Unit 9 – Liquids and Solutions
Properties of Liquids
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What Determines the Physical Properties of
Liquids
Physical Properties of Gases
•  According to the kinetic molecular theory, why are gases able to
spread far apart and take the shapes of their containers?
Liquids
•  How are liquids different from gases? Explain these differences by
applying what you know about bonding.
•  “The interplay between the disruptive motion of the particles in the
liquid phase and the attractions among the particles determines the
physical properties of the liquid.”
–  Liquids are more dense than gases
–  Increasing pressure hardly affects the volume of a liquid
(a.k.a “Liquids and solids are condensed states of matter”)
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Evaporation
How is evaporation different from boiling?
•  The conversion of a liquid to a gas or vapor is
called vaporization.
• Evaporation -
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1
Distribution of Thermal Energy
•  Only a small fraction of the molecules in a liquid have enough energy to
escape.
•  As the temperature increases, the fraction of the molecules with “escape
energy” increases.
•  The higher the temperature, __________________________________
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Evaporation in a Closed Container
•  In a closed container
molecules that escape the
liquid phase cannot escape
the container.
•  The molecules collect as a
vapor above the liquid.
Some molecules condense
back into the liquid phase.
•  The vapor particles exert a
pressure on top of the
surface of the liquid called
the vapor pressure.
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Vapor equilibrium in a closed container
•  What is a dynamic equilibrium?
•  At equilibrium, no net change occurs in the amount
of liquid or vapor, but system is dynamic on the
molecular level. Gas phase is saturated with vapor.
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2
Vapor Pressure
•  Vapor pressure
•  How do you think temperature affects vapor pressure
(VP)?
•  For a given liquid,
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Vapor Pressure
How do you think Intermolecular Forces affect VP?
•  Comparing two different liquids at the same
temperature, the stronger the intermolecular forces,
the harder it is to vaporize the liquid, so the lower
the VP.
–  Liquids with high VP are called volatile
–  Order these molecules from strongest to weakest
IMF.
H─O─H,
C2H5 ─O─C2H5,
CH3CH2─OH,
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Equilibrium Vapor Pressure
•  The curves show all conditions of P and T
where liquid and vapor are in equilibrium.
•  What conditions must exist for the liquid to
boil?
•  When VP =
•  If external P = 760 mm Hg, T of boiling is the
Equilibrium Vapor Pressure Curve Animation
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3
Why is the boiling point of water 100ºC?
Is it always the same?
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Boiling Point of Water = 100°C
•  When water boils, bubbles form in the interior of the liquid, rise
to the surface, and pop.
•  An air bubble acts as a birthplace for a bubble, and high-energy
water molecules enter.
•  The bubble expands only when the pressure inside the bubble
can oppose the atmospheric pressure pushing down on the
surface of the water.
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Affects of External Pressure
•  At a lower external pressure,
•  At a higher external pressure,
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4
The Heating/Cooling Curve of Water
•  If you’re continuously adding heat, why are there plateaus? Where
did the energy go (why doesn’t the temperature increase)?
•  Why are the flat portions of different length?
•  Why are the slopes different?
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Energy and Changes of State, Heating and
Melting Ice
• 
As energy flows into the ice, random vibrations increase as
the temp rises
• 
At 0°C, the molecules become so energetic that they break
loose from their lattice positions, and the ice changes to liquid
• 
At the MP, all the added energy is used to disrupt the ice
structure by
• 
• 
At constant pressure, heat transferred = ΔH = “enthalpy”
ΔHfusion =
ΔHvap
=
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Energy and Changes of State: Boiling water
•  Water has a different molar heat capacity than ice, so
the slopes (temperature increase vs. time) are
different
•  At 100°C liquid water reaches its boiling point
–  Temperature remains constant as added energy is used to
vaporize the liquid
–  The plateau at the boiling point is longer than the plateau at
the melting point because it takes more energy to vaporize
liquid water than to melt ice
–  At the water/vapor plateau, added energy goes into P.E.
•  When water is totally converted to steam, the
temperature rises again
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5
Example # 1 - How much energy is required to
convert 10.0 g H2O(s) at -10.0ºC to steam at
150.ºC?
Cp(ice) = 2.06 J/g ⋅ ºC
ΔHfusion = 334 J/g
Cp(water) = 4.18 J/g⋅ºC
ΔHvap = 2260 J/g
Cp(steam) = 2.02 J/g⋅ºC
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Phase Diagrams
Important Aspects:
1. Each colored area represents all the conditions of
temperature and pressure where that phase exists.
2. Lines represent conditions of T and P where two
phases are in equilibrium.
3. The Triple Point is the point
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Phase Diagram of Water
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Water Phase Diagram
Look closely…
•  The solid-liquid equilibrium line of most substances
has a positive slope.
•  What does an increase in pressure do to the volume
of a substance?
•  Most substances are more dense in the solid phase,
meaning that the volume is smaller for the same
mass of substance.
•  Water’s fusion curve has a negative slope. Why?
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CO2 Phase Diagram
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Critical point, critical T and P
As P and T increase, you finally reach the critical T and P
Pcritical
Note that line
goes straight up
.
High Pressure
• critical
point
LIQUID
Tcritical
GAS
High Temperature
Above critical T no liquid
exists no matter how high
the pressure. The
temperature is so high
that KE overcomes IMFs.
(There’s nothing “critical”
about the critical P—it’s
just the pressure
associated with critical
point)
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Unit 9 – Liquids and Solutions
Solution Formation and
Concentration
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Solubility Factors
Solubility – •  Factors that Affect Solubility
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Water – Ethanol Interactions
Dissolving a solute in a solvent is favored when the non-covalent interactions
between the solute and solvent molecules are stronger than the noncovalent solvent-solvent and solute-solute interactions.
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8
Oil is Insoluble in Water
For a substance to dissolve, the water-water hydrogen bonds must be
broken to make a “hole” for each solute particle. However, the waterwater interactions will break only if they are replaced by similar strong
interactions with the solute.
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Classification of Solutes in Aqueous Solution
Strong Electrolytes
Weak Electrolytes
Nonelectrolytes
Ionic Compounds
Weak acids: (e.g. HF,
HC2H3O2, H3PO4,
H2CO3)
(NH2)2CO (urea)
Strong acids:
(HCl,HBr,, HI, HNO3,
H2SO4, HClO4)
Strong bases: (Alkaline
and Alkaline Earth
bases(that are soluble).
Weak Bases: NH3
CH3OH (methanol)
H 2O
C6H12O6 (glucose)
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9
Decomposition versus dissociation
2 NaCl(l)
electricity
2 Na(l) + Cl2(g)
NaCl(s)
water
Na+(aq) + Cl– (aq)
CaI2 (s)
water
Ca2+ (aq) + 2 I– (aq)
Note that electrolysis of NaCl only works with molten NaCl.
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Net Ionic Equations
•  A complete ionic equation
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•  An ion that appears on both sides of an equation and is not
directly involved in the reaction is called a spectator ion.
•  The net ionic equation is
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Net Ionic Equations
1.  Write the balanced molecular equation.
2.  Write the ionic equation showing the strong
electrolytes completely dissociated into cations and
anions.
3.  Cancel the spectator ions on both sides of the ionic
equation
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Writing Net Ionic Equations
Write the net ionic equation for the reaction of Zn
with HCl(aq)
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Writing Net Ionic Equations
Write the net ionic equation for the reaction of aqueous
acetic acid with aqueous sodium hydroxide.
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Solubility Factors
Temperature of the Solvent
What do you think? How does temperature affect the
quantity of a solute that dissolves in a given solvent?
Talk with your group and write down your thoughts.
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How does temperature affect solubility?
•  The solubility of
•  The solvent and solute particles have more
KE. They move faster and collide with one
another more often. “Holes” are opened in
the solvent more easily. IMF between solute
particles are broken more easily.
•  This trend holds for water as a solvent
because there’s already a “hole” for solutes
(open structure of water).
Solubility in organic solvents does not
follow this trend.
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Solubility of a gas in water typically
Gas Solubility in Water
Solubility, mmol/L
CH4
O2
2.0
CO
1.0
N2
He
0.0
0
10
20
30
Temperature, °C
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12
Gas Solubility and Pressure:
Carbonated beverages are packed under
pressure in a chamber filled with
CO2(g), much of which dissolves in the
drink at this pressure.
When the bottle is opened, the partial
pressure of CO2 above the solution
drops to atmospheric pressure, which
causes the concentration of CO2 in
solution to drop, because the gas
bubbles out of the solution. Why?
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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).
Sg is the
Pg is the
k is a
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Example 1 – Henry’s Law
Calculate the concentration of CO2 in a soft drink that is
bottled with a partial pressure of CO2 of 4.0 atm over
the liquid at 25˚C. The Henry’s law constant for CO2
in water at this temperature is 3.1 x 10-2 mol/L-atm.
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Factors Affecting RATE of
Dissolution
How would agitation of the solvent affect the rate of
dissolution? Why?
Would a sugar cube dissolve more or less quickly
than a spoonful of sugar? Why?
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A saturated solution contains the
An unsaturated solution contains
A supersaturated solution contains
Sodium acetate crystals rapidly form when a seed crystal is
added to a supersaturated solution of sodium acetate.
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Solubility and Ionic Solutions
unsaturated
-
exactly saturated
K
ClCl -
+
K
+
K+
Cl -
Cl
Cl+
- K
l
C
K+
K+
K+
Cl -
K+
K+
Cl
Cl -
K+
KCl KCl
KCl KCl KCl
saturated solution
in equilibrium with
undissolved solute
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A large crystal of sugar is added to a sugar
solution that is unsaturated. Which
picture best represents the likely situation
after 15 minutes has passed?
a.
b.
c. (same size as
original crystal)
d.
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A large crystal of sugar is added to a sugar
solution that is saturated. Which picture
best represents the likely situation after 15
minutes has passed?
a.
b.
c. (same size as
original crystal)
d.
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Concentration Units
The concentration of a solution is the amount of solute present in a given
quantity of solvent or solution.
Percent by Mass
% by mass
= Since masses are additive, the mass of the solution is the sum of the masses of solute
and solvent
Mole Fraction (X)
XA = 45
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Mass Percent Example •  Parts of solute in every 100 parts solution
–  If a solution is 0.9% by mass, then there are 0.9
grams of solute in every 100 grams of solution
•  or 10 kg solute in every 100 kg solution
Example: Calculate the mass percent of a solution
containing 27.5 g of ethanol and 175 mL of H2O (assume
the density of H2O is 1.00 g/mL)
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Concentration Units Continued
Molarity (M)
moles of solute
M =
liters of solution
Molality (m)
m =
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Molality Calculation Example
Calculate the molality of a solution containing
17.2 g of ethylene glycol (C2H6O2) dissolved in
0.500 kg of water.
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Unit 9 – Liquids and Solutions
Colligative Properties
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Colligative Properties of Solutions
Colligative properties are properties that depend only on the
1. Vapor-Pressure Lowering
P1 = X1 × P 10
2. Boiling-Point Elevation
∆Tb = Kb × m × i
3. Freezing-Point Depression
∆ Tf = Kf × m × i
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1. Effect of nonvolatile solute on vapor pressure
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Raoult’s Law
Psolution= xsolvent P
o
solvent
Psolution =
Xsolvent =
P osolvent =
•  Nonvolatile solute particles •  Must break solute-solvent attractive forces to evaporate
solvent
•  Vapor pressure decrease is characteristic of the solvent
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Raoult’s Law - Example 1
Psolution = Xsolvent P
o
solvent
Assume the solution containing 1 mol of glucose in 250. g
of water is ideal. What is the vapor pressure of water over
the solution at 30oC in atm? (The VP of pure H2O is 31.8
mm Hg)
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Why does a nonvolatile solute reduce vapor pressure?
A pure solvent tends to
vaporize since the vapor
is more random (has
more possible
microstates).
When a nonvolatile solute is
present, the liquid phase has a
greater
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Calculating Boiling Point Elevation
ΔTbp = m Kbp i
•  Change in bp = •  Kbp depends on solvent
–  How many dissociated particles do each of the following
ionic compounds produce?
•  (NH4)2SO4
•  K2CO3
•  CaCl2
•  NaCl
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Example 2: Change in Boiling Point
Dissolve 1.00 mol glucose in 250. g of water. What is the BP
of the solution? (Kbp = 0.512 oC/molal for water)
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Kf and Kb depend on the solvent, not solute
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Freezing Point Depression
Ice in equilibrium with water.
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Example 3: Freezing Pt Depression
How much NaCl must be dissolved in 4.00 x103 g of water
to lower FP to -10.00 oC?
Calculate required molality
∆Tfp = Kfp • m • i
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Example 4. Using Freezing Pt Depression to
Calculate Molar Mass
•  8.02 g of a non-electrolyte solute in 861 g of
H2O lowers the f.p. to -0.430°C. Calculate
the molar mass of the solute.
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