Properties of Solutions

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Chapter 17

• Properties of Solutions

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Chapter 17 Properties of Solutions
17.1 Solution Composition 17.2 The
Thermodynamics of Solution Formation 17.3
Factors Affecting Solubility 17.4 The Vapor
Pressure of Solutions 17.5 Boiling-Point
Elevation and Freezing-Point Depression 17.6
Osmotic Pressure 17.7 Colligative Properties
of Electrolytic Solutions 17.8 Colloids
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The left beaker contains copper sulfate solution.
In the right beaker, ammonia is being added to
create a precipitate of copper(II) hydroxide.
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Definitions for Solutions
Solute - The smaller (in mass) of the components
in a solution, the material
dispersed into the solvent. Solvent - The major
component of the solution, the material that the
solute is dissolved
into. Solubility - The maximum amount that can
be dissolved into a
particular solvent to form a stable solution at a
specified temperature. Miscib
le - Substances that can dissolve in any
proportion, so that it is
difficult to tell which is the solvent or solute!
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Hydration shells around an aqueous
ion
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Solution Composition
Mass percent weight percent
Mass percent
x 100 Mole fraction symbolized by the Greek
letter, chi X mole fraction of component A
XA Molarity M (chapter 4)
M Molality m
m
grams of solute grams of solution
nA nA nB
moles of solute liter of solution
moles of Solute Kg of solvent
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Like Dissolves Like

• Polar molecules - dissolve best in Polar
  solvents.
• Polar molecules can hydrogen bond with polar
  solvents, such as water, hence increasing their
  solubility.
• Non-polar molecules - dissolve best in non -
  polar solvents.
• Hydrocarbons, non - polar molecules, do not
  dissolve, or mix with water!

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Like dissolves Like Solubility of methanol in
water
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Fig 13.4 (P 486) The structure and
function of a soap
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Predicting Relative Solubility's of Substances -I
Problem Predict which solvent will dissolve
more of the given solute. (a) Sodium Chloride
in methanol (CH3OH) or in propanol
(CH3CH2CH2OH). (b) Ethylene glycol
(HOCH2CH2OH) in water or in hexane

(CH3CH2CH2CH2CH2CH3). (c) Diethyl ether
(CH3CH2OCH2CH3) in ethanol (CH3CH2OH) or in


water. Plan Examine the formulas of each solute
and solvent to determine which forces will occur.
A solute tends to be more soluble in a
solvent which has the same type of forces binding
its molecules.
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Predicting Relative Solubility's of Substances -
II
Solution (a) Methanol - NaCl is an ionic
compound that dissolves through ion- dipole
forces. Both methanol and propanol contain a
polar hydroxyl group, and propanols longer
hydrocarbon chain would form only weak forces
with the ions, so it would be less effective at
replacing the ionic attractions of the
solvent. (b) Water Ethylene glycol molecules
have two -OH groups, and the molecules interact
with each other through H bonding. They would
be more soluble in water, whose H bonds can
replace solute H bonds better than can the
dispersion forces in hexane. (c) Ethanol Diethyl
ether molecules interact with each other
through dipole and dispersion forces and could
form H bonds to both water and ethanol. the
ether would be more soluble in ethanol because
the solvent can form H bonds and replace the
dispersion forces in the solute, whereas the H
bonds in water must be partly replaced with much
weaker dispersion forces.
19
An Energy Solution? Solar Ponds
Solar ponds are shallow bodies of salt water ( a
very high salt content) designed to collect
solar energy as it water the water in the ponds,
and then it can be used for heating, or
converted into other forms of energy. Sufficient
salt must be added to establish a salt gradient
in a pool 2-3 meters deep, with a dark bottom. A
salt gradient will be established in which the
upper layer, called the conductive layer, has a
salt content of about 2 by mass. The bottom
layer, called the heat storage layer has a salt
content of about 27. The middle layer called the
nonconvective layer has an intermediate salt
content, and acts as an insulator between the
two layers. The water in the deepest layer can
reach temperatures of between 90 and 100oC,
temperatures as high as 107oC have been
reported. A 52 acre pond near the Dead sea in
Israel can produce up to 5 mega watts of power.
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Three Steps in making a Solution
Step 1 Breaking up the solute into individual
components (expanding the
Solute) Step 2 Overcoming intermolecular
forces in the solvent to make room for the
solute (expanding the solvent) Step 3
Allowing the solvent and solute to interact
and form the solution.
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Figure 17.1 The formation of a liquid solution
can be divided into three steps
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Figure 17.2 (a) Enthalpy of solution Hsoln
has a negative sign (the process is exothermic)
if Step 3 releases more energy than is required
by Steps 1 and 2. (b) Hsoln has a positive
sign (the process is endothermic) if Steps 1 and
2 require more energy than is released in Step 3.
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H of solution for Sodium Chloride
NaCl(s) Na (g) Cl (g)
Ho1 786 kJ/mol H2O(l) Na(g) Cl
(g) Na(aq) Cl (aq)
Hohyd Ho2 Ho3
_____________ kJ/mol Hhyd enthalpy
(heat) of hydration Hosoln 786 kJ/mol
_______ kJ/mol _______ kJ/mol The dissolving
process is positive, requiring energy. Then why
is NaCl so soluble? The answer is in the Gibbs
free energy equation from chapter 10, G
H T S The entropy term T S
is Negative, and the result is that G becomes
negative, and as a Result, NaCl dissolves very
well in the polar solvent water.
24
French Navy ship L'Ailette equipped with vacuum
pumps approaches an oil slick.
Source AP/Wide World Photos
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Solution Cycle
Step 1 Solute separates into Particles -
overcoming attractions
Therefore -- Endothermic Step 2 Solvent
separates into Particles - overcoming
intermolecular attractions
Therefore -- Endothermic solvent
(aggregated) heat
solvent (separated)
Hsolvent gt 0 Step 3
Solute and Solvent Particles mix - Particles
attract each other
Therefore -- Exothermic solute (separated)
solvent (separated)
solution heat
Hmix lt 0
The Thermochemical Cycle Hsolution
Hsolute Hsolvent Hmix
If HEndothermic Rxn lt HExothermic Rxn
solution becomes warmer If
HEndothermic Rxn gt HExothermic Rxn
solution becomes colder
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Solution Cycles and the Enthalpy Components
ofthe Heat of Solution
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Multistage supercritical fluid extraction
apparatus
Source USDA
30
Chicken fat obtained by using supercritical
carbon dioxide as the extracting agent.
Source USDA
31
Figure 17.3 Vitamin A, C
A Fat Soluble Vitamin A
Water Soluble Vitamin A Hydrophobic Vitamin
A Hydrophilic Vitamin
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Carbonation in a bottle of soda
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Figure 17.4 (a) a gaseous solute in equilibrium
with a solution. (b) the piston is pushed in,
which increases the pressure of the gas and the
number of gas molecules per unit volume. (c)
greater gas
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Henrys Law of Gas solubilities in Liquids P
kHX P Partial pressure of dissolved
gas X mole fraction of dissolved gas kH
Henrys Law Constant
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Henrys Law of Gas Solubility
Problem The lowest level of oxygen gas
dissolved in water that will support life is
1.3 x 10 - 4 mol/L. At the normal atmospheric
pressure of oxygen is there adaquate oxygen to
support life? Plan We will use Henrys law and
the Henrys law constant for oxygen in water
with the partial pressure of O2 in the air to
calculate the amount. Solution
The Henrys law constant for oxygen in water is
1.3 x 10 -3 mol

liter atm and the partial pressure of oxygen
gas in the atmosphere is 21, or 0.21 atm.
.
Soxygen kH x PO2 1.3 x 10 -3 mol x ( 0.21
atm) liter
atm SOxygen
mol O2 / liter
.
This is adaquate to sustain life in water!
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Figure 17.5 The solubilities of several solids
as a function of temperature.
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Predicting the Effect of Temperature
on Solubility - I
Problem From the following information, predict
whether the solubility of each compound
increases or decreases with an increase
in temperature. (a) CsOH Hsoln -72
kJ/mol (b) When CsI dissolves in water the
water becomes cold (c) KF(s)
K(aq) F -(aq) 17.7 kJ Plan We use the
information to write a chemical reaction that
includes heat being absorbed (left) or released
(right). If heat is on the left, a temperature
shifts to the right, so more solute dissolves. If
heat is on the right, a temperature increase
shifts the system to the left, so less solute
dissolves. Solution (a) The negative H
indicates that the reaction is exothermic, so
when one mole of Cesium Hydroxide dissolves 72
kJ of heat is released.
H2O
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Predicting the Effect of Temperature
on Solubility - II
(a) continued
H2O
CsOH(s) Cs(aq)
OH -(aq) Heat
A higher temperature (more heat) decreases the
solubility of CsOH.
(b) When CsI dissolves, the solution becomes
cold, so heat is absorbed.
H2O
CsI(s) Heat
Cs(aq) I -(aq)
A higher temperature increases the solubility of
CsI.
(c) When KF dissolves, heat is on the product
side, and is given off so the reaction is
exothermic.
H2O
KF(s) K(aq) F
-(aq) 17.7 kJ
A higher temperature decreases the solubility of
KF
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Figure 17.6 The solubilities of several gases
in water as a function of temperature at a
constant pressure of 1 atm of gas above the
solution.
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Figure 17.7 Pipe with accumulated mineral
deposits (left) lengthwise section (right)
Source Visuals Unlimited
41
Lake Nyos in Cameroon
Source Corbis
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Figure 17.8 An aqueous solution and pure water
in a closed environment
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Figure 17.9 The presence of a nonvolatile
solute inhibits the escape of solvent molecules
from the liquid
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Figure 17.10 For a solution that obeys Raoults
law, a plot of Psoln versus xsolvent yields a
straight line.
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Fig. 13.15
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Vapor Pressure Lowering -I
Problem Calculate the vapor pressure lowering
when 175g of sucrose is
dissolved into 350.00 ml of water at 750C. The
vapor pressure of pure water
at 750C is 289.1 mm Hg, and its
density is 0.97489 g/ml. Plan Calculate the
change in pressure from Raoults law using the
vapor pressure of pure water at 750C. We
calculate the mole fraction of sugar
in solution using the molecular formula of
sucrose and density of water at
750C. Solution
molar mass of sucrose ( C12H22O11) 342.30 g/mol
175g sucrose 342.30g sucrose/mol
0.51125 mol sucrose
350.00 ml H2O x 0.97489g H2O 341.21g H2O
ml H2O

341.21 g H2O 18.02g H2O/mol
______ molH2O
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Vapor Pressure Lowering - II
mole sucrose moles of water moles
of sucrose
Xsucrose
0.51125 mole sucrose 18.935 mol H2O
0.51125 mol sucrose
Xsurose
0.2629
P Xsucrose x P 0H2O 0.2629 x 289.1 mm Hg
________ mm Hg
48
Like Example 17.1 (P 841-2)
A solution was prepared by adding 40.0g of
glycerol to 125.0g of water at 25.0oC, a
temperature at which pure water has a vapor
pressure of 23.76 torr. The observed vapor
pressure of the solution was found to be 22.36
torr. Calculate the molar mass of
glycerol! Solution Roults Law can be rearranged
to give XH2O
0.9411 mol H2O
6.94 mol H2O
0.9411
mol gly
0.4357 mol
Psoln PoH2O
22.36 torr 23.76 torr
mol H2O mol gly mol H2O
125.0 g 18.0 g/mol
6.94 mol mol gly 6.96 mol
6.94 mol (6.94 mol)(0.9411) 0.9411
40.0 g 0.4357 mol
g/mol
(MMglycerol 92.09 g/mol)
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Figure 17.11 Vapor pressure for a solution of
two volatile liquids.
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-H
H H H H H H H-C-C-C-C-C-C-H H H H H
H H Hexane H H
H-C-C-O-H Ethanol H
H
H-O
Acetone Water
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Figure 17.12 Phase diagrams for pure water (red
lines) and for an aqueous solution containing a
nonvolatile solution (blue lines).
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Like Example 17.2 (P 845-6)
A solution is prepared by dissolving 62g of
sucrose in 150.0g of water the resulting solution
was found to have a boiling point of
100.61oC. Calculate the molecular mass of
sucrose. Solution T
kbmsolute kb 0.51 T 100.61oC
100.00 oC 0.61oC msolute
1.20
mol/Kg Msolute
mol solute (0.150 kg)(1.2 mol/kg) mol
solute 0.18 mol MM
_________g/mol
oC Kg msolute
0.61oC
T kb
oC Kg msolute
0.51
mol solute kg solvent
62g 0.18 mol
(MMsucrose 342.18 g/mol)
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Figure 17.13 Ice in equilibrium with liquid
water.
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Like Example 17.3 (P847)
What mass of ethanol (C2H6O) must be added to
20.0 liters of water to keep it from freezing at
a temperature of -15.0oF? Solution
oC (oF 32)5/9 (-15 32)5/9 -26.1oC
T kf msolute msolute
14.0 mol/kg 14.0
mol/kg(20 kg H2O) 280 mol ethanol
Ethanol 2x12.01 6x 1.008 1x16.0 46.07
280 mol ethanol (46.07 g ethanol/mol)
___________ kg ethanol
T kf
-26.1oC 1.86
oC kg mol
58
Determining the Boiling Point Elevation and
Freezing Point Depression of an Aqueous
Solution
Problem We add 475g of sucrose (sugar) to 600g
of water. What will be the Freezing Point and
Boiling Points of the resultant solution? Plan
We find the molality of the sucrose solution by
calculating the moles of sucrose and dividing by
the mass of water in kg. We then apply the
equations for FP depression and BP elevation
using the constants from table 12.4. Solution
Sucrose C12H22O11 has a molar mass 342.30 g/mol
475g sucrose 342.30gsucrose/mol
1.388 mole sucrose
1.388 mole sucrose 0.600 kg H2O
molality
2.313 m
0.5120C m
Tb Kb x m (2.313 m) 1.180C BP
100.000C 1.180C
BP
101.180C
1.860C m
Tf Kf x m (2.313 m) 4.300C FP
0.000C - 4.300C -4.300C
59
Determining the Boiling Point Elevation and
Freezing Point Depression of a Non-Aqueous
Solution
Problem Calculate the effect on the Boiling
Point and Freezing Point of a chloroform solution
if to 500.00g of chloroform (CHCl3) 257g of
napthalene (C10H8, mothballs) is
dissolved. Plan We must first calculate the
molality of the cholorform solution
by calculating moles of each material, then we
can apply the FP and BP change equations and the
contants for chloroform. Solution napthalene
128.16g/mol chloroform 119.37g/mol
257g nap 128.16g/mol
molesnap 2.0053 mol nap
moles nap kg(CHCl3)
2.0053 mol 0.500 kg
molarity
4.01 m
3.630C m
Tb Kb m (4.01m) 14.560C
normal BP 61.70C

new BP ______0C
4.700C m
Tf Kf m (4.01m) 18.850C
normal FP - 63.50C

new FP _______0C
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Figure 17.14 A tub with a bulb on the end is
covered by a semipermeable membrane.
61
Figure 17.15 The normal flow of solvent into
the solution (osmosis) can be prevented by
applying an external pressure to the solution.
62
Figure 17.16 A pure solvent and its solution
(containing a nonvolatile solute) are separated
by a semipermeable membrane through which solvent
molecules (blue) can pass but solute molecules
(green) cannot.
63
Osmotic pressure calculation
Calculate the osmotic pressure generated by a
sugar solution made up of 5.00 lbs of sucrose per
5.00 pints of water. Solution
5.00 lbs ( ) 2.27 kg Molar
mass of sucrose 342.3 g/mol 2,270g
5.00 pints H2O ( )(
) 2.36 liters P MRT (
)(0.08206 )(298 K)
__________ atm
1 kg 2.205 lbs
6.63 mol sucrose
342.3g/mol
1.00 gallon 8 pints
3.7854 L 1.00 gallon
L atm mol K
6.63 mol 2.36 L
64
Like example 17.4 (P 848-9)
To determine the molar mass of a certain protein,
1.7 x 10-3g of the protein was dissolved in
enough water to make 1.00 ml of solution.
The osmotic pressure of this solution was
determined to be 1.28 torr at 25oC. Calculate the
molar mass of the protein. Solution
P 1.28 torr( ) 1.68 x 10-3
atm T 25 273 298 K M
6.87 x 10-5
mol/L 1.70 g xg 6.87 x
10-5mol mol x
_______________g/mol
1 atm 760 torr
1.68 x 10-3 atm 0.08206 L atm(298 K)
mol K

65
Determining Molar Mass from Osmotic
Pressure - I
Problem A physician studying a type of
hemoglobin formed during a fatal disease
dissolves 21.5 mg of the protein in water at
5.00C to make 1.5 ml of solution in order to
measure its osmotic pressure. At equilibrium,
the solution has an osmotic pressure of 3.61
torr. What is the molar mass(M) of the
hemoglobin? Plan We know the osmotic pressure
(?),R, and T. We convert ? from torr to atm and T
from 0C to K and use the osmotic pressure
equation to solve for molarity (M). Then we
calculate the moles of hemoglobin from the known
volume and use the known mass to find M. Solution
1 atm 760 torr
P 3.61 torr x 0.00475 atm
Temp 5.00C 273.15 278.15 K
66
Molar Mass from Osmotic Pressure - II
?? RT
0.00475 atm 0.082 L atm (278.2 K) mol K
M
2.08 x 10 - 4 M
Finding moles of solute
2.08 x 10 - 4 mol L soln
n M x V x 0.00150
L soln 3.12 x 10 - 7 mol
Calculating molar mass of Hemoglobin (after
changing mg to g)
0.0215 g 3.12 x 10-7 mol
M
________________ g/mol
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Figure 17.17 Representation of the functioning
of an artificial kidney
69
Example 17.5 (P 850)
What concentration of sodium chloride in water is
needed to produce an aqueous solution isotonic
with blood (p 7.70 atm at 25oC). Solution
P MRT M M
0.315 mol/L
Since sodium chloride gives two ions per
molecule, the concentration would be ½ that
value, or 0.158 M NaCl Na Cl-
P RT
7.70 atm (0.08206 L atm) (298 K)
mol K
70
Figure 17.18 Reverse osmosis
71
Reverse Osmosis for Removal of Ions
72
Family uses a commercially available desalinator,
similar to those developed by the Navy for life
rafts.
Source Recovery Engineering, Inc.
73
Figure 17.20 Residents of Catalina Island off
the coast of southern California are benefiting
from a desalination plant that can supply 132,000
gallons of drinkable water per day, one-third of
the island's daily needs.
74
Colligative Properties of Volatile Nonelectrolyte
Solutions
From Raoults law, we know that
Psolvent Xsolvent x P0solvent and Psolute
Xsolute x P0solute
Let us look at a solution made up of equal molar
quantities of acetone and chloroform. Xacetone
XCHCl3 0.500, at 350C the vapor pressure
of pure acetone 345 torr, and pure chloroform
293 torr. What is vapor pressure of the solution,
and the vapor pressure of each component.
What are the mole fractions of each component?
Pacetone Xacetone x P0acetone 0.500 x 345
torr 172.5 torr PCHCl3 XCHCl3 x P0CHCl3
0.500 x 293 torr 146.5 torr
PA PTotal
From Daltons law of partial pressures we know
that XA
Pacetone PTotal
172.5 torr 172.5 146.5 torr
Xacetone
0.541
Total Pressure 319.0 torr
PCHCl3 PTotal
146.5 torr 172.5 146.5 torr
XCHCl3
0.459
75
Colligative Properties
I ) Vapor Pressure Lowering - Raoults Law II
) Boiling Point Elvation III ) Freezing
Point Depression IV ) Osmotic Pressure
76
Colligative Properties of Ionic Solutions
For ionic solutions we must take into account the
number of ions present!
i vant Hoff factor ionic strength, or the
number of ions present
For vapor pressure lowering P i
XsoluteP 0solvent
For boiling point elevation Tb
i Kb m
For freezing point depression Tf
i Kf m
For osmotic pressure ?
i MRT
77
Figure 17.21 In a aqueous solution a few ions
aggregate, forming ion pairs that behave as a
unit.
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Like Example 17.6 (P 853)

• The observed osmotic pressure for a 0.10M
  solution of Na3PO4 at 25oC
• is 8.45 atm. Compare the expected and
  experimental values of i!
• Solution
• Tri sodium phosphate will produce
  4 ions in solution.
• Na3PO4
  3 Na PO4-3
• Thus i is expected to be 4, now to calculate the
  experimental value of i
• from the osmotic pressure equation.
• iMRT or i
• i ______ This is less than the value
  expected of 4 so there must be
• some ion paring
  occurring in the solution.

P MRT
8.45 atm (0.10 )(0.08206
)(298 K)
mol L
L atm mol K
80
Figure 17.22 The Tyndall effect
Source Stock Boston
81
Figure 17.23 Representation of two colloidal
particles
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Magnetic liquid seal
84
Figure 17.24 Cottrel precipitator installed in
a smokestack.
85
A Cottrell Precipitator for Removing Particulates
from Industrial Flue Gases
Fig. 13.24