Chapter 17
Properties of Solutions
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
The left beaker contains copper sulfate
solution. In the right beaker, ammonia is
being added to create a precipitate of
copper(II) hydroxide.
3
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.
Miscible - Substances that can dissolve in any proportion, so that it is
difficult to tell which is the solvent or solute!
Hydration
shells around
an aqueous
ion
Solution Composition
Mass percent:
weight percent
Mass percent =
grams of solute
x 100%
grams of solution
Mole fraction: symbolized by the Greek letter, chi = X
nA
mole fraction of component A = XA = n + n
A
B
Molarity: M ; (chapter 4)
moles of solute
M = liter of solution
Molality: m
moles of Solute
m = Kg of solvent
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!
Like dissolves Like: Solubility of methanol in water
Fig 13.4 (P 486)
The structure and function of a soap
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.
Predicting Relative Solubility's of Substances - II
Solution:
(a) Methanol - NaCl is an ionic compound that dissolves through iondipole forces. Both methanol and propanol contain a polar hydroxyl
group, and propanol’s 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.
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.
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.
Figure 17.1: The formation of a liquid
solution can be divided into three steps
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.
H of solution for Sodium Chloride
NaCl(s)
Na +(g) + Cl –(g)
H2O(l) + Na+(g) + Cl –(g)
Hohyd =
Ho1 = 786 kJ/mol
Na+(aq) + Cl –(aq)
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.
French Navy ship L'Ailette equipped with vacuum
pumps approaches an oil slick.
Source: AP/Wide World Photos
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 > 0
Step 3: Solute and Solvent Particles mix - Particles attract each other
Therefore -- Exothermic
solute (separated) + solvent (separated)
solution + heat
Hmix < 0
The Thermochemical Cycle
Hsolution =
Hsolute + Hsolvent + Hmix
If
HEndothermic Rxn <
HExothermic Rxn
solution becomes warmer
If
HEndothermic Rxn >
HExothermic Rxn
solution becomes colder
Solution Cycles
and the Enthalpy
Components of
the Heat of
Solution
Multistage
supercritical
fluid
extraction
apparatus
Source: USDA
Chicken fat obtained by using supercritical
carbon dioxide as the extracting agent.
Source: USDA
Figure 17.3: Vitamin A, C
A Fat – Soluble Vitamin
A Hydrophobic Vitamin
A Water – Soluble Vitamin
A Hydrophilic Vitamin
Carbonation in a bottle of soda
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
Henry’s Law of Gas solubilities in Liquids
P = kHX
P = Partial pressure
of dissolved gas
X = mole fraction
of dissolved gas
kH = Henry’s Law
Constant
Henry’s 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 Henry’s law and the Henry’s law constant for oxygen
in water with the partial pressure of O2 in the air to calculate the amount.
Solution:
The Henry’s 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!
Figure 17.5: The solubilities of several
solids as a function of temperature.
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) H2O 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.
Predicting the Effect of Temperature
on Solubility - II
(a) continued
CsOH(s)
H2O
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
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.
Figure 17.7: Pipe with accumulated mineral
deposits (left) lengthwise section (right)
Source: Visuals Unlimited
Lake Nyos in Cameroon
Source: Corbis
Figure 17.8: An aqueous solution and pure
water in a closed environment
Figure 17.9: The presence of a nonvolatile
solute inhibits the escape of solvent
molecules from the liquid
Figure 17.10: For a solution that obeys
Raoult’s law, a plot of Psoln versus xsolvent
yields a straight line.
Fig. 13.15
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 it’s
density is 0.97489 g/ml.
Plan: Calculate the change in pressure from Raoult’s 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 ( C H O ) = 342.30 g/mol
12
22
11
175g sucrose
= 0.51125 mol sucrose
342.30g sucrose/mol
350.00 ml H2O x 0.97489g H2O = 341.21g H2O
ml H2O
341.21 g H2O
= ______ molH2O
18.02g H2O/mol
Vapor Pressure Lowering - II
Xsucrose =
Xsurose =
mole sucrose
moles of water + moles of sucrose
0.51125 mole sucrose
= 0.2629
18.935 mol H2O + 0.51125 mol sucrose
P = Xsucrose x P 0H2O = 0.2629 x 289.1 mm Hg = ________ mm Hg
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:
Psoln
22.36 torr
XH2O = o
=
= 0.9411 =
P H2O
23.76 torr
125.0 g
mol H2O =
= 6.94 mol H2O
18.0 g/mol
mol H2O
mol gly + mol H2O
6.94 mol
0.9411 = mol gly + 6.96 mol
6.94 mol – (6.94 mol)(0.9411)
mol gly =
= 0.4357 mol
0.9411
40.0 g
=
g/mol (MMglycerol = 92.09 g/mol)
0.4357 mol
Figure 17.11: Vapor pressure for a solution
of two volatile liquids.
-
HHHHHH
H-C-C-C-C-C-C-H
HHHHHH
HH
H-C-C-O-H
HH
- -- -
HOH
H-C-C-C-H
H H
Acetone + Water
- - -
-
H-O
Hexane
+
Ethanol
Figure 17.12: Phase diagrams for pure water
(red lines) and for an aqueous solution
containing a nonvolatile solution (blue lines).
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
oC
Kg
kb = 0.51 m
solute
T = 100.61oC – 100.00 oC = 0.61oC
oC
0.61
T
msolute =
=
= 1.20 mol/Kg
oC Kg
kb
0.51 m
solute
mol solute
Msolute =
mol solute = (0.150 kg)(1.2 mol/kg)
kg solvent
62g
mol solute = 0.18 mol
MM = 0.18 mol = _________g/mol
(MMsucrose = 342.18 g/mol)
Figure 17.13: Ice in equilibrium with
liquid water.
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 =
T
kf
-26.1oC
= 1.86 oC kg = 14.0 mol/kg
mol
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
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
= 1.388 mole sucrose
342.30gsucrose/mol
molality = 1.388 mole sucrose = 2.313 m
0.600 kg H2O
0
Tb = Kb x m = 0.512 C(2.313 m)= 1.180C BP = 100.000C + 1.180C
m
BP = 101.180C
0C
1.86
Tf = Kf x m =
(2.313 m) = 4.300C FP = 0.000C - 4.300C = -4.300C
m
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
molesnap = 257g nap =2.0053 mol nap
128.16g/mol
molarity = moles nap =2.0053 mol= 4.01 m
kg(CHCl3)
0.500 kg
0C
3.63
Tb = Kb m =
(4.01m) = 14.560C
normal BP = 61.70C
m
new BP = ______0C
0C
4.70
Tf = Kf m =
(4.01m) =18.850C
m
normal FP = - 63.50C
new FP = _______0C
Figure 17.14: A tub with a bulb on the end is
covered by a semipermeable membrane.
Figure 17.15: The normal flow of solvent into
the solution (osmosis) can be prevented by
applying an external pressure to the solution.
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.
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:
1 kg
5.00 lbs (
) = 2.27 kg
2.205 lbs
Molar mass of sucrose = 342.3 g/mol
2,270g = 6.63 mol sucrose
342.3g/mol
1.00 gallon
5.00 pints H2O (
)( 3.7854 L ) = 2.36 liters
8 pints
1.00 gallon
L atm
6.63
mol
P = MRT = (
)(0.08206 mol K)(298 K) = __________ atm
2.36 L
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:
1 atm
P = 1.28 torr(
) = 1.68 x 10-3 atm
760 torr
T = 25 + 273 = 298 K
1.68 x 10-3 atm
M=
= 6.87 x 10-5 mol/L
0.08206 L atm(298 K)
mol K
1.70 g
xg
=
6.87 x 10-5mol
mol
x = _______________g/mol
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:
P = 3.61 torr x 1 atm = 0.00475 atm
760 torr
Temp = 5.00C + 273.15 = 278.15 K
Molar Mass from Osmotic Pressure - II
M =  =
RT
0.00475 atm
0.082 L atm (278.2 K)
mol K
= 2.08 x 10 - 4 M
Finding moles of solute:
2.08 x 10 - 4 mol
n=MxV=
x 0.00150 L soln = 3.12 x 10 - 7 mol
L soln
Calculating molar mass of Hemoglobin (after changing mg to g):
M = 0.0215 g -7
= ________________ g/mol
3.12 x 10 mol
Figure 17.17: Representation of the
functioning of an artificial kidney
Example 17.5 (P 850)
What concentration of sodium chloride in water is needed to produce
an aqueous solution isotonic with blood (= 7.70 atm at 25oC).
Solution:
P
P=MRT
M = RT
7.70 atm
M = (0.08206 L atm) (298 K)
mol K
= 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-
Figure 17.18: Reverse osmosis
Reverse Osmosis for Removal of Ions
Family uses a commercially available desalinator,
similar to those developed by the Navy for life rafts.
Source: Recovery Engineering, Inc.
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.
Colligative Properties of Volatile Nonelectrolyte Solutions
From Raoult’s 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
From Dalton’s law of partial pressures we know that XA = P
Total
Pacetone
Xacetone = P
= 172.5 torr
= 0.541
Total
172.5 + 146.5 torr
146.5 torr
XCHCl3 = PCHCl3 =
= 0.459
PTotal 172.5 + 146.5 torr
Total Pressure = 319.0 torr
Colligative Properties
I ) Vapor Pressure Lowering - Raoult’s Law
II ) Boiling Point Elvation
III ) Freezing Point Depression
IV ) Osmotic Pressure
Colligative Properties of Ionic Solutions
For ionic solutions we must take into account the number of ions present!
i = van’t 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:
For osmotic pressure:
Tf = i Kf m
 = i MRT
Figure 17.21: In a aqueous solution a few ions
aggregate, forming ion pairs that behave as a
unit.
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.
P= iMRT or i =
i = ______
8.45 atm
P
=
(0.10 mol )(0.08206 L atm )(298 K)
MRT
L
mol K
This is less than the value expected of 4 so there must be
some ion paring occurring in the solution.
Figure 17.22: The Tyndall effect
Source: Stock Boston
Figure 17.23: Representation of two
colloidal particles
Magnetic liquid seal
Figure 17.24:
Cottrel
precipitator
installed
in a smokestack.
A Cottrell Precipitator for Removing
Particulates from Industrial Flue Gases
Fig. 13.24