Chapter 13


3/22/2016
Solutions

Intermolecular Forces & Solubility

Liquid Solutions

Gas Solutions

Solid Solutions
The Solution Process

Heats of Solution

Heats of Hydration

Solution Process & Change in Entropy
1
Chapter 13


3/22/2016
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
2
Chapter 13


3/22/2016
Colligative Properties of Solutions

Nonvolatile Nonelectrolyte Solutions

Solute Molar Mass

Volatile Nonelectrolyte Solutions

Strong Electrolyte Solutions
Structure and Properties of Colloids
3
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


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Bonding (Intramolecular) forces

Attraction between Cations & Anions (ionic bonding)

Electron pairs (covalent bonding)

Metal Cations and Delocalized valence electrons
(metallic bonding)
Intermolecular forces

Attractions between molecules as a result of partial
charges

Attractions between ions & molecules
4
Intermolecular Forces

Bonding (Intramolecular)


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Relatively strong, larger charges close to each other
Intermolecular

Relatively weak 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) (weakest)
5
Intermolecular Forces
Bonding
(Intramolecular)
Forces
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NonBonding
(Intermolecular)
Forces
6
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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7
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., methanol) and
would be soluble in polar substances (solvents)
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8
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
crystal structure
9
Intermolecular Forces & Solubility
Intermolecular Forces

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:
Hydrogen Bonding (N, O, F)
:

:

H N
H O
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
(H2O) molecule
10
Intermolecular Forces & Solubility

Intermolecular Forces

Dipole-Dipole forces

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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)
11
Intermolecular Forces & Solubility

Intermolecular Forces

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Ion-induced Dipole forces

One of 2 types of charged-induced dipole forces
(next slide – dipole-induced dipole)

Rely on polarizability of components

Ion’s charge distorts electron cloud of nearby
nonpolar molecule

Ion increases magnitude of any nearby dipole;
thus contributes to solubility of salts in less polar
solvents – LiCl in ethanol
12
Intermolecular Forces & Solubility

Intermolecular Forces

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Dipole-induced dipole forces

Intermolecular attraction between a polar
molecule and the oppositely charged pole it
induces in a nearby molecule

Also based on polarizability, but are weaker than
ion-induced dipole forces because the magnitude
of charge is smaller – ions vs dipoles (coulombs
law)

Solubility of atmospheric gases (O2, N2, noble
gases) have limited solubility in water because of
these forces
13
Intermolecular Forces & Solubility

Intermolecular Forces

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Dispersion forces (Instantaneous Dipoles)

Contribute to solubility of all solutes in all
solvents.

Principal type of intermolecular force in solutions
of nonpolar substances, such as petroleum and
gasoline

Keep cellular macromolecules in their biologically
active shapes
14
Liquid Solutions



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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
 Oil/Hexane
– Dispersion forces
 Solid – Liquid
 Salts/Water
– Ionic forces
 Gas – Liquid
 (O/H2O)
– Weak intermolecular forces (low
solubility, but biologically
important)
15
Liquid Solutions

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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
16
Liquid Solutions

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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)
17
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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18
The Solution Process

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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 is absorbed when particles separate

Energy is released when particles mix and attract
each other

Thus, the solution process is accompanied by
changes in Enthalpy (sum of internal energy and
product of pressure & volume H  E  P V)
and Entropy (number of ways the energy of system
can be dispersed through motions of particles)
19
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 solvent is different
for 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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20
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
electronegative Oxygen,
Nitrogen or Fluorine
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Hydration Energy
attraction between
solvent and solute
Lattice Energy
solute-solute forces
21
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
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
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22
Practice Problem (con’t)
b. Ethylene Glycol (HOCH20CH20H) in Hexane
(CH3(CH2)4CH3) or in Water (H2O)
Ans: Very Soluble in Water
Ethylene Glycol molecules have two polar –OH groups
and the Molecules interact with each other
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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23
Practice Problem (con’t)
c. Diethyl Ether (CH3CH2OCH2CH3) in water 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 and its
hydrocarbon group (CH3CH2) can substitute for
the ether’s dispersion forces
Water, on the other hand, can form H bonds with
the ether, but it lacks any hydrocarbon portion, so
it forms much weaker dispersion forces with the
ether
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24
The Solution Process

Solute particles absorb heat (endothermic) and separate
from each other by overcoming intermolecular
attractions
Solute (aggreagted) + heat  solute (separated)

Solvent particles absorb heat (endothermic) and
separate from each other by overcoming intermolecular
attractions
Solvent (aggreagted) + heat  solvent (separated)

H
> 0
solute
H
> 0
solvent
Solute and Solvent particles mix (form a solution) by
attraction and release energy - an exothermic process
Solute (separated) + Solvent (separated)  Solution + Heat ΔH
< 0
mix

Total Enthalpy change is the “Heat of Solution” (∆Hsoln)
ΔH
= ΔH
soln
solute
Endothermic
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+ ΔH
solvent
Endothermic
+ ΔH
mix
Exothermic
25
The Solution Process

Solution cycles and the enthalpy components of the heat
of solution
If (Hsolute + Hsolvent)
If (Hsolute + Hsolvent)
Exothermic solution process
Solute Soluble
< Hmix
> Hmix
exothermic
endothermic
Endothermic solution process
Solute Insoluble
Note: Recall Born-Haber Process – Chap 6
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26
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:
ΔH
= ΔH
hydration
solvent
+ ΔH
mix

Since:
ΔH
= ΔH
solution
solute
+ ΔH

Then:
ΔH
+ ΔH
solution
=
ΔH
solute
solvent
+ ΔH
mix
hydration
Forming strong ion-dipole forces during hydration (Hmix <0)
more than compensates for the braking of water bonds; thus the
process of hydration is “exothermic”, i.e.,
ΔH
always < 0 (exothermic)
hydration
27
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
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)
ΔΗ
(always > 0 endothermic) = ΔΗ
solute
lattice


Thus, the heat of solution for ionic compounds in water
combines the lattice energy (always positive) and the
combined heats of hydration of cation and anion (always
negative)
Hsoln can be either positive or negative depending on
the net values of Hlattice and Hhydr
ΔH

3/22/2016
soln
= ΔH
lattice
+ ΔH
hydr 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
28
Charge Density - Heats of Hydration

Heat of Hydration is related to “Charge Density” of 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,,
i.e., the more soluble

Periodic Table

Going down Group


Charge density & Heat of Hydration decrease
Going across Period – greater charge, smaller size

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– same charge, increasing size
Charge density & Heat of Hydration increase
29
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
= ΔH
+ ΔH
soln
lattice
hydration
 Lattice Energy


Small size + high charge = high lattice E
For a given charge, Lattice energy (Hlattice) decreases as
the radius of ions increases, increasing solubility
 Energy of Hydration
 Energy of Hydration (Hhydr) 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) higher
solubility

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30
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
31
Solubility - Hhydr vs Hlattice

Relative Solubilities – Alkaline Earth Hydroxides
Mg(OH)2 < Ca(OH)2 < Sr(OH)2 < Ba(OH)2
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
Lattice energy decreases as the radius of alkaline
earth ion increases from Mg2+ to Ba2+

 Lattice energy alone would suggest solubility to
increase down the group from Magnesium Hydroxide
to Barium Hydroxide

Energy of Hydration is greatest for small ions (Mg2+)

 Energy of Hydration alone would suggest solubility
to decrease down the group from Mg(OH)2 to
Ba(OH)2
32
Solubility - Hhydr vs Hlattice

Relative Solubilities - Alkaline Earth Sulfates
MgSO4 > CaSO4 > SrSO4 > BaSO4
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
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 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
33
Solubility - Hhydr vs Hlattice
Mg++
Lattice
Energy
Solubility
(Hlattice)
Hydration
Energy
Solubility
(HHydration)
High
Low
Low
High
High
High
Solubility
Alkaline
Alkaline
Earth
Earth
Hydroxides
Sulfates
Low
High
Ionic
Size
Ba++
Low
Low


Down group, the lattice energy of alkaline
earth hydroxides decreases more rapidly
than does the energy of hydration

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 (Hhyd) dominates

MgSO4 is more soluble than BaSO4
 Lattice-energy factor (Hlattice) dominates
Ba(OH)2 more soluble than Mg(OH)2
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Low
Sulfates
Hydroxides

High
34
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)
 The resulting temperature of the solution is a function of the
relative value of (Hsoln)
Neg (Hot)
Pos (Cold)

• Hlattice (Hsolute) is always positive for ionic compounds
• Combined ionic heats of hydration (Hhydr ) are always negative
Hot
NaOH Hot
Hsoln = -44.5 kJ/mol
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NaCl Neutral
NH4NO3 Cold
Hsoln = 3.9 kJ/mol
Hsoln = 25.7 kJ/mol
35
Predicting Solubility
Compatible vs Incompatible Intermolecular Forces (IMFs)

C6H14 + C10H22
compatible; both dispersion forces &
non-polar molecule
(soluble)

C6H14 + H2O
incompatible; dispersion vs polar
and H-bonding
(insoluble)

H2O + CH3OH
compatible; both H-bonding

NaCl + H2O
compatible; ion-dipole

NaCl + C6H6
incompatible; ion vs dispersion (insoluble)

CH3Cl + CH3COOH compatible; both polar covalent
& dipole-dipole
(soluble)
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(soluble)
(soluble)
36
Practice Problem
Predicting Solubility
Which is more soluble in water?
3/22/2016
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
37
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)
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H2O
 C8H9NO2(aq)
38
Solution Process - Entropy

The heat of solution
ΔH
= ΔH
+ ΔH
solution
solute
hydration
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

Entropy (S) is a Thermodynamic variable directly related
to the number of ways a system can distribute its energy
Entropy will be discussed in more detail in Chapter 18

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Entropy is related to freedom of motion of the particles
in the system and the number of ways they can be
arranged
39
Solution Process - Entropy


Gas vs Liquid vs Solid
A liquid has a higher Entropy (S) than a solid because
the particles have higher freedom of motion
Sliquid > Ssolid

Similarly, a gas has higher entropy than a liquid
Sgas


3/22/2016
> Sliquid
The change in entropy when a liquid changes to a gas is
positive
Svap > 0
A solution usually has a higher entropy than a pure
solute and a pure solvent.
 The solution provides more ways to distribute energy
 Particles in a solution have a higher freedom of
motion with increased possibilities for interactions
between molecules
40
Solution Process - Entropy

Sodium Chloride Solution in Hexane vs Toluene in Hexane

The ion-dipole forces in NaCl are weak
and incompatible with the dispersion
forces in Hexane (Hydrocarbon – C6H14)

Hmix (exothermic) is much smaller than
Hsolute (endothermic)
Hsoln >> 0  NaCl insoluble

Solution does not form because the
entropy increase that would accompany
the mixing of solute & solvent is much
smaller than the enthalpy increase
required to separate

The dispersive forces in hexane and
toluene are compatible

Hmix dominates Hsolute & Hsolvent
 soluble
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41
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
(undissolved)
solute
(dissolved)

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
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42
Comparison of Unsaturated and
Saturated Solutions
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43
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
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44
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
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45
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 with Ce2(SO4)3

Hsoln is positive, heat is
absorbed to form solution

If heat is added, the rate of
solution should increase
Solute + Solvent + Heat
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saturated solution
46
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
3/22/2016
2
saturated solution
 CCO
2
47
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 = kHP
3/22/2016
S
= solubility of gas (mol/L)
Pg
= partial pressure of gas (atm)
KH
= Henry’s Law Constant (mol/L-atm)
48
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
3/22/2016
= K h × Pgas
g 1mol 

 2.79 L × 80g 
=

 0.24 atm 




Kh
 S gas 
= 
 P 
 gas 
Kh
= 0.145 mol / L • atm
49
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?
Sgas(1) = K h × Pgas(1)
 S gas(1) 
Kh = 

P
 gas(1) 
Sgas(2) = K h × Pgas(2)
 S gas(2) 
Kh = 

P
 gas(2) 
 S gas(2)   S gas(1) 

 = 

P
P
 gas(2)   gas(1) 
 Sgas(1) 
 0.32g / L 
S ga(2) = Pgas(2) 
=
0.0063atm



P
1atm


 gas(1) 
Sgas(2) = 0.0020 g / L
3/22/2016
50
Concentration
Definitions
Concentration Term
Molarity (M)
Molality (m)
Parts by mass
Parts by volume
Mole Fraction (X)
3/22/2016
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
amount (mol) of solute
moles of solute + moles of solvent
51
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
3/22/2016
52
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 
molality(m) = 

 kgsolvent 
1mol sucrose

1.75gsucrose

180.16gsucrose
 mol  
kgsolvent = 
=

 m   0.125mol / kg H2O


1000g
kg solvent = 0.0777kg H2 O ×
= 77.7g H 2 O
1kg
3/22/2016






53
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”
3/22/2016

Vapor Pressure

Boiling Point Elevation

Freezing Point Depression

Osmotic Pressure
54
Colligative Properties of Solutions
Phase Diagram for pure solvent and for Solution
Pure Liquid
Solvent
Triple Point
Of Solvent
Solution
Pure Solid
Solvent
Triple Point
Of Solution
3/22/2016
55
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
3/22/2016

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
56
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.
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57
Vapor Pressure



3/22/2016
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
 Entropy of solution is always greater than for solvent;
thus, it has less tendency to vaporize to gain entropy
58
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
(Raoult’s Law)
Psolvent = Xsolvent x 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 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
3/22/2016
solutions give good approximation of Raoult’s law
59
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
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60
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

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The presence of each volatile component lowers the
vapor pressure of the other by making each mole fraction
less than 1
61
Example Problem

Equi-molar solution of Benzene and Toluene, both
nonelectrolyte hydrocarbons
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 = 47.6torr
Ptol = Xtol × Ptol = 0.5 × 28.4 torr = 14.2 torr

Benzene lowers the vapor pressure of Toluene and
Toluene lowers the vapor pressure of Benzene

3/22/2016
Vapor composition vs Solution composition
Xben =
Pben
47.6 torr
=
= 0.770 torr
Ptot
47.6torr + 14.2torr
Xtol =
Pben
14.2 torr
=
= 0.230 torr
Ptot
47.6 torr + 14.2 torr
Mole fractions in
vapor are different
62
Boiling Point Elevation

Boiling Point (Tb) of a liquid is the temperature at which
its vapor pressure equals the external (atmospheric
usually) pressure

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
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63
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

3/22/2016
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
64
Freezing Point Depression
Recall: The vapor pressure of a solution is lower than
the vapor pressure of the pure solvent
 The lower vapor pressure of a solution is the result of
fewer solvent molecules being able to leave the liquid
phase because of the competition with solute molecules
for space at the surface
 In solutions with nonvolatile solutes, only solvent
molecules can vaporize; thus, only solvent molecules can
solidify (freeze)
 The formation of solid solvent particles leaves the
solution more concentrated (less solvent available)
 At the freezing point of a solution, there is an equilibrium
established between solid solvent and liquid solvent
 There is also an equilibrium between the number of
particles of solvent leaving the solution and the number
of particles returning to the solution

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65
Freezing Point Depression

The vapor pressure of a pure solvent at its freezing
point, i.e., an equilibrium solution of solid and liquid
solvent, is higher than the vapor pressure of a solution
of a solute in this solvent

The vapor pressure of a solvent in a solution at its
freezing point has been lowered from its value as a pure
solvent until it equals that of the solution

Since vapor pressure is a function of temperature, the
solution, with its lower vapor pressure, will freeze at a
lower temperature than the pure solvent
3/22/2016
66
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 / kg solvent )
K f = molal freezing point depression constant
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67
The van’t Hoff Factor

Colligative properties depend on the relative number of
solute to solvent “particles”

In strong electrolytes 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
i=
3/22/2016
measured value for electrolyte solution
expected value for nonelectrolyte solution
68
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
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T  5 K c
f
f m
69
Colligative Properties of Solutions
Colligative
Property
Mathematical
Relation
1. vapor pressure
(Raoult’s Law)
2. freezing pt depression
P
solvent
4. osmotic pressure
3/22/2016
solvent

f
X
T  i K m
f
3. boiling pt elevation

= i Po

T  i K m
b
b
solute
)
)
)
  i  MRT )
70
Practice Problem
What is the boiling point of 0.0075 m aqueous calcium
chloride, CaCl2?
 or ΔTb = iK bm
ΔTb
(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
 0.0075m + 100.0o C
m
TbCaCl = 100.1o C
2
3/22/2016
71
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 


Tfsolution = -0.46o C
3/22/2016
72
Practice Problem
What is the freezing point of 0.150 g of glycerol (C3H8O3) in
20.0 g of water?
ΔTf
= Tfsolvent - Tfsolution = iK f m
(i = 1)
-Tfsolution = K f m - Tfsolvent
Tfsolution = Tfsolvent - K f m
Tfsolution
 1mol 
0.150g
 92.094g 
 oC 


o
= 0.0 C - 1.86 
×

m
 1kg 


20.0g × 

1000g


Tfsolution


 oC 
mol
o
 × 0.0814
= 0.0 C - 1.86 
kg
 mol 
 kg 


Tfsolution = -0.151o C
3/22/2016
73
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?
ΔTf
= Tfsolvent - Tfsolution = iK f m
(i = 1)
 Tfsolvent - Tfsolution 
m=

iK
f


 5.455o C - 4.880o C 
m=
 = 0.1135 mol BHT / kg Benzene
o
 1× 5.065 C / m 
MmBHT = Molar MassBHT (g / mol)
m = 0.1135
mol BHT
kg Benzene
1mol BHT  
1mol BHT 

2.500
g
×
25.00
g


M mBHT (g)  
M mBHT (g) 
=

=
1kg  
1kg Benzene
 100 g

Benzene




1000g  


MmBHT  g )
25.00 g BHT
=
= 220.3g / mol
1 mol BHT
0.1135 mol BHT
3/22/2016
(Actual M m = 220.3 g / mol)
74
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

Colloids range in size from 1 – 1000 nm
3/22/2016
75
Colloids

Colloids have tremendous surface areas, which allows
many more interactions to exert an large total adhesive
force, which attracts other particles

Many useful applications

Colloids are classified according to whether the dispersed
and dispersing substances are gases, liquids, or solids
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76
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.

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
3/22/2016
77
Osmosis
3/22/2016
78
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
• Osmotic Pressure = pressure applied to just stop osmosis
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79
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

Osmotic Pressure () is proportional to the number of
solute particles in a given volume of solution, i.e.,
Molarity (M) (moles per liter of solution)
Π
nsolute
Vsolution
or
ΠM
The Proportionality Constant is "R" times the absolute temperature (T)
Π=
3/22/2016
nsolute
RT = MRT
Vsolution
80
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)
1 pascal = 7.5006 x 10-3 torr(mm)= 9.8692 x 10-6 atm
3/22/2016
81
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
3/22/2016
82
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?
Π=
nNaCl
RT
VNaCl Soln
VNaCl Soln
M NaCl soln =
nNaCl
VNaCl Soln
n NaCl = M NaCl soln × VNaClSoln
n NaCl
 M NaCl soln × VNaCl Soln
=
RT = 
Π
Π

VNaCl Soln

 RT

mol


0.55
×
1000
L


L • atm 
o
o
L
=
0.0821
[25
C
+
273.15]
K)



o
75 atm
mol • K 

 


VNaCl Soln = 179.5 L = 180 L
3/22/2016
83
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
3/22/2016
84
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 (increase temperature (VP) to match atmospheric
pressure
b. B (decrease temperature to reduce
vapor pressure of solvent)
c. A (The movement of solvent between
solutions is from the solution of
lower concentration to the solution
of higher concentration)
3/22/2016
85
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
b. A
c. Condense “A” by ½
3/22/2016
86
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



3
)
3
molescaffeine = 0.0042525 molcaffeine
Masscaffeine = molcaffeine ×
3/22/2016
194.14g
= 0.826g
molcaffeine
87
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) = mol solute / kgsolvent
1mol
Mm
0.0368 =
1kg
8.31gethanol
1000g
0.0653g
Mm =
0.0653g
mol
1kg
0.0368
× 8.31g ×
kg
1000g
Mm = 213g / mol = MolecularWeight
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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 (XH 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 mmHg - 22.8 mmHg = 1.00 mmHG
o
o
Psolvent
- Psolvent = ΔP = Xurea × Psolvent
= 0.0425× 23.8 mmHg = 1.01 mm Hg
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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 mmHg = 3.49 mm Hg
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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)
Molarity(M) =
mol solute
Lsoln
1mol 

0.0140
g

243.5 g 
=
 = 0.0022998 M
1L
 25 ml


1000 mL 

1atm
760 mmHg
 Π 
i=
=

 MRT  0.0022998  mol  × 0.0821 L • atm × (273.15 + 25.0)o K


 L 
mol •o K
119mmHg
i = 2.78 = 3
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Equation Summary

Component Enthalpies of Heat of Solution
ΔH
= ΔH
soln
solute
ΔH

soln
= ΔH
+ ΔH
+ ΔH
solute
solvent
+ ΔH
mix
hydration
Component Enthalpies of Ionic Compound Heat of Soln
ΔH
soln
= ΔH
lattice
+ ΔH
hydr of the ions

Relating Gas Solubility to its Partial Pressure (Henry’s
Law)
Sgas = k H Pgas

Relationship between vapor pressure & mole fraction
Psolvent = Xsolvent × PoSolvent
Psolute = Xsolute × PoSolute
Ptotal = Psolvent + Psolute = (Xsolvent × PoSolvent ) + (Xsolute × PoSolute )
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Equation Summary

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
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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
mass of solute
mass of solution
Parts by volume
Mole Fraction (X)
volume of solute
volume of solution
amount (mol) of solute
amount (mol) of solute + amount (mol) of solvent
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