Solutions
Chapter 14
Types of Solution
• A solution is a homogeneous mixture of two or
more substances.
– Solvent – medium in which the solute dissolves
• Usually a liquid
• Usually the most abundant species
– 10 grams of H2O(l) in 25 grams of CH3OH(l)
– Solute – substance that is dissolved
Discuss (review) homogeneous and heterogeneous
mixtures
Spontaneity of the Dissolution
Process
• Dissolving a substance in a solvent (usually a
liquid)
– Dissolving with a reaction
• 2Na(s) + 2H2O(l)  2Na+(aq) + 2OH-(aq) + H2(g)
– Dissolving without a reaction
H 2O
C 6 H 12 O 6 ( s) 
 C 6 H 12 O 6 ( aq )
We will focus on the latter type of dissolution.
Spontaneity of the Dissolution
Process
• Two major factors determine the dissolution of
solutes
– Change in energy, Hsolution
• An _____ process (a decrease in energy) favors dissolution.
• An _____ process (an increase in energy) does not favor
dissolution.
– Change in disorder or randomness, Smixing
• ______ in disorder favors dissolution.
• ______ in disorder does not favor dissolution.
– When a substance dissolves the disorder of the system almost
always increases.
Spontaneity of the Dissolution
Process
• Heat of solution, Hsolution, primarily depends on the strength
of ____________ between solute and solvent particles. A
negative value indicates that heat is _____ favoring
dissolution. The larger the magnitude of the value, the more
dissolution is favored.
• Factors affecting Hsolution
– Solute-solute attractions
• Weak attractions favor solubility (why?)
– Solvent-solvent attractions
• Weak attractions favor solubility (why?)
– Solvent-solute attractions
• Strong attractions favor solubility (why?)
Spontaneity of the Dissolution
Process
Input of energy is required to separate solute-solute (step a) and
solvent-solvent (step b) attractions. Energy is released due to solutesolvent attractions (step c). If the amount of energy released in step c
is greater than the energy absorbed in steps a and b the process is
exothermic and favored for dissolution. What if it is less?
Spontaneity of the Dissolution
Process
• Many solids, however, will dissolve in liquids by
endothermic processes. The increase in disorder of
the system upon mixing (magnitude of Smixing) is
enough to outweigh the endothermic process.
– The solute particles go from highly ordered in a
crystalline solid to highly disordered in solution.
– Most dissolving processes involve an overall increase in
disorder.
• Discuss mixing of gases
Dissolution of Solids in Liquids
• The crystal lattice energy is the energy change
accompanying the formation of one mole of formula
units in the crystalline state from constituent particles
in the gaseous state.
– The crystal lattice energy is always ______ (i.e. the process
is exothermic).
• M+(g) + X-(g)  MX(s) + energy
– The amount of energy released increases with the strength
of attraction between particles (e.g. ions in the solid)
• Energy released increases with increasing charge density. Why?
Dissolution of Solids in Liquids
• Solute-solute attractive interactions present
in the solid must be overcome for the solid
to dissolve in the liquid.
– This can be related to the crystal lattice energy
• MX(s) + energy  M+(g) + X-(g)
• If the magnitude of the crystal lattice energy is small
only a small amount of energy is needed to start the
dissolution process.
– This can be thought of as step a in Figure 14-1.
• Is this step exothermic or endothermic?
Dissolution of Solids in Liquids
• Energy is also required to break up or expand the
solvent molecules (i.e. solvent-solvent interactions).
This can be fairly large if the solvent is water. Why?
– This can be thought of as step b in Figure 14-1.
• Energy is released, however, due to attractive solutesolvent interactions.
– This can be thought of as step c in Figure 14-1.
– Solvation is the process by which solvent molecules
interact and surround solute particles or ions.
• Hydration refers to when the solvent molecules are water.
Dissolution of Solids in Liquids
• Solvation energy is the energy change involved in the
solvation of one mole of gaseous ions.
– Process is almost always exothermic.
– Equivalent to the sum of steps b and c in Figure 14-1.
– Hydration energy is the energy change when water is the
solvent.
Hydration is highly exothermic for ionic or polar covalent
compounds. Why?
Hsolution = (solvation energy) – (crystal lattice energy)
If Hsolution is negative the dissolution process is exothermic.
Dissolution of Solids in Liquids
• A nonpolar solid such as naphthalene, C10H8, does
not dissolve in polar solvents such as H2O. It does,
however, dissolve in nonpolar solvents. Why?
– “like dissolves like”
• Let’s look at the dissolving process of a soluble salt
such as NaCl (Figure 14-2).
H 2O
NaCl ( s) 
 Na  ( aq )  Cl  ( aq )
– The individual ions become solvated (hydrated if the
solvent is water) due to electrostatic interactions between
the ions and water molecules.
– Most cations (e.g. Na+) are surrounded by 4 to 9 H2O
molecules.
Dissolution of Solids in Liquids
• An increasing charge density (charge/radius
ratio) increases the hydration energy (heat
of hydration). Table 14-1
Ion
Radius (Å)
Chg/Radius
Heat of Hydration
K+
1.33
0.75
- 351 kJ/mol
Ca 2+
0.99
2.02
- 1650 kJ/mol
Cu 2+
0.72
2.78
- 2160 kJ/mol
Al 3+
0.50
6.00
- 4750 kJ/mol
Dissolution of Solids
• An increasing charge density, however, also increases
the magnitude of the crystal lattice energy.
• For many salts that possess low-charge species (e.g.
NaCl) the hydration energy and crystal lattice energy
nearly cancel each other.
– Demo: The dissolution of NH4NO3 is endothermic. What
can you say about the heat of solvation and crystal lattice
energy? The dissolution of Ca(CH3COO)2 is exothermic.
– With large charge densities the magnitude of the crystal
lattice energy increases more than the hydration energy.
• This is the reason why the dissolution process for many solids that
contain highly charged ions (e.g. Cr2O3) is endothermic. Many are
insoluble. Look at equation.
Dissolution of Liquids in Liquids
• Miscibility is the ability of one liquid to dissolve
in another. If two liquids are miscible one liquid
completely dissolves in the other.
• Types of attractive forces to consider when
determining miscibility
– Solute-solute attractions (step a)
– Solvent-solvent attractions (step b)
– Solute-solvent attractions (step c)
When will two liquids be the most miscible?
Dissolution of Liquids in Liquids
• ‘Like dissolves like’
determines miscibility
• Polar liquids tend to
dissolve in other polar
liquids
– H2O and CH3OH
Discuss in terms of the types
of interactions.
What are some other liquids
that will dissolve in water?
Dissolution of Liquids in Liquids
• Will hexane, C6H14, dissolve in water, H2O?
Why or why not? Explain in terms of types
of interactions.
• Will hexane dissolve in gasoline which is
nonpolar? Why or why not? What are the
strengths of attractive interactions?
Dissolution of Gases in Liquids
• ‘Like dissolves like’ holds fairly well for
gases dissolving in liquids.
– Polar gases dissolve in water
HF ( g)  HF ( aq )
H 2O
– Polar gases can also react with water
HBr ( g)  H 2 O ( l)  H 3 O  ( aq )  Br  ( aq )
Dissolution of Liquids in Liquids
• Nonpolar gases (e.g. O2) dissolves to a limited extent
in H2O due to dispersion forces
– Dissolved O2 is responsible for keeping fish alive
– CO2, a nonpolar gas, dissolves in water appreciable
because it reacts with H2O
CO2(g) + H2O(l)
H2CO3(aq)
H2CO3(aq)
H+(aq) + HCO3-(aq)
HCO3-(aq)
H+(aq) + CO32-(aq)
What happens to the acidity of water when CO2(g) is dissolved?
Because gases have such weak solute-solute attractions,
gases dissolve in liquids exothermically.
Rates of Dissolution and Saturation
• Rate of dissolution of a solid increases if the size of the
solid particles is decreased (e.g. ground to a powder).
Why?
– Sugar cubes versus granulated sugar
• The amount of solid in a liquid will increase until the
rate of dissolution equals the rate of crystallization.




crystallization
dissolution
solid
dissolved particles
– The opposing processes are in dynamic equilibrium. The
solution is saturated. It is holding all the solute it can at a
given temperature.
DEMO: NaCl crystals increasing in size in a saturated
solution.
Rates of Dissolution and Saturation
• Saturated solutions have an established equilibrium
between dissolved and undissolved particles.
– NaCl(s)
Na+(aq) + Cl-(aq)
– The forward and reverse rates are equal.
• Supersaturated solutions contain higher-than-saturated
concentrations of solute. How is this possible?
– A supersaturated solution is in a ‘metastable’ state. There
needs to be mechanism to start the crystallization.
• Hot packs, sodium acetate in H2O
Effect of Temperature on
Solubility
• LeChatelier’s Principle states that a chemical
system responds in a way that best relieves the
stress or disturbance.
• Exothermic dissolution
H 2O
LiBr ( s) 
 Li  ( aq )  Br  ( aq )  48 .8 kJ / mol
• Endothermic dissolution
H 2O
KMnO 4 ( s)  43 .6 kJ / mol 
 K  ( aq )  MnO 4 ( aq )
Effect of Temperature on Solubility
• What will happen to the solubility if the temperature is changed?
The system will respond according to LeChatelier’s Principle.
• Adding heat to an exothermic process ______ dissolution.
Why?
• Adding heat to an endothermic process _____ dissolution.
Why?
– For most solids dissolving in liquids the process is
endothermic.
• Most gases dissolve in liquids by exothermic processes. What
does this mean if the temperature is increased?
Demo: Dissolve NaCH3COO. Is this process endothermic or exothermic?
What will happen if the temp. is increased? Cool the mixture. What type
of solution results?
Effect of Temperature on
Solubility
• Increasing the
temperature increases the
solubility of most solids
in H2O. Are these
processes exothermic or
endothermic?
• Na2SO4 is the only
exothermic process
Effect of Pressure on Solubility
• Changing the pressure has no appreciable
effect on the solubilities of solids or liquids
in liquids.
• Pressure changes have large effects on the
solubilities of gases in liquids.
– Carbonated beverages
– Scuba divers get the ‘bends’
Effect of Pressure on Solubility
• Henry’s Law expresses that the concentration of
dissolved gas is directly related to the pressure
of the gas above the solution.
Pgas = kCgas
– Pgas = pressure of the gas above the sollution
– k is a constant for a particular gas and solvent at a
specific T
– Cgas = the concentration of dissolved gas (molarity)
Molality and Mole Fraction
• Previously, we covered molarity and weight
precent to express concentration. What are
they?
• Molality – number of moles of solute per
kilogram of solvent
molality 
number of moles solute
number of ki log rams solvent
– What is the molality of a solution that contains
142 grams of CH3OH in 315 grams of water?
Molality and Mole Fraction
• Calculate the molality and the molarity of an
aqueous solution that is 10.0% glucose, C6H12O6.
The density of the solution is 1.04 g/mL. 10.0%
glucose solution has several medical uses. 1 mol
C6H12O6 = 180 g
• Calculate the molality of a solution that contains
7.25 g of benzoic acid C6H5COOH, in 200 mL of
benzene, C6H6. The density of benzene is 0.879
g/mL. 1 mol C6H5COOH = 122 g
Molality and Mole Fraction
• Mole fraction is the number of moles of one
component per moles of all the components.
no . mol A
XA 
( two componens )
no . mol A  no . mol B
– The sum of the mole fractions (XA + XB) = 1
What are the mole fraction for methanol and
water in the previous problem?
What are the mole fractions of glucose and water
in a 10.0% glucose solution?
Colligative Properties of Solutions
• Colligative properties depend solely on the
number of particles dissolved in the solution
and not the kinds of particles dissolved.
– Physical property of solutions
• Four kinds that will be discussed
–
–
–
–
Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
Lowering of Vapor Pressure and
Raoult’s Law
• Addition of a nonvolatile solute to a solution lowers
the vapor pressure of the solution
– Fewer solvent molecules present at the surface since
some solute molecules occupy the space.
– As a result, molecules evaporate at a slower rate
• Raoult’s Law describes this effect in ideal solutions
0
Psolvent  XsolventPsolvent
0
– Where Xsolvent represents the mole fraction, Psolvent
is the
vapor pressure of the pure solvent, and Psolvent is the
vapor pressure of the solvent in the solution.
• Figure 14-9
Lowering of Vapor Pressure and
Raoult’s Law
• The change in vapor pressure of the solvent
in solution can be expressed in terms of the
solute mole fraction.
0
Psolvent  XsolutePsolvent
Where Psolvent is the change in vapor pressure of
the solvent in the solution.
This relationship assumes ideal solutions and that
the solute is nonvolatile (i.e. has not vapor
pressure). This is Raoult’s Law.
Lowering of Vapor Pressure and
Raoult’s Law
• Determine the vapor pressure of a solution,
at 25C, that is made by dissolving 5.00
grams of sucrose, C6H12O6, in 15.0 grams of
water. The vapor pressure of pure water at
25C is 23.8 torr (Appendix E).
• Determine the vapor pressure of a 5.25
molal aqueous sucrose solution at 47C.
Determining the Vapor Pressure
of a Two-Component System
• Both components are considered as volatile and
contribute to the total vapor pressure.
– A solution of hexane and heptane
• Each component behaves as if it were pure.
Therefore, the vapor pressure would be a sum of
its components.
Ptotal = PA + PB
0
0
P

X
P
and
P

X
P
From Raoult’s Law A
B
B B
A A
Therefore, PTotal  X A PA0  X B PB0
Determining the Vapor Pressure
when a Two-Component System
• The left-hand side
corresponds to pure B
(XB=1) and the right-hand
side to pure A (XA=1).
Which is more volatile?
• Notice that the black line
is always equal to the sum
of PA and PB.
Determining the Vapor Pressure
when a Two-Component System
• At 45C, the vapor pressure of pure heptane is 112
torr and the vapor pressure of pure octane is 36
torr. The solution contains two moles of heptane
and three moles of octane. Calculate the vapor
pressure of each component (heptane and octane)
and the total vapor pressure above the solution.
What is the composition (in mole fractions) of the
vapor above the solution?
• The volume above the solution is equal to 1.50 L.
Assuming that the gases behave ideally calculate
the amount of heptane and octane in the gas phase.
Also, do Example 14-5 for practice.
Boiling Point Elevation
• The boiling point of a liquid is the temperature at
which its _____ ______ equals the external pressure.
It was stated with Raoult’s law that the addition of a
nonvolatile solute decreases the vapor pressure.
Therefore, a _____ temperature must be acquired to
cause the liquid to boil (vapor pressure equals
atmospheric pressure).
• The amount of boiling point elevation depends on the
number of moles of solute particles dissolved in the
solvent.
Boiling Point Elevation
• The boiling point change can be expressed as
Tb = Kbm
– Tb is the change in boiling point (add)
– Kb is the molal boiling point constant (Table 14-2)
• Kb corresponds to the change in boiling point by a onemolal solution
– m is the molality of the solution
What is the normal boiling point of a 2.50 m
glucose, C6H12O6, solution?
Boiling Point Elevation
• Do this later. Predict the boiling point
elevation if 33.0 grams of NaCl is added to
100 grams of water? Note: Calculate
moles of solute particles not moles of
solute.
Freezing Point Depression
• Addition of a nonelectrolyte lowers the freezing
point according to the following expression
Tf=Kfm
– Tf is the change in the freezing point (subtract)
– Kf is the the molal freezing point depression constant
(Table 14-2)
– m is the molality of the solution
Upon freezing, the solvent solidifies as a pure substance.
Solute molecules make it more difficult for the solvent
molecules to come together and freeze. A lower
temperature must be acquired to freeze the solvent.
Freezing Point Depression
• Calculate the freezing point of a 2.50 m aqueous
glucose solution.
• CaCl2(s) is commonly added to melt snow on
roads at lower temperatures. What is the freezing
point when 20.0 grams of CaCl2 is dissolved in
100 grams of H2O at 20C? For comparative
purposes, calculate the lowering of the freezing
point when 20.0 grams of NaCl is dissolved in 100
grams of H2O at the same temperature.
• Antifreeze (HOCH2CH2OH) addition to water is
another example of freezing point depression.
Changes in Boiling and Freezing
Point Temperatures
• The relationships for
calculating the change in
freezing and boiling
points are very similar.
– It is essentially the same
effect. The difference is in
the ‘size’ or magnitude of
the effect which is
indicated by the constants,
Kf and Kb.
Determination of Molecular Weight
from Freezing Point Depression or
Boiling Point Elevation
• These colligative properties (especially
freezing point depression) can be used to
determine molecular weight of a solute.
– Kf and the amount of solvent in kilograms is
usually known. The freezing point lowering
was measured. Therefore, the molality can be
determined, and the moles of solute can be
calculated. The molecular weight can then be
determined.
Determination of Molecular Weight
from Freezing Point Depression or
Boiling Point Elevation
• A 37.0 g sample of a new covalent compound, a
nonelectrolyte, was dissolved in 200 g of water.
The resulting solution froze at -5.580C. What is
the molecular weight of the compound?
• Either ethylene glycol (C2H6O2) or propylene
glycol (C3H8O2) can be used to make an antifreeze
solution. A 15.0 gram sample of either ethylene
glycol or propylene glycol dissolved in 50.0 grams
of water lowers the freezing point to –7.33C.
Which glycol is added to the solution? The
solutions behave ideally at these concentrations.
Colligative Properties and
Dissociation of Electrolytes
• Since a colligative property only depends on the
‘number’ of solute particles in a given mass,
electrolytes have a larger effect on boiling point
elevation and freezing point depression.
– How many moles of solute particles will be
produced from one mole of sucrose (C6H12O6)?
How many from 1 mole of MgCl2?
• As a result, the boiling point elevation and freezing point
depression are larger for a given molar quantity of a
typical electrolyte.
Colligative Properties and
Dissociation of Electrolytes
• Solute particles, however, are
not randomly distributed in an
ionic solution. This causes the
boiling point elevation and
freezing point depression to be
not as large.
– Ionic solutions behave nonideally.
The ions start to undergo
association, and the number of
solute particles decreases. The
effective molality, therefore, is
reduced.
• Especially noticeable at higher
concentrations
Colligative Properties and
Dissociation of Electrolytes
• Due to association in ionic solutions, the
dissociation (or ionization) is reduced. The extent
of dissociation is measured by the van’t Hoff
factor, i.
i
Tf ( actual )
Tf ( if nonelectro lyte)
K f meffective meffective


K f mstated
mstated
– i has an ideal value of 2 for NaCl
– What is the ideal value for i with CaCl2?
Look at Table 14-3 at values for i and notice that the
actual values of i are closer to the ideal values of i at
lower concentrations.
Colligative Properties and
Dissociation of Electrolytes
• The freezing point of 0.0100 m NaCl
solution is -0.0360oC. Calculate the van’t
Hoff factor and apparent percent
dissociation of NaCl in this aqueous
solution.
• A 0.0500 m acetic acid solution freezes at 0.09480C. Calculate the percent ionization
of CH3COOH in this solution.
Osmostic Pressure
• Osmosis is the net flow of solvent between
two solutions separated by a semipermeable
membrane.
– Solvent will travel from lower concentration
solutions to higher concentration solutions
• Semipermeable membranges
– Skin, Saran wrap, and cells
Osmotic Pressure
Semipermeable membrane
sugar dissolved
HO
in water 2
H2O
H2O
H2O
H2O
H2O
H2O
net2solvent
flow
O
The solvent molecules will pass through the semipermeable
membrane into the more concentrated solution. The sugar
cannot pass through the membrane. The solvent passes through
the semipermeable membrane into the more concentrated
solution at a faster rate.
Osmotic Pressure
• Solvent molecules will pass through to the more
concentrated solution until the pressure is
sufficient to force molecules back through the
membrane at the same rate. The pressure exerted
to make the rates through the membrane equal is
called the osmotic pressure of the solution.
– Look at Figure 14-6a
• Osmotic pressure depends on the number, and not
the kind, of solute particles in solution.
– Colligative property
Osmotic Pressure
• Osmotic pressure follows the equation

–
–
–
–
nRT
or MRT
V
n = number of moles
R is the gas constant (0.082057 atmL/molK)
T is the temperature
M = molarity
• Osmotic pressure can be very significant.
– 1 M sugar solution has an osmotic pressure of
22.4 atm or 330 p.s.i.
Osmotic Pressure
• What osmotic pressure would a 1.25 molal sucrose
solution exhibit at 25C? The density of the
solution is 1.34 g/mL.
• A 1.00 g sample of a biological material was
dissolved in enough water to give 100 mL of
solution. The osmotic pressure of the solution was
2.80 torr at 250C. Calculate the molarity and
approximate molecular weight of the material.
Colloids
• In a solution, the solute particles do not settle upon
standing. Additionally, the mixture is at the molecular
or ionic level.
• Colloids (or colloidal dispersions) are mixtures that
have particle sizes between solutions and suspensions
(Table 14-5). The different components do not
separate upon standing.
– Fog, smoke, paint, and milk are examples
• The different components of a suspension do separate
upon standing.
Colloids
• The Tyndall effect
– Colloidal particles will scatter light due to their
size
• DEMO: Laser through a container filled with smoke
• The Adsorption Phenomenon
– Colloids have very large surface areas and
exhibit interesting surface chemistries
– They interact strongly with substances near their
surfaces.
Hydrophilic and Hydrophobic
Colloids
• Hydrophilic colloids (“water loving”)
– These colloids are polar (or possess a polar component)
that will enable them to be well dispersed in polar
solvents such as water.
– Examples are blood plasmas and proteins
• Hydrophobic colloids (“water hating”)
– These colloids are nonpolar and will not dissolve in polar
solvents such as water. These colloids require
‘emulsifiers’ which coat the colloids to enable dispersion
in a polar solvent such as water.
• Emulsifier agents coats the particles of a dispersed phase and
prevents coagulation (i.e. coming together of colloidal particles.
Hydrophobic Colloids and
Emulsifiers
• Most oils and greases are long-chain hydrocarbons
that are essentially nonpolar. Will they dissolve
water?
• Soaps (and detergents) are excellent emulsifying
agents. They are long-chain fatty acids that
contain a polar “head” and a nonpolar “tail”. The
nonpolar “tail” is attracted to the oil or grease and
the polar “head” is compatible with the polar
solvent (e.g. water).
Hydrophobic Colloids and
Emulsifiers
O
C C
nonpolar tail
hydrophobic portion
C C
O-Na+
C C
C C
C C
C C
C C
C C
C C
polar (ionic) head
hydrophilic portion
These emulsifers (here it is sodium stearate) coat or
surround the entire nonpolar particle so that it will
become dispersed in the polar solvent.
Hydrophobic Colloids and
Emulsifiers
Look at Figure 14-21 in book.
DEMO: Sulfur on surface of water. Add soap.