Physical Properties of Solutions
Chapter 13
1
Outline of the Chapter
• The fundamental properties of Solutions
– Types
– Energetics
• Working with solutions
– Concentration Units
• Saturated Solutions and Equilibrium
• Colligative Properties
– Freezing point depression
– Boiling point elevation
Definitions to Know
Solution
Solute
Solvent
Saturated, unsaturated, supersaturated solutions
Solubility
Dynamic Equilibrium
Types of solutions
3
Solutions
• Solutions are homogeneous mixtures of two or
more pure substances.
• Solute is dispersed uniformly throughout the
solvent ( we will deal with water
solutions).
• Liquids that mix in all proportions are called
miscible
4
12.1 Some Types Of Solutions
12.1
12.1 Types of Solutions
• Saturated
 Solvent holds as much solute as
is possible (maximum) at that
temperature.
 Dissolved solute is in dynamic
equilibrium with solid solute
particles.
Dynamic equilibrium: rate of
crystallization = rate of
dissolving
Solubility: concentration of
saturated solution.
6
12.1 Types of Solutions
• Unsaturated
 Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.
7
12.1 Types of Solutions
• Supersaturated
 Solvent holds more solute than is normally possible at that
temperature.
 Solutions are unstable; crystallization can usually be
stimulated by adding a “seed crystal” or scratching the side
of the flask.
• Sodium acetate crystals rapidly form when a seed
crystal is added to a supersaturated solution of
sodium acetate.
8
A Molecular View of the Solution
Process
Intermolecular Forces In Solution Formation
1. In the following processes differentiate
between physical process that leads to
solution formation and chemical reaction
a) Ni(s) + 2HCl → NiCl2(aq) + H2(g)
b) NaCl (s) → Na+(aq) + Cl-(aq)
c) CH3OH(l) → CH3OH(aq)
10
Intermolecular Forces In Solution
Solute-solute
Solute-solvent
Solvent-solvent
Formation Of Solution: Energy Considerations
Energy needed to break
the:
solute-solute interaction:
Δ H1 > 0
Solvent-solvent interactions:
Δ H2 > 0
Energy released when:
Solvent-solute interaction
Δ H3 < 0
DHsoln = DH1 + DH2 + DH3
12
Energy Changes in Solution
The enthalpy
change of the
overall process
depends on DH
for each of
these steps.
13
Why Do Endothermic Processes Occur?
Things do not tend to
occur spontaneously
(i.e., without outside
intervention) unless
the energy of the
system is lowered.
14
Why Do Endothermic Processes Occur?
Yet we know that in
some processes, like
the dissolution of
NH4NO3 in water,
heat is absorbed, not
released.
15
Enthalpy Is Only Part of the Picture
The reason is that
increasing the disorder
or randomness (known
as entropy) of a
system tends to lower
the energy of the
system.
16
Enthalpy Is Only Part of the Picture
So even though
enthalpy may increase,
the overall energy of
the system can still
decrease if the system
becomes more
disordered.
17
Intermolecular Forces In Solution Formation
• Ideal Solution: All intermolecular forces are of comparable
strength
Δ Hsoln = 0.
(N2 and O2; gasoline ; benzene + toluene)
Enthalpy driven dissolution:
• Intermolecular forces between solute and solvent molecules
are stronger than other intermolecular forces,
|Δ H3| > Δ H1 + Δ H2,
Δ Hsoln < 0 (exothermic)
MgSO4, ΔHsolution = -92. kJ/mol
Sometimes, volume of solution is smaller than volume of
individual components (C2H5OH + H2O)
Intermolecular Forces In Solution Formation
Entropy driven dissolution:
• Intermolecular forces between solute and solvent
molecules are weaker than other intermolecular forces,

|Δ H3| < Δ H1 + Δ H2,
 Δ Hsoln > 0
Endothermic. Entropy driven.
NH4NO3, ΔH = 26.4 kJ/mol
• Intermolecular forces between solute and solvent are
much weaker ( ΔH1 + ΔH2 << |ΔH3| ) than other
intermolecular forces, the solute does not dissolve in the
solvent. The compound is relatively insoluble in the
solvent.
19
Example
2. Predict the relative solubilities in the following
case:
a)
b)
c)
d)
e)
Bromine in benzene (μ = 0 D)
Bromine in water (μ = 1.87 D)
KCl in carbon tetrachloride (μ = 0 D)
KCl in liquid ammonia (μ = 1.46 D)
Formaldehyde (CH2O) in carbon disulfide CS2, μ = 0
D)
f) Formaldehyde (CH2O) in water (μ = 1.87 D)
g) Urea, (NH2)2CO in carbon disulfide
h) Urea in water
20
Factors Affecting Solubility
• Chemists use the axiom “like dissolves like”:
 Polar substances tend to dissolve in polar solvents.
 Nonpolar substances tend to dissolve in nonpolar
solvents: CCl4 in C6H6
21
Factors Affecting Solubility
Glucose (which has
hydrogen bonding) is
very soluble in water,
while cyclohexane
(which only has
dispersion forces) is
not.
22
Liquid-Liquid Solutions
• Miscible: completely soluble
• Immiscible: do not dissolve in each other
• Solvent: component with greater
concentration
• Like dissolves like
• Acetic acid dimers
23
Solutions of Solids in Liquids
Things to know:
• aqueous solutions
– hydration
• polar-polar solutions
– Solvation
• Crystallization
• Solubility of metals
24
Aqueous Solutions
• Two forces
- Break ionic attractions (positive and
negative ions) in the solid
- Form ion-dipole forces between solute
particles and solvent particles (form
hydration shell).
Solubility determined by the strength of the
solid-solid IMfs and solute-solvent
interactions
Network and metallic + water: does not happen
25
Ion-Dipole Forces in Dissolution
Hydration shell
12.4
26
Example
3. Predict whether each of the following is
likely to be a solution or a heterogeneous
mixture.
a) Ethanol, CH3OH, and water, HOH
b) Pentane, CH3(CH2)3CH3, and octane,
CH3(CH2)6CH3
c) Sodium chloride, NaCl, and carbon
tetrachloride, CCl4
d) 1-decanol, CH3(CH2)8CH2OH, and water,
HOH
12.4
27
Examples
4.
Some solution processes are endothermic and some are
endothermic. Provide a molecular interpretation for these
differences.
5.
Describe the factors that affect the solubility of a solid in a liquid.
What does it mean to say that two liquids are miscible? Give
examples.
Answers:
•
(IMFs: solid-solid; liqud-liquid, solid-liquid; Energy diagrams)
•
(miscible: completely soluble in all proportions; IMFs comparable
in strength, butane in octane; methanol in water, describe the
IMFs))
28
Ways of Expressing Concentrations
of Solutions
29
Concentration Units
•
•
•
•
•
•
•
•
Things to know:
Percent by mass
Mole fraction
Molarity
Molality
Comparison of Concentration Units
Parts per million (ppm)
Parts per billion (ppb)
30
Mass Percentage
mass of A in solution
 100
Mass % of A =
total mass of solution
31
Parts per Million and
Parts per Billion
Parts per Million (ppm), 1 ppm = 1 mg/L
ppm =
mass of A in solution
 106
total mass of solution
Parts per Billion (ppb), 1 ppb = 1 µg/L
mass of A in solution
 109
ppb =
total mass of solution
32
Mole Fraction (X)
moles of A
XA =
total moles in solution
• In some applications, one needs the mole
fraction of solvent, not solute—make sure
you find the quantity you need!
• xi < 1;
x1 + x2 + x3 + … = 1
33
Molarity (M)
M=
mol of solute
L of solution
• Volume is temperature dependent.
Therefore
• molarity can change with temperature.
34
Molality (m)
m=
mol of solute
kg of solvent
• Moles and mass do not change with
temperature.
Therefore,
• Molality (unlike molarity) is not
temperature dependent.
35
Concentration Units: Examples
6. A sample of 0.892 g KCl dissolved in 54.6 g
H2O. Calculate % by mass.
7. A solution is prepared by mixing 200.4 g
C2H5OH to 143.9 g of H2O. Calculate mole
fractions of both ingredients of the solution.
(Xalcohol 0.3539; xH2O = 0.647
8. Calculate molality of H2SO4 solutions
containing 24.4 g H2SO4 in 198 g of H2O.
(1.26 m)
36
Example
9. How many milliliters of water (d = 0.998 g/mL) are
required to dissolve 25.0 g of urea and thereby produce
a 1.65 m solution of urea, CO(NH2)2?
(253 mL of H2O)
10. The density of a 2.45 M aqueous methanol solution is
0.976 g/mL. What is the molality of the solution?
(2.73 m)
11. Calculate the molality of a 34.5 % (by mass) aqueous
solution of phosphoric acid, H3PO4. The molar mass of
phosphoric acid is 98.00 g/mol.
(5.37 m)
12. What is the molality of a 5.86 M ethanol
(C2H5OH) solution whose density is 0.927 g/mL?
(8.92 m)
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Assume 1 L of solution:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x 0.927 g/mL)
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = 0.657 kg
moles of solute
m =
mass of solvent (kg)
=
5.86 moles C2H5OH
= 8.92 m
0.657 kg solvent
38
Example (H@P, 516)
13. An aqueous solution of ethylene glycol used as an
automobile engine coolant is 40.0% HOCH2CH2OH by
mass and has a density of 1.05 g/mL. What are the
(a) molarity,
(b) molality, and
(c) mole fraction of HOCH2CH2OH.
(a. 6.77 M
b. 10.7 m
c. 0.162)
Changing Molarity to Molality
If we know the density
of the solution, we
can calculate the
molality from the
molarity, and vice
versa.
40
Effect of Temperature on Solubility
• Solubility: the amount of solute that will dissolve in a
given amount of solvent at a given temperature
• Solubilities of ionic compounds increase
significantly with increasing temperature (About
95% of compounds). Rest do not change or very
slightly (NaCl).
• A very few have solubilities that decrease with
increasing temperature.
• A supersaturated solution is created when a warm,
saturated solution is allowed to cool without the
precipitation of the excess solute.
12.6
Temperature and Solubility
Generally, the
solubility of solid
solutes in liquid
solvents increases
with increasing
temperature.
Exceptions?
42
Fractional Crystallization
When KNO3(s) is
crystallized from an
aqueous solution of
KNO3 containing
CuSO4 as an impurity,
CuSO4 remains in the
solution.
12.6
Fractional crystallization is the separation of a mixture of
substances into pure components on the basis of their differing
solubilities.
Suppose you have 90 g KNO3
contaminated with 10 g NaCl.
Fractional crystallization:
1. Dissolve sample in 100 mL of
water at 600C
2. Cool solution to 00C
3. All NaCl will stay in solution
(s = 34.2g/100g)
4. 78 g of PURE KNO3 will
precipitate (s = 12 g/100g).
90 g – 12 g = 78 g
12.6
44
Solubility: Examples
15.
The solubility of ammonium formate, NH4HCO2, in
100.0 g of water is 102.0 g at 0ºC and 546 g at 80ºC. A
solution is prepared by dissolving NH4HCO2 in 200.0 g of
water until no more can dissolve at 80ºC. The solution is
then cooled to 0ºC. What mass of ammonium formate
precipitates? (Assume that no water evaporates and the
solution is not supersaturated.
45
Gases in Solution
• In general, the
solubility of gases in
water increases with
increasing mass.
• Larger molecules
have stronger
dispersion forces.
46
Gas Solubility and Temperature
12.6
Effect of Pressure on Solubility of Gases
Things to know:
Henry’s law
Molecular Interpretation of Henry’s Law
Check online tutorial
Gases in Solution
• The solubility of
liquids and solids
does not change
appreciably with
pressure.
• The solubility of a gas
in a liquid is directly
proportional to its
pressure.
49
Pressure and Solubility of Gases
Explain the pictures.
50
Pressure and Solubility of Gases
HENRY’S LAW : the solubility of a gas in a liquid is
proportional to the pressure of the gas over the solution.
c is the concentration (mol/L) of the dissolved gas
c = kP
P is the pressure of the gas over the solution
k is a constant (mol/L•atm) that depends only
on temperature (for a specific solvent/solute)
12.7
low P
high P
low c
high c
51
Solubility of Gas in Water as a Function Of Gas
Pressure
12.7
Pressure and Solubility of Gases: Henry’s Law
16.
Calculate the concentration of CO2 in a soft drink that is bottled
with a partial pressure of CO2 of 4.00 atm over the liquid at 25.0
ºC. The Henry's Law constant for CO2 in water at this
temperature is 3.1 x 10-2 mol/L-atm.
(0.12 M)
17.
Calculate the concentration of CO2 in a soft drink after the bottle is
opened and it equilibrates at 25.0 ºC under a CO2 partial pressure
of 3.0 x 10-4 atm.
(9.3 x 10-6 M)
18.
The solubility of pure N2 at 25ºC and 1 atm is 6.8 x 10-4 mol/L.
What is the concentration of N2 dissolved in H2O under
atmospheric conditions? The partial pressure of nitrogen gas in
the atmosphere is 0.78 atm.
(5.3 x 10-4 mol/L)
•
NOTE: gases that interact (dissolve) with the solvent do not obey
Henry’s Law.
53
Colligative properties of
Nonelectrolyte Solutions
Colligative Properties
Colligative Properties – physical properties of
solutions that depend on the number of
solute particles present but not on the identity
(nature) of the solute.
•
•
•
•
Freezing point depression
Boiling point elevation
Vapor Pressure of a Solution
Osmosis
Vapor Pressure
Because of solutesolvent intermolecular
attraction, higher
concentrations of
nonvolatile solutes
make it harder for
solvent to escape to
the vapor phase.
56
Vapor Pressure
Therefore, the vapor
pressure of a solution
is lower than that of
the pure solvent.
57
Vapor Pressures of Solutions
• For a solution with a nonvolatile solute
( vapor pressure = 0), the vapor pressure of
its solution is always less than that of the
pure solvent.
• Raoult’s law: the partial pressure of the solvent
above a solution (P1) is the product of the vapor
pressure of the pure solvent (Po1) and the mole
fraction of the solvent in the solution (x1):
P1 = x1 . Po1
Raoult’s Law
Vapor-Pressure Lowering:
Raoult’s law
P 10 = vapor pressure of pure solvent
P1 = X1 P 10
P1 = vapor pressure of solution
X1 = mole fraction of the solvent
If the solution contains only one solute (nonvolatile):
0
11
X1 = 1 – X2
X2 = mole fraction of the solute
DP  P10 - P1  P10 - x 1P10  P10 (1 - x 1 )  x 2 P10
ΔP = X2 P10
X2 < 1
59
Example: Raoult’s Law
19. At 25 ºC, the vapor pressure of pure water is 23.76
mm Hg and that of aqueous urea solution is 22.98 mm
Hg. Estimate the molality of the solution.
(1.8 m)
20. The vapor pressure of glucose, C6H12O6 solution is
17.01 mm Hg at 20.0 ºC, while that of pure water is
17.25 mm Hg at the same temperature. Calculate the
molality of the solution.
(0.77 m)
21. How many grams of urea (H2NCONH2) must be added
to 450 g of water to give a solution with vapor pressure
2.50 mm Hg less then the pure water at 30.0 ºC? The
partial pressure of water at 30.0 ºC is 31.80 mm Hg.
(126 g)
60
Boiling Point Elevation and Freezing Point Depression
Nonvolatile solutesolvent
interactions also
cause solutions to
have higher
boiling points and
lower freezing
points than the
pure solvent.
61
Boiling Point Elevation
DTb = Tb – T b0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b0
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)
62
Freezing-Point Depression
DTf = T 0f – Tf
T
0
Tf
f
is the freezing point of
the pure solvent
is the freezing point of
the solution
T 0f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)
63
64
Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 10
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Molar Mass Calculations
Osmotic Pressure (p)
12.8
p = MRT
65
What is the freezing point of a solution containing
478 g of ethylene glycol (antifreeze) in 3202 g of
water? The molar mass of ethylene glycol is 62.01 g.
DTf = Kf m
Kf water = 1.86 0C/m
moles of solute
m =
mass of solvent (kg)
478 g x
1 mol
62.01 g
=
= 2.41 m
3.202 kg solvent
DTf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
DTf = T 0f – Tf
Tf = T 0f – DTf = 0.00 0C – 4.48 0C = -4.48 0C
12.8
66
Example: Colligative Properties
22. What mass of sucrose, C12H22O11, should be
added to 75.0 g H2O to raise the boiling point to
100.35 0C?
23. Automotive antifreeze consists of ethylene glycol,
C2H6O2, a nonvolatile electrolyte. Calculate the
b.p. and f.p. of a 25.0 mass percent solution of
ethylene glycol in water.
( 102.8 ºC; -10.0
ºC)
24. List the following aqueous solutions in order of
their expected freezing points (lowest to highest):
0.050 m CaCl2; 0.15 m NaCl; 0.10 m HCl; 0.050
m HC2H3O2; 0.10 m C12H22 O11
Determination of Molar Mass
25. A 7.85 g sample of a compound with empirical
formula C5H4 is dissolved in 301 g of benzene.
The freezing point of the solution is 1.05 ºC
below that of pure benzene. What are the molar
mass and molecular formula of this compound?
(127 g/mol, C10H8)
26. A solution of an unknown nonvolatile
compound was prepared by dissolving 0.250 g
of the substance in 40.0 g CCl4. The B.P of the
resultant solution was 0.357 ºC higher than of
the pure solvent. Calculate the molar mass of
the solute.
(88.0 g/mol)
68
Colligative Properties of Electrolyte Solutions
Colligative Properties of Electrolytes
Since these properties depend on the number of
particles dissolved, solutions of electrolytes (which
dissociate in solution) should show greater changes
than those of nonelectrolytes.
70
Colligative Properties of Electrolytes
However, a 1 M solution of NaCl does not show
twice the change in freezing point that a 1 M
solution of methanol does.
71
van’t Hoff Factor
One mole of NaCl in
water does not really
give rise to two moles
of ions.
72
van’t Hoff Factor
Some Na+ and Cl−
reassociate for a
short time, so the true
concentration of
particles is somewhat
less than two times
the concentration of
NaCl.
73
The van’t Hoff Factor
• Reassociation is more likely at higher concentration.
• Therefore, the number of particles present is
concentration dependent.
74
The van’t Hoff Factor
We modify the
previous equations
by multiplying by the
van’t Hoff factor, i
DTf = Kf  m  i
van’t Hoff factor (i) =
actual number of particles in soln after dissociation
number of formula units initially dissolved in
soln
75
Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iMRT
76
12.9
12.8 Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous
membrane from a dilute solution (high solvent concentration) to a more
concentrated one (low solvent concentration).
A semipermeable membrane allows the passage of solvent molecules but
blocks the passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.
dilute
more
concentrated
77
12.8
Osmotic Pressure (p)
High
P
Low
P
p=
n
V
RT = MRT
M is the molarity of the solution
R is the gas constant: 0.0821 L-atm/mol-K
T is the temperature (in K)
12.8
78
A cell in an:
Isotonic Solution:
same conc.
12.8
hypotonic
Less conc.
hypertonic
More conc.
79
Osmosis in Blood Cells
• If the solute
concentration outside
the cell is greater than
that inside the cell, the
solution is hypertonic.
• Water will flow out of
the cell, and crenation
results.
80
Osmosis in Cells
• If the solute
concentration outside
the cell is less than
that inside the cell, the
solution is hypotonic.
• Water will flow into the
cell, and hemolysis
results.
81
Osmotic Pressure: Example
27. The osmotic pressure of an aqueous
solution of a certain protein was measured
in order to determine its molar mass. The
solution contained 3.50 mg of protein
dissolved in sufficient water to form
5.00 ml solution. The osmotic pressure of
the solution at 25.0 ºC was found to be
1.54 torr. Calculate the molar mass of the
protein.
(8.45 x 103 g/mol)
82
A colloid is a dispersion of particles of one substance
throughout a dispersing medium of another substance.
Colloid versus solution
•
colloidal particles are much larger than solute molecules
•
colloidal suspension is not as homogeneous as a solution
83
12.8
The Cleansing Action of Soap
84
Tyndall Effect
• Colloidal suspensions
can scatter rays of light.
• This phenomenon is
known as the Tyndall
effect.
85
Colloids in Biological Systems
Some molecules have
a polar, hydrophilic
(water-loving) end and
a nonpolar,
hydrophobic (waterhating) end.
86
Colloids in Biological Systems
Sodium stearate
is one example of
such a molecule.
87
Colloids in Biological Systems
These molecules can
aid in the
emulsification of fats
and oils in aqueous
solutions.
88
The End