1
Chapter Twelve
Physical Properties of Solutions
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Chapter Twelve
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Some Types of Solutions
Solution: Solute dispersed in a solvent.
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Solution Concentration
Most concentration
units are expressed as:
Amount of solute
Amount of solvent or solution
• Molarity: moles of solute/liter of solution
• Percent by mass: grams of solute/grams of solution (then
multiplied by 100%)
• Percent by volume: milliliters of solute/milliliters of
solution (then multiplied by 100%)
• Mass/volume percent: grams of solute/milliliters of
solution (then multiplied by 100%)
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Example 12.1
How would you prepare 750 g of an aqueous solution that
is 2.5% NaOH by mass?
Example 12.2
At 20 °C, pure ethanol has a density of 0.789 g/mL and
USP ethanol has a density of 0.813 g/mL. What is the mass
percent ethanol in USP ethanol?
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Solution Concentration (cont’d)
Most concentration
units are expressed as:
Amount of solute
Amount of solvent or solution
• Parts per million (ppm): grams of solute/grams of
solution (then multiplied by 106 or 1 million)
• Parts per billion (ppb): grams of solute/grams of solution
(then multiplied by 109 or 1 billion)
• Parts per trillion (ppt): grams of solute/grams of solution
(then multiplied by 1012 or 1 trillion)
• ppm, ppb, ppt ordinarily are used when expressing
extremely low concentrations (a liter of water that is 1 ppm
fluoride contains only 1 mg F–!)
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Example 12.3
The maximum allowable level of nitrates in drinking water
in the United States is 45 mg NO3–/L. What is this level
expressed in parts per million (ppm)?
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Solution Concentration (cont’d)
Most concentration
units are expressed as:
Amount of solute
Amount of solvent or solution
Molality (m): moles of solute/kilograms of solvent.
• Molarity varies with temperature (expansion or
contraction of solution).
• Molality is based on mass of solvent (not solution!) and is
independent of temperature.
• We will use molality in describing certain properties of
solutions.
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Example 12.4
What is the molality of a solution prepared by
dissolving 5.05 g naphthalene [C10H8(s)] in 75.0 mL
of benzene, C6H6 (d = 0.879 g/mL)?
Example 12.5
How many grams of benzoic acid, C6H5COOH, must
be dissolved in 50.0 mL of benzene, C6H6 (d = 0.879
g/mL), to produce 0.150 m C6H5COOH?
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Example 12.6
An aqueous solution of ethylene glycol
HOCH2CH2OH 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 in this solution?
Example 12.7 An Estimation Example
Without doing detailed calculations, determine which
aqueous solution has the greatest mole fraction of
CH3OH: (a) 1.0 m CH3OH, (b)10.0% CH3OH by
mass, or (c) xCH3OH = 0.10.
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Solution Concentration (cont’d)
Concentration
expressed as:
Amount of solute
Amount of solvent or solution
• Mole fraction (xi): moles of component i per moles of all
components (the solution).
• The sum of the mole fractions of all components of a
solution is ____.
• Mole percent: mole fraction times 100%.
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Chapter Twelve
11
Enthalpy of Solution
Solution formation can be considered to take place in three
steps:
1. Move the molecules of solvent apart to make room for
the solute molecules. DH1 > 0 (endothermic)
2. Separate the molecules of solute to the distances found
between them in the solution. DH2 > 0 (endothermic)
3. Allow the separated solute and solvent molecules to mix
randomly. DH3 < 0 (exothermic)
DHsoln = DH1 + DH2 + DH3
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Visualizing Enthalpy of Solution
For dissolving to
occur, the magnitudes
of DH1 + DH2 and of
DH3 must be roughly
comparable.
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Intermolecular Forces
in Solution Formation
• An ideal solution exists when all intermolecular forces are
of comparable strength, DHsoln = 0.
• When solute–solvent intermolecular forces are somewhat
stronger than other intermolecular forces, DHsoln < 0.
• When solute–solvent intermolecular forces are somewhat
weaker than other intermolecular forces, DHsoln > 0.
• When solute–solvent intermolecular forces are much
weaker than other intermolecular forces, the solute does
not dissolve in the solvent.
– Energy released by solute–solvent interactions is
insufficient to separate solute particles or solvent
particles.
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Intermolecular Forces in Solution
For a solute to dissolve,
the strength of solvent–
solvent forces …
… must be
comparable to solute–
solvent forces.
… and solute–
solute forces …
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Non-Ideal Solutions
… when
mixed, give
less than 100
mL of solution.
50 mL of
ethanol …
… and 50 mL
of water …
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In this solution, forces
between ethanol and
water are _____er
than other
intermolecular forces.
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Chapter Twelve
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Aqueous Solutions
of Ionic Compounds
• The forces causing an ionic solid to dissolve in water are
ion–dipole forces, the attraction of water dipoles for
cations and anions.
• The attractions of water dipoles for ions pulls the ions out
of the crystalline lattice and into aqueous solution.
• The extent to which an ionic solid dissolves in water is
determined largely by the competition between:
– interionic attractions that hold ions in a crystal, and
– ion–dipole attractions that pull ions into solution.
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Ion–Dipole Forces in Dissolution
Negative ends of dipoles
attracted to cations.
Positive ends of dipoles
attracted to anions.
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Example 12.8
Predict whether each combination is likely to be a
solution or a heterogeneous mixture:
(a) methanol, 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
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Some Solubility Terms
• Liquids that mix in all proportions are called miscible.
• When there is a dynamic equilibrium between an
undissolved solute and a solution, the solution is saturated.
• The concentration of the solute in a saturated solution is
the solubility of the solute.
• A solution which contains less solute than can be held at
equilibrium is unsaturated.
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Formation of a Saturated Solution
Solid begins
to dissolve.
Eventually, the rates of dissolving
and of crystallization are equal; no
more solute appears to dissolve.
As solid dissolves,
some dissolved solute
begins to crystallize.
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Longer standing does
not change the amount
of dissolved solute.
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Solubility as a Function
of Temperature
• Most ionic compounds have aqueous solubilities that
increase significantly with increasing temperature.
• A few have solubilities that change little with temperature.
• A very few have solubilities that decrease with increasing
temperature.
• If solubility increases with temperature, a hot, saturated
solution may be cooled (carefully!) without precipitation of
the excess solute. This creates a supersaturated solution.
• Supersaturated solutions ordinarily are unstable …
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A Supersaturated Solution
A single “seed
crystal” of
solute is added.
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Solute immediately
begins to crystallize …
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… until all of the
excess solute has
precipitated.
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Some
Solubility
Curves
What is the (approx.)
solubility of KNO3
per 100 g water at
90 °C? At 20 °C?
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Selective Crystallization
When KNO3(s) is
crystallized from an
aqueous solution of
KNO3 containing
CuSO4 as an impurity,
CuSO4 (blue) remains
in the solution.
KNO3 crystallized from
a hot, saturated solution
is virtually pure.
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The Solubilities of Gases
• Most gases become less soluble in liquids as the
temperature increases. (Why?)
• At a constant temperature, the solubility (S) of a gas is
directly proportional to the pressure of the gas (Pgas) in
equilibrium with the solution.
S = k Pgas
The value of k depends on the particular gas and the solvent.
• The effect of pressure on the solubility of a gas is known
as Henry’s law.
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Effect of Temperature on Solubility
of Gases
Thermal pollution: as
river/lake water is warmed
(when used by industry for
cooling), less oxygen dissolves,
and fish no longer thrive.
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Pressure and Solubility of Gases
Higher partial pressure
means more molecules of
gas per unit volume …
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… thus more
frequent collisions
of gas molecules
with the surface …
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… giving a higher
concentration of
dissolved gas.
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Example 12.9
A 225-g sample of pure water is shaken with air under a pressure of
0.95 atm at 20 °C. How many milligrams of Ar(g) will be present
in the water when solubility equilibrium is reached? Use data from
Figure 12.14 and the fact that the mole fraction of Ar in air is
0.00934.
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Colligative Properties of Solutions
• Colligative properties of a solution depend only on the
concentration of solute particles, and not on the nature of
the solute.
• Non-colligative properties include: color, odor, density,
viscosity, toxicity, reactivity, etc.
• We will examine four colligative properties of solutions:
– Vapor pressure (of the solvent)
– Freezing point depression
– Boiling point elevation
– Osmotic pressure
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Vapor Pressure of a Solution
• The vapor pressure of solvent above a solution is less than
the vapor pressure above the pure solvent.
• Raoult’s law: the vapor pressure of the solvent above a
solution (Psolv) is the product of the vapor pressure of the
pure solvent (P°solv) and the mole fraction of the solvent in
the solution (xsolv):
Psolv = xsolv ·P°solv
• The vapor in equilibrium with an ideal solution of two
volatile components has a higher mole fraction of the more
volatile component than is found in the liquid.
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Example 12.10
The vapor pressure of pure water at 20.0 °C is 17.5 mmHg.
What is the vapor pressure at 20.0 °C above a solution that
has 0.250 mol sucrose (C12H22O11) and 75.0 g urea
[CO(NH2)2] dissolved per kilogram of water?
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Example 12.11
At 25 °C, the vapor pressures of pure benzene (C6H6) and
pure toluene (C7H8) are 95.1 and 28.4 mmHg, respectively.
A solution is prepared that has equal mole fractions of
C7H8 and C6H6. Determine the vapor pressures of C7H8
and C6H6 and the total vapor pressure above this solution.
Consider the solution to be ideal.
Example 12.12
What is the composition, expressed as mole fractions, of
the vapor in equilibrium with the benzene–toluene solution
of Example 12.11?
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Fractional
Distillation
The vapor here …
… is richer in the more
volatile component than
the original liquid here …
… so the liquid that
condenses here will also
be richer in the more
volatile component.
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Example 12.13 A Conceptual Example
Figure 12.16 (below) shows two different aqueous
solutions placed in the same enclosure. After a time, the
solution level has risen in container A and dropped in
container B. Explain how and why this happens.
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Vapor Pressure Lowering by a
Nonvolatile Solute
… the vapor
pressure from the
pure solvent.
Raoult’s Law: the vapor pressure
from a solution (nonvolatile
solute) is lower than …
Result: the boiling
point of the solution
increases by DTb.
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Freezing Point Depression and
Boiling Point Elevation
DTf = –Kf × m
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DTb = Kb × m
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Example 12.14
What is the freezing point of an aqueous sucrose solution
that has 25.0 g C12H22O11 per 100.0 g H2O?
Example 12.15
Sorbitol is a sweet substance found in fruits and berries
and sometimes used as a sugar substitute. An aqueous
solution containing 1.00 g sorbitol in 100.0 g water is
found to have a freezing point of –0.102 °C. Elemental
analysis indicates that sorbitol consists of 39.56% C,
7.75% H, and 52.70% O by mass. What are the (a) molar
mass and (b) molecular formula of sorbitol?
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Osmotic Pressure
• A semipermeable membrane has microscopic pores,
through which small solvent molecules can pass but larger
solute molecules cannot.
• During osmosis, there is a net flow of solvent molecules
through a semipermeable membrane, from a region of
lower concentration to a region of higher concentration.
• The pressure required to stop osmosis is called the osmotic
pressure (p) of the solution.
p = (nRT/V) = (n/V)RT = M RT
This equation should
look familiar …
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Osmosis and Osmotic Pressure
The solution
increases in
volume until …
… the height of
solution exerts the
osmotic pressure (π)
of the solution.
Net flow of water
from the outside
(pure H2O) to the
solution.
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Example 12.16
An aqueous solution is prepared by dissolving 1.50 g
of hemocyanin, a protein obtained from crabs, in
0.250 L of water. The solution has an osmotic pressure
of 0.00342 atm at 277 K. (a) What is the molar mass
of hemocyanin? (b) What should the freezing point of
the solution be?
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Practical Applications of Osmosis
Ordinarily a patient must be given intravenous fluids that are
isotonic—have the same osmotic pressure as blood.
External solution is
hypertonic; produces
osmotic pressure > πint.
Net flow of water out
of the cell.
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Red blood cell in
isotonic solution
remains the same
size.
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External solution is
hypotonic; produces
osmotic pressure <
πint. Net flow of water
into the cell.
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42
Practical
Applications of
Osmosis (cont’d)
• Reverse osmosis (RO):
reversing the normal net flow
of solvent molecules through
a semipermeable membrane.
• Pressure that exceeds the
osmotic pressure is applied to
the solution.
• RO is used for water
purification.
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Pressure greater than π is
applied here …
… water flows from the more
concentrated solution, through
the membrane.
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Chapter Twelve
43
Solutions of Electrolytes
• Whereas electrolytes dissociate, the number of solute
particles ordinarily is greater than the number of formula
units dissolved. One mole of NaCl dissolved in water
produces more than one mole of solute particles.
• The van’t Hoff factor (i) is used to modify the colligativeproperty equations for electrolytes:
DTf = i × (–Kf) × m
DTb = i × Kb × m
p = i × M RT
• For nonelectrolyte solutes, i = 1.
• For electrolytes, we expect i to be equal to the number of
ions into which a substance dissociates into in solution.
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At very low
concentrations, the
“theoretical” values
of i are reached.
At higher concentrations, the
values of i are significantly lower
than the theoretical values; ion
pairs form in solution.
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Example 12.17
(a)
(b)
(c)
(d)
A Conceptual Example
Without doing detailed calculations, place the following
solutions in order of decreasing osmotic pressure:
0.01 M C12H22O11(aq) at 25 °C
0.01 M CH3CH2COOH(aq) at 37 °C
0.01 m KNO3(aq) at 25 °C
a solution of 1.00 g polystyrene (molar mass: 3.5 × 105
g/mol) in 100 mL of benzene at 25 °C
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Colloids
• In a solution, dispersed particles are molecules, atoms, or
ions (roughly 0.1 nm in size). Solute particles do not
“settle out” of solution.
• In a suspension (e.g., sand in water) the dispersed
particles are relatively large, and will settle from
suspension.
• In a colloid, the dispersed particles are on the order of
1–1000 nm in size.
• Although they are larger than molecules/atoms/ions,
colloidal particles are small enough to remain dispersed
indefinitely.
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Why are there no gasin-gas colloids?
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The Tyndall Effect
Light scattered by the
(larger) colloidal
particles of Fe2O3
makes the beam visible.
Fe3+
The dissolved
ions are not large
enough to scatter
light; the beam is
virtually invisible.
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A Suspension and a Colloid
Suspended SiO2 (sand)
settles very quickly.
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Each colloidal particle of
SiO2 (Ludox®) attains a
(–) charge, which repels
other colloidal particles.
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Formation and Coagulation
of a Colloid
When a strong electrolyte is
added to colloidal iron
oxide, the charge on the
surface of each particle is
partially neutralized …
… and the colloidal
particles coalesce
into a suspension
that quickly settles.
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Chapter Twelve
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Cumulative Example
A 375-mL sample of hexane vapor in equilibrium with
liquid hexane C6H14 (d = 0.6548 g/mL), at 25.0 °C is
dissolved in 50.0 mL of liquid cyclohexane, C6H12, at
25.0 °C (d = 0.7739 g/mL, vp = 97.58 Torr). Use
information found elsewhere in the text (such as
Example 11.3) to calculate the total vapor pressure
above the solution at 25.0 °C. How reliable is this
calculation?
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Chapter Twelve