1 chapter 13 physical properties of solutions insert picture from first page of chapter
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Chapter 13
Physical Properties of Solutions
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13.1 Types of Solutions
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13.1 Types of Solutions
Solution classifications based on amount of solute dissolved relative to the maximum:
• Saturated – maximum amount at a given temperature (This amount is termed the solubility of the solute.)
• Unsaturated – less than the maximum
• Supersaturated – more than a saturated solution but is an unstable condition
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unsaturated
supersaturated
saturated
heat
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Conversion of a Supersaturated Solution to a Saturated Solution
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13.2 A Molecular View of the Solution Process
Factors that determine solubility
• Intermolecular forces present in the formation of a solution– Solute-solute interactions– Solvent-solvent interactions– Solute-solvent interactions
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321soln HHHH
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321soln HHHH
0soln321 HHHH
0soln321 HHHH
endothermic
exothermic
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Solubility can be predicted based on “like dissolves like” in terms of intermolecular forces.For example: water and methanol are both polar
and dissolve in each other
• Two liquids that are soluble in each other in all proportions are term miscible.
• Ions readily dissolve in polar solvent due to solvation by the solvent molecules.
O
HH
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Predict whether Vitamin B6 is water soluble
or fat soluble.
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Water soluble due to the presence of polar
groups.
polar groups
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• Energy and entropy – Exothermic processes are generally more
favorable than endothermic processes– Solutions do form when the overall process is
endothermic– Entropy (randomness or disorder) contributes
to the solution process
• Entropy tends to increase for all process
• A solution is more disordered than the isolated solute and solvent
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13.3 Concentration Units
• Molality (m) – number of moles of solute dissolved in 1 kg (1000 g) of solvent
• Percent by Mass – ratio of mass of solute to the mass of the solution times 100
kg)(insolutionofmass
soluteofmolesmolality m
x100solventofmasssoluteofmass
soluteofmassmassby percent
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Units analogous to percent by mass (part per hundred) to express very small concentrations
• Parts per million (ppm)
• Parts per billion (ppb)
*masses must be expressed in the same units
6x10solutionof*mass
soluteof*massppm
9x10solutionof*mass
soluteof*massppb
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Choice of units depends on the purpose of the experiment
• Mole fraction – used for gases and vapor pressures of solutions
• Molarity – commonly used since volumes of solutions are easy to measure
• Molality – temperature independent• Percent by Mass – temperature independent
and need not know molar masses• Conversion between units requires the use of
density if any mass to volume or volume to mass conversion in needed.
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Determine the a) molality, b) percent by
mass and c) the ppm concentrations of a
solution prepared by dissolving 15.56 g of
glucose (C6H12O6) in 255 g of water.
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a) molality
Molar mass of glucose = 180.2 g/mol
mol0.086349g180.2
molxg15.56glucosemol
mm 0.339kg
g10x
g 255
1 x mol0.086349
3
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b) percent by mass
c) ppm
glucose5.751% x100g 270.56
g 15.56
glucose5.751x10 x10g 270.56
g 15.56 46 ppm
(original slide typo!)
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13.4 Factors that Affect Solubility
• Temperature– The solubility of solids may increase,
decrease or remain relatively constant with increasing temperature
– The solubility of gases decreases with increasing temperature• Thermal pollution – a consequence of
the relation between gas solubility and temperature
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Temperature Dependence of the Solubility of Selected Solids
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• Pressure – significantly affects only the solubility of gases– Henry’s law – the solubility of a gas, c, is
directly proportional to the pressure of the gas, P, over the solution
where c, is in mol/L, k is Henry’s law
constant with units of mol/L. atm and P is in atm.
Pc kPc
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Molecular View at Different Pressures
P1 P2
P1< P2
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13.5 Colligative Properties
Colligative properties depend only on the number of solute particles in solution and not the nature of the solute• Vapor-Pressure lowering
– Nonvolatile solute (no appreciable pressure)• Vapor pressure of the solvent is decreased
– Volatile solute (exhibit appreciable pressure)• Vapor pressure is the sum of the individual
pressures
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– Raoult’s law – quantitative expression of the solution vapor pressure
• Nonvolatile solute (P1 is the solution vapor pressure, P0 is the vapor pressure of the pure substance, and is the mole fraction.)
• Volatile solute (PT is the solution vapor pressure, and are vapor pressures of pure solution components)
0111 PP
0BB
0AAT PPP
0AP 0
BP Volatile solute: not exam material
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Calculate the vapor pressure of a solutionmade by dissolving 115 g of urea, a nonvolatilesolute, [(NH2)2CO; molar mass = 60.06 g/mol]in 485 g of water at 25°C. (At 25°C, .)mmHg23.8OH2
P
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0OHOHOH 222
PP
0.9336mol1.915mol26.91
mol26.91OH2
mol26.91g18.02
molgx485mol OH2
mol1.915g60.06
mol xg115molurea
mmHg22.2mmHg23.8 x 0.9392OH2P
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• Boiling-Point Elevation (Tb)
– The boiling point of a solution, Tb, of a nonvolatile solute will be higher than that of the pure solvent, .
– Elevation is directly proportional to the molal concentration.
– Kb is the molal boiling-point elevation constant.
0bT
mKT bb
0bbb TTT
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Effect of Vapor Pressure Lowering
Effect on Boiling Point
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• Freezing-Point Depression (Tf)
– The freezing point of a solution, Tf, will be lower than that of the pure solvent, .
– Depression is directly proportional to the molal concentration.
– Kf is the molal freezing-point elevation constant.
0fT
mKT ff
f0
ff TTT
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Effect of Vapor Pressure Lowering
Effect on Freezing Point
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Calculate a) the freezing point and b) the
boiling point of a solution containing 268 g of
ethylene glycol and 1015 g of water. (The
molar mass of ethylene glycol (C2H6O2) is
62.07 g/mol. Kb and Kf for water are 0.512°C/m
and 1.86°C/m, respectively.)
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a) freezing point
C7.914.254C1.86 o
o
f mxm
T
mol4.318g62.07
molxg268glycolethylenemol
mm 4.254kg
g10x
g 1015
1 x mol3184
3
.
foo C0.00C7.91 T
C7.91of T
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b) boiling point
C100.00C2.18 ob
o T
C2.184.254C0.512 o
o
b mxm
T
C102.18ob T
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Osmosis - Selective passage of solvent molecules through a semipermeable membrane
solvent solution solvent solution
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• Osmotic Pressure () – Pressure required to stop osmosis
– Directly proportional to molar concentration, M
= MRT
where R is 0.08206 L.atm/mol . K and T is in
kelvins
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• Solutions of Electrolytes
– Dissociation of strong and weak electrolytes affects the number of particles in a solution
– van’t Hoft factor (i) – accounts for the effect of dissociation
solutionindissolvedinitiallyunitsformulaofnumber
ondissociatiaftersolutioninparticlesofnumberactuali
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Modified Equations for Colligative Properties
miKT ff
miKT bb
iRTM
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Formation of ion pairs affects colligative properties
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The freezing-point depression of a 0.100 m
MgSO4 solution is 0.225°C. Determine the
experimental van’t Hoff factor of MgSO4 at
this concentration.
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miKT ff
One approach:
mxm
i 0.100C1.86
C0.225o
o
1.21i
Note, at this concentration the dissociation of MgSO4 is not complete.
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Ideal freezing point depression
C0.1860.100x C1.86 o
o
f mm
T
Compare ideal and real freezing point depression
211C0.186
C0.225o
o
.i
Another approach to the same problem:
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A solution made by dissolving 25.0 mg ofinsulin in 5.00 mL of water has an osmoticpressure of 15.5 mmHg at 25°C. Calculatethe molar mass of insulin. (Assume that there is no change in volume when theinsulin is added to the water and that insulinis a nondissociating solute.)
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K298
1
atmL0.08206
Kmolxatm2.039x10 2 x
RTM
Calculate the M of the solution
atm2.039x10mmHg760
atmxmmHg15.5 2
L
mol10x8.33810x8.338
44
MM
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Calculate the moles of insulin
mol x104.169mL
L10xmL5.00x
L
mol10x8.340mol 6
34
Molar mass is ratio of grams to moles
mol x104.169
1x
mg
g10xmg25.0
6
3
M
mol
g10x6.00 3
M
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13.7 ColloidsA colloid is a dispersion of particles of onesubstance throughout another substance.• Intermediate between a homogenous and heterogeneous
mixture• Range of particle size: 103 – 106 pm• Categories
– Aerosols – Foams– Emulsions– Sols– Gels
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Aerogel
• Lowest density solids known
• Good thermal, electric, sound insulators
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A 2.5-kg brick supported by a 2-g piece of aerogel
(source: Wikipedia)
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• Exhibit the Tyndall effect
colloid solution
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Fog – A Familiar Manifestation of the Tyndall Effect
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• Many important colloids are aqueous and can be further classified.
– Hydrophilic (water loving)
– Hydrophobic (water fearing)
Hydrophilic groups on the surface of a protein stabilize the molecule in water
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Stabilization of Hydrophobic Colloidal Particles by Ion Adsorption
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Removal of Grease by Soap (Sodium Stearate)
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Emulsification – • stabilization of an unstable colloid• accomplished by the addition of an emulsifier or emulsifying agent
Sodium glycoholate – a bile salt or biological emulsifying agent for ingested fats
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Lecithin is a mixture of substances found in natural sources such as egg yolk. It contains molecules such asphosphatidylcholine (shown right) that have both hydrophobic (nonpolar) and hydrophilic (polar; charged) regions. Lecithin acts as an emulsifier.
This is why egg yolk is used to make mayonnaise. The egg yolk is typically whisked with a small amount of vinegar or lemon juice (the aqueous phase). Vegetable oil is then added gradually while whisking. Mayonnaise is the resulting emulsion of vegetable oil and aqueous acid.