Chapter
11
Solutions and Their Properties
Chemistry 4th Edition
McMurry/Fay
Dr. Paul Charlesworth
Michigan Technological University
Solution Formation
01
••
Saturated: Contains
Contains the
the maximum
maximum amount
amount of
of
Saturated:
solute that
that will
will dissolve
dissolve in
in a
a given
given solvent.
solvent.
solute
••
Unsaturated: Contains
Contains less
less solute
solute than
than a
a solvent
solvent
Unsaturated:
has the
the capacity
capacity to
to dissolve.
dissolve.
has
••
Supersaturated: Contains
Contains more
more solute
solute than
than would
would
Supersaturated:
be present
present in
in a
a saturated
saturated solution.
solution.
be
••
Crystallization: The
The process
process in
in which
which dissolved
dissolved
Crystallization:
solute comes
comes out
out of
of the
the solution
solution and
and forms
forms crystals.
crystals.
solute
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Chapter 11
Slide 2
Solution Formation
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02
Slide 3
Solution Formation
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02
Slide 4
Solution Formation
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03
Slide 5
Solution Formation
•
Exothermic ∆Hsoln:
•
The solute–solvent
interactions are
stronger than solute–
solute or solvent–
solvent.
•
Favorable process.
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Chapter 11
04
Slide 6
Solution Formation
•
Endothermic ∆Hsoln:
•
The solute–solvent
interactions are weaker
than solute–solute or
solvent–solvent.
•
Unfavorable process.
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Chapter 11
05
Slide 7
Solution Formation
06
•
Solubility: A measure of how much solute will
dissolve in a solvent at a specific temperature.
•
Miscible: Two (or more) liquids that are completely
soluble in each other in all proportions.
•
Solvation: The process in which an ion or a
molecule is surrounded by solvent molecules
arranged in a specific manner.
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Chapter 11
Slide 8
Solution Formation
1.
07
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)?
3.
Which would have the largest (most negative)
hydration energy and which should have the
smallest? Al3+, Mg2+, Na+
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Chapter 11
Slide 9
Concentration Units
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Chapter 11
01
Slide 10
Concentration Units
01
•
Concentration: The amount of solute present in a
given amount of solution.
•
Percent by Mass (weight percent): The ratio of
the mass of a solute to the mass of a solution,
multiplied by 100%.
mass of solute
% by massof solute =
100%

mass of solution
mass of solution =mass of solute +mass of solvent
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Chapter 11
Slide 11
Concentration Units
•
02
Parts per Million:
Mass of component
6
x
10
• Parts per million (ppm) =
Total mass of solution
= % mass x 104
•
One ppm gives 1 gram of solute per 1,000,000 g or
one mg per kg of solution. For dilute aqueous
solutions this is about 1 mg per liter of solution.
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Chapter 11
Slide 12
Concentration Units
•
03
A sample of 0.892 g of potassium chloride (KCl) is
dissolved in 54.6 g of water. What is the percent by
mass of KCl in this solution?
•
An aqueous solution is 5.50% H2SO4. How many
moles of sulfuric acid (MM = 98.08 g/mol) are
dissolved in 250.0 g of the solution?
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Chapter 11
Slide 13
Concentration Units
•
•
•
Mole Fraction (X): X A 
04
Moles of A
Total number of moles
Molarity (M):
Moles of solute
Molarity 
Liters of SOLUTION
Molality (m):
Moles of solute
Molality =
Kilograms of SOLVENT
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Chapter 11
Slide 14
Concentration Units
06
•
Molality from Mass: Calculate the molality of a
sulfuric acid solution containing 24.4 g of sulfuric
acid in 198 g of water. The molar mass of sulfuric
acid is 98.08 g.
•
Molality from Molarity: Calculate the molality of a
5.86 M ethanol (C2H5OH) solution whose density is
0.927 g/ml.
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Chapter 11
Slide 15
Concentration Units
07
•
Molality from Mass %: Assuming that seawater is
a 3.50 mass % aqueous solution of NaCl, what is
the molality of seawater?
•
Molarity from Molality: The density at 20°C of a
0.258 m solution of glucose in water is 1.0173
g/mL, and the molar mass of glucose is 180.2 g.
What is the molarity of the solution?
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Chapter 11
Slide 16
Concentration Units
08
•
Mole Fraction from Molality: An aqueous solution
is 0.258 m in glucose (MM = 180.2 g/mol). What is
the mole fraction of the glucose?
•
Mass from Molality: What mass (in grams) of a
0.500 m aqueous solution of urea [(NH2)2CO, MM
= 60.1 g/mol] would you use to obtain 0.150 mole
of urea?
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Chapter 11
Slide 17
Effect of Temperature
on Solubility
•
01
Solids:
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Chapter 11
Slide 18
Effect of Temperature
on Solubility
•
02
Gases:
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Slide 19
The Effect of Pressure on the
Solubility of Gases
01
Henry’s Law:
• The solubility of a
gas is proportional
to the pressure of
the gas over the
solution.
•
cP
c = k·P
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c  kP
Chapter 11
Slide 20
The Effect of Pressure on the
Solubility of Gases
02
Flash Animation - Click to Continue
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Chapter 11
Slide 21
The Effect of Pressure on the
Solubility of Gases
•
02
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).
•
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)?
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Chapter 11
Slide 22
Colligative Properties
of Nonvolatile Solutes
01
•
Colligative Properties: Depend only on the
number of solute particles in solution. These affect
properties of the solvent.
•
There are four main colligative properties:
1.
2.
3.
4.
Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
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Chapter 11
Slide 23
Colligative Properties
of Nonvolatile Solutes
•
02
When solute molecules displace solvent molecules
at the surface, the vapor pressure drops since
fewer gas molecules are needed to equalize the
escape rate and capture rates at the liquid surface.
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Chapter 11
Slide 24
Colligative Properties
of Nonvolatile Solutes
•
03
Raoult’s Law: P1 = x1P°1 where x1 is the solvent mole
fraction.
•
For a single solute solution, x1= 1 – x2 , where x2 is the
solute mole fraction.
•
We can obtain an expression for the change in vapor
pressure of the solvent (the vapor pressure lowering).
P = P°1 – P1 = P°1 – x1 P°1 = P°1 – (1 – x2 ) P°1
∆P = x2 P°1
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Chapter 11
Slide 25
Colligative Properties
of Nonvolatile Solutes
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Chapter 11
04
Slide 26
Colligative Properties
of Nonvolatile Solutes
05
•
The vapor pressure of a glucose (C6H12O6) solution
is 17.01 mm Hg at 20°C, while that of pure water is
17.25 mm Hg at the same temperature. Estimate the
molality of the solution.
•
How many grams of NaBr must be added to 250 g
of water to lower the vapor pressure by 1.30 mm Hg
at 40°C? The vapor pressure of water at 40°C is
55.3 mm Hg.
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Chapter 11
Slide 27
Colligative Properties of a
Mixture of Two Volatile Liquids
•
What happens if both components are volatile
(have measurable vapor pressures)?
•
The vapor pressure has a value intermediate
between the vapor pressures of the two liquids.
01
PT = PA + PB = XAP°A + XBP°B = XAP°A + (1 – XA)P°B
PT = P°B + (P°A – P°B)XA = straight line = a + bXA
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Chapter 11
Slide 28
Colligative Properties of a
Mixture of Two Volatile Liquids
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Chapter 11
02
Slide 29
Colligative Properties of a
Mixture of Two Volatile Liquids
•
03
The following diagram shows a close-up view of
part of the vapor–
pressure curves
for two pure liquids
and a mixture of
the two. Which
curves represent
pure liquids, and
which the mixture?
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Chapter 11
Slide 30
Boiling-Point Elevation and
Freezing-Point Depression
•
01
Boiling-Point Elevation (∆Tb): The boiling point of
the solution (Tb) minus the boiling point of the pure
solvent (T°b):
∆Tb = Tb – T°b
∆Tb is proportional to concentration:
∆Tb = Kb m
Kb = molal boiling-point elevation constant.
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Chapter 11
Slide 31
Boiling-Point Elevation and
Freezing-Point Depression
•
02
Freezing-Point Depression (∆Tf): The freezing point
of the pure solvent (T°f) minus the freezing point of
the solution (Tf).
∆Tf = T°f – Tf
∆Tf is proportional to concentration:
∆Tf = Kf m
Kf = molal freezing-point depression constant.
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Chapter 11
Slide 32
Boiling-Point Elevation and
Freezing-Point Depression
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Chapter 11
03
Slide 33
Boiling-Point Elevation and
Freezing-Point Depression
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Chapter 11
04
Slide 34
Boiling-Point Elevation and
Freezing-Point Depression
•
05
The phase diagram shows a close-up of the liquid–vapor
phase transition boundaries for pure chloroform.
a)
Estimate the boiling
point of pure
chloroform.
a)
Estimate the molal
concentration of the
nonvolatile solute.
(See Table 11.4 for Kb).
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Chapter 11
Slide 35
Boiling-Point Elevation and
Freezing-Point Depression
•
•
06
van’t Hoff Factor, i: This factor equals the number
of ions produced from each molecule of a
compound upon dissolving.
i = 1 for CH3OH
i = 3 for CaCl2
i = 2 for NaCl
i = 5 for Ca3(PO4)2
For compounds that dissociate on dissolving, use:
∆Tb = iKb m
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∆Tf = iKf m
Chapter 11
∆P = ix2 P°1
Slide 36
Boiling-Point Elevation and
Freezing-Point Depression
•
07
How many grams of ethylene glycol antifreeze,
CH2(OH)CH2(OH), must you dissolve in one liter of
water to get a freezing point of –20.0°C. The molar
mass of ethylene glycol is 62.01 g. For water, Kf =
1.86 (°C·kg)/mol. What will be the boiling point?
Prentice Hall ©2004
Chapter 11
Slide 37
Boiling-Point Elevation and
Freezing-Point Depression
•
08
What is the molality of an aqueous solution of KBr
whose freezing point is –2.95°C? Kf for water is
1.86 (°C·kg)/mol.
•
What is the freezing point (in °C) of a solution
prepared by dissolving 7.40 g of K2SO4 in 110 g of
water? The value of Kf for water is 1.86
(°C·kg)/mol.
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Chapter 11
Slide 38
Osmosis and Osmotic Pressure
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Chapter 11
01
Slide 39
Osmosis and Osmotic Pressure
•
Osmosis: The selective passage of solvent
molecules through a porous membrane from a
dilute solution to a more concentrated one.
•
Osmotic pressure (π or ∏): The pressure
required to stop osmosis.
01
π = iMRT
R = 0.08206 (Latm)/(molK)
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Chapter 11
Slide 40
Osmosis and Osmotic Pressure
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Chapter 11
02
Slide 41
Osmosis and Osmotic Pressure
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03
Slide 42
Osmosis and Osmotic Pressure
04
•
Isotonic: Solutions have equal concentration of
solute, and so equal osmotic pressure.
•
Hypertonic: Solution with higher concentration of
solute.
•
Hypotonic: Solution with lower concentration of
solute.
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Chapter 11
Slide 43
Osmosis and Osmotic Pressure
•
05
The average osmotic pressure of seawater is about
30.0 atm at 25°C. Calculate the molar
concentration of an aqueous solution of urea
[(NH2)2CO] that is isotonic with seawater.
•
What is the osmotic pressure (in atm) of a 0.884 M
sucrose solution at 16°C?
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Chapter 11
Slide 44
Uses of Colligative Properties
•
01
Desalination:
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Chapter 11
Slide 45
Uses of Colligative Properties
•
02
A 7.85 g sample of a compound with the
empirical formula C5H4 is dissolved in 301 g of
benzene. The freezing point of the solution is
1.05°C below that of pure benzene. What are
the molar mass and molecular formula of this
compound?
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Chapter 11
Slide 46
Uses of Colligative Properties
03
•
A 202 ml benzene solution containing 2.47 g of an
organic polymer has an osmotic pressure of 8.63
mm Hg at 21°C. Calculate the molar mass of the
polymer.
•
What is the molar mass of sucrose if a solution of
0.822 g of sucrose in 300.0 mL of water has an
osmotic pressure of 149 mm Hg at 298 K?
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Chapter 11
Slide 47
Uses of Colligative Properties
•
04
Fractional Distillation is the separation of volatile
liquid mixtures into fractions of different
composition.
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Chapter 11
Slide 48
Uses of Colligative Properties
•
05
Fractional distillation can be represented on a
phase diagram by plotting temperature against
composition.
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Chapter 11
Slide 49
Uses of Colligative Properties
•
06
Two miscible liquids, A and B, have vapor
pressures of 250 mm Hg and 450 mm Hg,
respectively. They were mixed in equal molar
amounts. What is the total vapor pressure of the
mixture and what are their mole fractions in the
vapor phase?
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Chapter 11
Slide 50