14
Solutions
Endothermic
Exothermic
2
Chapter Goals
The Dissolution Process
Spontaneity of the Dissolution Process
• Dissolution of Solids in Liquids
• Dissolution of Liquids in Liquids (Miscibility)
• Dissolution of Gases in Liquids
• Rates of Dissolution and Saturation
• Effect of Temperature on Solubility
• Effect of Pressure on Solubility
• Molality and Mole Fraction
3
Chapter Goals
•
•
•
•
•
Colligative Properties of Solutions
Lowering of Vapor Pressure and Raoult’s Law
Fractional Distillation
Boiling Point Elevation
Freezing Point Depression
Determination of Molecular Weight by Freezing
Point Depression or Boiling Point Elevation
• Colligative Properties and Dissociation of
Electrolytes
• Osmotic Pressure
4
Chapter Goals
Colloids
•
•
•
The Tyndall Effect
The Adsorption Phenomena
Hydrophilic and Hydrophobic Colloids
5
The Dissolution Process 溶解過程
• Solutions are homogeneous mixtures of two or more
substances.
– Dissolving medium is called the solvent.
– Dissolved species are called the solute.
• There are three states of matter (solid, liquid, and
gas) which when mixed two at a time gives nine
different kinds of mixtures.
– Seven of the possibilities can be homogeneous.
– Two of the possibilities must be heterogeneous.
6
The Dissolution Process
•
•
•
•
•
•
•
Solute
Solid
Liquid
Gas
Liquid
Solid
Gas
Gas
• Solid
• Liquid
Seven Homogeneous Possibilities
Solvent
Example
Liquid
salt water
Liquid
mixed drinks
Liquid
carbonated beverages
Solid
dental amalgams
Solid
alloys
Solid
metal pipes
Gas
air
Two Heterogeneous Possibilities
Gas
dust in air
Gas
clouds, fog
7
Spontaneity of the Dissolution
Process
Many solid do dissolve in liquids by endothermic processes.
A large increase in disorder of the solute during the
8
dissolution process
Spontaneity of the Dissolution
Process
•As an example of dissolution, let’s assume that
the solvent is a liquid.
–Two major factors affect dissolution of solutes
1.Change of energy content or enthalpy of solution,
Hsolution
– If Hsolution is exothermic (< 0) dissolution is favored
– If Hsolution is endothermic (> 0) dissolution is not
favored
2.Change in disorder, or randomness, of the solution
Smixing
– If Smixing increases (> 0) dissolution is favored
– If Smixing decreases (< 0) dissolution is not favored
9
Spontaneity of the Dissolution
Process
• Thus the best conditions for dissolution are :
– For the solution process to be exothermic .
•Hsolution < 0
– For the solution to become more disordered.
•Smixing > 0
10
Spontaneity of the Dissolution
Process
• Disorder in mixing a solution is very common.
– Smixing is almost always > 0.
• What factors affect Hsolution?
– There is a competition between several different
attractions.
• Weak solute-solute attractions favor solubility
• Weak solvent-solvent attractions favor solubility
• Strong solvent-solute attractions favor solubility
• Solute-solute attractions such as ion-ion attraction,
dipole-dipole, etc.
– Breaking the solute-solute attraction requires an
absorption of Energy.
11
Dissolution of Solids in Liquids
• The energy released (exothermic) when a mole of
formula units of a solid is formed from its constituent
ions (molecules or atoms for nonionic solids) in the
gas phase is called the crystal lattice energy.
(always negative)
Ionic solid:
M+(g)+ X-(g)  M+X-(s) + crystal lattice energy
•The crystal lattice energy is a measure of the
attractive forces in a solid.
•These attractions are strong, a large amount of
energy is released as the solid forms
•The crystal lattice energy increases as the
charge density increases.
12
Dissolution of Solids in Liquids
MX(s) + energy M+(g)+ X-(g)
•This process can be considered the
hypothetical first step in forming a solution of
a solid in a liquid
•It always endothermic
•The smaller the magnitude of the crystal
lattice energy, the more readily dissolution
occurs
13
Spontaneity of the Dissolution
Process
• Solvent-solvent attractions such as hydrogen bonding
in water.
– This also requires an absorption of Energy.
• Solvent-solute attractions, solvation, releases energy.
– If solvation energy is greater than the sum of the
solute-solute and solvent-solvent attractions, the
dissolution is exothermic, Hsolution < 0.
– If solvation energy is less than the sum of the
solute-solute and solvent-solvent attractions, the
dissolution is endothermic, Hsolution > 0.
14
Spontaneity of the Dissolution Process
• Solvent-solute attractions, solvation, releases energy.
– When the solvent is water, the more specific term is
hydration
– Hydration energy: the energy change involved in
the hydration of one mole of gaseous ions
Mn+(g)+ xH2O(l) M(OH2)xn++ energy (for cation)
Xy-(g)+ rH2O(l) X(H2O)ry-+ energy (for anion)
– Hydration is usually highly exothermic for ionic or
polar covalent compounds, because the polar
water molecules interact very strongly with ions
and polar molecules
– Nonpolar solids do not dissolve appreciably in
polar solvent, because they do not attract each
other
15
Spontaneity of the Dissolution
Process
亂度增加
16
Dissolution of Solids in Liquids
• Dissolution is a competition
between:
1. Solute -solute attractions
• crystal lattice energy for ionic solids
2. Solvent-solvent attractions
• H-bonding for water
3. Solute-solvent attractions
• Solvation or hydration (the solvent is
water) energy
17
Dissolution of Solids in Liquids
• Solvation is directed by the water to ion attractions
as shown in these electrostatic potentials.
18
Dissolution of Solids in Liquids
• In an exothermic dissolution, energy is released
when solute particles are dissolved.
– This energy is called the energy of solvation or
the hydration energy (if solvent is water).
• Let’s look at the dissolution of CaCl2.
19
Dissolution of Solids in Liquids
CaCl2 (s) 
Ca(OH 2 ) 6   2Cl H 2 O x
2
H 2O
-
where x is approximat ely 7 or 8
OH2
2+
Ca
H2O
H
OH2
H2O
OH2
OH2
O
O
H
H
H
H
ClH
O
H
O
H
20
21
Dissolution of Solids in Liquids
• The energy absorbed when one mole of formula
units becomes hydrated is the molar energy of
hydration.
M n + (g) + x H 2 O  M(OH 2 ) x   hydration E for M n +
n
X ( g )  n H 2 O  X(H 2 O)n   hydration E for X y y-
y
22
23
Dissolution of Liquids in
Liquids (Miscibility)
• Most polar liquids are miscible in other
polar liquids.
• In general, liquids obey the “like dissolves
like” rule.
– Polar molecules are soluble in polar solvents.
– Nonpolar molecules are soluble in nonpolar
solvents.
• For example, methanol, CH3OH, is very
soluble in water
24
50ml H2O
50ml H2SO4
Sulfuric acid is always
diluted by adding the
acid slowly and
carefully to water.
Water should never be
added to the acid
50ml H2SO4 +50ml H2O
25
Dissolution of Liquids in
Liquids (Miscibility)
• Nonpolar molecules essentially “slide” in between
each other.
– For example, carbon tetrachloride and benzene
are very miscible.
H
Cl
Cl C
Cl
H
Cl
H
C
C
C
C
C
C
H
H
H
26
Dissolution of Gases in Liquids
• Polar gases are more soluble in water than nonpolar
gases.
– This is the “like dissolves like” rule in action.
• Polar gases can hydrogen bond with water
• Some polar gases enhance their solubility by
reacting with water.
27
Dissolution of Gases in Liquids
• A few nonpolar gases are soluble in water
because they react with water.






CO2  g   H 2O  H 2CO3 
 H 3O aq   HCO 3 aq 
H 2O
very wea k acid
• Because gases have very weak solute-solute
interactions, gases dissolve in liquids in
exothermic processes.
28
Rates of Dissolution and
Saturation
• Finely divided solids dissolve more rapidly than
large crystals.
– Compare the dissolution of granulated sugar and
sugar cubes in cold water.
• The reason is simple, look at a single cube of NaCl.
Breaks
NaCl
up
many smaller crystals
• The enormous increase in surface area
helps the solid to dissolve faster.
29
Rates of Dissolution and
Saturation
• Saturated solutions have established an
equilibrium between dissolved and undissolved
solutes
– Examples of saturated solutions include:
• Air that has 100% humidity.
• Some solids dissolved in liquids.
30
Rates of Dissolution and
Saturation
• Symbolically this equilibrium is written
as:


MX s   M aq   X aq 
• In an equilibrium reaction, the forward rate
of reaction is equal to the reverse rate of
reaction.
31
Rates of Dissolution and
Saturation
• The solubilities of many solids
increase at higher temperature
Supersaturated solutions have
higher-than-saturated
concentrations of dissolved solutes.
– Metastable (temporarily stable)
32
33
Effect of Temperature on
Solubility
•According to LeChatelier’s Principle
–when stress is applied to a system at equilibrium,
the system responds in a way that best relieves the
stress.
–Since saturated solutions are at equilibrium,
LeChatelier’s principle applies to them.
•Possible stresses to chemical systems include:
1.Heating or cooling the system.
2.Changing the pressure of the system.
3.Changing the concentrations of reactants or
products.
34
Effect of Temperature on Solubility
• What will be the effect of heating or cooling the
water in which we wish to dissolve a solid?
– It depends on whether the dissolution is exo- or
endothermic.
• For an exothermic dissolution, heat can be considered as
a product.
LiBr s   
 Li  aq   Br  aq   48.8 kJ / mol
H 2O
+
-
– Warming the water will decrease solubility and
cooling the water will increase the solubility.
• Predict the effect on an endothermic dissolution
like this one.
KMnO 4 s   43.6 kJ / mol  
 K  aq   MnO 4  aq 
H 2O
+
-
35
Effect of Temperature on
Solubility
• For ionic solids that dissolve endothermically
dissolution is enhanced by heating.
• For ionic solids that dissolve exothermically dissolution
is enhanced by cooling.
• Be sure you understand these trends.
36
Effect of Temperature on
Solubility
37
Effect of Pressure on Solubility
• Pressure changes have little or no effect on solubility
of liquids and solids in liquids.
– Liquids and solids are not compressible.
• Pressure changes have large effects on the
solubility of gases in liquids.
– Sudden pressure change is why carbonated
drinks fizz when opened.
– It is also the cause of several scuba diving
related problems including the “bends”.
38
Effect of Pressure on Solubility
• The effect of pressure on the solubility of gases
in liquids is described by Henry’s Law.
Cgas = kPgas
Where Cgas =the concentration of gas
k= is constant for a particular gas and
solvent at a particular temperature
Pgas= the pressure of the gas
39
Effect of Pressure on Solubility
40
Molality重量莫耳濃度and Mole
Fraction莫耳分率
•In Chapter 3 we introduced two
important concentration units.
1.% by mass of solute
mass of solute
% w/w =
 100%
mass of solution
41
Molality and Mole Fraction
2.Molarity
moles of solute
M =
Liters of solution
• We must introduce two new
concentration units in this chapter.
42
Molality and Mole Fraction
• Molality is a concentration unit based on the
number of moles of solute per kilogram of solvent.
m=
moles of solute
kg of solvent
In dilute aqueous solutions molarity and
molality are nearly equal
43
Example 14-1: Molality
What is the molality of a solution that contain 128g of CH3OH in
108g of water?
The molecule weight of CH3OH is 32
? mole CH3OH=128/32=4 mole
4mole/108x10-3 kg =37m CH3OH
Example 14-2: Molality
How many grams of H2O must be used to dissolve 50.0g of
sucrose to prepare a 1.25m solution of sucrose, C12H22O11?
The molecule weight of C12H22O11is 342
1.25m=(50/342)/xkg
x=0.117kg H2O
44
Molality and Mole Fraction
Example 14-1: Calculate the molarity and the molality
of an aqueous solution that is 10.0% glucose,
C6H12O6. The density of the solution is 1.04 g/mL.
10.0% glucose solution has several medical uses. 1
mol C6H12O6 = 180 g
10% means: 10g C6H12O6 in 90g H2O
10.0g C6H12O6
?mol C6H12O6
180.0g C6H12O6
=0.617 m C6H12O6
=
m=
-3
90.0gx10 H2O
kg H2O
This is the concentration in molality
10.0g C6H12O6
?mol C6H12O6
180.0g C6H12O6
=
M=
L solution
(100g/1.04)x10-3 sol’n
=0.587 M C6H12O6
This is the concentration in molarity
45
Molality and Mole Fraction
Example 14-2: Calculate the molality of a solution that
contains 7.25 g of benzoic acid C6H5COOH, in 2.00
x 102 mL of benzene, C6H6. The density of benzene
is 0.879 g/mL. 1 mol C6H5COOH = 122 g
7.25g C6H5COOH
?mol C6H5COOH
122g C6H5COOH
=
kg C6H6
2.0x102 ml x0.879 C6H6
1000
=0.338 m C6H5COOH
46
Molality and Mole Fraction
• Mole fraction is the number of moles of one
component divided by the moles of all the
components of the solution
– Mole fraction is literally a fraction using moles
of one component as the numerator and moles
of all the components as the denominator.
• In a two component solution, the mole
fraction of one component, A, has the
symbol XA.
number of moles of A
XA 
number of moles of A + number of moles of B
47
Molality and Mole Fraction
• The mole fraction of component B - XB
number of moles of B
XB 
number of moles of A + number of moles of B
Note that X A  X B  1
The sum of all the mole fractions must equal 1.00.
48
Example 14-3: Mole Fraction
What are the fractions of CH3OH and H2O in the solution
described in Example 14-1? It contains 128g of CH3OH and
108g of H2O?
? mole CH3OH=128/32= 4 mol
? Mole H2O=108/18= 6 mol
XCH3OH= 4mol/(4mol+6mol) = 0.4
XH2O= 6mol/(4mol+6mol) = 0.6
49
Molality and Mole Fraction
• Example 14-3: What are the mole fractions of
glucose and water in a 10.0% glucose solution
(Example 14-1)?
Let the solution equal 100g
10.0g of glucose, 90.0g of water
?mole C6H12O6 = 10.0g/180=0.0556 mol C6H12O6
?mole H2O = 90.0g/18=5.0 mol H2O
5.0 mole H2O
XH2O =
= 0.989
5.0 mole H2O + 0.0556 mol C6H12O6
0.0556 mole C6H12O6
= 0.011
XC6H12O6 =
5.0 mole H2O + 0.0556 mol C6H12O6
50
Molality and Mole Fraction
The sulfuric acid in a car battery has a density of 1.225
g/cm3 and is 3.75 M. What is the molality, mole
fraction and percentage of sulfuric acid by mass in
this solution?
Let the solution volume is 1.0L
3.75 M means that 3.75 moles of H2SO4
Mass of solute
?g H2SO4=3.75 mole x 98=367.5 g H2SO4
?g solution=1.00L x 103 x 1.225g/ml = 1225 g solution
Mass of solution
Mass of solvent = 1225-367.5 = 857.5g water
m= moles of solute / kg of solvent
% = g of solute / g of solution
= 3.75 moles/0.8575kg
= 367.5 g/ 1225 g
= 4.37 m
= 30%
3.75 mol H2SO4
= 0.073
XH2SO4 = (857.5g/18) mole H O + 3.75 mol H SO
2
2
4
51
Colligative Properties of
Solutions
• Colligative properties are properties of solutions that
depend solely on the number of particles dissolved
in the solution.
– Colligative properties do not depend on the
kinds of particles dissolved.
• Colligative properties are a physical property of
solutions.
52
Colligative Properties of
Solutions
• There are four common types of colligative
properties:
1.Vapor pressure lowering
2.Freezing point depression
3.Boiling point elevation
4.Osmotic pressure
Vapor pressure lowering is the key to all four of the
colligative properties.
53
Lowering of Vapor Pressure
and Raoult’s Law
Lowering of vapor pressure
54
Lowering of Vapor Pressure
and Raoult’s Law
• Addition of a nonvolatile solute to a solution lowers
the vapor pressure of the solution.
– The effect is simply due to fewer solvent
molecules at the solution’s surface.
– The solute molecules occupy some of the spaces
that would normally be occupied by solvent.
• Raoult’s Law models this effect in ideal solutions .
• 溶液的組成決定了溶液的蒸汽壓。
• 根據拉午耳定律，溶質為非揮發性非電解質的
稀薄溶液，其蒸氣壓和溶劑的莫耳分率成正比。
55
Lowering of Vapor Pressure
and Raoult’s Law
• Derivation of Raoult’s Law.
Psolvent = Xsolvent P0solvent
Where psolvent = vapor pressure of solvent in solution
P0solvent = vapor pressure of pure solvent
Xsolvent = mole fraction of solvent in solution
Lowering of vapor pressure, Psolvent, is defined as:
Psolvent = P0solvent - Psolvent
= P0solvent –(Xsolvent)(P0solvent)
= (1-Xsolvent) P0solvent
Xsolute = 1- Xsolvent
Psolvent = XsoluteP0solvent
56
Lowering of Vapor Pressure
and Raoult’s Law
Example 14-4: Vapor Pressure of a Solution of
Nonvolatile Solute
Sucrose is a nonvolatile, nonionizing solute in water.
Determine the vapor lowering, at 25oC, of 1.25 m sucrose
solution in Example 14-2. Assume that the solution
behaves ideally. The vapor pressure of pure water at 25oC
is 23.8 torr.
1.25m sucrose= 1.25 mol/1kg solvent
Xsucrose= 1.25mol/(1.25mol+(1000/18)mol)
= 0.0220
Psolvent = XsoluteP0solvent
= (0.022)(23.8 torr)
= 0.524 torr
57
•若溶質為揮發性高蒸氣壓物質
–溶液的蒸汽壓等於溶液中每一成份的蒸汽壓的總和(包含
溶質和溶劑)
–每一成份的蒸汽壓等於此純物質的蒸汽壓和它在溶液中的
莫耳分率(mole fraction)的乘積
•溶液的蒸汽壓符合拉午耳定律的溶液稱為「理想溶液
(Ideal solution)」
–理想溶液中不論是溶質－溶質、溶劑－溶劑或溶質－溶劑
間的分子之間的引力相似
–此時溶液的體積具有加成性，符合1 + 1 = 2原則
–因為分子間的引力相近，混合時無能量改變，理想溶液在
混合時沒有能量的變化，不會吸熱或放熱。譬如苯
(Benzene)與甲苯(Toluene)混合溶液，兩者的分子大小相
近、化學結構和極性相似，可形成理想溶液。
58
volatile Solute
• When a solution consists of two components that
are very similar
– Such as heptane C7H16 and Octane C8H18
– Experiences nearly the same intermolecule force
– Ptotal =PA+ PB
Ptotal =XAP0A+ XBP0B
59
Example 14-5: Vapor Pressure of a Solution of
volatile Components
At 40oC, the vapor pressure of pure heptane 92.0 torr and the
vapor pressure of pure octane is 31.0 torr. Consider a
solution that contains 1.0 mole of hepatne and 4.0 mole of
octane. Calculate the vapor pressure of each component
and the total vapor pressure above the solution.
Xheptane= 1.0mol/(1.0mol+ 4.0mol) = 0.2
Xcctane= 1- 0.2 = 0.8
Pheptane = XheptaneP0heptane= (0.2)(92.0)=18.4 torr
Poctane = XoctaneP0octane= (0.8)(31.0)=24.8 torr
Ptotal = Pheptane + Poctane= 18.4 +24.8 =43.2 torr
60
Example 14-6: Composition of Vapor
Calculate the mole fractions of heptane and octane in the
vapor that is equilibrium with the solution in Example 14-5.
The mole fraction of a component in a gaseous mixture equals
the ratio of its partial pressure to the total pressure
Xheptane = Pheptane/Ptotal= 18.4/43.2= 0.426
Xoctane = Pocptane/Ptotal= 24.8/43.2= 0.574
61
• Some solution do not behave ideally over the entire
concentration range.
Positive deviation
Negative deviation
62
Boiling Point Elevation
• Addition of a nonvolatile solute to a solution raises
the boiling point of the solution above that of the
pure solvent.
– This effect is because the solution’s vapor
pressure is lowered as described by Raoult’s law.
– The solution’s temperature must be raised to
make the solution’s vapor pressure equal to the
atmospheric pressure.
• The amount that the temperature is elevated is
determined by the number of moles of solute
dissolved in the solution.
63
Boiling Point Elevation
• Boiling point elevation relationship is:
Tb = Kbm
Where Tb = boiling point elevation
m= molality concentration of solute
Kb = boiling point elevation constant for solvent
• Example 14-4: What is the normal boiling point of a 2.50
m glucose, C6H12O6, solution?
Tb = Kbm
= (0.512 oC/m)(2.5m)
=1.28 oC
Boiling point of the solution
=100+1.28
=101.28oC
64
65
66
Freezing Point Depression
• Addition of a nonvolatile solute to a solution lowers
the freezing point of the solution relative to the pure
solvent.
• See table 14-2 for a compilation of boiling point and
freezing point elevation constants.
67
68
Freezing Point Depression
• Relationship for freezing point depression is:
Tf = Kfm
Where Tf = freezing point depression of solvent
m= molality concentration of solute
Kb = freezing point depression constant for solvent
Example 14-5: Calculate the freezing point of a 2.50 m
aqueous glucose solution.
Tf = Kfm
= (1.86 oC/m)(2.5m)
=4.65 oC
Freezing point of the solution
=0-4.65
=-4.65oC
69
Freezing Point Depression
• Notice the similarity of the two relationships for
freezing point depression and boiling point
elevation.
 Tf  K f m vs.  Tb  K b m
• Fundamentally, freezing point depression and
boiling point elevation are the same
phenomenon.
– The only differences are the size of the effect which
is reflected in the sizes of the constants, Kf & Kb.
• This is easily seen on a phase diagram for a
solution.
70
Freezing Point Depression
Example 14-6: Calculate the freezing point of a solution
that contains 8.50 g of benzoic acid (C6H5COOH, MW =
122) in 75.0 g of benzene, C6H6.
molality = mole of benzoic acid / kg of benzene
= (8.50/122)/(75.0/1000)
= 0.929 m
Tf = Kfm
= (5.12 oC/m)(0.929m)
=4.76 oC
Freezing point of the solution
=5.48-4.76
=0.72oC
71
Determination of Molecular Weight by
Freezing Point Depression
•The size of the freezing point depression
depends on two things:
1.The size of the Kf for a given solvent, which
are well known.
2.And the molality concentration of the
solution which depends on the number of
moles of solute and the kg of solvent.
• If Kf and kg of solvent are known, as is often
the case in an experiment, then we can
determine # of moles of solute and use it to
determine the molecular weight.
72
Determination of Molecular Weight by
Freezing Point Depression
Example 14-7: A 37.0 g sample of a new covalent compound,
a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The
resulting solution froze at -5.58oC. What is the molecular
weight of the compound? (Kf =1.86)
Tf = Kfm  m = Tf /Kf
= (0-(5.58))(1.86)
=3.0m
3.0=mole of compound/0.2kg solvent
 Mole of compound = 0.6 mole
0.6mole =37/molecular weight
 The compound molecular weight = 37/0.6
=61.7 g/mol
73
Colligative Properties and
Dissociation of Electrolytes
• Electrolytes have larger effects on boiling point
elevation and freezing point depression than
nonelectrolytes.
– This is because the number of particles released in
solution is greater for electrolytes
• One mole of sugar dissolves in water to produce one
mole of aqueous sugar molecules.
• One mole of NaCl dissolves in water to produce two
moles of aqueous ions:
– 1 mole of Na+ and 1 mole of Cl- ions
74
Colligative Properties and
Dissociation of Electrolytes
• Remember colligative properties depend on the
number of dissolved particles.
– Since NaCl has twice the number of particles we
can expect twice the effect for NaCl than for sugar.
• The table of observed freezing point depressions in the
lecture outline shows this effect.
75
Colligative Properties and
Dissociation of Electrolytes
• Ion pairing or association of
ions prevents the effect from
being exactly equal to the
number of dissociated ions
76
Colligative Properties and
Dissociation of Electrolytes
• The van’t Hoff factor, symbol i, is used to introduce
this effect into the calculations.
• i is a measure of the extent of ionization or
dissociation of the electrolyte in the solution.
i =
Tf (actual)
Tf (if nonelectrolyte)
77
Colligative Properties and
Dissociation of Electrolytes
• i has an ideal value of 2 for 1:1 electrolytes like
NaCl, KI, LiBr, etc.
2O
Na + Cl- H

 Na +aq   Cl-aq  2 ions
formula unit
• i has an ideal value of 3 for 2:1 electrolytes
like K2SO4, CaCl2, SrI2, etc.
2O
Ca 2 + Cl-2 H

 Ca 2aq+   2 Cl-aq  3 ions
formula unit
78
79
Colligative Properties and
Dissociation of Electrolytes
• Example 14-8: The freezing point of 0.0100 m NaCl solution
is -0.0360oC. Calculate the van’t Hoff factor and apparent
percent dissociation of NaCl in this aqueous solution.
meffective = total number of moles of solute particles/kg solvent
First let’s calculate the i factor.
i =
Tf (actual)
Tf (if nonelectrolyte)
Tf (actual)= Kf meffective
meffective= 0.0194m
Kfm effective
m effective
=
= m
stated
Kfm stated
0.0360oC
Tf (actual)
=
meffective =
Kf
1.86oC/m
m effective
0.0194m =1.94
=
i = m
0.0100m
stated
80
Colligative Properties and
Dissociation of Electrolytes
• Next, we will calculate the apparent percent
dissociation.
• Let x = mNaCl that is apparently dissociated.
H2O
+
-
NaCl
Na + Cl
(0.01-x)m
xm
xm
meffective= [0.01-x+x+]m
=[0.01+x]m
=0.0194m
x=0.094m
mapp diss
x100%
Apparent % dissociation = m
stated
0.0094m
x100%
=
0.0100m
= 94%
81
Colligative Properties and
Dissociation of Electrolytes
• Example 14-9: A 0.0500 m acetic acid solution
freezes at -0.0948oC. Calculate the percent
ionization of CH3COOH in this solution.
CH3COOH
H+ + CH3COO-
(0.05-x)m
xm
xm
meff= [0.05-x+x+x]m=[0.05+x]m
0.0948oC
Tf
=0.051m
meff =
=
Tf = Kf meff
o
Kf
1.86 C/m
meff= [0.05+x]m =0.051m
x= 0.001m m
ionized
x100%
% ionized = m
original
0.0010m
x100%
=
0.0500m
= 2.0% ionized and 98.0% unionized 82
Osmotic Pressure
• Osmosis is the net flow of a solvent between
two solutions separated by a semipermeable
membrane.
– The solvent passes from the lower concentration
solution into the higher concentration solution.
• Examples of semipermeable membranes
include:
1. Cellophane and saran wrap
2. skin
3. cell membranes
83
Osmotic Pressure
84
Osmotic Pressure
net solvent flow
85
Osmotic Pressure
• Osmosis is a rate controlled phenomenon.
– The solvent is passing from the dilute solution into
the concentrated solution at a faster rate than in
opposite direction, i.e. establishing an
equilibrium.
• The osmotic pressure is the pressure exerted by a
column of the solvent in an osmosis experiment.
  MRT
where:  = osmotic pressure in atm
M = molar concentration of solution
L atm
R = 0.0821
mol K
T = absolute temperature
86
Osmotic Pressure
 For very dilute aqueous solutions, molarity and
molality are nearly equal.
 Mm
  mRT
for dilute aqueous solutions only
87
Osmotic Pressure
• Osmotic pressures can be very large.
– For example, a 1 M sugar solution has an osmotic pressure of
22.4 atm or 330 p.s.i.
• Since this is a large effect, the osmotic pressure
measurements can be used to determine the
molar masses of very large molecules such as:
1.Polymers
2.Biomolecules like
• proteins
• ribonucleotides
88
Osmotic Pressure
Example 14-18: A 1.00 g sample of a biological material was
dissolved in enough water to give 1.00 x 102 mL of solution.
The osmotic pressure of the solution was 2.80 torr at 25oC.
Calculate the molarity and approximate molecular weight
of the material.
  MRT  M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
4
M =

150
.

10
M
L atm
0.0821 mol K 298 K
89
Osmotic Pressure
  MRT  M 

RT
1 atm
? atm = 2.80 torr 
 0.00368 atm = 
760 torr
0.00368 atm
4
M =

150
.

10
M
L atm
298 K 
0.0821 mol
K 
?g
1.00 g
1L
4g


 6.67  10 mol
4
mol 0.100 L 150
.  10 M
typical of small proteins
90
Osmotic Pressure
Water Purification by Reverse Osmosis
• If we apply enough external pressure to an osmotic
system to overcome the osmotic pressure, the
semipermeable membrane becomes an efficient
filter for salt and other dissolved solutes.
– Ft. Myers, FL gets it drinking water from the Gulf of
Mexico using reverse osmosis.
– US Navy submarines do as well.
– Dialysis is another example of this phenomenon.
91
Osmotic Pressure
92
Osmotic Pressure
Hypotonic solution Isotonic solution
Hypertonic solution
93
Colloids
•Colloids are an intermediate type of mixture
that has a particle size between those of true
solutions and suspensions.
–The particles do not settle out of the solution but
they make the solution cloudy or opaque.
•Examples of colloids include:
1.Fog
2.Smoke
3.Paint
4.Milk
5.Mayonnaise
6.Shaving cream
7.Clouds
94
95
96
The Tyndall Effect
• Colloids scatter light when it is shined upon them.
– Why they appear cloudy or opaque.
– This is also why we use low beams on cars when
driving in fog.
Distilled water
Tap water 500ml distilled water
1 drop milk
97
The Adsorption Phenomenon
• Colloids have very large surface areas
– They interact strongly with substances near their
surfaces.
• One of the reasons why rivers can carry so much
suspended silt in the water.
98



2  Fe 2+  3 Cl -   y  3 H 2 O  Fe 2 O 3  y H 2 O + 6 H + + Cl -

A colloidal particle contains many Fe 2 O 3  y H 2 O units with Fe 3+ ions
bound to its surface. The + charged particles repel each other and
99
keep the colloid from precipitating.
Hydrophilic and Hydrophobic
Colloids
• Hydrophilic colloids like water and are water
soluble.
– Examples include many biological proteins like blood
plasma.
• Hydrophobic colloids dislike water and are water
insoluble.
– Hydrophobic colloids require emulsifying agents to
stabilize in water.
• Homogenized milk is a hydrophobic colloid.
– Milk is an emulsion of butterfat and protein particles
dispersed in water
– The protein casein is the emulsifying agent.
100
Hydrophilic and Hydrophobic
Colloids
• Mayonnaise is also a hydrophobic colloid.
– Mayonnaise is vegetable oil and eggs in a colloidal
suspension with water.
– The protein lecithin from egg yolk is the emulsifying
agent.
• Soaps and detergents are excellent emulsifying
agents.
– Soaps are the Na or K salts of long chain fatty acids.
– Sodium stearate is an example of a typical soap.
101
Hydrophilic and Hydrophobic
Colloids
• Sodium stearate
102
Hydrophilic and Hydrophobic
Colloids
micelle
103
Hydrophilic and Hydrophobic
Colloids
• So called “hard water” contains Fe3+, Ca2+, and/or
Mg2+ ions
– These ions come primarily from minerals that are
dissolved in the water.
• These metal ions react with soap anions and
precipitate forming bathtub scum and ring around
the collar.
2
Ca  soap anion  Ca(soap anion) 2(s)
insoluble scum
• Synthetic detergents were designed as soap substitutes
that do not precipitate in hard water
• Detergents are good emulsifying agents.
• Chemically, we can replace COO- on soaps with sulfonate
104
or sulfate groups
Hydrophilic and Hydrophobic
Colloids
• The use of detergents containing phosphates is now
discouraged because they cause eutrophication in
rivers
105
Hydrophilic and Hydrophobic
Colloids
• Linear alkylbenzenesulfonates are good detergents.
106