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13-2
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
13-4
Figure 13.1
The major types of
intermolecular forces in solutions
(from Chapter 12)
(energies in parenthesis)
13-5
“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 solvent-solvent 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.
13-6
A major factor that determines whether a solution forms:
The relative strengths of the intermolecular forces within andbetween solute and solvent molecules
13-7
Some Definitions
Solvent: the most abundant component of a given solutionSolute: 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 presenceof excess solute)
Dilute and concentrated solutions: qualitative terms
13-8
Figure 13.2
Hydration shells
around an aqueous ion
Formation of ion-dipoleforces when a salt dissolves
in water
13-9
Liquid Solutions
Liquid-Liquid Gas-Liquid
Gas and Solid Solutions
Gas-Gas Gas-Solid Solid-Solid
13-11
Figure 13.3
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)
13-12
(c) Diethyl ether can interact through dipole and dispersion forces. Ethanolcan provide both while water can only H-bond.
(b) Hexane has no dipoles to interact with the OH groups of ethylene glycol. Water can H-bond to ethylene glycol.
SAMPLE PROBLEM 13.1 Predicting relative solubilities of substances
SOLUTION:
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 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.
(a) NaCl is ionic and forms ion-dipoles with the OH groups of bothmethanol and propanol. However, propanol is subject to greaterdispersion forces (more CH bonds than methanol).
13-13
Figure B13.1
Structure-Function Correlations: A Soap
Soap: the salt form of a long-chain fatty acid; is amphipathicin character (has polar and non-polar components)
13-14
Figure B13.2
The mode of action of the
antibiotic, Gramicidin A
Destroys the Na+/K+ ionconcentration gradients
in the cell
13-15
QuickTime™ and aPhoto - JPEG decompressor
are needed to see this picture.
Gas-Liquid Solutions
Non-polar gas solubility in water is directly related to the boiling point of the gas.
important toaquatic life
13-16
Gas-gas solutions: All gases are infinitely soluble in one another.
Gas-solid solutions: The gas molecules occupy the spacesbetween the closely packed particles of the solid.
Solid-solid solutions: alloys (substitutional or interstitial)
13-18
Heats of solution and solution cycles
1. Solute particles separate from each other - endothermic
solute (aggregated) + heat solute (separated) Hsolute > 0
2. Solvent particles separate from each other - endothermic
3. Separate solute and solvent particles mix - exothermic
solvent (aggregated) + heat solvent (separated) Hsolvent > 0
solute (separated) + solvent (separated) solution + heat Hmix < 0
Dissolution of a solid: breaking down the process into three steps
13-19
Hsoln = Hsolute + Hsolvent + Hmix
Calculating the heat of solution, Hsoln
The total enthalpy change that occurs when a solution forms by dissolving a solute into a solvent.
A thermochemical solution cycle
13-21
Heats of Hydration
The solvation of ions by water is always exothermic.
M+ (g) [or X- (g)] M+ (aq) [or X- (aq)] Hhydr of the ion < 0H2O
Hhydr is related to the charge density of the ion, that is, both coulombic charge and ion size are important.
Lattice energy is the H involved in the formation of an ionic solid from its gaseous ions.
M+ (g) + X- (g) MX(s) Hlattice is always (-)
Thus, Hsoln = -Hlattice + Hhydr
(for 1 mole of gaseous ions)
13-22
Heats of Hydration and Ionic Character
For a given size, greater charge leads to a more (-) Hhydr
For a given charge, smaller size leads to a more (-) Hhydr
13-23
Table 13.4 Trends in Ionic Heats of Hydration
ion ionic radius (pm) Hhydr (kJ/mol)
Group 1A
Group 2A
Group 7A
Li+
Na+
K+
Rb+
Cs+
Mg2+
Ca2+
Sr2+
Ba2+
F-
Cl-
Br-
I-
76102138152167
-510-410-336-315-282
72100118
133
-1903-1591-1424
-431181 -313196 -284220 -247
135 -1317
13-24
Figure 13.6
Enthalpy Diagrams for Dissolving Three Different Ionic Compounds in Water
NaOH
NH4NO3NaCl
13-25
Entropy Considerations
The natural tendency of most systems is to become moredisordered; entropy increases.
Dissolution: involves a change in enthalpy and a change inentropy.
Entropy always favors the formation of solutions.
13-26
Figure 13.7
Enthalpy diagrams for dissolving NaCl and octane in hexane
In this case, dissolution isentropy-driven!
NaCl in insoluble in hexane!
13-27
When excess undissolved solute is in equilibrium withthe 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 morethan the equilibrium amount of dissolved solute
More Definitions
13-29
Figure 13.9
Sodium acetate crystallizing from a supersaturated solution
nucleation a saturated solutionresults
13-30
Solubility and Temperature
Most solids are more soluble at higher temperatures.
The sign of the heat of solution, however, does not predict reliably the effectof temperature on solubility; e.g., NaOH and NH4NO3 have
Hsoln of opposite signs, yet their solubility in H2O increaseswith temperature.
13-32
Gas Solubility in Water: Temperature Effects
For all gases, Hsolute = 0, Hhydr < 0; thus, Hsoln < 0
Implications: gas solubility in water decreases with increasing temperature
solute(g) + water(l) saturated solution(aq) + heat
13-34
Pressure Effects on Solubility
Essentially zero for solids and liquids, but substantial for gases!
gas + solvent saturated solution
13-35
Figure 13.12
The effect of pressure on gas solubility
gas volume is reduced;pressure (concentration!)increases; more collisionsoccur with liquid surface
13-36
Henry’s Law
Sgas = kH x Pgas
The solubility of a gas (Sgas) is directly
proportional to the partial pressure of the gas (Pgas) above the
solution.
A quantitative relationshipbetween gas solubility and
pressure
kH = Henry’s law constant
for a gas; units of mol/L.atm
Implications for scuba diving!
13-37
SAMPLE PROBLEM 13.2 Using Henry’s Law to calculate gas solubility
PLAN:
SOLUTION:
PROBLEM: 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.
0.1 mol / LS = (3.3 x 10-2 mol/L.atm)(4 atm) =CO2
13-38
Table 13.5 Concentration Definitions
concentration term ratio
molarity (M)amount (mol) of solutevolume (L) of solution
molality (m)amount (mol) of solute
mass (kg) of solvent
parts by massmass of solute
mass of solution
parts by volumevolume of solute
volume of solution
mole fraction amount (mol) of solute
amount (mol) of solute + amount (mol) of solvent
13-39
SAMPLE PROBLEM 13.3 Calculating molality
PLAN:
SOLUTION:
PROBLEM: What is the molality of a solution prepared by dissolving 32.0 g of CaCl2 in 271 g of water?
Convert grams of CaCl2 into moles and grams of water to kg. Then substitute into the equation for molality.
molality =
= 0.288 mole CaCl2
271 g H2O
0.288 mole CaCl2
kg
103 gx
= 1.06 m CaCl2
32.0 g CaCl2mole CaCl2
110.98 g CaCl2x
13-40
Figure 13.13
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 andtrap specific insects (Japanese beetles)
13-41
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
13-42
SAMPLE PROBLEM 13.4 Expressing concentration in parts by mass, parts by volume, and mole fraction
PLAN:
PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50 g pill thatcontains 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 alcoholdoes the wine contain?
(c) A sample of rubbing alcohol contains 142 g of isopropylalcohol (C3H7OH) and 58.0 g of water. What are the
mole fractions of alcohol and water?
(a) Convert mg to g of Ca, find the ratio of g Ca to g pill, and multiplyby 106.
(b) Knowing the % alcohol and the total volume, the volume ofalcohol can be calculated.
(c) Convert g of solute and solvent to moles, and find the ratios ofeach part to the total.
13-43
SAMPLE PROBLEM 13.4
SOLUTION:
(continued)
(a)
3.5 g
103 mg
g40.5 mg Ca x
106x = 1.16 x 104 ppm Ca
(b) 11.5 L alcohol
100. L chianti0.750 L chianti x = 0.0862 L alcohol
(c) moles isopropyl alcohol = 142 gmole
60.09 g = 2.36 mol C3H7OH
moles water = 58.0 gmole
18.02 g = 3.22 mol H2O
2.36 mol C3H7OH
2.36 mol C3H7OH + 3.22 mol H2O
3.22 mol H2O
2.36 mol C3H7OH + 3.22 mol H2O
= 0.423 C3H7
OH
= 0.577 H2O
x
x
13-44
SAMPLE PROBLEM 13.5 Converting concentration units
PLAN:
SOLUTION:
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
(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.
(a) g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O
molality =
30.0 g H2O234.02 g H2O2
mol H2O2
70.0 g H2Okg H2O103 g
= 12.6 m H2O2
x
x
13-45
SAMPLE PROBLEM 13.5 (continued)
(b)
0.882 mol H2O2
70.0 g H2Omol H2O
18.02 g H2O= 3.88 mol H2O
0.882 mol H2O2 + 3.88 mol H2O= 0.185 = of H2O2
(c)100.0 g solution
mL
1.11 g= 90.1 mL solution
0.882 mol H2O2
90.1 mL solution L
103 mL
= 9.79 M H2O2
x
x
x
13-46
Colligative Properties
Physical properties of solutions dictated by the number ofsolute particles present. Their chemical structures are
not factors in determining these properties!
vapor pressure lowering boiling point elevation freezing point depression osmotic pressure
13-47
Figure 13.14
Three types of electrolytes
strong
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
weak
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
non-electrolyte
Copyright © The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
13-48
Vapor Pressure Lowering
The vapor pressure of a solution of a nonvolatile nonelectrolyteis always lower than the vapor pressure of the pure solvent.
Figure 13.15
An entropy argument!
13-49
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
Posolvent - Psolvent = P = solute x Po
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)
(change in VP is proportional to the mole fraction of solute)
13-50
SAMPLE PROBLEM 13.6 Using Raoult’s Law to find the vapor pressure lowering
SOLUTION:
PROBLEM: Calculate the vapor pressure lowering, P, 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.
PLAN: Find the mol fraction, , of glycerol in solution and multiply by the vapor pressure of water.
10.0 mL C3H8O3
1.26 g C3H8O3
mL C3H8O3
mol C3H8O3
92.09 g C3H8O3= 0.137 mol C3H8O3
500.0 mL H2O0.988 g H2O
mL H2O
mol H2O
18.02 g H2O= 27.4 mol H2O
P = 0.137 mol C3H8O3
0.137 mol C3H8O3 + 27.4 mol H2O92.5 torrx
x
x
= 0.461 torr
= 0.00498
x
x
13-51
Boiling Point Elevation
A solution boils at a higher temperature than the pure solvent.
This effect is explained by differences between the VP ofthe solution and VP of the pure solvent at a given temperature.
13-52
Figure 13.16
Superimposed phase diagrams of solvent and solution
aqueous solution:dashed lines
pure water:solid lines
13-53
(m = solution molality, Kb = molal BP elevation constant, Tb = BP elevation)
Quantitative Treatment of BP Elevation
Tb = Kbm
The magnitude of the effect is proportional to solute concentration.
Tb = Tb (solution) - Tb (solvent)
13-54
(m = solution molality, Kf = molal FP depression constant, Tf = FP depression)
Quantitative Treatment of FP Depression
Tf = Kfm
The magnitude of the effect is proportional to solute concentration.
Tf = Tf (solvent) - Tf (solution)
13-55
Table 13.6 Molal Boiling Point Elevation and Freezing Point Depression Constants of Several Solvents
solventboiling
point (oC)* Kb (oC/m) Kf (
oC/m)
melting
point (oC)
acetic acid
benzene
carbon disulfide
carbon tetrachloride
chloroform
diethyl ether
ethanol
water
117.9
80.1
46.2
76.5
61.7
34.5
78.5
100.0
3.07 16.6 3.90
2.53 5.5 4.90
2.34 -111.5 3.83
5.03 -23 30.
3.63 -63.5 4.70
2.02 -116.2 1.79
1.22 -117.3 1.99
0.512 0.0 1.86
*at 1 atm.
13-56
SAMPLE PROBLEM 13.7 Determining the boiling point elevation and freezing point depression of a solution
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?
PLAN: 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.
1.00 x 103 g C2H6O2
mol C2H6O2
62.07 g C2H6O2
= 16.1 mol C2H6O2
Tb = 0.512 oC/m
16.1 mol C2H6O2
4.450 kg H2O= 3.62 m C2H6O2
3.62 mx = 1.85 oC
BP = 101.85 oC
Tf = 1.86 oC/m 3.62 mx
FP = -6.73 oC
x
13-57
Osmotic Pressure
Applies only to aqueous solutions!
Two solutions of different concentrations are separated by a semi-permeable membrane (allows water but notsolute to pass through)
13-58 Figure 13.17
The development of osmotic pressure
pure solvent
solution
applied pressure needed to prevent volume
increase; equal to
the osmotic pressure
osmotic pressure
semipermeablemembrane
13-59
T is the Kelvin temperature
Quantitative Treatment of Osmotic Pressure ()
OP is proportional to the number of solute particles in a givenvolume of solution (to M).
nsolute/Vsoln or M
The constant of proportionality = RT, so = M x R x T
13-60
Underlying Principle of Colligative Properties
Each property stems from an inability of solute particles tocross between two phases.
13-61
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).
13-62
SAMPLE PROBLEM 13.8 Determining molar mass from osmotic pressure
SOLUTION:
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?
PLAN: We know as well as R and T. Convert to atm and T to Kelvin. Use the equation to find the molarity M and then the amount and volume of the sample to calculate M.
M =
RT= 3.61 torr
atm
760 torr
(0.0821 L . atm/mol . K)(278.15 K)
= 2.08 x 10-4 M
2.08 x 10-4 molL 1.50 mL
103 mL
L= 3.12 x 10-7 mol
21.5 mgg
103 mg
1
3.12 x 10-7 mol= 6.89 x 104 g/mol
x
x x
xx
# mol = g/M
13-63
Fractional Distillation of VolatileNonelectrolytes
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)
13-64
Figure 13.18
The process of fractional distillation
Gas
Gasoline 38 oCKerosene 150 oC
Heating oil 260 oC
Lubricating oil 315 oC - 370 oC
Crude oil vapors from heater
Steam
Residue (asphalt, tar)
Condenser
Gasoline vapors
13-65
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, = i (MRT).
For ideal behavior, i = mol particles in solution / mol dissolved solute
But solutions are not ideal; for example, for BP elevation of NaClsolutions, i = 1.9, not 2!
Data suggest that the ions are not behaving as independent particles!
13-66
Figure 13.19
Non-ideal behavior of electrolyte solutions
Observed values of i are lessthan the predicted (expected)
values.
13-67
Figure 13.20
An ionic atmosphere model for non-ideal
behavior of electrolyte solutions
ionic atmospheres
Concept of effectiveconcentration
13-72
Figure 13.21
Light scattering and the Tyndall effect
Photo by C.A.Bailey, CalPoly SLO (Myanmar)
13-73 Figure 13.22
A Cottrell precipitator for removing particulates from industrial smokestack gases