Transcript Document

PowerPoint Lecture Presentation
by
J. David Robertson
University of Missouri
Physical Properties
of Solutions
Chapter 12
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
LEARNING OBJECTIVES
• To understand the concept of
solution & solution process;
• To be able to describe the
concentration of a solution
in the most appropriate
way;
A solution is a homogenous mixture of 2 or
more substances
The solute is(are) the substance(s) present in the
smaller amount(s)
The solvent is the substance present in the larger
amount
12.1
In the process of making solutions, intermolecular
forces are rearranged.
The H-bonds in water have to be interrupted.
Ion-dipole forces form: Na+ … -OH2 , Cl- … +H2O
A saturated solution contains the maximum amount of a
solute that will dissolve in a given solvent at a specific
temperature.
An unsaturated solution contains less solute than the
solvent has the capacity to dissolve at a specific
temperature.
A supersaturated solution contains more solute than is
present in a saturated solution at a specific temperature.
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
12.1
Three types of interactions in the solution process:
• solvent-solvent interaction
• solute-solute interaction
• solvent-solute interaction
DH1 = separation
of solvent
molecules
DH2 = separation
of solute
molecules
DH3 = formation
of solute-solvent
interaction
DHsoln = DH1 + DH2 + DH3
DHsoln can either be positive or negative depending
on the intermolecular forces.
12.2
Revision question: What is the strongest type of
intermolecular forces between solute and solvent in each of
the following solution?
Answer:
Ion-dipole forces
a) CsCl (s) in H2O (ℓ)
b) CH3OH (ℓ) in CCl4 (ℓ)
Dipole-induced dipole forces
c) CH3Cl (g) in CH3OCH3 (g) Dipole-dipole forces
d) CH3OCH3 (g) in H2O (ℓ)
Hydrogen bonding
e) Br2 (ℓ) in CCl4 (ℓ)
Dispersion forces
Energy Changes and Solution Formation
Breaking intermolecular forces is always endothermic.
Forming new intermolecular forces is always
exothermic.
To determine whether DHsoln is positive or negative, we
consider the strengths of all solute-solute and solutesolvent interactions:
1.
2.
3.
DH1 and DH2 are both positive.
DH3 is always negative.
It is possible to have either:
DH3 > (DH1 + DH2)
or
DH3 < (DH1 + DH2).
DHsoln negative
CaCl2 (s) : DHsoln = -81.3 kJ/mol
MgSO4 (s) : DHsoln = -91.2 kJ/mol
DHsoln positive.
NH4NO3 (s)
DHsoln = +25.7 kJ/mol
Concentration Units
The concentration of a solution is the amount of solute
present in a given quantity of solvent or solution.
Percent by Mass (w/w %)
mass of solute
x 100%
% by mass =
mass of solute + mass of solvent
mass of solute x 100%
=
mass of solution
Mole Fraction (X)
moles of A
XA =
sum of moles of all components
12.3
Concentration Units (cont.)
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
12.3
Concentration Units (cont.)
Parts per million (ppm): This unit of concentration may be
expressed in a number of ways. It is often used to express
the concentration of very dilute solutions. The "technical"
definition of parts per million is:
ppm =
1 g solute
1000 000 g solution
Parts per million may also be expressed in the following two
ways:
Concentration Units (cont.)
Parts per billion (ppb): This concentration unit is also
used for very dilute solutions. The "technical"
definition is as follows:
1 g solute
ppb =
1000 g solution
Owing to the dilute nature of the solution, once again,
the density of the solution will be about the same as the
density of the solvent. Thus, we may also express parts
per billion as:
Question #1: Sterile saline solutions containing NaCl in
water are often used in medicine. What is the percent
by mass of NaCl in a solution made by dissolving 4.6 g
NaCl in 500 g of pure water?
Answer:
% by mass =
mass of solute
mass of solute + mass of solvent
4.6 g
=
4.6 g + 500 g
= 0.91 %
x 100%
x 100%
Question #2: Calculate the molality of a solution
containing 88.4 g glycine (NH2CH2COOH) dissolved in
1.250 kg H2O
Answer:
Mole glycine = 88.4 g  1 mol = 1.178 mol
75.07 g
1.178 mol
Moles solute
= 0.9424 molal
=
molality =
1.250 kg
Mass solvent (kg)
Question #3: A sample of water is found to contain 0.010
ppm lead ions (Pb2+).
a) What is the mass of lead ions per liter of this solution?
Assume density of the solution is 1.0 g/mL.
b) What is the lead concentration in ppb?
Answer:
0.010 g Pb2+
1 g solution
a)
x
x
1000 000 g solution
1 mL solution
1000 mL
1L
= 1 x 10–5 g Pb2+/L solution
b)
1 x 10–5 g Pb2+
1 L solution
x
106 g
= 10 g Pb2+/L solution
1g
= 10 ppb Pb2+
Question #4: Hydrochloric acid is sold as a concentrated
aqueous solution of HCl with a density of 1.18 g/mL. The
concentrated acid contains 38% HCl by mass. Calculate
the molarity of HCl in this solution.
Answer:
38% HCl by mass means that 100 g solution contains 38 g HCl
Mole HCl = 38 g 
1 mol HCl
36.5 g HCl
Volume of HCl = 100 g 
molarity =
1 mL
1L
= 0.0847 L

1.18 g
1000 mL
Moles solute
L solvent
= 1.041 mol HCl
=
1.041 mol
0.0847 L
= 12.3 M
Question #5: What is the molality of a 5.86 M
ethanol (C2H5OH) solution whose density is
0.927 g/mL?
moles of solute
moles of solute
m =
M =
mass of solvent (kg)
liters of solution
Answer:
Assume 1 L of solution
Moles of ethanol = 5.86 mol
Mass of ethanol = 270 g
Mass of solution = 0.927 g/mL  1000 mL = 927 g
Mass solvent = mass of solution – mass of solute
= 927 g – 270 g
= 657 g
molality =
Moles solute
Mass solvent (kg)
=
5.86 mol
= 8.92 molal
0.657 kg
12.3
Question #6: Vodka is essentially a solution of pure
ethanol in water. A typical vodka is sold as “80
proof”, which means that it contains 40.0% ethanol
by volume. The density of pure ethanol is 0.789
g/mL at 20oC. If we assume that the volume of the
solution is the sum of the volumes of the
components, what are
(a) the mass percent
(b) the mole fraction
(c) the molarity
(d) the molality
of ethanol in “80 proof” vodka?
Assume 1 L of solution:
Volume of ethanol = 40.0% x 1000 mL = 400 mL
Volume of water = 1000 mL – 400 mL = 600 mL
Mass of ethanol = 40.0% x 0.789 g/mL = 315.6 g
Mass of water = 600 mL x 1.0 g/mL = 600 g
(a) the mass percent of ethanol in vodka:
% ethanol =
315.6 g
x 100% = 34.5 %
315.6 g + 600 g
(b) the mole fraction
Mole ethanol = 315.6 g 
Mole water =
600 g 
1 mol
46.07 g
= 6.85 mol
1 mol
18.02 g
= 33.3 mol
Mole fraction of ethanol in vodka
XEtOH =
c) molarity =
6.85
= 0.171
6.85 mol + 33.3 mol
6.85 mol
= 6.85 M
1L
d) molality =
6.85 mol
0.6 kg
= 11.4 m
LEARNING OBJECTIVES
• To understand the relationship
between solubility and molecular
structure;
• To understand the
relationships between
temperature, pressure
and solubility;
Factors Affecting Solubility:
(1) Molecular structure
Like-Dissolves-Like
Two substances with similar intermolecular forces are likely
to be soluble in each other.
•
non-polar solutes are soluble in non-polar solvents
CCl4 in C6H6
•
polar solutes are soluble in polar solvents
C2H5OH in H2O
•
ionic compounds are more soluble in polar solvents
NaCl in H2O or NH3 (l)
If both solute and solvent are in liquid forms, solubility
can be interpreted as ‘miscible’ or ‘immiscible’.
12.2
e.g. of like-dissolves-like : soaps & detergents
Polar head group
(hydrophilic)
Non-polar tail
(hydrophobic)
Head group dissolves in water and tail dissolves in grease.
Tails sink into blob of grease up as far as head groups &
head groups remain on surface.
Head groups interact favourably with water and the whole
blob can be dissolved in water and be washed away.
Action of soap on oil
QUESTION #7: Predict which of these substances
will be most soluble and which will be the least
soluble in water:
LiCl ;
Benzoic acid (C6H5COOH) ;
Napthalene
Answer:
LiCl is an ionic compound, and it is highly soluble in water
(polar).
Benzoic acid has polar carboxylic group and a non-polar
aromatic ring. The polar group is soluble in water (polar).
Napthalene is a non-polar compound, it is
not soluble in water (polar).
Napthalene < benzoic acid < LiCl
Increasing solubility
11.2
QUESTION #8: Arrange the following compounds
in order of their expected increasing solubility in
water:
Br2
KBr
Toluene (C7H8, a constituent of gasoline).
Answer:
Toluene and Br2 are both non-polar and
toluene has a larger size than Br2.
KBr is ionic and easily dissolve in water.
Toluene < Br2 < KBr
Increasing solubility
Factors Affecting Solubility:
(2) Pressure effect (for gases in any solvent)
Solubility of a gas in a solvent is a function of
the pressure of the gas.
The higher the pressure, the more molecules of
gas becoming closer to the solvent and the
greater the chance of a gas molecule to strike
the surface and entering the solution.
Therefore, the higher the pressure, the
greater the solubility of gas.
The lower the pressure, the fewer
molecules of gas are close to the
solvent and the lower the solubility.
The concentration of dissolved gas depends on the partial
pressure of the gas. The partial pressure controls the number
of gas molecule collisions with the surface of the solution. If
the partial pressure is doubled the number of collisions with
the surface will double. The increased number of collisions
produce more dissolved gas.
Low pressure
Low concentration
Double the pressure
Double the concentration
Pressure and Solubility of Gases
The solubility of a gas in a liquid is proportional to the
pressure of the gas over the solution (Henry’s law).
c is the concentration (M) 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
low P
high P
low c
high c
12.5
Henry’s Law explains why soft drinks “fizz” and then go “flat”
after being opened.
Bubbles of CO2 form as soon
as a carbonated beverage is
opened because the drink was
bottled under CO2 at a pressure
higher than 1 atm.
When the bottle is opened, the pressure
of CO2 above the solution drops rapidly,
and some of the dissolved gas escapes
from the solution as bubbles.
Factors Affecting Solubility:
(3) Temperature effect (for gases in any
solvent)
Gas solubility and temperature
solubility usually
decreases with
increasing temperature
12.4
Experience tells us that carbonated beverages
go flat as they get warm.
Gases are less soluble at higher temperature.
Increased temperature
causes an increase in
kinetic energy.
The higher kinetic energy
causes more motion in
molecules which break
intermolecular bonds and
escape from solution.
What is Thermal Pollution?
The discharge of heated liquid
or air into lakes, rivers, etc., as
by an industry or nuclear power
plant, causing such a rise in the
water temperature as to affect
the life cycles within the water
and disrupt the ecological
balance.
If lakes get too warm, CO2
and O2 become less soluble
and are not available for
plants or animals.
Factors Affecting Solubility:
(3) Temperature effect (for solids in any solvent)
Solid solubility and
temperature:
solubility increases with
increasing temperature
solubility decreases
with increasing
temperature
12.4
Ionic
compounds
Experience tells us
that sugar
dissolves better in
warm water than
cold water.
Generally, as
temperature increases,
solubility of solids
increases.
But sometimes,
solubility decreases
as temperature
increases.
Although the solubility of a substance generally increases
with increasing temperature, there is no simple relationship
between the structure of a substance and the temperature
dependence of its solubility.
In chemistry, fractional crystallization is a method of
refining substances based on differences in solubility. If two
or more substances are dissolved in a solvent, they will
crystallize out of solution (precipitate) at different rates.
Crystallization can be induced by changes in concentration,
temperature or other means.
This technique is often used in chemical engineering to
obtain very pure substances, or to recover sellable
products from waste solutions.
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
12.4
LEARNING OBJECTIVES
• To understand the
colligative properties
of a solution.
• To understand the
effect of
intermolecular
forces on the
colligative properties
of a solution.
Physical Behavior of Solutions:
Colligative Properties
When, comparing a pure solvent and a
solution, the solution’s
• vapour pressure is lower;
• boiling point is elevated;
• freezing point is lower;
than the pure solvent.
And, osmosis occurs from
solvent to solution when
separated by a membrane.
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
0
1
Raoult’s law
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = DP = X2 P 10
X2 = mole fraction of the solute
12.6
Ideal Solution
PA = XA P A0
PB = XB P 0B
PT = PA + PB
PT = XA P A0 + XB P 0B
12.6
QUESTION #9: The vapor pressures of pure benzene
and toluene at 25°C are 95.1 and 28.4 mm Hg,
respectively. A solution is prepared in which the
mole fractions of benzene and toluene are both
0.500.
(a) What are the partial pressures of the benzene and
toluene above this solution?
(b) What is the total vapor pressure?
Answer:
a) PBenzene
PToluene
= XBenzene x PoBenzene
= 0.500 x 95.1 mmHg
= 47.6 mmHg
= XToluene x PoBenzene
= 0.500 x 28.4 mmHg
= 14.2 mmHg
b) PTotal = PBenzene + PToluene = 61.8 mmHg
Solutions that obey Raoult’s Law are called
‘ideal solution’.
Most real solutions exhibit positive or negative
deviation from Raoult’s Law, just like most real
gases that do not obey the ‘Ideal gas Law’.
Positive deviation or
negative deviation can
be distinguished
depending on the
intermolecular forces
between solute-solvent,
solvent-solvent and
solute-solute.
Negative deviation:
Methanol in acetone:
Methanol contains Hbonding and acetone is a
polar substance.
PT is less than
predicted by Raoults’s law
Acetone can forms acetonemethanol (A-B) interactions
that are stronger than acetoneacetone (A-A) or methanolmethanol (B-B) interactions.
The A-B interactions
effectively stabilize the
solution thus, lowering the
vapour pressure.
Force
Force
Force
> A-A & B-B
A-B
DH soln = -ve
12.6
PT is greater than
predicted by Raoults’s law
Force
Force
Force
< A-A & B-B
A-B
DH soln = +ve
Positive deviation:
Cyclohexane in ethanol:
Cyclohexane is a non-polar
substance and ethanol
contains H-bonding.
The non-polar cyclohexane
cannot interact favourably with
the polar ethanol, therefore the
cyclohexane-ethanol
interactions (A-B) is weaker
than both ethanol-ethanol and
cyclohexane-cyclohexane.
Both cyclohexane & ethanol have
the tendency to escape from the
solution, thus, increasing the
vapour pressure.
12.6
Fractional Distillation Apparatus
A procedure to
separate liquid
components
based on their
different boiling
points.
12.6
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)
12.6
QUESTION #10: Calculate the boiling point of a
30.2% aqueous solution of ethylene glycol (EG).
Given: Kb (water) = 0.51 oC/m.
Answer:
30.2% aqueous solution of EG means 30.2 g of
EG in 100 g solution
mass water = 100 g – 30.2 g = 69.8 g
30.2 g EG x 1 mol EG
62.01 g EG
Molality of EG =
0.0698 kg
= 6.98 m
∆Tb = Kb m = (0.51 oC/m)(6.98 m) = 3.6oC
Boiling point of pure water = 100oC
Boiling point of EG solution = 100oC + 3.6oC = 103.6oC
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)
12.6
Question #11: 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
Answer:
478 g EG x
Molality of EG =
1 mol EG
62.01 g EG
3.202 kg
= 2.41 m
∆Tf = Kf m = (1.86 0C/m)(2.41 m) = 4.48oC
Freezing point of pure water = 0oC
Freezing point of EG solution = 0oC – 4.48oC = – 4.48oC
12.6
12.6
Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a
porous membrane from a dilute solution to a more concentrated one.
A semi-permeable 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
12.6
Osmotic Pressure (p)
High
P
Low
P
p = MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K)
12.6
Question #12: Assume that the fluids inside a
sausage are approximately 0.80 M in dissolved
particles due to the salt and sodium nitrite used to
prepare them. Calculate the osmotic pressure
inside the sausage at 100oC to learn why
experienced cooks pierced the semi-permeable
skin of sausages before boiling them.
Answer:
p = MRT
p = (0.80 mol/L)(0.0821 L.atm/mol.K)(373 K)
p = 24.5 atm
Reverse Osmosis : Desalination
• Application of a
pressure to the
solution (that is equal
to or greater than the
Osmotic Pressure)
and the solvent flows
from the more
concentrated side to
the other one.
• This process is used
to obtain pure water
from salt water.
A cell in an:
Isotonic
solution
Equal concentration,
same osmotic
pressure, the cell
remains unchanged.
Hypotonic
solution
Dilute solution
outside the cell, the
cell swells.
Hypertonic
solution
Concentrated
solution outside
the cell, the cell
shrinks.
12.6
Example of Osmosis:
Cucumber placed in NaCl
solution loses water to shrivel
up and become a pickle.
Limp carrot placed in water
becomes firm because water
enters via osmosis.
‘Salt added to meat’ or ‘sugar
added to fruit’ prevents bacterial
infection (a bacterium placed on
the salt will lose water through
osmosis and die).
Colligative Properties of
Non-electrolyte 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
Osmotic Pressure (p)
p = MRT
12.6
Colligative Properties of Electrolyte Solutions
0.1 m Na+ ions & 0.1 m Cl- ions
0.1 m NaCl solution
Colligative properties are properties that depend only on the
number of solute particles in solution and not on the nature of
the solute particles.
0.1 m NaCl solution
Van’t Hoff =
factor, i
0.2 m ions in solution
actual number of particles
in solution after dissociation
number of formula units initially
dissolved in solution
i should be
nonelectrolytes
NaCl
CaCl2
1
2
3
12.7
Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = i MRT
12.7
Question #13: Calculate the van’t Hoff factor for a
0.050 m aqueous solution of MgCl2 that has a
measured freezing point of –0.25oC.
Given Kf (H2O) = 1.86 oC/m.
Answer:
DTf = i Kf m
–
0.25oC
= (i)(1.86
oC/m)(0.050
m)
i (measured)
 i = 2.7
MgCl2  Mg2+ + 2 Cl–
i =3
i (calculated)
i (measured) is lower than i (calculated) because of
the formation of ion-pair :– electrostatic force
between cation(s) and anion(s).