Transcript Document 7565894

CHAPTER 13

Solutions
1
Chapter Goals
1.
2.
3.
4.
5.
6.
7.
8.
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
2
Chapter Goals
9.
10.
11.
12.
13.
14.
15.
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
3
Chapter Goals
16.
17.
18.
Colloids
The Tyndall Effect
The Adsorption Phenomena
Hydrophilic and Hydrophobic Colloids
4
The Dissolution Process

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.
5
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
6
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
Change of energy content or enthalpy of
solution, Hsolution


1.


If Hsolution is exothermic (< 0) dissolution is favored.
If Hsolution is endothermic (> 0) dissolution is not
favored.
7
Spontaneity of the Dissolution Process
Change in disorder, or randomness, of the
solution Smixing
2.


If Smixing increases (> 0) dissolution is favored.
If Smixing decreases (< 0) dissolution is not favored.
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
8
Spontaneity of the Dissolution Process

Disorder in mixing a solution is very common.


What factors affect Hsolution?


Smixing is almost always > 0.
There is a competition between several different
attractions.
Solute-solute attractions such as ion-ion
attraction, dipole-dipole, etc.

Breaking the solute-solute attraction requires an
absorption of E.
9
Spontaneity of the Dissolution Process

Solvent-solvent attractions such as hydrogen
bonding in water.


This also requires an absorption of E.
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.
10
Spontaneity of the Dissolution Process
11
Dissolution of Solids in Liquids

M
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.
+
(g)
 X (g)  M X (s)  crystal lattice energy
-
+
-
•The crystal lattice energy is a measure of
the attractive forces in a solid.
•The crystal lattice energy increases as the
charge density increases.
12
Dissolution of Solids in Liquids
Dissolution is a competition between:

Solute -solute attractions
1.

crystal lattice energy for ionic solids
Solvent-solvent attractions
2.

H-bonding for water
Solute-solvent attractions


Solvation or hydration energy
13
Dissolution of Solids in Liquids

Solvation is directed by the water to ion
attractions as shown in these electrostatic
potentials.
14
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.
15
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+
H
OH2
H2O
H
H
H
Ca
H2O
O
Cl-
O
OH2
H
H
H
OH2
O
H
O
16
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 yy-
y
17
Dissolution of Solids in Liquids
Hydration energy increases with increasing charge
density
Ion Radius(Å) Charge/radius
Heat of Hydration
K+
1.33
0.75
-351 kJ/mol
Ca2+ 0.99
2.02
-1650 kJ/mol
Cu2+ 0.72
2.78
-2160 kJ/mol
Al3+
0.50
6.00
-4750 kJ/mol

18
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
19
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
C
H
Cl
H
H
C
C
C
C
C
H
H
20
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.
+
HBr + H 2 O  H 3O aq   Br aq 
strong acid
SO 2 + H 2 O  H 2SO 3aq 
H 2SO 3(aq)
+





 H 3O aq   HSO 3 aq 
H 2O
weak acid
21
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   HCO3 aq 
H 2O
very wea k acid

Because gases have very weak solute-solute
interactions, gases dissolve in liquids in exothermic
processes.
22
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
up
many smaller crystals
NaCl

The enormous increase in surface area helps the solid to
dissolve faster.
23
Rates of Dissolution and Saturation

Saturated solutions have established an
equilbrium between dissolved and undissolved
solutes

Examples of saturated solutions include:


Air that has 100% humidity.
Some solids dissolved in liquids.
24
Rates of Dissolution and Saturation

Symbolically this equilibrium is written as:

In an equilibrium reaction, the forward rate of
reaction is equal to the reverse rate of reaction.


MX s   Maq   Xaq 
25
Rates of Dissolution and Saturation

Supersaturated solutions have higher-thansaturated concentrations of dissolved solutes.
26
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.
2.
3.
Heating or cooling the system.
Changing the pressure of the system.
Changing the concentrations of reactants or products.
27
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 .8kJ/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
+
-
28
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.
29
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”.
30
Effect of Pressure on Solubility

The effect of pressure on the solubility of gases in
liquids is described by Henry’s Law.
 k M gasgas
Pgas
gas  k M
where M
M gas
 molar concentrat ion of gas
where
gas  molar concentrat ion of gas
k = Henry' s Law constant, unique number for each
gas - liquid combinatio n
Pgas = partial pressure of gas
31
Molality and Mole Fraction

1.
In Chapter 3 we introduced two important
concentration units.
% by mass of solute
mass of solute
% w/w =
100%
mass of solution
32
Molality and Mole Fraction
2.
Molarity
moles of solute
M =
Liters of solution

We must introduce two new concentration
units in this chapter.
33
Molality and Mole Fraction

Molality is a concentration unit based on the
number of moles of solute per kilogram of
solvent.
moles of solute
m
kg of solvent
in dilute aqueous solutions molarity and
molality are nearly equal
34
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
C66 H1210
O.60 g C
106 .H
012gOC66 H121000
O 6 g 1000
? mol C?6mol
H12O
H 2 Og H 2 O
 .0 g C 6 H 12 O6



? mol C 6 H 12O 6 10

1 kg H12 O
kg H 2 O
gH
kg H 2 O 90.0
90.0
kg H 2 O
2O g H 2O
kg H 2 O
90.0 g H 2 O
1 mol C6 H12O 6
 0.617m C 6 H12O 6
180 g C 6 H12O 6
This is the concentrat ion in molality .
35
Molality and Mole Fraction

Example 14-1: Calculate the molality and the
molarity 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
You calculate the molarity!
? mol C 6 H12O 6 10.0 g C 6 H12O 6 1.04 g sol' n



L H 2O
100.0 g sol' n
mL sol' n
1000 mL 1 mol C 6 H12O 6

 0.578 M C 6 H12O 6
1L
180 g C 6 H12O 6
36
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
You do it!
? mol C 6 H 5COOH 7.25g C 6 H 5COOH
1 mL C 6 H 6


kg C 6 H 6
200.0 mL C 6 H 6
0.879 g C 6 H 6
1000 g C 6 H 6 1 mol C 6 H 5COOH

 0.338 m C 6 H 5COOH
1 kg C 6 H 6
122 g C 6 H 5COOH
37
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
38
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.
39
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)?
You do it!
In 1.00 10 g of this solution t here are
10.0 g of glucose and 90.0 g of water.
? mol C 6 H12O 6  10.0 g C 6 H12O 6 
2
1 mol C 6 H12O 6
 0.0556 mol C 6 H12O 6
180 g C 6 H12O 6
40
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)?
1 mol H 2O
? mol H 2O  90.0 g H 2O 
 5.00 mol H 2O
18 g H 2O
41
Molality and Mole Fraction

Now we can calculate the mole fractions.
X H 2O
5.00 mol H 2 O

5.00 mol H 2 O + 0.0556 mol C 6 H12O 6
 0.989
X C 6 H1 2O 6
0.0556 mol C 6 H12O 6

5.00 mol H 2 O + 0.0556 mol C 6 H12O 6
 0.011
1.00  0.989  0.011
42
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.
43
Colligative Properties of Solutions
There are four common types of colligative
properties:

1.
2.
3.
4.

Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
Vapor pressure lowering is the key to all
four of the colligative properties.
44
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.
45
Lowering of Vapor Pressure and
Raoult’s Law

Derivation of Raoult’s Law.
0
Psolvent  X solvent Psolvent
where Psolvent  vapor pressure of solvent in solution
0
Psolvent
 vapor pressure of pure solvent
X solvent  mole fraction of solvent in solution
46
Lowering of Vapor Pressure and
Raoult’s Law

Lowering of vapor pressure, Psolvent, is defined as:
Psolvent  P
0
solvent
 Psolvent
0
0
 Psolvent
- ( X solvent)( Psolvent
)
 (1  X solvent)P
0
solvent
47
Lowering of Vapor Pressure and
Raoult’s Law


Remember that the sum of the mole fractions
must equal 1.
Thus Xsolvent + Xsolute = 1, which we can
substitute into our expression.
X solute  1 - X solvent
Psolvent  X solute P
0
solvent
which is Raoult' s Law
48
Lowering of Vapor Pressure and
Raoult’s Law

This graph shows how the solution’s vapor pressure
is changed by the mole fraction of the solute, which
is Raoult’s law.
49
Fractional Distillation


Distillation is a technique used to separate solutions
that have two or more volatile components with
differing boiling points.
A simple distillation has a single distilling column.


Simple distillations give reasonable separations.
A fractional distillation gives increased separations
because of the increased surface area.

Commonly, glass beads or steel wool are inserted into the
distilling column.
50
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.
51
Boiling Point Elevation

Boiling point elevation relationship is:
Tb  K b m
where : Tb  boiling point elevation
m  molal concentrat ion of solution
K b  molal boiling point elevation constant
for the solvent
52
Boiling Point Elevation

Example 14-4: What is the normal boiling
point of a 2.50 m glucose, C6H12O6, solution?
Tb  K b m
Tb  (0.512 0 C/m)( 2.50m)
Tb  1.280 C
Boiling Point of the solution = 100.0 0 C + 1.280 C = 101.280 C
53
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.
54
Freezing Point Depression

Relationship for freezing point depression is:
Tf  K f m
where: Tf  freezing point depression of solvent
m  molal concentration of soltuion
K f  freezing point depression constant for solvent
55
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.
56
Freezing Point Depression
57
Freezing Point Depression

Example 14-5: Calculate the freezing point of
a 2.50 m aqueous glucose solution.
Tf  K f m
Tf  (1.86 0 C/m)( 2.50m)
Tf  4.650 C
Freezing Point of solution = 0.000 C - 4.650 C = - 4.650 C
58
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.
You do it!
59
Freezing Point Depression
1. Calculate molality!
? mol C 6 H 5 COOH 8.50 g C 6 H 5 COOH


kg C 6 H 6
0.0750 kg C 6 H 6
1 mol C 6 H 5 COOH
 0.929m
122 g C 6 H 5 COOH
2. Calculate the depression for this solution.
Tf  K f m
Tf  (5.12 0 C/m)( 0.929m)  4.76 0 C
F.P. = 5.480 C - 4.76 0 C = 0.72 0 C
60
Determination of Molecular Weight
by Freezing Point Depression
The size of the freezing point depression
depends on two things:

1.
2.

The size of the Kf for a given solvent, which are
well known.
And the molal 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.
61
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?
62
Determination of Molecular Weight
by Freezing Point Depression
Tf  K f m thus the
Tf 5.580 C
m

 3.00m
0
K f 1.86 C
In this problem there are
200 mL  0.200 kg of water.
? mol compound in 0.200 kg H 2 O = 3.00 m  0.200 kg
 0.600 mol compound
37 g
Thus the molar mass is
 61.7 g/mol
0.600 mol
63
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
64
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.
65
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
66
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
67
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
68
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.
Tf  actual 
K f meffective meffective
i


Tf  if nonelectrolyte 
K f mstated
mstated
69
Colligative Properties and Dissociation
of Electrolytes
Tf  actual 
K f meffective meffective
i


Tf  if nonelectrolyte 
K f mstated
mstated
Tf  actual 
0.0360 0 C
Tf  actual   K f meffective  meffective 

Kf
1.86 0 C m
meffective 0.0194 m
meffective  0.0194 m  i 

 194
.
mstated
0.0100 m
70
Colligative Properties and Dissociation
of Electrolytes


Next, we will calculate the apparent percent
dissociation.
Let x = mNaCl that is apparently dissociated.
71
Colligative Properties and Dissociation
of Electrolytes
NaCl 
 Na + Cl
(0.0100  x)m
xm
xm
H 2O
+
-
72
Colligative Properties and Dissociation
of Electrolytes
NaCl 
 Na + Cl
H 2O
meffective
+
(0.0100  x)m
xm
 0.0100  x  x  x  m
-
xm
 0.0100  x  m  0.0194 m
x  0.0094 m
73
Colligative Properties and Dissociation
of Electrolytes
apparent % dissociati on =
mapp diss
mstated
100%
0.0094m

100%
0.0100m
 94%
74
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.
You do it!
75
Colligative Properties and Dissociation
of Electrolytes
 H + + CH COOCH 3COOH 
3
0.0500  x m x m
xm
meff  0.0500  x   x  x  m  0.0500  x  m
Tf  K f  meff
meff
Tf
0.09480 C


 0.0510 m
0
Kf
1.86 C m
meff  0.0500  x  m  0.0510 m
x  0.0010 m
% ionized =
mionized
0.0010 m
 100% 
 100%
moriginal
0.0500 m
 2.0% ionized and 98.0% unionized
76
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.
2.
3.
cellophane and saran wrap
skin
cell membranes
77
Osmotic Pressure
semipermeable membrane
sugar dissolved
H2O
in water
H2O
H2O
H2O
H2O
H2O
H2O
net2solvent
flow
O
78
Osmotic Pressure
79
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
80
Osmotic Pressure
 For very dilute aqueous solutions, molarity
and molality are nearly equal.
 Mm
  mRT
for dilute aqueous solutions only
81
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:

Polymers
Biomolecules like
1.
2.


proteins
ribonucleotides
82
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.
You do it!
83
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
0.0821 mol K 298 K
84
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
85
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.
86
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.
2.
3.
4.
5.
6.
7.
Fog
Smoke
Paint
Milk
Mayonnaise
Shaving cream
Clouds
87
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.

See Figure 14-18 in Textbook.
88
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.




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 Fe3+ ions
bound to its surface. The + charged particles repel each other and
keep the colloid from precipitating.
89
Hydrophilic and Hydrophobic Colloids

Hydrophilic colloids like water and are water soluble.


Hydrophobic colloids dislike water and are water insoluble.


Examples include many biological proteins like blood plasma.
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.
90
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.
91
Hydrophilic and Hydrophobic Colloids

Sodium stearate
O
nonpolar tail
hydrophobic portion
CH2 C
O-Na+
CH2 CH2
CH2 CH2
H3C CH2
CH2 CH2
CH2 CH2
CH2 CH2
CH2 CH2
CH2 CH2
polar (ionic) head
hydrophilic portion
92
Hydrophilic and Hydrophobic Colloids
93
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.
Ca
2
 soap anion  Ca(soap anion)2(s)
insoluble scum
94
Hydrophilic and Hydrophobic Colloids

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 or sulfate groups
95
Hydrophilic and Hydrophobic Colloids

Linear alkylbenzenesulfonates are good
detergents.
C H2
C H3
C H2
C H2
C H2
C H2
C H2
C H2
C H2
C H2
C H2
C H2
O
S
O- N a +
O
96
Synthesis Question

The world’s record for altitude in flying gliders
was 60,000 feet for many years. It was set
by a pilot in Texas who flew into an updraft in
front of an approaching storm. The pilot had
to fly out of the updraft and head home not
because he was out of air, there was still
plenty in the bottle of compressed air on
board, but because he did not have a
pressurized suit on. What would have
happened to this pilot’s blood if he had
continued to fly higher?
97
Synthesis Question

As the pilot flew higher, the atmospheric
pressure became less and less. With the
lower atmospheric pressure, eventually the
blood in the pilot’s veins would have begun to
boil. This is a deadly phenomenon which the
pilot wisely recognized.
98
Group Question

Medicines that are injected into humans,
intravenous fluids and/or shots, must be at
the same concentration as the existing
chemical compounds in blood. For example,
if the medicine contains potassium ions, they
must be at the same concentration as the
potassium ions in our blood. Such solutions
are called isotonic. Why must medicines be
formulated in this fashion?
99
End of Chapter 14

Human Beings are solution chemistry in
action!
100