Chapter 12: Solutions and Their
Properties
Renee Y. Becker
Valencia Community College
Introduction
1. A mixture is any intimate combination of two or more pure
substances
2. Can be classified as heterogeneous or homogeneous
Heterogeneous
-The mixing of components is visually
nonuniform and have regions of different
composition
Homogenous
-Mixing is uniform, same composition
throughout
-Can be classified according to the size of their
particles as either solutions or colloids
Solutions
Solution
1. Homogeneous mixtures
2. Contain particles with diameters in the range of
0.1–2 nm
3. Transparent but may be colored
4. Do not separate on standing
Colloids
1. Milk & fog
2. Diameters 2-500 nm
3. Do not separate on standing
Types of Solutions
Solution Formation
Solute
• Dissolved substance, or smaller quantity substance
Solvent
• Liquid dissolved in, larger quantity substance
Saturated solution
• Contains the maximum amount of solute that will
dissolve in a given solvent.
Solution Formation
Unsaturated
Contains less solute than a solvent has the capacity
to dissolve.
Supersaturated
Contains more solute than would be present in a
saturated solution.
Crystallization
The process in which dissolved solute comes out of
the solution and forms crystals.
Energy Changes and the Solution Process
Three Types of interactions
1.
Solvent-solvent
2.
Solvent-solute
3.
Solute-solute
“Like dissolves like”
solutions will form when three types of
interactions are similar in kind and magnitude
Energy Changes and the Solution Process
Example NaCl and water:
Ionic solid NaCl dissolve in polar solvents like water
because the strong ion-dipole attractions between
Na+ and Cl- ions and polar water molecules are
similar in magnitude to the strong dipole-dipole
attractions between water molecules and to the
strong ion-ion attractions between Na+ and Cl- ions
Example Oil and water
Oil does not dissolve in water because the two liquids
have different kinds of intermolecular forces. Oil is
not polar or an ionic solvent
Energy Changes and the Solution Process
NaCl in H2O
1. Ions that are less tightly held because of their position at a corner or an
edge of the crystal are exposed to water molecules
2. Water molecules will collide with the NaCl until an ion breaks free
3. More water molecules then cluster around the ion, stabilizing it by iondipole attractions
4. The water molecules attack the weak part of the crystal until it is dissolved
5. Ions in solution are said to be solvated they are surrounded and stabilized
by an ordered shell of solvent molecules
Energy Changes and the Solution Process
G, Free energy change
1. If G is negative the process is spontaneous, and the
substance is dissolved
2. If G is positive the process is non-spontaneous, the
substance is not dissolved
3. G = H -TS
H, enthalpy, heat flow in or out of the system,
Hsoln heat of solution
S, entropy, disorder, Ssoln entropy of solution
Energy Changes and the Solution Process
Energy Changes and the Solution Process
Ssoln Entropy of Solution
Usually a positive number because when you dissolve
something you are increasing disorder
Hsoln Heat of Solution
1. Harder to predict because it could be exothermic
(- Hsoln) or endothermic (+Hsoln)
2. The value of the heat of solution for a substance
results from an interplay of the three kinds of
interactions
Energy Changes and the Solution Process
1) Solvent-solvent interactions: Energy is required
(endothermic) to overcome intermolecular forces
between solvent molecules because the molecules
must be separated and pushed apart to make room
for solute particles
2) Solute-solute interactions: Energy is required
(endothermic) to overcome interactions holding
solute particles together in a crystal. For an ionic
solid, this is the lattice energy. Substances with
higher lattice energies therefore tend to be less
soluble than substances with lower lattice energies.
Energy Changes and the Solution Process
3)
Solvent-solute interactions: Energy is released
(exothermic) when solvent molecules cluster around
solute particles and solvate them. For ionic
substances in water, the amount of hydration energy
released is generally greater for smaller cations than
for larger ones because water molecules can
approach the positive nuclei of smaller ions more
closely and thus bind more tightly. Hydration
energy generally increases as the charge on the ion
increases.
Energy Changes and the Solution Process
Exothermic -Hsoln
The solute–solvent
interactions are stronger
than solute–solute or
solvent–solvent.
Favorable process
Exothermic rxn.
Energy Changes and the Solution Process
Endothermic +Hsoln
The solute–solvent
interactions are
weaker than
solute–solute or
solvent–solvent.
Unfavorable
process.
Endothermic rxn
Energy Changes and the Solution Process
• Hydration
– The attraction of ions for water molecules
• Hydration Energy
– The energy associated with the attraction
between ions and water molecules
• Lattice energy
– The energy holding ions together in a crystal
lattice
Example 1
Arrange the following in order of their
expected increasing solubility in water:
Br2, KBr, C7H8
Example 2
Which would you expect to have the larger (more
negative) hydration energy?
1. Na+
1. Mg2+
2. Cs+
2. Na+
3. Li+
3. Li+
4. Rb+
Units of Concentration
Molarity (M)
M = mole of solute / Liter of solution
Molality (m)
m = moles of solute/mass of solvent (kg)
Mole Fraction (x)
X = mole of component / total moles
Units of Concentration
Mass Percent (mass %)
Mass % =
(mass of component / total mass of sol’n) x 100%
Parts per million, ppm =
(mass of component / total mass of solution) x 106
Parts per billion, ppb =
(mass of component / total mass of solution) x 109
Example 3
What is the mass % concentration of a saline sol’n
prepared by dissolving 1.00 mol of NaCl in 1.00
L of water? DensityH2O=1.00 g/mL MMNaCl =
58.443 g/mol
Example 4
Assuming that seawater is an aqueous solution of
NaCl what is its molarity? The density of
seawater is 1.025 g/mL at 20C and the NaCl
concentration is 3.50 mass %
• Assume 1 L to make easier, 1000 mL
Example 5
What is the molality of a solution prepared by
dissolving 0.385 g of cholesterol, C27H46O in 40.0
g of chloroform, CHCl3? What is the mole
fraction of cholesterol in the solution?
Example 6
What mass in grams of a 0.500 m solution of sodium acetate,
CH3CO2Na, in water would you use to obtain 0.150 mol
of sodium acetate?
• Assume 1 kg
Example 7
The density at 20°C of a 0.258 m solution of
glucose in water is 1.0173 g/mL and the molar
mass of glucose is 180.2 g/mol. What is the
molarity of the solution?
• Assume 1 kg
Example 8
The density at 20°C of a 0.500 M solution of acetic
acid in water is 1.0042 g/mL. What is the
concentration of this solution in molality? The
molar mass of acetic acid, CH3CO2H, is 60.05
g/mol.
• Assume 1 L
Some Factors Affecting Solubility
Solubility
The amount of solute per unit of solvent needed to form
a saturated solution
Miscible
Mutually soluble in all proportions
Effect of Temperature on Solubility
1. Most solid substances become more soluble as
temperature rises
2. Most gases become less soluble as temperature rises
Some Factors Affecting Solubility
Effect of Pressure on Solubility
1. No effect on liquids or solids
2. The solubility of a gas in a liquid at a given
temperature is directly proportional to the partial
pressure of the gas over the solution, @ 25°C
Henry’s Law
solubility = k x P
k = constant characteristic of specific gas, mol/L atm
P = partial pressure of the gas over the sol’n
Some Factors Affecting Solubility
a) Equal numbers of gas molecules escaping liquid and
returning to liquid
b) Increase pressure, increase # of gas molecules
returning to liquid, solubility increases
c) A new equilibrium is reached, where the #’s of
escaping = # of returning
Example 9
Which of the following will become less
soluble in water as the temperature is
increased?
1) NaOH(s)
2) CO2(g)
Example 10
The solubility of CO2 in water is 3.2 x 10-2 M @
25°C and 1 atm pressure. What is the Henry’sLaw constant for CO2 in mol/L atm?
solubility = k x P
Physical Behavior of Solutions: Colligative Properties
Colligative properties
Properties that depend on the amount of a
dissolved solute but not its chemical identity
There are four main colligative properties:
1. Vapor pressure lowering
2. Freezing point depression
3. Boiling point elevation
4. Osmotic pressure
Physical Behavior of Solutions: Colligative Properties
In comparing the properties of a pure solvent with those
of a solution…
1. Vapor pressure of sol’n is lower
2. Boiling point of sol’n is higher
3. Freezing point of sol’n is lower
4. Osmosis, the migration of solvent molecules through
a semipermeable membrane, occurs when solvent
and solution are separated by the membrane
Vapor-pressure Lowering of Solutions: Raoult’s Law
1.
A liquid in a closed container is in equilibrium with its vapor
and that the amount of pressure exerted by the vapor is called
the vapor pressure.
2.
When you compare the vapor pressure of a pure solvent with
that of a solution at the same temperature the two values are
different.
3.
If the solute is nonvolatile and has no appreciable vapor
pressure of its own (solid dissolved) the vapor pressure of the
solution is always lower that that of the pure solvent.
4.
If the solute is volatile and has a significant vapor pressure (2
liquids) the vapor pressure of the mixture is intermediate
between the vapor pressures of the two pure liquids.
Solutions with a Nonvolatile Solute
When solute molecules displace solvent molecules at the
surface, the vapor pressure drops since fewer gas
molecules are needed to equalize the escape rate and
capture rates at the liquid surface.
Solutions with a Nonvolatile Solute!!!
Raoult’s Law
Psoln = Psolv · Xsolv
Psoln = vapor pressure of the solution
Psolv = vapor pressure of the pure solvent
Xsolv = mole fraction of the solvent in the
solution
Raoult’s Law applies to only Ideal solutions
1.
Law works best when solute concentrations are low and
when solute and solvent particles have similar
intermolecular forces.
2.
If intermolecular forces between solute particles and
solvent molecules are weaker than solvent molecules
alone, solvent molecules are less tightly held, vapor
pressure is higher than Raoult predicts
3.
If intermolecular forces between solute and solvent are
stronger than solvent alone, solvent molecules are more
tightly held and the vapor pressure is lower than
predicted
4.
No Van’t Hoff factor!!!
Example 11
What is the vapor pressure (in mm Hg) of a solution
prepared by dissolving 5.00 g of benzoic acid
(C7H6O2) in 100.00 g of ethyl alcohol (C2H6O) at
35°C? The vapor pressure of the pure ethyl
alcohol at 35°C is 100.5 mm Hg
Psoln = Psolv · Xsolv
Example 11
Solutions with a Nonvolatile Solute
Close-up view of part of the vapor pressure curve for
a pure solvent and a solution of a nonvolatile
solute. Which curve represents the pure solvent,
and which the solution?
Why?
• The lower vapor pressure of a sol’n relative to that of a
pure solvent is due to the difference in their entropies of
vaporization, Svap. Because the entropy of the solvent
in a sol’n is higher to begin with, Svap is smaller for the
sol’n than for the pure solvent. As a result vaporization
of the solvent from the sol’n is less favored (less
negative Gvap), and the vapor pressure of the solution is
lower.
Solutions with a Volatile Solute!!
Ptotal = PA + PB
Ptotal = (P°A · XA) + (P°B · XB)
P°A = vapor pressure of pure A
XA = mole fraction of A
P°B = vapor pressure of pure B
XB = mole fraction of B
Ptotal should be intermediate to A & B
Close-up view of part of the vapor pressure curves for
two pure liquids and a mixture of the two. Which
curves represent the mixture?
1) Red
2) Green
3) blue
Example 12
What is the vapor pressure ( in mm Hg) of a sol’n
prepared by dissolving 25.0 g of ethyl alcohol
(C2H5OH) in 100.0 g of water at 25°C? The vapor
pressure of pure water is 23.8 mm Hg and the
vapor pressure of ethyl alcohol is 61.2 mm Hg at
25°C
Example 13
The following phase
diagram shows part of the
vapor pressure curves for
a pure liquid (green
curve) and a solution (red
curve) of the first liquid
with a second volatile
liquid (not shown)
a)
b)
Is the boiling point of the second pure liquid higher or lower
than that of the first liquid?
On the diagram where is the approximate position of the
second pure liquid?
Boiling Point Elevation and Freezing Point Depression of Solutions
Boiling Point Elevation and Freezing Point Depression of Solutions
1. Red line is pure solvent
2. Green line solution of nonvolatile solute
3. Vapor pressure of sol’n is lower
4. Temp at which vapor pressure = 1 atm for sol’n is higher
5. Boiling point of sol’n is higher by Tb
6. Liquid/vapor phase transition line is lower for sol’n
7. Triple point temp is lower for sol’n
8. Solid/liquid phase transition has shifted to a lower temp.
9. The freezing point of the sol’n is lower by Tf
Boiling Point Elevation and Freezing Point Depression of Solutions
 Tb = K b · m
 Tf = K f · m
Kb = molal boiling-point elevation constant
Kf = molal freezing-point depression constant
m = molality
NO Van’t Hoff Factor!!
Boiling Point Elevation and Freezing Point Depression of Solutions
The higher boiling point of a solution relative to that of a pure
solvent is due to a difference in their entropies of
vaporization, Svap. Because the solvent in a solution has a
higher entropy to begin with, Svap is smaller for the solution
than for the pure solvent. As a result, the boiling point of the
solution Tb is higher than that of the pure solvent.
Boiling Point Elevation and Freezing Point Depression of Solutions
The lower freezing point of a solution relative to that of a
pure solvent is due to a difference in their entropies of
fusion, Sfusion. Because the solvent in a solution has a
higher entropy level to begin with, Sfusion is larger for the
solution than for the pure solvent. As a result the freezing
point of the solution Tf is lower than that of the pure
solvent.
Example 14
What is the normal boiling point in °C of a solution
prepared by dissolving 1.50 g of aspirin (C9H8O4)
in 75.00 g of chloroform (CHCl3)? The normal
boiling point of chloroform is 61.7 °C and Kb of
chloroform is 3.63 °C kg/mol
Osmosis and Osmotic Pressure
1. Semipermeable membranes allow water or other
small molecules to pass through, but they block
the passage of large solute molecules or ions.
2. When a solution and a pure solvent are separated
by the right kind of semipermeable membrane,
solvent molecules pass through the membrane in
a process known as osmosis.
3.
Passage of solvent through the membrane takes
place in both directions
Osmosis and Osmotic Pressure
4.
Passage from the pure solvent side to the
solution side is more favored and faster.
5. The amount of liquid on the pure solvent side
decreases
6. The amount of liquid on the solution side
increases
7. The concentration of the solution decreases
Osmosis and Osmotic Pressure
Osmosis and Osmotic Pressure
Osmotic Pressure
1. The amount of pressure necessary to achieve
equilibrium
2.  = MRT
 = osmotic pressure
M = molarity
R = gas constant, .08206 L atm/K mol
T = temperature in kelvins
Example 15
What osmotic pressure in atm would you expect for
a solution of 0.125 M C6H12O6 that is separated
from pure water by a semipermeable membrane at
310 K?
 = MRT
Example 16
A solution of unknown substance in water at
300 K gives rise to an osmotic pressure of
3.85 atm. What is the molarity of the
solution?
 = MRT
M = /RT
Some uses of Colligative Properties
1. The most important use of colligative properties
in the laboratory is for determining the
molecular mass of an unknown substance.
2. Any of the four colligative properties can be
used but using osmotic pressure gives the most
accurate results
Example 17
• What is the molar mass of sucrose if a
solution prepared by dissolving 0.822 g of
sucrose in water and diluting to a volume of
300.0 mL has an osmotic pressure of 149
mm Hg at 298 K?
If you have to calculate the molar mass of a
compound which of the following will give you
the most accurate results?
1)
2)
3)
4)
Osmotic pressure
Vapor pressure lowering
Boiling point elevation
Freezing point depression