Chapter 13
Solutions and Their Properties
1. Important Definitions
2. Expressing Concentrations of Solutions
3. The Thermodynamics of the Dissolving Process
4. Dynamic Equilibrium
5. The Solubility of Solids, Liquids, and Gases
6. Colligative Properties
1
Important Definitions
Solution - A homogeneous mixture consisting of one or
more substances uniformly dispersed as separate
atoms, molecules, or ions in another substance.
Solvent - The component of a solution that is the
“dissolving”
dissolving” medium. The solvent determines the
physical state of the solution (solid,
(solid, liquid, or gas).
gas).
Solute - The components of a solution that are “disdissolved”
solved” by the medium.
Aqueous Solution - A solution wherein water is the
solvent.
When the solute and solvent are both in the same physical
state, the one in the largest quantity is the solvent.
2
1
Table 13.1 Some Common
Solutions
3
13-2 Solution Concentration.
Concentration - The amount of solute present in a given
quantity of solvent or solution.
solution.
Mass percent.
Volume percent.
Mass/volume percent.
(m/m)
(v/v)
(m/v)
Isotonic saline is prepared by
dissolving 0.9 g of NaCl in 100 mL of
water and is said to be:
0.9% NaCl (mass/volume)
4
2
ppm, ppb and ppt
Very low solute concentrations are expressed
as:
ppm: parts per million
ppb: parts per billion
ppt: parts per trillion
(µg/g, mg/L)
(ng/g, µg/L)
(pg/g, ng/L)
note that 1.0 L × 1.0 g/mL = 1000 g
ppm, ppb, and ppt are properly m/m or v/v.
5
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%
6
3
Molarity and Molality
Molarity (M) =
Amount of solute (in moles)
Volume of solution (in liters)
Amount of solute (in moles)
Molality (m) =
Mass of solvent (in kilograms)
7
8
4
9
10
5
Expressing Concentration
Concentration - The amount of solute present in a given
quantity of solvent or solution.
solution.
1.
mass solute
35.7 g NaCl
volume solvent
100 mL H2O
(Typical way to express solubilities)
2. Percent by Mass
mass solute (100 %)
mass solution
What is the percent by mass of sugar in a solution concontaining 50.0 g of sugar and 65.0 g H2O?
11
(mass of solute/mass of solution)(100%) =
(50.0 g sugar)/(50.0 g sugar + 65.0 g H2) (100.0%) =
43.5 % by mass
3. Molarity (M) (Temperature Dependent)
moles solute
Molarity =
liter solution
Practice Problem: A 650650-mL solution contains 1.22 mol
of KNO3. Calculate the molarity of the solution.
12
6
1.22 mol KNO3
650 mL soln
1000 mL soln
L soln
= 1.88 M KNO3
Practice Problem: Calculate the molarity of a solution
that contains 5.25 g AgNO3 in 175 mL of solution.
A. Calculate the formula mass of AgNO3.
1 Ag@ 107.868 u = 107.868 u
1 N@ 14.0067 u = 14.0067 u
3 O@ 15.9994 u = 47.9982 u
1 AgNO3
= 169.873 u
B. Determine the moles of AgNO3 in 5.25 g AgNO3.
(5.25 g AgNO3)(1 mol AgNO3/169.873 g AgNO3) =
13
0.030 91 mol AgNO3
C. From the moles of AgNO3 and the mL of solution,
calculate the molarity of the solution.
0.030 91 mol AgNO3
175 mL soln
1000 mL soln
L soln
= 0.177 M AgNO3
(“When in doubt, calculate moles”)
# of Particles
Avogadro’
Avogadro’s
#
Moles
Mass
Molar Mass
Molarity
VOLUME
14
7
4. Molality (m) (Temperature Independent)
moles solute
molality =
kg solvent
Practice Problem: What is the molality of a saturated
solution of NaCl in water at 0oC? NaCl’
NaCl’s solubility is
35.7 g/100 mL at 0oC. Water’
Water’s density is 1.000 g/mL at
o
0 C.
35.7 g NaCl 1 mL H2O 103 g H2O 1 mol NaCl
100 mL H2O 1.000 g H2O kg H2O 58.44 g NaCl
6.11 mol NaCl = 6.11 m
kg H2O
15
5. Mole Fraction (Χ
(Χ) (Temperature Independent)
moles solute
=
moles solute + moles solvent
moles solute
total moles
Practice Problem: If 26.2 g of H2O (g) is mixed with
43.7 g of O2 (g), what is the mole fraction of water vapor
in this gaseous solution?
26.2 g H2O
1 mol H2O = 1.454 mol H O
2
18.015 g H2O
43.7 g O2
1 mol O2
= 1.366 mol O2
31.9988 g O2
16
8
Χwater =
Χwater =
1.454 mol H2O
1.454 mol H2O + 1.366 mol O2
1.454 mol H2O
= 0.516 mol fraction
2.818 mol H2O
17
Solubility - The amount of solute that will dissolve in a
given quantity of solvent at a given temperature.
temperature.
Saturated Solution - A solution that contains an amount
of solute that is equal to its solubility.
The dissolved solute in the solution is
in dynamic equilibrium with the unundissolved solute (the precipitate).
precipitate).
Dissolved solute
Precipitate
The rate of dissolving is equal to the rate of precipitation.
This is a dynamic equilibrium.
equilibrium.
18
9
Unsaturated Solution - A solution that contains an
amount of solute that is less than its solubility.
All of the solute present is dissolved in
an unsaturated solution.
Supersaturated Solution - A solution that contains an
amount of solute that is more than its solubility. This
is a “metastable”
metastable” state (“
(“without stability”
stability”).
If disturbed in anyway, the excess solute
will precipitate out of solution and a
saturated solution will result.
19
Solubility Curves
Saturated solution
along the line
Supersaturated above this line
Unsaturated below this line
20
10
The Thermodynamics of Dissolving
Why is n-hexane soluble in CCl4 but water is not??
Stay Tuned!!
21
The Thermodynamics of Dissolving
The Dissolving Process: In order to predict whether or
not a material will dissolve in another material, one
must know the enthalpy change (∆
∆H) and the entropy
change (∆
∆S)that will occur upon dissolving and
the temperature.
Entropy - the disorder of a system. Most dissolving
processes involve an increase in entropy. That is, the
disorder present in the separate solute and solvent is
lower than it is in the solution. - ∆Ssolution > 0! This
favors dissolving.
Enthalpy - the total energy of a system. Some dissolving
processes are endothermic.
endothermic That is, the solution has a
higher enthalpy than the separate solute and solvent 22
∆H > 0! These processes do NOT favor dissolving.
11
The Thermodynamics of Dissolving
Enthalpy - Some dissolving processes are exothermic.
exothermic
That is, the solution has a lower enthalpy than the
separate solute and solvent - ∆Hsolution < 0! These
processes do favor dissolving.
Summary: To predict whether or not a solute will
dissolve spontaneously in a given solvent, one must evaluate both the enthalpy change that will occur and the
entropy change that will occur during dissolving. The
temperature must also be taken into consideration.
Since the entropy almost always increases during dissolving, we can concentrate, for now, on the enthalpy
change.
23
The Thermodynamics of Dissolving
The Dissolving Process:
Solvent
∆ Ha
∆ Hb
∆ Hc
∆Ha - Endothermic
Endothermic
Solute
∆Hb - Endothermic
Endothermic
Solution
∆Hc - Exothermic
Exothermic
24
12
13-3 Intermolecular Forces and
the Solution Process
∆Hc
∆Hb
∆Ha
25
The Thermodynamics of Dissolving
??
∆Hsolution = ∆Ha + ∆Hb + ∆Hc
Always endothermic !
Always exothermic !
26
13
The Thermodynamics of Dissolving
∆Hsolution = ∆Ha + ∆Hb + ∆Hc
If ∆Hc > ∆Ha + ∆Hb then ∆Hsolution < 0
If ∆Hc < ∆Ha + ∆Hb then ∆Hsolution > 0
If ∆Hc = ∆Ha + ∆Hb then ∆Hsolution = 0
If the solvent is a liquid, then ∆Ha is the heat of
vaporization for the liquid.
If the solute is a nonionic (covalent) solid, then ∆Hb is the
heat of sublimation of the solid.
If the solute is an
an ionic solid, then ∆Hb is the negative of
lattice energy of the solid.
27
How could you determine the value of ∆Hc?
Intermolecular Forces in Mixtures
Magnitude of ∆Ha, ∆Hb, and
∆Hc depend on
intermolecular forces.
Ideal solution
are similar between
all combinations of
components.
Cohesive
Forces
Adhesive
Forces
Forces
∆Hsoln = 0
Adhesive forces (forces between unlike molecules) are equal to
Cohesive forces (forces between like molecules) in magnitude
28
14
Increasing Energy
Fig. 13.1
29
Non-ideal Solutions
Adhesive forces
greater than cohesive
forces.
∆Hsoln < 0
30
15
Non-ideal Solutions
Adhesive forces are less
than cohesive forces.
(Acetone-Carbon disulfide)
∆Hsoln > 0
At the limit these solutions
are heterogeneous.
31
Ionic Solutions
32
16
Hydration Energy
NaCl(s) → Na+(g) + Cl-(g)
-∆Hlattice > 0
Na+(g) + xs H2O(l) → Na+(aq)
∆Hhydration < 0
Cl-(g) + xs H2O(l) → Cl-(aq)
∆Hhydration < 0
------------------------------------------------------------Overall:
NaCl(s) → Na+(aq) + Cl-(aq)
∆Hsoln > 0 but ∆Gsolution < 0
33
The Thermodynamics of Dissolving
The Role of Intermolecular Attractive Forces:
For a dissolving process to be exothermic,
exothermic, the solventsolventsolute attractive forces must be greater than the solutesolutesolute and the solventsolvent-solvent attractive forces combined.
Can a substance dissolve if the process is endothermic?
endothermic?
Substance
∆Hsolution
CO2 (g)
-19.41 kJ/mol
LiCl (s)
- 37.03 kJ/mol
KCl (s)
+ 17.22 kJ/mol
NaCl (s)
+ 3.88 kJ/mol
H2SO4 (l)
- 95.28 kJ/mol
34
17
The Thermodynamics of Dissolving
Some Useful Thermodynamic Concepts - Read
Chapter 19
1. The enthalpy change accompanying the dissolving
process can be thought of as the change in total
energy.
energy. The total energy consists of the energy
available for doing work (Free Energy, ∆G) and
the energy NOT available for doing work (entropy,
T∆S).
S).
∆Hsolution = ∆Gsolution + T∆
T∆Ssolution
∆Gsolution = ∆Hsolution - T∆Ssolution
For a change to take place spontaneously, ∆G of
the process must be less than 0.
0.
35
The Thermodynamics of Dissolving
∆Hsolution = ∆Gsolution + T∆
T∆Ssolution
∆Gsolution = ∆Hsolution - T∆Ssolution
If ∆Gsolution > 0 at temperature T, the solute will NOT
spontaneously dissolve.
If ∆Gsolution < 0 at temperature T, the solute will
spontaneously dissolve.
Case 1:
(-)
(-)
(+)
∆Gsolution = ∆Hsolution - T∆Ssolution
Spontaneous at
all temperatures
exothermic
more disorder
36
18
The Thermodynamics of Dissolving
Case 2:
(+)
∆Gsolution
NonNon-spontaneous
at all temperatures
Case 3:
∆Gsolution
(+)
(-)
= ∆Hsolution - T∆Ssolution
endothermic
less disorder
(-)
(-)
= ∆Hsolution - T∆Ssolution
exothermic
less disorder
At higher temperatures, the entropy term
dominates and the dissolving process is nonnonspontaneous.
spontaneous. At lower temperatures, the enthalpy
term dominates and the dissolving is spontaneous.
spontaneous.
Thus, this type of system would have higher solubility
at lower temperatures.
37
The Thermodynamics of Dissolving
Case 4:
∆Gsolution
(+)
(+)
= ∆Hsolution - T∆Ssolution
endothermic more disorder
At higher temperatures, the entropy term
dominates and the dissolving process is spontaneous.
spontaneous.
At lower temperatures, the enthalpy term dominates
and the dissolving is nonnon-spontaneous.
spontaneous.
Dynamic Equilibrium: When two opposing processes
are occurring at exactly the same rate, a state of dynamic
equilibrium has been achieved.
38
19
The Thermodynamics of Dissolving
1. Ethyl alcohol (ethanol)
(ethanol) is miscible with both water
and n-hexane. However, water is not miscible with
n-hexane.
Why??
H2O - H2O interaction - H-bonding and London Forces
CH3CH2OH - CH3CH2OH - H-bonding and London Forces
H2O - CH3CH2OH - H-bonding and London Forces
The interactions are all of similar strength. Thus,
entropy is the predominating factor in solubility.
39
The Thermodynamics of Dissolving
2. AgCl is Not very soluble in water.
Why??
H2O - H2O interaction - H-bonding and London Forces
Ag + - Cl - interaction - Very strong ion-ion attraction
(Crystal Lattice Energy)
H2O - Ag +
and interactions - Relatively weak ion-dipole
H2O - Cl attractions
The water-ion attractions are not strong enough to overcome the water-water and ion-ion attractions. Thus,
enthalpy is the predominating factor in solubility.
40
20
The Thermodynamics of Dissolving
3. n-Heptane and n-hexane are completely miscible
in each other.
Why??
heptane - heptane interactions - London Forces
hexane - hexane interactions - London Forces
heptane - hexane interactions - London Forces
The attractive forces are very similar. Thus entropy is
the predominant factor.
41
The Thermodynamics of Dissolving
4. n-Hexane and water are immiscible (not
(not miscible).
Why??
H2O - H2O interaction - H-bonding and London Forces
hexane - hexane interactions - London Forces
H2O - hexane interactions - London Forces
The water-hexane interactions are much weaker than
the water-water interactions. Thus enthalpy (which
which is
very positive - endothermic
endothermic) is the predominant factor.
42
21
The Thermodynamics of Dissolving
The previous four examples demonstrate the useful
principle that “like dissolves like”
like”. In other words,
materials that have similar structures and polarity
such as n-heptane and n-hexane or water and ethanol
will tend to be soluble or miscible in one another.
CH3CH2CH2CH2CH2CH3
n-hexane
CH3CH2CH2CH2CH2CH2CH3
n-heptane
H-OH
water
CH3CH2-OH
ethanol
43
Dynamic Equilibrium
Examples of Dynamic Equilibrium The rate of water flowing into
the tub is equal to the rate of
water flowing out of the tub.
Thus, the water level in the tub
is not changing.
Rate of Input = Rate of Output
What will happen if we increase the rate of water
going into the tub? The rate of water flowing out of the
tub will increase until the two rates are again equal.
44
22
Dynamic Equilibrium
Open System
(No Equilibrium)
Evaporation
Evaporation
Liquid
Gas
(No Equilibrium)
Liquid
Gas
(No Equilibrium)
Liquid
Gas
(Equilibrium)
45
Dynamic Equilibrium
Ag + + Cl -
AgCl (s)
Chemical
Equilibrium
Ag + Cl Cl - Ag +
AgCl (s)
Rate of Precipitation = Rate of Dissolving
HC2H3O2 (aq)
H + + C2H3O2 -
Rate of dissociation (ionization) = Rate of Association
HC2H3O2
H+
+
H
C2H3O2 C2H3O2 HC2H3O2
46
23
13-4 Solution Formation
and Equilibrium
Saturated
Rate of crystallization≈ 0
Rate of dissolution>
Rate of crystallization
unsaturated
Rate of
dissolution =
Rate of
crystallization
47
Dynamic Equilibrium
The net result of a dynamic equilibrium is that no
change in the system is evident.
Le Chatelier’
Chatelier’s Principle - If a change is made in a
system at equilibrium, the equilibrium will shift in such
a way so as to reduce the effect of the change.
Apply
Pressure
Pressure applied to the system at equilibrium caused it to
shift until a new equilibrium was established.
48
24
Solubility of Solids, Liquids, and Gases
Using Le Chatelier’
Chatelier’s Principle to predict the
temperature effect on solubility:
1. If a solute has an endothermic
endothermic enthalpy of solution
(∆Hsolution > 0),
0), it will TEND to have a higher
solubility at higher temperatures.
Thermal
Energy
+ solute + solvent
solution
49
Solubility of Solids, Liquids, and Gases
2. If a solute has an exothermic
exothermic enthalpy of solution
(∆Hsolution < 0),
0), it will TEND to have a lower
solubility at higher temperatures.
solute + solvent
solution + Thermal
Energy
Note: The enthalpy of solution is somewhat dependent
on the concentration of the solute. Therefore, the effect
of temperature on the ∆Hsolution is measured at the
saturation point.
50
25
Solubility of Solids and Liquids
All solid and liquid solutes that dissolve exothermically
exothermically
should be less soluble at higher temperatures, according
to Le Chatelier’
Chatelier’s Principle.
BUT...
If the solution process is exothermic
exothermic AND produces a
higher state of disorder (higher entropy), an increase in
temperature will favor dissolving!
(-)
(-)
(+)
∆Gsolution = ∆Hsolution - T∆Ssolution
Spontaneous
Higher temperatures make the solution process MORE
spontaneous!
51
Solubility of Solids, Liquids, and Gases
The Effects of Temperature on Solubility - Most but
not all liquids and solids show increase in solubility as the
temperature is increased.
All gases, on the other hand, decrease in solubility with
increasing temperature. Why???
* g gas
100 g H2O
Gas
0oC
25oC
50oC
CO2
0.33*
0.145*
0.076*
O2
0.0069*
0.0039 *
0.001 22*
SO2
23*
9.4*
4.3*
52
26
13-5 Solubility of Gases
Most gases are
less soluble in
water as
temperature
increases.
In organic solvents
the reverse is often
true.
53
Solubility of Solids, Liquids, and Gases
Le Chatelier’
Chatelier’s Principle is useful for predicting the
effects of temperature on the solubility of gases in
liquids because these are all (or nearly all) exothermic
exothermic
changes.
Solute (g) + solvent (l)
solution + Thermal
Energy
Le Chatelier’
Chatelier’s Principle is not useful for predicting the
effects of temperature on the solubility of solids and
liquids.
liquids.
54
27
Solubility of Gases
Then if the dissolving of gases is exothermic, why do their
solubilities ALWAYS DECREASE with increasing
temperature???
Because when gases dissolve, the entropy actually
decreases!
(+)
(-)
(-)
∆Gsolution = ∆Hsolution - T∆Ssolution
Thus, at higher temperatures, the entropy term
dominates and ∆Gsolution becomes more positive!
This is why Le Chatelier’
Chatelier’s Principle works for the
solubility of gases in liquids.
55
Solubility of Gases
Recall the thermodynamics of dissolving:
Very minimal
in the case of
gases!!
∆S < 0
56
28
Solubility of Gases
The Effects of Pressure on the Solubility of Gases:
Henry’
Henry’s Law - The solubility of a sparingly soluble gas
is directly proportional to the partial pressure of the gas
above the liquid.
S α P
S = kP
Le Chatelier’
Chatelier’s Principle can be used to predict the effects
of pressure on the solubility of a gas in a liquid.
Solute (g) + Solvent (l)
Solution
57
Solubility (g /100 g of H20)
Solubility of Gases
Henry’
Henry’s Law
O2
N2
He
Pressure
58
29
Henry’s Law
Solubility of a gas
increases with
increasing pressure.
S= k Pgas
S: Solubility of a gas
in a particular solvent
Pgas:Partial Pressure of the gas
above the solution
k=
S
=
Pgas
23.54 mL
1.00 atm
= 23.54 ml N2/atm
To increase the solubility to 100 mL
S
100 mL
= 4.25 atm
Pgas =
=
k
23.54 ml N2/atm
Henry’s law fails at high P or if the gas ionizes or reacts in the solvent
Applies only to equilibrium between molecules of a gas and the same
59
molecules in solution
Henry’s Law
What happens when you open a soft drink?
In a saturated solution,
Rate of vaporization =
Rate of Condensation
These rates are dependent on
number of molecules per unit
volume.
As P increases, # of molecules in gas phase/volume increases,
the # of molecules in the liquid phase must also increase to
reach a new equilibrium ( through an increase in concentration)
60
30
Colligative Properties of Solutions
Definition: A property of solutions that depends only
on the concentration (number) of solute particles and
not on the nature of the solute.
The same number of particles of different solutes in a
given solvent will produce the same change in a
colligative property.
property.
h
h
h
h
Vapor Pressure Lowering (VPL)
Boiling Point Elevation (∆ Tbp)
Freezing Point Depression (∆ Tfp)
Osmotic Pressure (Π)
61
Colligative Properties of Solutions
Ideal Solution: An imaginary solution in which the
component molecules are subject to forces that are
Identical to those they would experience in the pure
state.
A. The volume of an ideal solution is the sum of its
pure component volumes; there is no expansion or
contraction on mixing.
B. The ∆ Hsolution = 0; mixing is neither exothermic
exothermic nor
endothermic.
endothermic.
C. The vapor pressure above an ideal solution is
given by Dalton’
Dalton’s Law of Partial Pressures.
62
31
Colligative Properties of Solutions
Raoult’
Raoult’s Law: The vapor pressure of a solution that concontains a nonnon-volatile solute is directly proportional to the
mole fraction of the solvent in the solution.
o
PA α ΧA
PA = ΧAPA
“Ideal solutions”
solutions” containing more than one volatile
component also obey Raoult’
Raoult’s Law.
o
PA = ΧAPA
o
PB = ΧBPB
o
o
Ptotal = PA + PB = ΧAPA + ΧBPB
63
13-6 Vapor Pressures of
Solutions
Roault, 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.
Applies to ideal solutions and to all volatile components of
ideal solutions. ( Also to very dilute non-ideal solutions
PB = χB P°
B
P° = Vapor pressure of B
B
Mixture:Solvent B and Solute A
Mole fractions: χA+ χB = 1 χB<1 PB< Po
B
64
32
65
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
66
33
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:
χvapbenzene = Pbenzene/Ptotal = 47.6 mm Hg/61.89 mm Hg = 0.770
χvaptoluene = Ptoluene/Ptotal = 14.2 mm Hg/61.89 mm Hg = 0.230
67
Liquid-Vapor Equilibrium
χliqbenzene= 0.5
χliqtoluene= 0.5
61.8 mm Hg
χvapbenzene = Pbenzene/Ptotal =
47.6 mm Hg/61.89 mm Hg = 0.770
47.6 mm Hg
χvaptoluene = Ptoluene/Ptotal =
14.2 mm Hg/61.89 mm Hg = 0.230
14.2 mm Hg
Benzene is more volatile
(higher vapor pressure)
The vapor is richer in the more
volatile component than is the liquid
phase
68
34
Fractional Distillation
At P= 1 atm Plot of Normal bp of solutions vs solution and vapor concentration
69
Fractional Distillation
70
35
Non-ideal behavior
Azeotrope=
A mixture boiling to
Yield a vapor with
constant composition all
the time as long as
enough of both
components are present
Đn the liquid phase
71
13-7 Osmotic Pressure
Direction of
net flow of H2O,
Osmosis
Semipermeable
membrane
(permeable to
only water)
Applying pressure to the sucrose solution slows down the net flow of water.
If sufficient pressure (called the osmotic pressure) is applied to the solution, the
osmotic flow of water can be stopped.
72
For 20% sucrose solution osmotic pressure is about 15 atm
36
The Development of Osmotic Pressure
73
Fig. 13.18
Osmotic Pressure
For dilute solutions of electrolytes:
osmotic
pressure (π)
πVsolution = nsoluteRT
π=
n
RT = M RT
V
RT
π = (m/MM)
V
Osmotic pressure can be used to determine molar mases (MM)
of solutes
74
37
Reverse Osmosis Desalination
75
14-8 Freezing-Point Depression
and Boiling Point Elevation of
Nonelectrolyte Solutions
Vapor pressure is lowered when a solute is
present.
This
results in boiling point elevation.
Freezing point is also effected and is lowered.
Colligative properties.
Depends
on the number of particles present.
76
38
Vapor Pressure
Lowering
∆Tf = -Kf x m
∆Tb = Kb x m
77
Colligative Properties of Solutions
Boiling Point Elevation: Because of VPL, a solution
containing a nonnon-volatile solute will boil at a higher
temperature than the pure solvent.
∆ Tbp α molality of the solute
∆ Tbp = kbpm
kbp - boiling point elevation constant
Solvent
H2 O
Ethanol
Cyclohexane
HC2H3O2
Kbp
0.512 oC/molal
1.22 oC/molal
2.79 oC/molal
3.07 oC/molal
78
39
Colligative Properties of Solutions
Practice Problem: Calculate the boiling point of a
solution containing 0.600 kg of CHCl3 and 42.0 g of
eucalyptol, C10H18O. Chloroforms b.p. = 61.2oC and
its kbp = 3.63 oC/molal.
1. What is the solvent?
CHCl3
2. Calculate the molar mass of the solute?
10 C @ 12.011 u = 120.11 u
18 H @ 1.00794 u = 18.1429 u
1 O @ 15.9994 u = 15.9994 u
1 C10H18O
= 154.25 u
79
Colligative Properties of Solutions
3. Calculate the molality of the solute.
42.0 g C10H18O
0.600 kg CHCl3
1 mol C10H18O
154.25 g C10H18O
= 0.4538 m
4. Calculate the ∆Tbp.
∆Tbp = kbpm = (3.63 oC/m)(0.4538 m) = 1.647 oC
5. Calculate the boiling point of the solution.
Tbp solution = Tbp solvent + ∆Tbp = 61.2 oC + 1.647 oC
= 62.8oC
80
40
Colligative Properties of Solutions
Freezing Point Depression: A solution containing a
nonnon-volatile solute will freeze at a lower temperature
than the pure solvent.
∆ Tfp α molality of the solute
∆ Tfp = kfpm
kfp - freezing point depression constant
Solvent
H2 O
Ethanol
Cyclohexane
HC2H3O2
Kfp
1.86 oC/molal
1.99 oC/molal
20.4 oC/molal
3.90 oC/molal
81
Colligative Properties of Solutions
Practice Problem: A 2.00 m solution of sugar in water
was found to have a freezing point of -3.70oC. What
is the kfp for water based on these data?
∆Tfp = kfp m
kfp = ∆Tfp / m
kfp = [0.00 oC + ((-3.70oC)]/2.00 m
kfp = -3.70oC/2.00 m = -1.85oC/m
What would the kfp be for a 1.23 m solution of sugar
C/m. The kfp is constant
in water? It would be -1. 85oC/m.
for a given solvent.
82
41
Colligative Properties of Solutions
If kbp or kfp of a solvent is known and the ∆Tfp or ∆Tbp
is measured for a solution containing a known mass of
solute, the molar mass of the solute can be calculated.
∆Tfp = kfp m = kfp
g solute
molar mass of solute kg solvent
molar mass of solute =
(kfp) (g solute)
(kg solvent)(∆
solvent)(∆Tfp)
“When in doubt, calculate moles!”
moles!”
83
Practical Applications
84
42
13-9 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)
∆Tf = -Kf x m = -1.86°C m-1 x 0.0100 m = -0.0186°C
Freezing point depression for NaCl is -0.0361°C.
85
van’t Hoff
i=
measured ∆Tf
expected ∆Tf
=
0.0361°C
= 1.98
0.0186°C
π = i x M x RT
∆Tf = -i x Kf x m
∆Tb = i x Kb x m
86
43
Interionic Interactions
Arrhenius theory does not correctly predict
the conductivity of concentrated electrolytes.
87
1923
Debye and Hückel
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.
88
44