George Mason University
General Chemistry 212
Chapter 13
Properties of Mixtures: Solutions & Colloids
Acknowledgements
Course Text: Chemistry: the Molecular Nature of Matter and
Change, 7th edition, 2011, McGraw-Hill
Martin S. Silberberg & Patricia Amateis
The Chemistry 211/212 General Chemistry courses taught at George
Mason are intended for those students enrolled in a science
/engineering oriented curricula, with particular emphasis on chemistry,
biochemistry, and biology The material on these slides is taken
primarily from the course text but the instructor has modified,
condensed, or otherwise reorganized selected material.
Additional material from other sources may also be included.
Interpretation of course material to clarify concepts and solutions to
problems is the sole responsibility of this instructor.
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1
Chap 13: Properties of Mixtures

Intermolecular Forces & Solubility
 (Intramolecular) (Bonding)
● Ionic
● Covalent
● Metallic
 Intermolecular
● Ion-Dipole (strongest)
● Hydrogen bonding
● Dipole-Dipole
● Ion-Induced Dipole
● Dipole-Induced Dipole
● Dispersion (London) (weakest
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2
Chap 13: Properties of Mixtures


Types of Solutions

Liquid Solutions

Gas Solutions

Solid Solutions
The Solution Process

Solvation /Hydration

Solute

Solvent

Mixing
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3
Chap 13: Properties of Mixtures
 Thermodynamic Solution Cycle, an application of “Hess’s” Law
The Enthalpy Change Of An Overall Process Is The
Sum Of The Enthalpy Changes Of The Individual Steps

Heat of Solution
ΔHsolution = ΔHsolute + ΔHsolvent + ΔHmix

Heat of Hydration
ΔHhydration =
ΔHsolution = ΔHsolute

ΔHsolvent + ΔHmix
+ ΔHhydration
Lattice Energy (Hsolute)
ΔHsolution = ΔH lattice + ΔHhydration of the ions

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Solution Process & Change in Entropy
ΔSgas > ΔSliquid > ΔSsolid
> 0
4
Chap 13: Properties of Mixtures
Solution as an Equilibrium Process
 Effect of Temperature
 Effect of Pressure
 Concentration
 Molarity & Molality
 Parts of Solute by Parts of Solution
 Interconversion of Concentration Terms
 Colligative Properties of Solutions
 Vapor Pressure
 Boiling Point Elevation
 Freezing Point Depression
 Osmotic Pressure
 Structure and Properties of Colloids

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Intermolecular Forces
Phases and Phase Changes (Chapter 12) are due to forces
among the molecules
 Bonding (Intramolecular) forces and Intermolecuar forces
arise from electrostatic attractions between opposite
charges
 Intramolecular Forces - Attractive forces within
molecules:
 Cations & Anions (ionic bonding)
 Electron pairs (covalent bonding)
 Metal Cations and Delocalized Valence Electrons
(metallic bonding)
 Intermolecular Forces - Nonbonding Attractive and
Repulsive forces involving partial charges among particles:
 Molecules
 Atoms
 ions
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
6
Intermolecular Forces


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Intramolecular (Bonding within molecules)
 Relatively strong, larger charges close to each other
 Ionic
 Covalent
 Metallic
Intermolecular (Nonbonding Attractions between
molecules)
 Relatively weak attractions involving smaller charges
that are further apart
 Types (in decreasing order of bond energy)
 Ion-Dipole
(strongest)
 Hydrogen bonding
 Dipole-Dipole
 Ion-Induced Dipole
 Dipole-Induced Dipole
 Dispersion (London, instantaneous dipoles)
(weakest)
7
Intramolecular Forces
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Intermolecular Forces
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Intermolecular Forces & Solubility

Solutions – Homogeneous mixtures consisting of a
solute dissolved in a solvent through the actions
of intermolecular forces

Solute dissolves in Solvent

Distinction between Solute & Solvent not always
clear

Solubility – Maximum amount of solute that
dissolves in a fixed quantity of solvent at a
specified temperature

“Dilute” & “Concentrated” are qualitative (relative)
terms
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Intermolecular Forces & Solubility

Substances with similar types of intermolecular
forces (IMFs) dissolve in each other
“Like Dissolves Like”

Nonpolar substances (e.g., hydrocarbons) are
dominated by dispersion force IMFs and are
soluble in other nonpolar substances

Polar substances have Hydrogen Bonding, ionic,
and dipole driven IMFs (e.g., Water, Alcohols) and
would be soluble in polar substances (solvents)
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Intermolecular Forces & Solubility

Intermolecular Forces

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Ion-Dipole forces

Principal forces involved in solubility of ionic
compounds in water

Each ion on crystal surface attracts oppositely
charged end of Water dipole

These attractive forces overcome attractive forces
between crystal ions leading to breakdown of
Water
crystal structure
12
Intermolecular Forces & Solubility

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:
:
Intermolecular Forces
 Hydrogen Bonding – Dipole-Dipole force that arises between
molecules that have an H atom bonded to a small highly
electronegative atom with lone electrons pairs (N, O, F)
:

H O
H N
H F
Especially important in aqueous (water) solutions.
 Oxygen- and nitrogen-containing organic and biological
compounds, such as alcohols, sugars, amines, amino
acids.
 O & N are small and electronegative, so their bound H
atoms are partially positive and can get close to negative
O in the dipole water (H O) molecule
13
Intermolecular Forces & Solubility

Intermolecular Forces

Dipole-Dipole forces

In the absence of H bonding, Dipole-Dipole forces
account for the solubility of polar organic molecules,
such as Aldehydes (R-CHO) in polar solvents such
as Chloroform (CHCl3)
-
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+
14
Intermolecular Forces & Solubility

Intermolecular Forces
 Ion-induced Dipole forces
 An ion-induced dipole attraction is a weak attraction that
results when the approach of an ion induces a dipole in
an atom or nonpolar molecule by disturbing the
arrangement of electrons in the nonpolar species
 The Ion increases the magnitude of any nearby dipole;
thus contributes to the solubility of salts in less polar
solvents
Ex. LiCl in Ethanol
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Intermolecular Forces & Solubility

Intermolecular Forces
 Dipole-Induced Dipole forces
 Intermolecular attraction between a polar molecule
and the oppositely charged pole it induces in a nearby
non-polar molecule
 Also based on polarizability, but are weaker than ioninduced dipole forces because the magnitude of
charge is smaller – ions vs dipoles (coulombs law)

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Solubility of atmospheric gases (O2, N2, noble gases)
have limited solubility in water because of these forces
16
Intermolecular Forces & Solubility

Intermolecular Forces
 Dispersion (or London or Instantaneous Dipole) forces
 These very weak forces are due to the non-zero
instantaneous dipole moments of all atoms and
molecules
 Such polarization can be induced either by a polar
molecule or by the repulsion of negatively charged
electron clouds in non-polar molecules
 Thus, London interactions are caused by random
fluctuations in electron density in an electron cloud
 Contribute to the solubility of all solutes in all solvents
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Liquid Solutions
Solutions can be:
Liquid
Gaseous
Solid
 Physical state of Solvent determines physical state of
solution
 Liquid solutions – forces created between solute and
solvent comparable in strength (“like” dissolves “like”)
 Liquid – Liquid
 Alcohol/water – H bonds between OH & H2O



Salts/Water
– Ionic forces
Gas – Liquid

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– Dispersion forces
Solid – Liquid


Oil/Hexane
(O2/H2O)
– Weak intermolecular forces (low
solubility, but biologically important) 18
Liquid Solutions

Gas Solutions
 Gas – Gas Solutions – All gases are infinitely
soluble in one another
 Gas – Solid Solutions – Gases dissolve into solids
by occupying the spaces between the closely
packed particles
 Utility – Hydrogen (small size) can be purified by
passing an impure sample through a solid metal
like palladium (forms Pd-H covalent bonds)
 Disadvantage – Conductivity of Copper is
reduced by the presence of Oxygen in crystal
structure (copper metal is transformed to
Cu(I)2O
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Liquid Solutions

Solid Solutions
 Solid – Solid Solutions
 Alloys – A mixture with metallic properties that
consists of solid phases of two or more pure
elements, solid-solid solution, or distinct
intermediate (heterogeneous) phases
 Alloys Types
 Substitutional alloys – Brass (copper & zinc),
Sterling Silver(silver & copper), etc. substitute
for some of main element atoms
 Interstitial alloys – atoms of another elements
(usually a nonmetal) fill interstitial spaces
between atoms of main element, e.g., carbon
steel (C & Fe)
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Practice Problem
A solution is ____________
a. A heterogeneous mixture of 2 or more substances
b. A homogeneous mixture of 2 or more substances
c. Any two liquids mixed together
d. very unstable
e. the answer to a complex problem
Ans: b
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The Solution Process

Elements of Solution Process
 Solute particles must separate from each other
 Some solvent particles must separate to make room
for solute particles
 Solute and Solvent particles must mix together
 Energy must be absorbed to separate particles
 Energy is released when particles mix and attract each
other
 Thus, the solution process is accompanied by changes
in Enthalpy (∆H), representing the heat energy tied
up in chemical bonds and Entropy (S), the number of
ways the energy of a system can be dispersed through
motions of particles)
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2. Entropy (S)
22
Solution Process - Solvation

Solvation – Process of surrounding a
solute particle with solvent particles

Hydrated ions in solution are
surrounded by solvent molecules in
defined geometric shapes

Orientation of combined
solute/solvent is different for solvated
cations and anions

Cations attract the negative charge
of the solvent

Anions attract the positive charge of
the solvent

Hydration # for Na+ = 6

Hydration # for Cl- = 5
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Solution Process
Solute v. Solvent
Coulombic
- attraction of oppositely charged particles
van der Waals forces:
a. dispersion (London)
b. dipole-dipole
c. ion-dipole
Hydrogen-bonding - attractive interaction of a
Hydrogen atom and a strongly
electronegative Element:
Hydration Energy
attraction between solvent
and solute
Lattice Energy
solute-solute forces
Oxygen, Nitrogen or Fluorine
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Practice Problem
Predict which solvent will dissolve more of the given solute:
a. Sodium Chloride (solute) in Methanol or in 1-Propanol
Ans: Methanol
A solute tends to be more soluble in a solvent
whose intermolecular forces are similar to those of
the solute
NaCl is an ionic solid that dissolves through
ion-dipole forces
Both Methanol & 1-Propanol contain a “Polar” -OH
group
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The Hydrocarbon group in 1-Propanol is longer
and can form only weak dispersive forces with the
ions; thus it is less effective at substituting for the
ionic attractions in the solute
25
Practice Problem (con’t)
Predict which solvent will dissolve more of the given solute:
b. Ethylene Glycol (HOCH2CH20H) in
Hexane (CH3(CH2)4CH3) or in Water (H2O)
Ans: Very Soluble in Water
Ethylene Glycol molecules have two polar –OH groups
and they interact with the Dipole water molecule
through H-Bonding
The Water H-bonds can substitute for the Ethylene
Glycol H-Bonds better than they can with the weaker
dispersive forces in Hexane (no H-Bonds)
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Practice Problem (con’t)
Predict which solvent will dissolve more of the given solute:
c.
Diethyl Ether (CH3CH2OCH2CH3) in
Water (H2O) or in Ethanol (CH3CH2OH)
Ans: Ethanol
Diethyl Ether molecules interact with each other
through dipole-dipole and dispersive forces and can
form H-Bonds to both H2O and Ethanol
The Ether is more soluble in Ethanol because Ethanol
can form H-Bonds with the Ether Oxygen and the
dispersion forces of the Hydrocarbon group (CH3CH2)
are compatible with the dispersion forces of the
Ether’s Hydrocarbon group
Water, on the other hand, can form H bonds with the
Ether, but it lacks any Hydrocarbon portion, so it
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The Solution Process

Solute particles absorb heat (Endothermic) and
separate from each other by overcoming
intermolecular attractions
Solute (aggregated) + Heat  Solute (separated) ΔHsolute > 0

Solvent particles absorb heat (Endothermic) and
separate from each other by overcoming
intermolecular attractions
Solvent (aggregated) + Heat  Solvent (separated)

ΔHsolvent > 0
Solute and Solvent particles mix (form a solution) by
attraction and release of energy - an Exothermic
process
Solute (separated) + Solvent (separated)  Solution + Heat ΔHmix < 0
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The Solution Process

Total Enthalpy change is called the:
“Heat of Solution” (∆Hsoln)
This process, called a “Thermodynamic Solution
Cycle”, is an application of “Hess’s” Law
Enthalpy Change is the Sum of the
Enthalpy Changes of Its Individual Steps
ΔHsoln = ΔHsolute
Endothermic
H > 0
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+ ΔHsolvent
Endothermic
H > 0
+ ΔHmix
Exothermic
H < 0
29
The Solution Process

Solution Cycles, Enthalpy, and the Heat of Solution
In an Exothermic Solution Process the
Solute is Soluble
If (Hsolute + Hsolvent) < Hmix (Hsoln < 0)
(+)
(+)
(–)
Exothermic
In an Endothermic Solution Process the
Solute is Insoluble
If (Hsolute + Hsolvent) > Hmix
(+)
(+)
(Hsoln > 0)
(–)
Endothermic
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The Solution Process

In an Exothermic Solution Process, the Solute is Soluble
ΔH < 0
Note: Recall Born-Haber Process – Chap 6
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The Solution Process

In an Endothermic Solution Process, the solute is Insoluble
ΔH > 0
Note: Recall Born-Haber Process – Chap 6
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32
Heats of Hydration

Hsolvent & Hmix are difficult to measure individually

Combined, these terms represent the Enthalpy change during
“Solvation” – The process of surrounding a solute particle with
solvent particles

Solvation in “Water” is called “Hydration”

Thus:
ΔHhydration = ΔHsolvent + ΔHmix
Since:
ΔHsolution = ΔHsolute
+ ΔHsolvent + ΔHmix
Then:
ΔHsolution = ΔHsolute
+ ΔHhydration
Forming strong ion-dipole forces during Hydration (Hmix < 0)
more than compensates for the braking of water bonds
(Hsolvent > 0); thus the process of Hydration is “Exothermic”, i.e.,
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ΔHhydration always < 0 (Exothermic)
33
Heat of Hydration & Lattice Energy

The energy required to separate an ionic solute
(Hsolute) into gaseous ions is:
The Lattice Energy (Hlattice)

This process requires energy input (Endothermic)
M X (s) + Heat  M (g) + X (g)
+
-
ΔHsolute (always > 0)

+
-
= ΔHlattice
Thus, the Heat of Solution for ionic compounds in
water combines the Lattice Energy (always
positive) and the combined Heats of Hydration of
Cations and Anions (always negative)
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Heat of Hydration & Lattice Energy

Hsoln can be either positive or negative
depending on the net values of:
Hlattice
(always >0)
and Hhydration (always <0)
Recall:
ΔHsoln = ΔHsolute + ΔHhydration of the ions
ΔHsoln = ΔH lattice + ΔHhydration of the ions

If the Lattice Energy is large relative to the
Hydration energy, the ions are likely to
remain together and the compound will tend
to be more insoluble
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Charge Density - Heats of Hydration

Heat of Hydration is related to the
“Charge Density” of the ion

Charge Density – ratio of ion’s charge to its volume

Coulombs Law – the greater the ion’s charge and
the closer the ion can approach the oppositely
charged end of the water molecule, the stronger
the attraction and, thus, the greater the charge
density

The higher the charge density, the more negative
Hhydr (always <0), i.e., the greater the solubility
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Charge Density - Heats of Hydration

Periodic Table


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Going down Group

Same Charge, Increasing Size

Charge Density & Heat of Hydration Decrease
Going across Period

Greater Charge, Smaller Size

Charge Density & Heat of Hydration Increase
37
Heat of Hydration & Lattice Energy
Hydration vs. Lattice Energies

The solubility of an ionic solid depends on the
relative contribution of both the energy of
Hydration (Hhydr) and the Lattice energy (Hlattice)
ΔH

soln
= ΔH
lattice
+ ΔH
hydration
Lattice Energy, Hlattice (always >0)

Small size + high charge = high Hlattice

For a given charge, lattice energy (Hlattice)
decreases (less positive) as the radius of ions
increases, increasing solubility
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Heat of Hydration & Lattice Energy

Energy of Hydration (Hhydr)

Energy of Hydration also depends on ionic radius

Small ions, such as Na1+ or Mg2+ have
concentrated electric charge and a strong
electric field that attracts water molecules

High Charge Density (more negative Hhydr)
produces higher solubility
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Heats of Hydration
Cation Radius (pm)
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Hhydr
(kJ/Mol)
Anion Radius (pm)
Hhydr
(kJ/Mol)
Na+
102
-410
F-
119
-431
K+
138
-336
OH-
119
-460
Rb+
152
-315
NO31-
165
-314
Cs+
167
-282
Cl-
167
-313
Mg2+
72
-1903
CN1-
177
-342
Ca2+
100
-1591
N31-
181
-298
Sr2+
118
-1424
Br-
182
-284
Ba2+
135
-1317
SH1-
193
-336
I-
206
-247
BF41-
215
-223
ClO41-
226
-235
CO32-
164
-1314
S2-
170
-1372
SO42-
244
-1059
40
Solubility – Solution Temperatures

Solubilities and Solution temperatures for 3 salts
NaCl

NaOH
NH4NO3
The interplay between (Hlattice) and (Hhydr) leads
to different values of (Hsoln)
ΔH
soln
= ΔH
lattice
+ ΔH
hydration
● Hlattice (Hsolute) for ionic compounds is always
positive
● Combined ionic heats of hydration (Hhydr) are
always negative
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41
Solubility – Solution Temperatures

The resulting temperature of the solution is a
function of the relative value of (Hsoln)

If the mixture solution becomes colder, the water,
representing the “surroundings,” has lost energy
to the system, i.e. an Endothermic process
(Hlattice dominates & Hsoln > 0)

If the solution becomes warmer, the surroundings
have gained energy, i.e., an Exothermic process
Hhydr dominates & Hsoln < 0
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Solubility – Solution Temperatures

Born-Haber diagram for the dissolution of 3 salts

(Hsoln) is positive for solutions that become “Colder”

(Hsoln) is negative for solutions that become “Warmer”
Hot
NaOH
Hot
Hsoln = - 44.5
kJ/mol
(Exothermic)
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Neutral NaCl
NH4NO3 Cold
Hsoln = 3.9 kJ/mol
Hsoln = 25.7 kJ/mol
(Endotherimc)
43
Solution Process – Entropy Change
The Heat of Solution (Hsoln) is one of two factors
that determine whether a solute dissolves in a
solvent
 The other factor concerns the natural tendency of
a system to spread out, distribute, or disperse its
energy in as many ways as possible
 The thermodynamic variable for this process is
called:

Entropy (S, S)
Directly related to the number of ways that a
system can distribute its energy
 Closely related to the freedom of motion of the
particles and the number of ways they can be
arranged

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Solution Process – Entropy Change

Solids vs. Liquids vs. Gases
 In a solid, particles are relatively fixed in their
positions
 In a liquid the particles are free to move around
each other
 This greater freedom of motion allows the
particles to distribute their kinetic energy in
more ways
 Entropy increases with increased freedom of
motion; thus, a material in its liquid phase will
have a higher Entropy than when it is a solid
Sliquid > Ssolid
A gas, in turn, would have a higher Entropy than
its liquid phase
Sgas > Sliquid > Ssolid
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Solution Process – Entropy Change

Relative change in Entropy (∆S)

The change in Entropy from solid to liquid to gas
is always positive
ΔSgas > ΔSliquid > ΔSsolid

> 0
The change from a liquid to a gas is called:
“Vaporization,”
thus,
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∆Svap > 0
46
Solution Process – Entropy Change

The change from a liquid to a solid is called:
“Fusion”
(freezing)
thus, Entropy change would be negative
After the liquid has frozen there is less freedom
of motion, thus the Entropy is less,
∆Sfusion
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< 0
47
Solution Process – Entropy Change

Entropy and Solutions
 There are far more interactions between particles in a
solution than in either the pure solvent or the pure
solute
 Thus, the Entropy of a solution is higher than the
sum of the solute and solvent Entropies
Ssoln > (Ssolute + Ssolvent )


or
ΔS > 0
The solution process involves the interplay between the:
Change in Enthalpy (∆H)
Change in Entropy (∆S)
Systems tend toward
“State of Lower Enthalpy (∆H)”
Dissolution
“State of Higher Entropy (∆S)”

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
48
Enthalpy Change vs Entropy Change

Does Sodium Chloride dissolve in Hexane?

NaCl and Hexane have very dissimilar
intermolecular forces:
Ionic (NaCl) vs Dispersive (Hexane)

Separating the solvent (Hexane) is easy because
intermolecular forces are weak ∆Hsep > 0

Separating NaCl requires supplying Lattice Energy
∆Hsep >> 0
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49
Enthalpy Change vs Entropy Change

Mixing releases very little heat because ioninduced dipole attractions between Na+ & Clions and Hexane are weak ∆Hmix  0

The sum of the Endothermic term (∆Hsep) is
much larger than the Exothermic term (∆Hmix),
thus ∆Hsoln is highly positive; thus “no mixing”
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50
Enthalpy Change vs Entropy Change
Solution does not form because the Entropy increase that
would accompany the mixing of the solute and solvent is
much smaller than the Enthalpy increase required to
separate the ions
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51
Enthalpy Change vs Entropy Change

Does Octane dissolve in Hexane?
Octane
Hexane
 Both compounds (Hydrocarbons with no
polarized functional groups) consist of nonpolar
molecules held together by Dispersive
Intermolecular Forces (IMF’s), which are
relatively weak
 Thus, these similar intermolecular forms would
suggest solubility
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52
Enthalpy Change vs Entropy Change
 Note ∆Hsoln is close to 0, suggesting little heat is released
 With little Enthalpy change in the solution process, Octane
dissolves readily in Hexane because the Entropy increase
(∆S) greatly exceeds the relatively small Enthalpy change
as a result of the ∆Hmix process
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53
Practice Problem

Water is added to a flask containing solid
Ammonium Chloride (NH4Cl). As the salt dissolves,
the solution becomes colder.
a. Is the dissolving of NH4Cl Exothermic or
Endothermic?
Ans: The solution process results in a cold
solution, i.e. – the system has taken energy from
the surroundings (water loses energy becoming
colder), thus,
an Endothermic process (∆Hsoln > 0)
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54
Practice Problem
b. Is the magnitude of ∆Hlattice of HN4Cl larger or smaller
than the combined ∆Hhyd of the ions?
Ans: Since ∆Hsoln > 0 and ∆Hhyd is always < 0; then
∆Hlattice must be much greater than ∆Hhyd
ΔH
soln
= ΔH
lattice
+ ΔH
hydration
c. Given the answer to (a), why does NH4Cl dissolve in
water?
Ans: The increase in Enthalpy (∆Hsoln > 0) would suggest
the solute would be insoluble
Since a solution does form, the increase in Entropy
must outweigh the change in Enthalpy
3/12/2016
Thus, Ammonium Chloride dissolves
55
Solubility - Hhydr vs Hlattice

Relative Solubilities – Alkaline Earth Hydroxides
Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2

Lattice energy decreases (becomes less positive) as
the radius of alkaline earth ion increases down the
group from Mg2+ to Ba2+

3/12/2016
Lattice energy decreases down a group suggesting
solubility to increase from Magnesium Hydroxide
to Barium Hydroxide
56
Solubility - Hhydr vs Hlattice

3/12/2016
Energy of Hydration becomes more negative
(Exothermic) as the ions become smaller

Since the atom and ion size increase down a
group, Energy of Hydration becomes more
positive (less Exothermic) down the group
suggesting a decrease in solubility down the
group from Mg(OH)2 to Ba(OH)2

However, in the case of Alkaline Earth Hydroxides,
Lattice energy (Hlattice) dominates and Ba(OH)2 is
the more soluble
57
Solubility - Hhydr vs Hlattice

Relative Solubilities - Alkaline Earth Sulfates
MgSO4 > CaSO4 > SrSO4 > BaSO4

Lattice energy depends on the sum of the
anion/cation radii

Since the Sulfate (SO42-) ion is much larger than the
Hydroxide ion (244 pm vs 119 pm), the percent
change in Lattice Energy (Hlattice) going from
Magnesium to Barium in the Sulfates is smaller than
for the Hydroxides, i.e., the solubility change is small
3/12/2016
58
Solubility - Hhydr vs Hlattice

The Energy of Hydration (HHydr) for the cations
decreases (becomes more positive) by a greater
amount going from Mg to Ba in the Sulfates relative
to the Hydroxides

Thus in the case of Alkaline Earth Sulfates, the
energy of Hydration (HHydr) dominates and Mg(SO4)
is more soluble than Ba(SO4)
3/12/2016
59
Solubility - Hhydr vs Hlattice
Lattice Energy (Hlattice)
Always Positive
(Endothermic)
More Pos
Less Pos
Solubility
Low
High
Solubility
Mg++
Hydration Energy (Hhyd)
Always Negative
(Exothermic)
More Neg
Solubility
High
Less Neg
Ba++
Low
Alkaline Earth Hydroxides
Down group
●
Lattice energy becomes less positive
(more soluble)
●
Hydration energy becomes less
negative (less soluble)
 Lattice-energy (Hlattice) dominates
Ba(OH)2 more soluble than Mg(OH)2
3/12/2016
Ionic
Size
●
●
●
Alkaline
Earth
Hydroxides
Alkaline
Earth
Sulfates
Solubility
Lower
Solubility
Higher
Higher
Lower
Alkaline Earth Sulfates
The Sulfate (SO42-) ion is much larger than
the Hydroxide ion (244 pm vs 119 pm)
The percent change in lattice energy going
from Magnesium to Barium in the Sulfates
is smaller than for the Hydroxides
The energy of hydration for the cations
decreases by a greater amount going from
Mg to Ba in the Sulfates relative to the
Hydroxides
 Hydration Energy (Hhydr) dominates
MgSO4 is more soluble than BaSO4
60
Predicting Solubility
Compatible vs Incompatible Intermolecular Forces (IMFs)

C6H14 + C10H22
compatible: both display dispersion forces
non-polar molecules (soluble)

C6H14 + H2O
incompatible: dispersion vs polar H-bonding
(insoluble)

H2O + CH3OH
compatible; both H-bonding
(soluble)

NaCl + H2O
compatible; ion-dipole
(soluble)

NaCl + C6H6
incompatible; ion vs dispersion
(insoluble)

CH3Cl + CH3COOH compatible; both polar covalent
(dipole-dipole)
3/12/2016
(soluble)
61
Practice Problem
Predicting Solubility
Which is more soluble in water?
a. NaCl
vs
C6H6
Ans: NaCl
b. CH3OH
vs
C6H6
Ans: CH3OH
c. CH3OH
vs
NaI
Ans: NaI
d. CH3OH
vs
CH3CH2CH2OH
Ans: CH3OH
e. BaSO4
vs
MgSO4
Ans: Mg(SO4)
f. BaF
vs
MgF
Ans: BaF
3/12/2016
62
Writing Solution Equations
1. Write out chemical equation for the dissolution of NaCl(s)
in water
NaCl(s)  Na+(aq) + Cl-(aq)
H2O
2. Write out chemical equation for the dissolution of Tylenol
(C8H9NO2) in water
C8H9NO2(s)
3/12/2016
H2O
 C8H9NO2(aq)
63
Solubility as Equilibrium Process

Ions disaggregate and become dispersed in solvent

Some undissolved solids collide with undissolved
solute and recrystallize

Equilibrium is reached when rate of dissolution
equals rate or recrystallization
Solute
3/12/2016
(undissolved)
€
Solute
(dissolved)
64
Solubility as Equilibrium Process

Saturation – Solution contains the maximum
amount of dissolved solute at a given temperature

Unsaturation – Solution contains less than
maximum amount of solute – more solute could be
dissolved!

Supersaturation – Solution contains more dissolved
solute than equilibrium amount
Supersaturation can be produced by
slowly cooling a heated equilibrium solution
3/12/2016
65
Comparison of Unsaturated and
Saturated Solutions
3/12/2016
66
Solubility Equilibrium

Solubility Equilibrium
A(s)  A(aq)
A(s) = solid lattice form
A(aq) = solvated solution form
Rate of dissolution = rate of
crystallization
Amount of solid remains constant
Dynamic Equilibrium
As many particles going into
solution as coming out of solution
3/12/2016
67
Gas Solubility – Temperature
Gas particles (solute) are already separated;
thus, little heat required for disaggregation
Hsolute  0
Heat of Hydration of a gaseous species is
always Exothermic (Hhydration < 0)
 Hsoln = Hsolute + Hhydration < 0
solute(g) + solvent(l)  sat’d soln + heat
Gas molecules have weak intermolecular
forces and the intermolecular forces between
a gas and solvent are also weak
As temperature rises, the kinetic energy of
the gas molecules also increases allowing
them to escape, lowering concentration, i.e.
Solubility of a gas decreases with increasing temperature
3/12/2016
68
Salt Solubility – Temperature

Most solids are more soluble at
higher temperatures

Most ionic solids have a
positive Hsoln because (Hlattice)
is greater than (HHydr)
Note exception - Ce2(SO4)3

Hsoln is positive
Heat must be absorbed
to form solution

If heat is added, the rate of
solution should increase
Solute + Solvent + Heat
3/12/2016
saturated solution
69
Effect of Pressure on Gas Solubility

At a given pressure, the same number of gas
molecules enter and leave a solution per unit time –
equilibrium

As partial pressure (P) of gas above solution increases,
the concentration of the molecules of gas increases in
solution
Gas + Solvent
PCO
3/12/2016
2
Saturated Solution
 CCO
2
70
Effect of Pressure on Solutions

Pressure changes do not effect solubility of liquids and
solids significantly aside from extreme P changes
(vacuum or very high pressure)

Gases are compressible fluids and therefore solubilities
in liquids are Pressure dependent

Solubility of a gas in a liquid is proportional ( to the
partial pressure of the gas above the solution

Quantitatively defined in terms of Henry’s Law
SP
3/12/2016
S = kHP
S
= solubility of gas (mol/L)
Pg
= partial pressure of gas (atm)
KH
= Henry’s Law Constant (mol/Latm)
71
Practice Problem

The partial pressure of Carbon Dioxide (CO2) gas in
a bottle of cola is 4 atm at 25oC
What is the Solubility of CO2?
Henry’s Law constant for CO2 dissolved in water is:
3.3 x 10-2 mol/Latm
Ans:
Solubility = k H × PCO
2
SCO = 3.3×10-2 mol / L • atm × 4 atm = 0.132 = 0.1 mol / L
2
3/12/2016
72
Practice Problem
The solubility of gas A (Mm = 80 g/mol) in water is
2.79 g/L at a partial pressure of 0.24 atm. What is
the Henry’s law constant for A (atm/M)?
M mgas = 80g / mol
Sgas
3/12/2016
= K h × Pgas
g 1 mol 

 2.79 L × 80 g 
=

 0.24 atm 




Kh
 S gas 
= 
 P 
 gas 
Kh
= 0.145 mol / L • atm
73
Practice Problem
If the solubility of gas Z in water is 0.32 g/L at a pressure of
1.0 atm, what is the solubility (g/L) of Z in water at a pressure
of 0.0063 atm?
 S gas(1) 
Kh = 
Sgas(1) = K h × Pgas(1)

P
 gas(1) 
Sgas(2) = K h × Pgas(2)
 S gas(2) 
Kh = 

P
 gas(2) 
 S gas(2)   S gas(1) 

 = 

 Pgas(2)   Pgas(1) 
Sga(2)
3/12/2016
 Sgas(1) 
 0.32g / L 
= Pgas(2) 
=
0.0063atm



P
1atm


 gas(1) 
Sgas(2) = 0.0020 g / L
74
Concentration
Definitions
Concentration Term
Molarity (M)
Molality (m)
Parts by mass
Parts by volume
Ratio
amount (mol) of solute
volume (L) of solution
amount (mol) of solute
mass (kg) of solvent
mass of solute
mass of solution
volume of solute
volume of solution
Mole Fraction (Xsolvent)
amount (mol) of solute
moles of solute + moles of solvent
Mole Fraction (Xsolute)
(i)amount (mol) of solute
(i)moles of solute + moles of solvent
3/12/2016
75
Practice Problem
The molality of a solution is defined as _______
a. moles of solute per liter of solution
b. grams of solute per liter of solution
c. moles of solute per kilogram of solution
d. moles of solute per kilogram of solvent
e. the gram molecular weight of solute per kilogram
of solvent
Ans: d
3/12/2016
76
Practice Problem

What is the molality of a solution prepared by dissolving
32.0 g of CaCl2 in 271 g of water
Ans:
amount (mol) solute
Molality =
mass (kg) solvent
Step 1 - Convert Mass to Moles
1 mol CaCl 2
Moles CaCl 2 = 32.0 g CaCl 2 ×
= 0.280 mol CaCl 2
110.98 g CaCl 2
Step 2 – Compute Molality
0.288 mol CaCl 2
mol solute
Molality =
=
= 1.06 m CaCl 2
1 kg
kg solvent
271 g × 3
10 g
3/12/2016
77
Practice Problem
Fructose, C6H12O6 (FW = 180.16 g/mol, is a sugar
occurring in honey and fruits. The sweetest sugar, it is
nearly twice as sweet as sucrose (cane or beet sugar). How
much water should be added to 1.75 g of fructose to give a
0.125 m solution?
 mol solute 
Molality(m) = 

 kgsolvent 
kg solvent
 mol solute
= 
 m
1 mol sucrose

 1.75 g sucrose  180.16 g
 
sucrose
 = 
0.125 mol / kg H2O



kgsolvent = 0.0777 kg H2O
3/12/2016






1000 g
×
= 77.7 g H2O
1 kg
78
Colligative Properties of Solutions

Presence of Solute particles in a solution changes
the physical properties of the solution

The number of particles dissolved in a solvent also
makes a difference in four (4) specific properties of
the solution known as:
“Colligative Properties”

Vapor Pressure

Boiling Point Elevation

Freezing Point Depression

Osmotic Pressure
3/12/2016
79
Colligative Properties of Solutions
Phase Diagram showing various phases of a substance and
the conditions under which each phase exists
Triple Point
Of Solution
3/12/2016
80
Colligative Properties of Solutions

Colligative properties deal with the nature of a solute
in aqueous solution and the extent of the dissociation
into ions

Electrolyte – Solute dissociates into ions and solution is
capable of conducting an electric current

Strong Electrolyte – Soluble salts, strong acids, and
strong bases dissociate completely; thus, the
solution is a good conductor

Weak Electrolyte – polar covalent compounds, weak
acids, weak bases dissociate weakly and are poor
conductors

Nonelectrolyte – Compounds that do not dissociate
at all into ions (sugar, alcohol, hydrocarbons, etc.)
are nonconductors
3/12/2016
81
Colligative Properties of Solutions

Prediction of the magnitude of a colligative
property
 Solute Formula
 Each mole of a nonelectrolyte yields 1 mole of
particles in the solution
Ex. 0.35 M glucose contains 0.35 moles of
solute particles (glucose molecules) per liter
 Each mole of strong electrolyte dissociates into
the number of moles of ions in the formula
unit
Ex. 0.4 M Na2SO4 contains 0.8 mol of Na+ ions
and 0.4 mol of SO42- ions (total 1.2 mol of
particles) per liter of solution
3/12/2016
82
Vapor Pressure

Vapor pressure (Equilibrium Vapor Pressure):
The pressure exerted by a vapor at equilibrium
with its liquid in a closed system

Vapor Pressure increases with increasing temperature

nonvolatile nonelectrolyte (ex. sugar) & pure solvent

The vapor pressure of a solution of a nonvolatile
nonelectrolyte (solute) is always lower than the
vapor pressure of the pure solvent

Presence of solute particles reduces the number of
solvent vapor particles at surface that can vaporize

At equilibrium, the number of solvent particles
leaving solution are fewer than for pure solvent,
thus, the vapor pressure is less
3/12/2016
83
Vapor Pressure

The vapor pressure of the solvent above the solution
(Psolvent) equals the Mole Fraction of solvent in the
solution (Xsolvent) times the vapor pressure of the pure
solvent (Posolvent)
(Raoult’s Law)
Psolvent = Xsolvent  Posolvent
In a solution, the mole fraction of the solvent (Xsolvent)
is always less than 1; thus the partial pressure of the
solvent above the solution (Psolvent) is always less than
the partial pressure of the pure solvent Posolvent
Ideal Solution – An ideal solution would follow
Raoult’s law for any solution concentration

Most gases in solution deviate from ideality

Dilute solutions give good approximation of Raoult’s
law
3/12/2016
84
Vapor Pressure

A solution consists of Solute & Solvent

The sum of their mole fractions equals 1
Xsolvent + Xsolute = 1
Xsolvent = 1 - Xsolute
From Raoult's law
o
o
Psolvent = Xsolvent Psolvent
= (1 - Xsolute ) × Psolvent
o
o
Psolvent = Psolvent
- Xsolute × Psolvent
Rearranging
o
o
Psolvent
- Psolvent = ΔP = Xsolute × Psolvent

The magnitude of P (vapor pressure lowering)
equals the mole fraction of the solute times the
vapor pressure of the pure solvent
3/12/2016
85
Vapor Pressure
Vapor Pressure & Volatile Nonelectrolyte Solutions
 The vapor now contains particles of both a volatile
solute and the volatile solvent

o
Psolvent = Xsolvent  Psolvent

o
and Psolute = Xsolute  Psolute
From Dalton’s Law of Partial Pressures
o
o
Ptotal = Psolvent + Psolute = Xsolvent  Psolvent
+ Xsolute  Psolute

The presence of each volatile component lowers
the vapor pressure of the other by making each
mole fraction less than 1
3/12/2016
86
Example Problem

Given: Equi-molar solution of Benzene (XB = 0.5) and
Toluene (XT = 0.5) (Both nonelectrolytes)
Xben = 0.5
o
Pben
= 95.1 torr @ 25o C
Xtol = 0.5
o
Ptol
= 28.4 torr @ 25o C
o
Pben = Xben × Pben
= 0.5 × 95.1 torr = 47.6 torr
Ptol


= Xtol × Po tol = 0.5 × 28.4 torr = 14.2 torr
Benzene lowers the vapor pressure of Toluene and
Toluene lowers the vapor pressure of Benzene
Compare Vapor composition vs Solution composition
Xben =
Xtol =
3/12/2016
Pben
47.6 torr
=
= 0.770 torr
Ptot
47.6torr + 14.2torr
Ptol
14.2 torr
=
= 0.230 torr
Ptot
47.6 torr + 14.2 torr
Mole fractions in
vapor are different
87
Boiling Point Elevation

Boiling Point (Tb) of a liquid is the temperature at which its
vapor pressure equals the external pressure (Atm Press)

The vapor pressure of a solution (solvent + solute) is lower
than that of the pure solvent at any temperature

Thus, the difference between the external (atmospheric)
pressure and the vapor pressure of the solution is greater
than the difference between external pressure and the
solvent vapor pressure - Psoln > Psolvent

The boiling point of the solution will be higher than the
solvent because additional energy must be added to the
solution to raise the vapor pressure of the solvent (now
lowered) to the point where it again matches the external
pressure

The boiling point of a concentrated solution is greater than
the boiling point of a dilute solution - Antifreeze in your car!!
3/12/2016
88
Boiling Point Elevation

The magnitude of the boiling point elevation is
proportional to the Molal concentration of the solute
particles
ΔTb  m or ΔTb = K bm
where : ΔTb = Tbsolution - Tbsolvent = boiling point elevation
m = solution molality (mol solute / kg solvent )

K b = molal boiling point elevation constant
Note the use of “Molality”
 Molality is related to mole fraction, thus particles
of solute
 Molality also involves “Mass of Solvent”; thus, not
affected by temperature
3/12/2016
89
Freezing Point Depression

Recall: The addition of a solute to a solvent dilutes the
number of solvent particles in the solution

As the solute concentration increases there is
increased competition between solute and solvent
particles for space at the surface preventing solvent
molecules from escaping to the vapor phase, thus the
vapor pressure of the solution is lower than the vapor
pressure of the pure solvent

Recall: The vapor pressure lowering, ΔP, of a solution
is a function of the mole fraction of solute and vapor
pressure of pure solvent, that is:
Raoults Law
o
o
Psolvent
- Psolvent = ΔP = Xsolute × Psolvent
3/12/2016
90
Freezing Point Depression

The Freezing Point of a substance (solvent) is the
temperature at which an equilibrium is established
between the number of solid particles coming out of
solution and the number of particles dissolving

Note: In solutions with nonvolatile solutes, only solvent
molecules can vaporize; thus, only solvent molecules
can solidify (freeze)

The vapor pressure of solid and liquid solvent particles
in equilibrium is same

Since vapor pressure is a function of temperature, the
lowering of the vapor pressure of the solution means a
lower temperature at which equilibrium exists between
solid & liquid solvent particles, i.e. the freezing point
3/12/2016
91
Freezing Point Depression

Thus, as the concentration of the solute in the
solution increases, the freezing point of the
solution is lowered
3/12/2016
92
Freezing Point Depression

The Freezing Point depression has the magnitude
proportional to the Molal concentration of the solute
ΔTf  m or ΔTf = K f m
where : ΔTf = Tfsolvent - Tfsolution = freezing point depression
m = solution molality (mol solute / kgsolvent )
K f = molal freezing point depression constant
Molal Boiling Point Elevation and Freezing Point Depression
Constants of Several Solvents
3/12/2016
93
The van’t Hoff Factor

From Slide 82, colligative properties depend on the relative
number of solute to solvent “particles”

In strong electrolyte solutions, the solute formula specifies
the number of particles affecting the colligative property

Ex. The BP elevation of a 0.5 m NaCl soln would be twice
that of a 0.5 m Glucose soln because NaCL dissociates into
“2” particles per formula unit, where glucose produces “1”
particle per formula unit

The van’t Hoff factor (i) is the ratio of the measured value
of the colligative property, e.g. BP elevation, in the
electrolyte solution to the expected value for a
nonelectrolyte solution
measured value for electrolyte solution
i=
expected value for nonelectrolyte solution
3/12/2016
94
The van’t Hoff Factor

To calculate the colligative properties of strong electrolyte
solutions, incorporate the van’t Hoff factor into the equation
CH3OH(l)
 CH3OH(aq)
1:1, i = 1
NaCl(s)
 Na+(aq) + Cl-(aq)
2:1, i = 2
CaCl2(s)
 Ca2+(aq) + 2Cl-(aq)
3:1, i = 3
Ca3(PO4)2(s)  3 Ca2+(aq) + 2 PO43-(aq)
5:1, i = 5
Ex. Freezing Point Depression with van’t Hoff factor for
Ca3(PO4)2(s)
T  iK c
f
f m
3/12/2016
T  5 K c
f
f m
95
Colligative Properties of Solutions
Colligative
Property
Mathematical
Relation
1. Vapor Pressure
(Raoult’s Law)
2. Freezing Point Depression
3. Boiling Point Elevation
4. Osmotic Pressure
3/12/2016
P
solvent

solvent

f
= i Po
X
T  i K m
f

T  i K m
b
b
solute



  i  MRT 
96
Practice Problem
What is the boiling point of 0.0075 m aqueous calcium
chloride, CaCl2?
CaCl2 = 3 particles (1 Ca & 2 Cl)  i = 3
ΔTb  m or ΔTb = iK bm
(i = 3)
where : ΔTb = Tbsolution - Tbsolvent = boiling point elevation
Tbsolution - Tbsolvent = 3  K bm
TbCaCl = 3  K bm + TbH O
2
2
o
TbCaCl
2
C
= 3  0.512
 0.0075 m + 100.0o C
m
TbCaCl = 100.1o C
2
3/12/2016
97
Practice Problem
What is the freezing point of a 0.25 m solution of glucose in
water (Kf for water is 1.86°C/m)?
a. 0.93°C b. –0.93°C
Ans: d
ΔTf
c. 0.46°C d. –0.46°C e. 0.23°C
= Tfsolvent - Tfsolution = iK f m
( i  1)
-Tfsolution = K f m - Tfsolvent
Tfsolution = Tfsolvent - K f m
T
fsolution
 oC 
o
 × 0.25 m
= 0.0 C - 1.86 
 m 


o
Tfsolution = -0.46 C
3/12/2016
98
Practice Problem
What is the freezing point of 0.150 g of glycerol (C3H8O3) in
20.0 g of water? (Kf for water is 1.86°C/m)
ΔTf
= Tfsolvent - Tfsolution = iK f m
(i = 1)
-Tfsolution = K f m - Tfsolvent
Tfsolution = Tfsolvent - K f m
Tfsolution
Tfsolution
Tfsolution
3/12/2016
 1mol 
0.150g 

o


92.094g
C


= 0.0o C - 1.86 
×

m
 1kg 


20.0g × 

1000g




 oC 
mol
o
 × 0.0814
= 0.0 C - 1.86 
mol
kg


 kg 


o
= -0.151 C
99
Practice Problem
What is the molar mass (Mm) of Butylated Hydroxytoluene
(BHT) if a solution of 2.500 g of BHT in 100.0 g of Benzene
(Kf = 5.065 oC/m; Tf = 5.455 oC) had a freezing point of 4.880
oC?
ΔT
=T
- T
= iK m (i = 1)
f
fsolvent
fsolution
f
 Tfsolvent - Tfsolution 
m=

iK
f

o
o
 5.455 C - 4.880 C 
m=
 = 0.1135 mol BHT / kg Benzene
o
 1× 5.065 C / m 
MmBHT = Molar MassBHT (g / mol)
1mol BHT  
1mol BHT 

2.500
g
×
25.00
g

mol BHT 
M mBHT (g)  
M mBHT (g) 
=

m = 0.1135
=
1kg  
kg Benzene  100 g
1kg Benzene

Benzene


1000g  


MmBHT  g 
25.00 g BHT
=
= 220.3g / mol (Actual M m = 220.3 g / mol)
1 mol BHT
0.1135 mol BHT
3/12/2016
100
Colloids

Suspensions vs Mixtures vs Colloids

A Heterogeneous mixture – fine sand suspended in
water – consists of particles large enough to be seen
by the naked eye, clearly distinct from surrounding
fluid

A Homogeneous mixture – sugar in water – forms a
solution consisting of molecules distributed throughout
and indistinguishable from the surrounding fluid

Between these extremes is a large group of mixtures
called colloidal dispersions – Colloids

Colloid particles are larger than simple molecules, but
small enough to remain in suspension and not settle
out
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101
Colloids

Colloids have tremendous surface areas, which allows many
more interactions to exert a large total adhesive force,
which attracts other particles

Particle Size & Surface Area


Diameter – 1 to 1000 nm (10-9 to 10-6)

Single macromolecule or aggregate of many atoms,
ions, or molecules

Very large surface area

Large surface area attracts other particles through
various intermolecular forces (IMF)
Surface Area

3/12/2016
A cube with 1 cm sides (SA – 6 cm2) if divided into
1012 cubes (size of large colloidal particles) would
have a total surface area of 60,000 cm2
102
Colloids

Colloid Classifications:

Colloids are classified according to whether the
dispersed and dispersing substances are gases,
liquids, or solids
Types of Colloids
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103
Colloids
Tyndall Effect
 Light passing through a colloid is scattered
randomly because the dispersed particles have
sizes similar to wavelengths of visible light
 The scattered light beams appears broader than
one passing through a solution
 Brownian Motion
 Observed erratic change of speed and direction
resulting from collisions with molecules of the
dispersing medium
 Einstein’s explanation of Brownian motion further
enhanced the concept of the molecular nature of
matter

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104
Colloids
Stabilizing & Destabilizing Colloids
 Colloidal particles dispersed in water have charged
surfaces
 Two major classes:
 Hydrophilic (Stabilized) Colloids
 London force attractions between charges on
dispersed phase (colloid) and the partial
charges on continuous phase (water)
 Appear like normal solutions, no settling out
o Gelatin (protein solution) in water
o Lipids & Soap form spherical micelles with
charged heads on exterior and hydrocarbon
tails
o Oily particles can be dispersed by adding
ions, which are absorbed on the particle
surface
3/12/2016

105
Colloids

3/12/2016
Hydrophobic (Unstable) Colloids
 Several methods exist to overcome the
attractions between the dispersed and
continuous phases causing the particles to
aggregate and settle out
 At the mouths of rivers, where salt water
meets fresh water, the ions in salt water are
absorbed onto the surfaces of clay particles
causing them to aggregate. The settling out
process produces deltas.
 In smokestack gases, electrolyte ions are
absorbed onto uncharged colloidal particles,
which are then attracted to charged plates
removing them from the effluent
106
Osmosis

Osmosis, a colligative property, is the movement of
solvent particles through a semipermeable
membrane separating a solution of one
concentration from a solution of another
concentration, while the solute particles stay within
their respective solutions

Many organisms have semipermeable membranes
that regulate internal concentrations
3/12/2016
107
Osmosis

Water can be purified by “reverse osmosis”

The direction of flow of the solvent is from the
solution of lower concentration to the solution of
higher concentration increasing its volume,
decreasing concentation

Osmotic pressure is defined as the amount of
pressure that must be applied to the higher
concentration solution to prevent the dilution and
change in volume
3/12/2016
108
Osmosis
3/12/2016
109
Osmosis (2-Compartment Chamber)
1










Semipermeable membrane
















2



= solute molecule
= solvent molecule
• Osmosis = solvent flow across membrane
• Net movement of solvent is from 2 to 1; pressure increases in 1
• PA,1 < PA,2, osmosis follows solvent vapor pressure
3/12/2016
• Osmotic Pressure = pressure applied to just stop osmosis
110
Osmotic Pressure

At equilibrium the rate of solvent movement into
the more concentrated solution is balanced by the
rate at which the solvent returns to the less
concentrated solution.

The pressure difference at equilibrium is the
Osmotic Pressure (), which is defined as the
applied pressure required to prevent the net
movement of solvent from the less concentrated
solution to the more concentrated solution
3/12/2016
111
Osmotic Pressure

Osmotic Pressure () is proportional to the
number of solute particles (moles) in a given
volume of solution, i.e.,
Molarity (M) (moles per liter of solution)
nsolute
Π 
Vsolution
or
Π  M
The Proportionality Constant is :
"R" times the absolute temperature (T)
nsolute
Π =
RT = MRT
Vsolution
3/12/2016
112
Osmotic Pressure

Osmotic Pressure / Solution Properties

The volume of the concentrated solution will
increase as the solvent molecules pass through the
membrane into the solution decreasing the
concentration

This change in volume is represented as a change in
height of the liquid in a tube open to the externally
applied pressure

This height change (h), the solution density (d), and
the acceleration of gravity (g) can also be used to
compute the Osmotic Pressure ()
 = gdh
 = g(m/s2)d(kg/m3)h(m) = pascals(kg/ms2)
-3 torr(mm)= 9.8692 x 10-6 atm
1
pascal
=
7.5006
x
10
3/12/2016
113
Reverse Osmosis

Membranes with pore sizes of 0.1 – 10 m are used to filter
colloids and microorganisms out of drinking water

In reverse osmosis, membranes with pore sizes of 0.0001 –
0.01 m are used to remove dissolved ions

A pressure greater than the osmotic pressure is applied to
the concentrated solution, forcing the water back through
the membrane, filtering out the ions

Removing heavy metals from domestic water supplies and
desalination of sea water are common uses of reverse
osmosis
3/12/2016
114
Practice Problem
How much potable water can be formed per 1,000 L of
seawater (~0.55 M NaCl) in a reverse osmosis plant that
operates with a hydrostatic pressure of 75 atm at 25 oC?
nNaCl
Π=
RT
VNaCl Soln
MNaCl soln
nNaCl
=
VNaCl Soln
nNaCl = MNaCl soln × VNaClSoln
VNaCl Soln
n NaCl
 M NaCl soln × VNaCl Soln
=
RT = 
Π
Π


 RT

VNaCl Soln
mol


0.55
×
1000
L


L • atm 
o
o
L
=
[25
C
+
273.15]
K

  0.0821

o
75 atm
mol • K 

 


VNaCl Soln = 179.5 L = 180 L
3/12/2016
115
Practice Problem
What is the osmotic pressure (mm Hg) associated with a
0.0075 M aqueous calcium chloride, CaCl2, solution at
25 oC?
a. 419 mm Hg
b. 140 mm Hg
d. 279 mm Hg
e. 371 mm Hg
Π = i (MRT)
c. 89 mm Hg
i=3
mol
L • atm
o
Π = 3 × 0.0075
× 0.0821
×
[25
+
273.15]
K

o
L
mol • K
760 mm
Π = 0.5507576 atm
= 418.5758 = 419 mm Hg
1 atm
3/12/2016
116
Practice Problem

Consider the following dilute NaCl(aq) solutions
a. Which one will boil at a higher temperature?
b. Which one will freeze at a lower temperature?
c. If the solutions were separated by a semi permeable
membrane that allowed only water to pass, which solution
would you expect to show an increase in the
concentration of NaCl?
Ans: a. B vapor pressure of more concentrated solution is lower; thus,
a higher temperature required to reach VP at boiling point
b. B vapor pressure of more concentrated solution is lower; thus,
temperature at equilibrium (freezing) point is lower
c. A The movement of solvent between
solutions is from the solution of
lower concentration to the solution
of higher concentration), thus less
solvent in beaker A resulting in
higher concentration
3/12/2016
117
Practice Problem

Consider the following three beakers that contain water and
a non-volatile solute. The solute is represented by the
orange spheres.
a. Which solution would have the highest vapor pressure?
 b. Which solution would have the lowest boiling point?
 c. What could you do in the laboratory to make each
solution have the same freezing point?
a. A More solvent particles can

escape because of fewer
solute particles present
b. A Requires less change in
vapor pressure to reach
VP at boiling point
c.
Condense “A” by 1/2
3/12/2016
118
Practice Problem
Caffeine, C8H10N4O2 (FW = 194.14 g/mol), is a stimulant
found in tea and coffee. A Sample of the substance was
dissolved in 45.0 g of chloroform, CHCl3, to give a 0.0946 m
solution.
How many grams of caffeine were in the sample?
Ans: molality(m) = molescaffeine
kgCHCl
3
molescaffeine = m ×kgCHCl
molescaffeine
3
 mol

1kg
caffeine
= 0.0945 
 × 45.0gCHCl
 kgCHCl 
1000g


= 0.0042525 molcaffeine

3

3
molescaffeine
Masscaffeine
3/12/2016
194.14g
= molcaffeine ×
= 0.826g
molcaffeine
119
Practice Problem
A solution contains 0.0653 g of a molecular compound in 8.31
g of Ethanol. The molality of the solution is 0.0368 m.
Calculate the molecular weight of the compound
molality(m) = molsolute / kgsolvent
1mol
0.0653g
Mm
0.0368 =
1kg
8.31gethanol
1000g
0.0653g
Mm =
mol
1kg
0.0368
× 8.31g ×
kg
1000g
Mm = 213g / mol = MolecularWeight
3/12/2016
120
Practice Problem
What is the vapor pressure (mm Hg) of a solution of 0.500 g of
urea [(NH2)2CO, FW = 60.0 g/mol] in 3.00 g of water at 25 oC?
What is the vapor pressure lowering of the solution?
The vapor pressure of water at 25 oC is 23.8 mm Hg
molurea = 0.500gurea
1molurea
= 0.00833 molurea
60.0g urea
mol H O = 3.00 g H O
Mole Fraction Urea (Xurea ) =
0.00833
= 0.0425
0.00833 + 0.187
Mole Fraction Water (X H O ) =
0.187
= 0.957
0.00833 + 0.187
2
2
2
1mol H O
2
18.016 g H O
= 0.187 mol H O
2
2
Partial Pressure of Solution, i.e., the solvent containing non - volatile electrolyte

o
Psolvent = i Psolvent
Xsolvent

Psolvent = 1×(23.8 mmHg × 0.957) = 22.8 mmHg
VaporPressure lowering
o
Psolvent
- Psolvent = ΔP = 23.8 mm Hg - 22.8 mmHg = 1.00 mm HG
or
o
o
Psolvent
- Psolvent = ΔP = Xurea × Psolvent
= 0.0425 × 23.8 mm Hg = 1.01 mm Hg
3/12/2016
121
Practice Problem
What is the vapor pressure lowering in a solution formed by
adding 25.7 g of NaCl (FW = 58.44 g/mol) to 100. g of water
at 25 oC? (The vapor pressure of pure water at 25 oC is 23.8
mm Hg)
mol NaCl = 25.7g NaCl
mol H O = 100. g H O
2
2
1mol NaCl
= 0.4398 mol NaCl
58.44g NaCl
1mol H O
2
18.016 g H O
= 5.5506 mol H O
2
2
Mole Fraction NaCl (X NaCl ) =
0.4398
= 0.07342
0.4398 + 5.5506
Vapor Pressure lowering
o
ΔP = i(XNaCl × Psolvent
) = 2  0.07432 × 23.8 mm Hg = 3.49 mm Hg
3/12/2016
122
Practice Problem
A 0.0140-g sample of an ionic compound with the formula
Cr(NH3)5Cl3 (FW = 243.5 g/mol) was dissolved in water to
give 25.0 mL of solution at 25 oC.
The osmotic pressure was determined to be 119 mm Hg.
How many ions are obtained from each formula unit when the
compound is dissolved in water?
Π = i(MRT)
1mol 

0.0140
g
molsolute 
243.5 g 
Molarity(M) =
=
 = 0.0022998 M
1L
Lsoln
 25 ml


1000 mL 

1atm
119mmHg
760 mmHg
 Π 
i=
=

 MRT  0.0022998  mol  × 0.0821 L • atm × (273.15 + 25.0)o K


 L 
mol •o K
The number of particles produced by the dissolution
i = 2.78 = 3
3/12/2016
of a "complex" molecule may not be readily descernible
by physical inspection of the compond formula
123
Equation Summary

Component Enthalpies of Heat of Solution
ΔH
= ΔH
+ ΔH
soln
solute
solvent
ΔH

= ΔH
solute
+ ΔH
mix
hydration
Component Enthalpies of Ionic Compound Heat of Soln
ΔH

soln
+ ΔH
soln
= ΔH
lattice
+ ΔH
hydr of the ions
Relating Gas Solubility to its Partial Pressure (Henry’s Law)
Sgas = k H Pgas
3/12/2016
124
Equation Summary

Relationship between vapor pressure & mole fraction
Psolvent = Xsolvent × P oSolvent
Psolute = Xsolute × P oSolute
Ptotal = Psolvent + Psolute = (Xsolvent × P oSolvent ) + (Xsolute × P oSolute )

Vapor Pressure Lowering
Po
- P
= ΔP = i (X
× Po
)
solvent
solvent
solute
solvent

Boiling Point Elevation
ΔTb = Tbsolution - Tbsolvent = iK bm

Freezing Point Depression
ΔTf = Tfsolvent - Tfsolution = iK f m
Osmotic Pressure
nsolute
Π = i
RT = iMRT
Vsolution
3/12/2016

125
Equation Summary
Concentration Term
Ratio
Molarity (M)
amount (mol) of solute
volume (L) of solution
Molality (m)
amount (mol) of solute
mass (kg) of solvent
Parts by mass
Parts by volume
Mole Fraction (Xsolute)
Mole Fraction (Xsolvent)
3/12/2016
mass of solute
mass of solution
volume of solute
volume of solution
(i)amount (mol) of solute
(i)moles of solute + moles of solvent
amount (mol) of solvent
(i)moles of solute + moles of solvent
126