3 4 5 6 Other Vapor Pressures of Solutions
•  Raoult, 1880s.
–  Dissolved solute lowers vapor pressure of solvent.
–  The partial pressure exerted by solvent vapor
above an ideal solution is the product of the mole
fraction of solvent in the solution and the vapor
pressure of the pure solvent at a given
temperature.
PA = χA P°A
PA = partial pressure above a solution
PoA = vapour pressure of a pure solvent
χA = mole fraction of solvent in solution
Slide 20
10
Example
Predicting vapor pressure of ideal solutions.
The vapor pressures of pure benzene and toluene at 25°C are
95.1 and 28.4 mm Hg, respectively. A solution is prepared in
which the mole fractions of benzene and toluene are both
0.500. What are the partial pressures of the benzene and
toluene above this solution? What is the total vapor pressure?
Balanced Chemical Equation:
Pbenzene = χbenzene P°benzene = (0.500)(96.1 mm Hg) = 47.6 mm Hg
Ptoluene = χtoluene P°toluene = (0.500)(28.4 mm Hg) = 14.2 mm Hg
Ptotal = Pbenzene + Ptoluene = 61.8 mm Hg
Slide 21
Example
Calculating the Composition of Vapor in Equilibrium with
a Liquid Solution.
What is the composition of the vapor in equilibrium with
the benzene-toluene solution?
Partial pressure and mole fraction:
χbenzene = Pbenzene/Ptotal = 47.6 mm Hg/61.89 mm Hg = 0.770
χtoluene = Ptoluene/Ptotal = 14.2 mm Hg/61.89 mm Hg = 0.230
Slide 22
11
Osmotic Pressure
Volatile solvent,
non-volatile solute
Raoult’s Law still applies
Water flows from the low concentration
of solute side to the high side to
reduce the solute concentration
Osmotic pressure - Depends on the
number of solute particles, not the
nature of the solute
Semipermeable membrane
Colligative property (depends on
number of particles present)
Slide 23
Osmotic Pressure
For dilute solutions of electrolytes, osmotic pressure,
π, obeys the following relationship (looks familiar!):
πV = nRT
π=
n
RT = M RT
V
Where M is molarity in mol/L
Slide 24
12
Osmotic Pressure – Practical Applications
Biology – electrolyte (saline) solutions to
treat dehydration.
Red Blood Cells will shrink in too salty a
solution but burst in pure water
Need to balance salinity with that of
blood (isotonic)
Hypertonic > 0.92% m/V
crenation
Isotonic Saline = 0.92% m/V
Hypotonic < 0.92% m/V
rupture
Slide 25
Reverse Osmosis - Desalination
Apply a pressure to the salt
water side B to obtain pure
water on side A
Force water through
membrane but leave salts
behind
Reverse osmosis pumps can
be found in emergency kits on
life boats and small open water
vessels to provide safe
drinking water
Slide 26
13
Two aqueous soluKons of equal volume, one containing 10 g of lacKc acid (C3H6O3) and one containing 10 g of glucose (C6H12O6) are separated by a semi permeable membrane (allowing only H2O to pass). The net flow of water through the membrane is? A. 
B. 
C. 
D. 
10 g C3H6O3 10 g C6H12O6 LeR Right There is no net flow cannot tell without knowing the exact volume of the soluKons Freezing-Point Depression and Boiling
Point Elevation of Nonelectrolyte
Solutions
•  Vapor pressure is lowered when a solute is
present.
–  Boiling point elevation (salt-water boils at > 100oC).
–  Freezing point depression (salt-ice melts at < 0oC).
–  The solution stays liquid over a wider T range.
•  Colligative properties.
–  Depends on the number of particles present.
Slide 28
14
Vapor Pressure Lowering
Phase Diagram
fp0, bp0 are normal freezing
and boiling pts of pure
solvent
fp and bp are observed
freezing and boiling pts of
solutions
m is molality (mol solute
per kg solvent)
Kf (freezing pt depression
constant) depends on mp,
ΔHfus, molar mass of
solvent
ΔTf = -Kf × m
ΔTb = Kb × m
Kb (boiling pt elevation
constant) depends on bp,
ΔHvap, molar mass of
solvent
Slide 29
Solutions of Electrolytes
•  Svante Arrhenius
–  Nobel Prize 1903.
–  Ions form when electrolytes dissolve in solution.
–  Explained anomalous colligative properties
Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq)
ΔTf = -Kf × m = -1.86°C m-1 × 0.0100 m = -0.0186°C
Freezing point depression for NaCl is -0.0361°C.
Why should it be different if the concentration is the same?
Slide 30
15
van’t Hoff Factor
Effect of ions! Ionic solutes (electrolytes) dissociate (more than
one particle per formula unit). Non-electrolytes do not.
NaCl yields two ions in solution, Na+ and Cli=
measured ΔTf
expected ΔTf
=
0.0361°C
0.0186°C
= 1.98
For urea (non-electrolyte) i = 1, for NaCl i = 2, for MgCl2 i = 3
Therefore, previous equations should be changed to:
π = i × M × RT
ΔTf = -i × Kf × m
ΔTb = i × Kb × m
Slide 31
Determine the freezing point of an anKfreeze soluKon prepared by dissolving 250 g of ethylene glycol (C2H6O2, MM=66.1 g mol-­‐1) in 900. g of water. The freezing point depression constant for water is 1.86 oC m-­‐1. ΔTf = Kf × m
A. 
B. 
C. 
D. 
E. 
F. 
7.8 oC -­‐7.8 oC 6.3oC -­‐6.3 oC 6.1oC -­‐6.1 oC 16
Which of the following aqueous soluKons will have the highest boiling point? A. 
B. 
C. 
D. 
3 molal glucose 4 molal ethanol 2.5 molal NaCl 2.5 molal CaCl2 Interionic Attractions - Debye and Hückel
•  Arrhenius theory does not correctly predict the conductivity
of concentrated electrolytes.
•  1923
–  Ions in solution do not
behave independently.
–  Each ion is surrounded by
others of opposite charge.
–  Ion mobility is reduced by
the drag of the ionic
atmosphere (interionic
attractions).
–  i is not exactly 2.00 for NaCl
Slide 34
17
Problems
1. 
Calculate the vapour pressure lowering, ΔP, when 10.0 mL of glycerol (C3H8O3) is
added to 500.0 mL of water at 50.0 oC. At this temperature, the vapour pressure of
pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26
g/mL. (Aside: glycerol is thought to be produced in fish living in cold water
conditions to lower the freezing point of their blood).
2. 
You add 1.00 kg of ethylene glycol (C2H6O2) antifreeze to your car radiator, which
contains 4450 g of water. What are the boiling and freezing points of the solution?
Kb water is 0.512 oC/m, Kf is 1.86 oC/m.
3. 
A chemist dissolves 21.5 mg of hemoglobin protein in water at 5.0 oC to make 1.5
mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar
mass of the hemoglobin?
Slide 35
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