Transcript Document 7728119

Chapter 13
Properties of Mixtures: Solutions
13-1
Properties of Mixtures: Solutions
13.1 Types of solutions: Intermolecular forces and
predicting solubility
13.2 Energy changes in the solution process
13.3 Solubility as an equilibrium process
13.4 Quantitative ways of expressing concentration
13.5 Colligative properties of solutions
13-2
13-3
The major
types of
intermolecular
forces in
solutions
(from Chapter 12)
Figure 13.1
13-4
(energies in parenthesis)
“LIKE DISSOLVES LIKE”
Substances with similar types of
intermolecular forces dissolve in each other.
When a solute dissolves in a solvent, solute-solute interactions and solventsolvent interactions are partly replaced with solute-solvent interactions.
The new forces created between solute and solvent must be comparable in
strength to the forces destroyed within the solute and the solvent.
13-5
A major factor that determines
whether a solution forms:
The relative strengths of the intermolecular forces within and
between solute and solvent molecules
13-6
Some Definitions
Solvent: the most abundant component of a given solution
Solute: component dissolved in the solvent
Solubility (S): the maximum amount of solute that dissolves in a
fixed quantity of solvent at a given temperature (in the presence
of excess solute)
Dilute and concentrated solutions: qualitative terms
13-7
Hydration
shells around
an aqueous
ion
Formation of ion-dipole
forces when a salt dissolves
in water
Figure 13.2
13-8
Liquid Solutions


Liquid-Liquid
Gas-Liquid
Gas and Solid Solutions



13-9
Gas-Gas
Gas-Solid
Solid-Solid
hexane =
CH3(CH2)4CH3
Competition
between
H-bonding
and dispersion
forces
13-10
Molecular Basis for the Solubility of CH3OH in H2O
H-bonding: CH3OH can serve as a donor and acceptor
(maximum number of three H-bonds / molecule)
Figure 13.3
13-11
SAMPLE PROBLEM 13.1
PROBLEM:
Predicting relative solubilities of substances
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 hexane (CH3CH2CH2CH2CH2CH3)
or in water.
(c) Diethyl ether (CH3CH2OCH2CH3) in water or in ethanol (CH3CH2OH)
PLAN:
Consider the intermolecular forces that exist between solute
molecules and consider whether the new solvent-solute interactions
can substitute for them.
SOLUTION:
(a) NaCl is ionic and forms ion-dipoles with the OH groups of both
methanol and propanol. However, propanol is subject to greater
dispersion forces (more CH bonds than methanol).
(b) Hexane has no dipoles to interact with the OH groups of ethylene glycol.
Water can H-bond to ethylene glycol.
(c) Diethyl ether can interact through dipole and dispersion forces. Ethanol
can provide both while water can only H-bond.
13-12
Structure-Function Correlations: A Soap
Soap: the salt form of a long-chain fatty acid; is amphipathic
in character (has polar and non-polar components)
Figure B13.1
13-13
The mode of
action of the
antibiotic,
Gramicidin A
Destroys the Na+/K+ ion
concentration gradients
in the cell
Figure B13.2
13-14
Gas-Liquid Solutions
QuickTime™ and a
Photo - JPEG decompressor
are needed to see this picture.
Non-polar gas solubility in
water is directly related to
the boiling point of the gas.
important to
aquatic life
13-15
Gas-gas solutions: All gases are infinitely soluble in one another.
Gas-solid solutions: The gas molecules occupy the spaces
between the closely packed particles of the solid.
Solid-solid solutions: alloys (substitutional or interstitial)
13-16
The arrangement of atoms in two types of alloys
Figure 13.4
13-17
Heats of solution and solution cycles
Dissolution of a solid: breaking down the process into three steps
1. Solute particles separate from each other - endothermic
solute (aggregated) + heat
solute (separated)
DHsolute > 0
2. Solvent particles separate from each other - endothermic
solvent (aggregated) + heat
solvent (separated)
DHsolvent > 0
3. Separate solute and solvent particles mix - exothermic
solute (separated) + solvent (separated)
13-18
solution + heat
DHmix < 0
Calculating the heat of solution, DHsoln
The total enthalpy change that occurs when a solution forms by
dissolving a solute into a solvent.
DHsoln = DHsolute + DHsolvent + DHmix
A thermochemical solution cycle
13-19
Solution cycles and the enthalpy components of the heat of solution
Figure 13.5
13-20
Heats of Hydration
The solvation of ions by water is always exothermic.
M+
(g) [or
X- (g)]
H2O
M+ (aq) [or X- (aq)]
DHhydr of the ion < 0
(for 1 mole of gaseous ions)
DHhydr is related to the charge density of the ion, that is,
both coulombic charge and ion size are important.
Lattice energy is the DH involved in the
formation of an ionic solid from its gaseous ions.
M+ (g) + X- (g)
MX(s)
DHlattice is always (-)
Thus, DHsoln = -DHlattice + DHhydr
13-21
Heats of Hydration and Ionic Character


13-22
For a given size, greater charge leads to a more (-) DHhydr
For a given charge, smaller size leads to a more (-) DHhydr
Table 13.4
ion
Trends in Ionic Heats of Hydration
ionic radius (pm)
DHhydr (kJ/mol)
Group 1A
Li+
Na+
K+
Rb+
Cs+
76
102
138
152
167
-510
-410
-336
-315
-282
72
100
118
135
-1903
-1591
-1424
-1317
133
181
196
220
-431
-313
-284
-247
Group 2A
Mg2+
Ca2+
Sr2+
Ba2+
Group 7A
13-23
FClBrI-
Enthalpy Diagrams for Dissolving Three
Different Ionic Compounds in Water
NaCl
Figure 13.6
13-24
NH4NO3
NaOH
Entropy Considerations
The natural tendency of most systems is to become more
disordered; entropy increases.
Entropy always favors the formation of solutions.
Dissolution: involves a change in enthalpy and a change in
entropy.
13-25
Enthalpy diagrams for dissolving NaCl and octane in hexane
NaCl in insoluble in hexane!
Figure 13.7
13-26
In this case, dissolution is
entropy-driven!
More Definitions
When excess undissolved solute is in equilibrium with
the dissolved solute: a saturated solution
An unsaturated solution: more solute can be dissolved,
ultimately producing a saturated solution
A supersaturated solution: a solution that contains more
than the equilibrium amount of dissolved solute
13-27
Equilibrium in a saturated solution
Figure 13.8
13-28
solute (undissolved)
solute (dissolved)
Sodium acetate crystallizing from a supersaturated solution
nucleation
Figure 13.9
13-29
a saturated solution
results
Solubility and Temperature
Most solids are more soluble at higher temperatures.
The sign of the heat of solution, however, does not predict reliably the effect
of temperature on solubility; e.g., NaOH and NH4NO3 have
DHsoln of opposite signs, yet their solubility in H2O increases
with temperature.
13-30
The relation
between
solubility and
temperature for
several ionic
compounds
Figure 13.10
13-31
Gas Solubility in Water: Temperature Effects
For all gases, DHsolute = 0, DHhydr < 0; thus, DHsoln < 0
solute(g) + water(l)
saturated solution(aq) + heat
Implications: gas solubility in water decreases with
increasing temperature
13-32
Thermal
Pollution
Leads to O2
deprivation in aquatic
systems
Figure 13.11
13-33
Pressure Effects on Solubility
Essentially zero for solids and liquids, but substantial for gases!
gas + solvent
13-34
saturated solution
The effect of pressure on gas solubility
Figure 13.12
13-35
gas volume is reduced;
pressure (concentration!)
increases; more collisions
occur with liquid surface
Henry’s Law
A quantitative relationship
between gas solubility and
pressure
Sgas = kH x Pgas
The solubility of a gas
(Sgas) is directly
proportional to the
partial pressure of the
gas (Pgas) above the
solution.
kH = Henry’s law constant
for a gas; units of mol/L.atm
13-36
Implications for scuba diving!
SAMPLE PROBLEM 13.2
PROBLEM:
PLAN:
The partial pressure of carbon dioxide gas inside a bottle of
cola is 4 atm at 25 oC. What is the solubility of CO2? The
Henry’s law constant for CO2 dissolved in water is 3.3 x 10-2
mol/L.atm at 25 oC.
Knowing kH and Pgas, we can substitute into the Henry’s Law
equation.
SOLUTION:
S
CO2
13-37
Using Henry’s Law to calculate gas solubility
= (3.3 x 10-2 mol/L.atm)(4 atm) =
0.1 mol / L
Table 13.5
concentration term
molarity (M)
molality (m)
Concentration Definitions
ratio
amount (mol) of solute
volume (L) of solution
amount (mol) of solute
mass (kg) of solvent
parts by mass
parts by volume
mass of solute
mass of solution
volume of solute
volume of solution
mole fraction 
amount (mol) of solute
amount (mol) of solute + amount (mol) of solvent
13-38
SAMPLE PROBLEM 13.3
PROBLEM:
Calculating molality
What is the molality of a solution prepared by dissolving 32.0 g
of CaCl2 in 271 g of water?
PLAN: Convert grams of CaCl2 into moles and grams of water to kg. Then
substitute into the equation for molality.
SOLUTION:
32.0 g CaCl2 x
mole CaCl2
110.98 g CaCl2
0.288 mole CaCl2
molality =
kg
271 g H2O x
13-39
103 g
= 0.288 mole CaCl2
= 1.06 m CaCl2
The Sex Attractant of the Gypsy Moth
Potent at Extremely Low Concentrations!
100-300 molecules/mL air
100 parts per quadrillion by volume!
Practical Implications: a strategy used to target and
trap specific insects (Japanese beetles)
Figure 13.13
13-40
Other Expressions of Concentration
mass percent (% w/w) = mass solute / mass of solution x 100
(related to parts per million (ppm) or parts per billion (ppb))
volume percent (% (v/v) = volume solute / volume of solution x 100
% (w/v) = solute mass / solution volume x 100
mole percent (mol%) = mole fraction x 100
13-41
SAMPLE PROBLEM 13.4
Expressing concentration in parts by mass,
parts by volume, and mole fraction
PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50 g pill that
contains 40.5 mg of Ca.
(b) The label on a 0.750 liter bottle of Italian chianti indicates
“11.5% alcohol by volume”. How many liters of alcohol
does the wine contain?
(c) A sample of rubbing alcohol contains 142 g of isopropyl
alcohol (C3H7OH) and 58.0 g of water. What are the
mole fractions of alcohol and water?
PLAN:
13-42
(a) Convert mg to g of Ca, find the ratio of g Ca to g pill, and multiply
by 106.
(b) Knowing the % alcohol and the total volume, the volume of
alcohol can be calculated.
(c) Convert g of solute and solvent to moles, and find the ratios of
each part to the total.
SAMPLE PROBLEM 13.4
(continued)
g
SOLUTION:
(a)
40.5 mg Ca x
103 mg
3.5 g
(b)
(c)
11.5 L alcohol
0.750 L chianti x
moles water = 58.0 g x
2.36 mol C3H7OH
= 0.0862 L alcohol
mole
18.02 g
mole
60.09 g
= 2.36 mol C3H7OH
= 3.22 mol H2O
3.22 mol H2O
2.36 mol C3H7OH + 3.22 mol H2O
13-43
= 1.16 x 104 ppm Ca
100. L chianti
moles isopropyl alcohol = 142 g x
= 0.423
x 106
C3H7OH
2.36 mol C3H7OH + 3.22 mol H2O
= 0.577
H2O
SAMPLE PROBLEM 13.5
Converting concentration units
PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in
concentrated solution in rocket fuels and in dilute solution as a
hair bleach. An aqueous solution of H2O2 is 30.0% by mass
and has a density of 1.11 g/mL. Calculate its:
(a) molality
(b) mole fraction
(c) molarity
PLAN: (a) To find the mass of solvent, assume the % is per 100 g of solution.
Take the difference in the mass of the solute and solution to determine
the mass of solvent.
(b) Convert g of solute and solvent to moles before finding .
(c) Use the density to find the volume of the solution.
SOLUTION:
(a)
g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O
30.0 g H2O2
x
molality =
70.0 g H2O x
13-44
mol
H O2
34.022g H
2O2
kg H2O
103 g
= 12.6 m H2O2
SAMPLE PROBLEM 13.5
(b)
70.0 g H2O
x
(continued)
mol H2O
18.02 g H2O
0.882 mol H2O2
= 3.88 mol H2O
= 0.185 =  of H2O2
0.882 mol H2O2 + 3.88 mol H2O
(c)
mL
100.0 g solution x
= 90.1 mL solution
1.11 g
0.882 mol H2O2
90.1 mL solution x
13-45
L
103 mL
= 9.79 M H2O2
Colligative Properties
Physical properties of solutions dictated by the number of
solute particles present. Their chemical structures are
not factors in determining these properties!
vapor pressure lowering
 boiling point elevation
 freezing point depression
 osmotic pressure

13-46
Three types of electrolytes
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
Copyright © The McGraw-Hill Companies, Inc.
Permission required for reproduction or display.
non-electrolyte
strong
Figure 13.14
13-47
weak
Vapor Pressure Lowering
The vapor pressure of a solution of a nonvolatile nonelectrolyte
is always lower than the vapor pressure of the pure solvent.
An entropy argument!
Figure 13.15
13-48
Quantitative Treatment of VP Lowering
Raoult’s Law (vapor pressure of a solvent above a solution, Psolvent)
Psolvent = solvent x Posolvent
where Posolvent = vapor pressure of the pure solvent
How does the amount of solute affect the magnitude of the VP lowering?
( substitute 1- solute for solvent in the above equation and rearrange)
Posolvent - Psolvent = DP = solute x Posolvent
(change in VP is proportional to the mole fraction of solute)
13-49
SAMPLE PROBLEM 13.6
PROBLEM:
PLAN:
Using Raoult’s Law to find the vapor pressure
lowering
Calculate the vapor pressure lowering, DP, when 10.0 mL of
glycerol (C3H8O3) is added to 500. mL of water at 50. oC. At this
temperature, the vapor pressure of pure water is 92.5 torr and its
density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.
Find the mol fraction, , of glycerol in solution and multiply by the
vapor pressure of water.
SOLUTION:
10.0 mL C3H8O3 x
500.0 mL H2O x
1.26 g C3H8O3
x
mL C3H8O3
0.988 g H2O
mL H2O
x
mol C3H8O3
= 0.137 mol C3H8O3
92.09 g C3H8O3
mol H2O
18.02 g H2O
= 27.4 mol H2O
 = 0.00498
DP =
0.137 mol C3H8O3
0.137 mol C3H8O3 + 27.4 mol H2O
13-50
x
92.5 torr
= 0.461 torr
Boiling Point Elevation
A solution boils at a higher temperature than the pure solvent.
This effect is explained by differences between the VP of
the solution and VP of the pure solvent at a given temperature.
13-51
Superimposed phase diagrams of solvent and solution
aqueous solution:
dashed lines
pure water:
solid lines
Figure 13.16
13-52
Quantitative Treatment of BP Elevation
The magnitude of the effect is proportional to solute concentration.
DTb = Kbm
(m = solution molality, Kb = molal BP elevation constant, DTb = BP elevation)
DTb = Tb (solution) - Tb (solvent)
13-53
Quantitative Treatment of FP Depression
The magnitude of the effect is proportional to solute concentration.
DTf = Kfm
(m = solution molality, Kf = molal FP depression constant, DTf = FP depression)
DTf = Tf (solvent) - Tf (solution)
13-54
Table 13.6
Molal Boiling Point Elevation and Freezing Point
Depression Constants of Several Solvents
Kb (oC/m)
melting
point (oC)
Kf (oC/m)
117.9
3.07
16.6
3.90
benzene
80.1
2.53
5.5
4.90
carbon disulfide
46.2
2.34
-111.5
3.83
carbon tetrachloride
76.5
5.03
-23
chloroform
61.7
3.63
-63.5
4.70
diethyl ether
34.5
2.02
-116.2
1.79
ethanol
78.5
1.22
-117.3
1.99
100.0
0.512
0.0
1.86
solvent
boiling
point (oC)*
acetic acid
water
*at 1 atm.
13-55
30.
SAMPLE PROBLEM 13.7
Determining the boiling point elevation and
freezing point depression of a solution
PROBLEM: You add 1.00 kg of ethylene glycol antifreeze (C2H6O2) to your car
radiator, which contains 4450 g of water. What are the boiling and
freezing points of the resulting solution?
Find the number of mols of ethylene glycol and m of the solution;
multiply by the boiling or freezing point constant; add or subtract,
respectively, the changes from the boiling point and freezing point
of water.
SOLUTION:
mol C2H6O2
1.00 x 103 g C2H6O2 x
= 16.1 mol C2H6O2
62.07 g C2H6O2
PLAN:
16.1 mol C2H6O2
4.450 kg H2O
= 3.62 m C2H6O2
DTb = 0.512 oC/m x 3.62 m = 1.85 oC
BP = 101.85 oC
13-56
DTf = 1.86 oC/m x 3.62 m
FP = -6.73 oC
Osmotic Pressure
13-57

Applies only to aqueous solutions!

Two solutions of different concentrations are separated by
a semi-permeable membrane (allows water but not
solute to pass through)
The development of osmotic pressure
osmotic
pressure
pure
solvent
semipermeable
membrane
13-58
solution
applied
pressure
needed to
prevent
volume
increase;
equal to
the
osmotic
pressure
Figure 13.17
Quantitative Treatment of Osmotic Pressure (P)
OP is proportional to the number of solute particles in a given
volume of solution (to M).
Pansolute/Vsoln
or
Pa M
The constant of proportionality = RT, so P = M x R x T
T is the Kelvin temperature
13-59
Underlying Principle of Colligative Properties
Each property stems from an inability of solute particles to
cross between two phases.
13-60
Determination of Solute Molar Mass by Exploiting Colligative Properties
In principle, any colligative property can be used, but OP gives the
most accurate results (better dynamic range).
13-61
SAMPLE PROBLEM 13.8
Determining molar mass from osmotic pressure
PROBLEM: Biochemists have discovered more than 400 mutant varieties of
hemoglobin (Hb), the blood protein that carries oxygen
throughout the body. A physician studying a form of Hb
associated with a fatal disease first finds its molar mass (M).
She dissolves 21.5 mg of the protein in water at 5.0 oC to make
1.50 mL of solution and measures an osmotic pressure of 3.61
torr. What is the molar mass of this Hb mutant?
We know Pas well as R and T. Convert P to atm and T to Kelvin.
Use the P equation to find the molarity M and then the amount and
volume of the sample to calculate M.
atm
P
SOLUTION:
3.61 torr x
M=
=
760 torr
RT
= 2.08 x 10-4 M
(0.0821 L . atm/mol . K)(278.15 K)
PLAN:
2.08 x 10-4 mol
x 1.50 mL
L
# mol = g/M
21.5 mg
13-62
x
g
103 mg
x
L
x
103 mL
1
3.12 x 10-7 mol
= 3.12 x 10-7 mol
= 6.89 x 104 g/mol
Fractional Distillation of Volatile
Nonelectrolytes
The presence of each volatile component
lowers the vapor pressure of the other.
partial pressure = mole fraction x vapor pressure of pure gas
For vapor: mole fraction = partial pressure / total pressure
(thus, the vapor has a higher mole fraction of the more volatile
solution component)
13-63
Gasoline vapors
Condenser
Gas
The process of
fractional
distillation
Gasoline 38 oC
Kerosene 150 oC
Heating oil 260 oC
Lubricating oil 315 oC - 370 oC
Crude oil vapors
from heater
Steam
Figure 13.18
13-64
Residue (asphalt, tar)
Colligative Properties of Electrolyte Solutions
Must consider the full dissociation into ions!
van’t Hoff factor (i) = measured value for electrolyte solution
expected value for nonelectrolyte solution
This factor is multiplied into the appropriate equations; for example,
P = i (MRT).
For ideal behavior, i = mol particles in solution / mol dissolved solute
But solutions are not ideal; for example, for BP elevation of NaCl
solutions, i = 1.9, not 2!
Data suggest that the ions are not behaving
as independent particles!
13-65
Non-ideal
behavior of
electrolyte
solutions
Observed values of i are less
than the predicted (expected)
values.
Figure 13.19
13-66
An ionic
atmosphere
model for nonideal behavior of
electrolyte
solutions
ionic atmospheres
Concept of effective
concentration
Figure 13.20
13-67
Some Practical Applications

ion-exchange (water softeners)

13-68
water purification
Ion exchange for
removal of hardwater cations
Use of ion-exchange
resins
Figure B13.4
13-69
Reverse osmosis for the removal of ions
Desalination Process
Figure B13.5
13-70
End of Assigned Material
13-71
Light scattering and the Tyndall effect
Figure 13.21
13-72
Photo by C.A.Bailey, CalPoly SLO (Myanmar)
A Cottrell precipitator for removing particulates from
industrial smokestack gases
13-73
Figure 13.22
The steps in a typical municipal water treatment plant
Figure B13.3
13-74