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Chapter 11: Properties of Solutions Their Concentrations and Colligative Properties
Chapter Outline
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11.1 Energy Changes when Substances Dissolve
11.2 Vapor Pressure
11.3 Mixtures of Volatile Substances
11.4 Colligative Properties of Solutions
11.5 Osmosis and Osmotic Pressure
11.6 Using Colligative Properties to Determine Molar Mass
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Enthalpy of Solution
 Dissolution of Ionic Solids:
• Enthalpy of solution (∆Hsoln) depends on:
» Energies holding solute ions in crystal lattice
» Attractive force holding solvent molecules together
» Interactions between solute ions and solvent
molecules
• ∆Hsoln = ∆Hion-ion + ∆Hdipole-dipole + ∆H ion-dipole
• When solvent is water:
» ∆Hsoln = ∆Hion-ion + ∆Hhydration
Enthalpy of Solution for NaCl
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Calculating Lattice Energies Using the Born-Haber Cycle
Lattice energy (U) is the energy required to completely separate one mole of
a solid ionic compound into gaseous ions. It is always endothermic.
e.g. MgF2(s)  Mg2+(g) + 2F-(g)
𝐸𝑒𝑙 = 2.31 𝑥 10−19 𝐽 ∙ 𝑛𝑚
𝑄1 𝑥𝑄2
𝑑
Q1 is the charge on the cation
Lattice energy (E) increases as Q
increases and/or as r decreases.
Compound
Lattice
Energy (U)
MgF2
2957
Q= +2,-1
MgO
3938
Q= +2,-2
LiF
LiCl
1036
853
radius F <
radius Cl
Q2 is the charge on the anion
d is the distance between the ions
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Born-Haber Cycles – Calculating Ulattice
Hess’ Law is used to calculate Ulattice
The calculation is broken down into a series of steps (cycles)
Steps:
1. sublimation of 1 mole of the metal
= ΔHsub
2. if present, breaking bond of a diatomic
= ½ ΔHBE
3. ionization of the metal(g) atom
= IE1 + IE2 etc
4. electron affinity of the nonmetal atom
= EA1 + EA2 etc
5. formation of 1 mole of the salt from ions(g) = ΔHlast = -Ulattice
In General for M(s) + ½ X2(g)  MX(s)
M+(g) +
M+(g) +
X(g)
X(g)
EA1 + EA2 etc
IE1 + IE2 etc
M+(g) +
M(g) +
X(g)
M(g) +
½ X2(g)
M(s) +
½ X2(g)
X-(g)
½ ΔHBE
ΔHlast = -Ulattice
ΔHsub
ΔHf
MX(s)
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Calculating Ulattice
ΔHf° = ΔHsub + ½ ΔHBE +  IE +  EA + ΔHlast
Rearranging and solving for ΔHlast -
ΔHlast = ΔHf − ΔHsub − ½ ΔHBE −  IE −  EA
Ulattice = -ΔHlast
Calculating Ulattice for NaCl(s)
Na+(g) +
Na+(g) +
Cl(g)
Cl(g)
EA1
IE1
Na+(g) +
Na(g) +
Cl-(g)
Cl(g)
½ ΔHBE
Na(g) +
½ Cl2(g)
Na(s) +
½ Cl2(g)
ΔHlast = -Ulattice
ΔHsub
ΔHf
NaCl(s)
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Calculating Ulattice for NaCl(s)
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ΔHlast = ΔHf − ΔHsub − ½ ΔHBE − IE1 − EA1
Sample Exercise 11.1
Calculating Lattice Energy
Calcium fluoride occurs in nature as the mineral fluorite, which is the
principle source of the world’s supply of fluorine. Use the following data to
calculate the lattice energy of CaF2.
ΔHsub,Ca = 168 kJ/mol
BEF2 = 155 kJ/mol
EAF = -328 kJ/mol
IE1,Ca = 590 kJ/mol
IE2,Ca = 1145 kJ/mol
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Lattice Energies and Solubility
𝐸𝑒𝑙 = 2.31 𝑥 10−19 𝐽 ∙ 𝑛𝑚
𝑄1 𝑥𝑄2
𝑑
Q1 is the charge on the cation
Q2 is the charge on the anion
d is the distance between the ions
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Chapter Outline
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
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11.1 Energy Changes when Substances Dissolve
11.2 Vapor Pressure
11.3 Mixtures of Volatile Substances
11.4 Colligative Properties of Solutions
11.5 Osmosis and Osmotic Pressure
11.6 Using Colligative Properties to Determine Molar Mass
Vapor Pressure
Pressure exerted by a gas in equilibrium with its liquid
Where are we going with this?
Colligative Properties – boiling
point elevation, freezing point
depression, osmosis
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Vapor Pressure of Solutions
• Vapor Pressure:
– Pressure exerted by a gas
in equilibrium with liquid
– Rates of
evaporation/condensation
are equal
The tendency for a liquid to evaporate
increases as 1. the temperature rises
2. the surface area
increases
3. the intermolecular
forces decrease
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Boiling Point
“If you have a beaker of water open to the
atmosphere, the mass of the atmosphere is
pressing down on the surface. As heat is
added, more and more water evaporates,
pushing the molecules of the atmosphere
aside (wsystem < 0). If enough heat is added,
a temperature is eventually reached at which
the vapor pressure of the liquid equals the
atmospheric pressure, and the liquid boils.”
Normal boiling point
the temperature at which the vapor pressure of a liquid is equal to the
external atmospheric pressure of 1 atm.
Increasing the external atmospheric pressure increases the boiling point
Decreasing the external atmospheric pressure decreases the boiling point
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Location
San Francisco
Salt Lake City
Denver
Mt. Everest
Elevation (ft)
b.p H2O (oC)
sea level
4400
5280
29,028
100.0
95.6
95.0
76.5
Intermolecular Forces and Vapor Pressure
Diethyl ether - dipole-dipole
interactions
Water - hydrogen
bonding (stronger)
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Clausius-Clapeyron Equation
Vapor Pressure vs Temperature
Pressure 
ln Pvap   
H vap  1 
   C
R
T 
liquid
solid
gas
Temperature 
Clausius-Clapeyron Equation
Vapor Pressure vs Temperature
Plot of ln(P) vs 1/T yields
straight line:
• Slope = −ΔHvap/R
• Intercept = constant
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Sample Exercise 11.2:
Calculating Vapor Pressure Using Hvap
The compound with the common name isooctane has the structure
shown below. Its normal boiling point is 99 oC, and Hvap = 35.2 kJ/mol.
What is the vapor pressure at 25 oC?
Practice Exercise, p. 473
Pentane (C5H12) gas is used to blow the bubbles in molten polystyrene that turn it into
Styrofoam, which is used in coffee cups and other products that have good thermal
insulation properties. The normal boiling point of pentane is 36 oC; its vapor pressure
at 25 oC is 505 torr. What is the enthalpy of vaporization of pentane?
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Chapter Outline
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11.1 Energy Changes when Substances Dissolve
11.2 Vapor Pressure
11.3 Mixtures of Volatile Substances
11.4 Colligative Properties of Solutions
11.5 Osmosis and Osmotic Pressure
11.6 Using Colligative Properties to Determine Molar Mass
Fractional Distillation
 A method of separating a
mixture of compounds
based on their different
boiling points
 Vapor phase enriched in
the more volatile
component
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Solutions of Volatile Compounds
 Raoult’s Law:
• Total vapor pressure of an ideal solution
depends on how much the partial pressure of
each volatile component contributes to total
vapor pressure of solution
• Ptotal = 1P1 + 2P2 + 3P3 + …
where i = mole fraction of component i, and
Pi = equilibrium vapor pressure of pure
volatile component at a given temperature
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Sample Exercise 11.3 (simplified):
Calculating the Vapor Pressure of a Solution
What is the vapor pressure of a solution prepared by dissolving 13 g
of heptane (C7H16) in 87 g of octane (C8H18) at 25 oC?
Real vs. Ideal Solutions
 Deviations from Raoult’s Law:
Due to differences in solute–solvent and
solvent–solvent interactions
(dashed lines = ideal behavior)
Negative deviations
Positive deviations
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Concept Test, p. 478
Which of the following solutions is least likely to follow
Raoult’s Law: (a) acetone/ethanol, (b) pentane/hexane, (c)
pentanol/water?
Chapter Outline
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11.1 Energy Changes when Substances Dissolve
11.2 Vapor Pressure
11.3 Mixtures of Volatile Substances
11.4 Colligative Properties of Solutions
11.5 Osmosis and Osmotic Pressure
11.6 Using Colligative Properties to Determine Molar
Mass
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Colligative Properties of Solutions
• Colligative Properties:
– Set of properties of a solution relative to the pure
solvent
– Due to solute–solvent interactions
– Depend on concentration of solute particles, not
the identity of particles
– Include lowering of vapor pressure, boiling point
elevation, freezing point depression, osmosis
and osmotic pressure
Vapor Pressure of Solutions
• Raoult’s Law:
• Vapor pressure of solution is equal to the vapor
pressure of the pure solvent multiplied by the
mole fraction of solvent in the solution
• Psolution = solvent·P solvent
• Vapor pressure lowering:
• A colligative property of solutions (Section 11.5)
• Ideal Solution:
• One that obeys Raoult’s law
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Sample Exercise 11.4: Calculating the Vapor
Pressure of a Solution of One or More Nonvolatile
Substances
Most of the world’s supply of maple syrup is produced in quebec, Ontario,
where the sap from maple trees is evaporated until the concentration of sugar
(mostly sucrose) in the sap reaches at least 66% by weight. What is the vapor
pressure in atm of a 66% aqueous solution of sucrose (MW=342) at 100 oC?
Assume the solution obeys Raoults’s Law.
Colligative Properties of Solutions
• Colligative Properties:
Vapor pressure
lowering
Consequences:
 in b.p. and f.p. of
solution relative to
pure solvent
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Solute Concentration: Molality
• Changes in boiling point/freezing point
of solutions depends on molality:
m
nsolute
kg of solvent
– Preferred concentration unit for properties
involving temperature changes because it
is independent of temperature
– For typical solutions: molality > molarity
Calculating Molality
Starting with:
a) Mass of solute
b) Volume of solvent, or
c) Volume of solution
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Sample Exercise 11.5: Calculating the
Molality of a Solution
A popular recipe for preparing corned beef calls for preparing a seasoned brine
(salt solution) that contains 0.50 lbs of kosher salt (NaCl) dissolved in 3.0 qt of
water. Assuming the density of water = 1.00 g/mL, what is the molality of NaCl in
this solution for “corning” beef?
Chapter Outline
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11.1 Energy Changes when Substances Dissolve
11.2 Vapor Pressure
11.3 Mixtures of Volatile Substances
11.4 Colligative Properties of Solutions
11.5 Osmosis and Osmotic Pressure
11.6 Using Colligative Properties to Determine Molar
Mass
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Colligative Properties of Solutions
 Colligative Properties:
• Solution properties that depend on concentration
of solute particles, not the identity of particles.
Previous example:
vapor pressure
lowering.
Consequences:
change in b.p.
and f.p. of solution.
© 2012 by W. W. Norton & Company
Boiling-Point Elevation
and Freezing-Point Depression
 Boiling Point Elevation (ΔTb):
• ΔTb = Kb∙m
• Kb = boiling point elevation
constant of solvent; m = molality.
 Freezing Point Depression (ΔTf):
• ΔTf = Kf∙m
• Kf = freezing-point depression
constant; m = molality.
© 2012 by W. W. Norton & Company
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Sample Exercise 11.6: Calculating the Boiling
Point Elevation of an Aqueous Solution
What is the boiling point of seawater? Assume that the
average concentration of sea salt is equivalent to 0.560 m
(molal) NaCl, and that the average density of seawater is
1.025 g/mL.
Sample Exercise 11.7: Calculating the
Freezing Point Elevation of a Solution
What is the freezing point of an automobile radiator fluid
prepared by mixing equal volumes of ethylene glycol
(MW=62.07) and water at a temperature where the density of
ethylene glycol is 1.114 g/mL and the density of water is 1.000
g/mL? The freezing point depression constant of water, kf, is
1.86 oC/m.
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The van’t Hoff Factor
 Solutions of Electrolytes:
• Need to correct for number of
particles formed when ionic
substance dissolves.
 van’t Hoff Factor (i):
•
number of ions in formula unit.
•
e.g., NaCl, i = 2
 ΔTb = i∙Kb∙m & ΔTf = i∙Kf∙m
 Deviations from theoretical
value due to ion pair formation.
© 2012 by W. W. Norton & Company
Values of van’t Hoff Factors
© 2012 by W. W. Norton & Company
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Sample Exercise 11.9:
Using the van’t Hoff Factor
Some Thanksgiving dinner chefs “brine” their turkeys prior to
roasting them to help the birds retain moisture and to season
the meat. Brining involves completely immersing a turkey fro
about 6 hours in a brine (salt solution) prepared by dissolving
2.0 pounds of salt (NaCl, MW=58.44) in 2.0 gallons of water.
Suppose a turkey soaking in such a solution is left for 6 hours
on an unheated porch on a day when the air temperature is
18 oF (-8 oC). Are the brine (and turkey) in danger of
freezing? The kf of water is 1.86 oC/m; assume that i=1.85 for
this NaCl solution.
Chapter Outline
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11.1 Energy Changes when Substances Dissolve
11.2 Vapor Pressure
11.3 Mixtures of Volatile Substances
11.4 Colligative Properties of Solutions
11.5 Osmosis and Osmotic Pressure
11.6 Using Colligative Properties to Determine
Molar Mass
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Molar Mass from Colligative Properties
ΔTf = Kfm and ΔTb = Kbm
Step 1:
solve for molality
ΔTf or ΔTb
m = ΔTf/Kf
kg of solvent
Step 2:
solve for moles
Step 3:
calc. MW
molality
moles
MW = g sample/moles
Sample Exercise 11.12: Using Freezing
Point Depression to Determine Molar Mass
Eicosene is a hydrocarbon used as an additive in the highpressure fracturing of rock layers to enhance recovery of
petroleum and natural gas, a process sometimes called
“fracking”. The freezing point of a solution prepared by
dissolving 100 mg of eicosene in 1.00 g of benzene is 1.75 oC
lower that the freezing point of pure benzene. What is the molar
mass of eicosene? (kf for benzene is 4.90 oC/m)
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