We just finished exploring the properties
of pure gases, liquids, and solids
 However, the matter that we encounter
in our daily lives are frequently mixtures
 Now, we examine homogeneous
mixtures, also referred to as solutions


Solutions can be solids, liquids, or gases
› Solid solutions – dental filings, 14K gold,
sterling silver
› Liquid solutions – saline, vinegar, sugar water
› Gas solutions – the atmosphere, anesthesia
gases

Liquids are most common, so we will
focus our attention on them

A solution consists of two component types:
› Solvent - component in the greater
concentration
› Solute- component in the lesser amount
(there may be more than one)

In a solution:
› The solute can’t be filtered out
› The solute always stays mixed
› Particles are always in motion
› Volumes may not be additive
› A solution will have different properties than the solvent

Molarity
M=
amount mol solute
volume L of solution

Parts by Mass
mass of solute
mass of solution

Mass percent
mass of solute
× 100
mass of solution

Mole fraction
amount mol of solute
amount mol of solute + amount mol of solvent

Molality

Parts by Volume
m=
amount mol solute
amount kg of solvent
volume of solute
volume of solution

The ability of substances to form solutions
depends on two factors:
› The natural tendency of substances to mix
and spread into larger volumes when not
restrained in some way
› Types of intermolecular interactions involved
in the solution process

Three kinds of intermolecular interactions are involved in
solution formation:
› Solute-solute
 Occur between solute particles
 Must be overcome to disperse the solute particles through the solvent
› Solvent-solvent
 Occur between solvent particles
 Must be overcome to make room for the solute particles in the solvent
› Solvent-solute
 Occur as the particles mix

The extent to which one substance is able to dissolve in
another depends on the relative magnitudes of these three
types of interactions
› Solutions form when the magnitudes of the solvent-solute
interactions are either comparable to or greater than the solutesolute and solvent-solvent interactions

Cardinal rule of solubility:
LIKE DISSOLVES LIKE
Substances with similar types of intermolecular forces
dissolve in each other


Polar solvent is used to dissolve polar or ionic solute
Nonpolar solvent is used to dissolve a nonpolar solute
WHY is this the case?
Let’s take a more in-depth look!

Assume that the formation of a liquid
solution takes place in three distinct
steps:
› Separation of solute molecules
› Separation of solvent molecules
› Formation of solute-solvent interactions

Each step is associated with an enthalpy change:
› Separation of solute molecules
ΔHsolute → ENDOTHERMIC
› Separation of solvent molecules
ΔHsolvent → ENDOTHERMIC
› These steps are always endothermic because attractive
forces must be overcome
› Formation of solute-solvent interactions
ΔHmix → EXOTHERMIC


This step is always exothermic because attractive forces
are forming between the solute and solvent particles
We can define the enthalpy (heat) change in the
solution process as:
∆Hsoln = ∆𝐇𝐬𝐨𝐥𝐮𝐭𝐞 + ∆𝐇𝐬𝐨𝐥𝐯𝐞𝐧𝐭 + ∆𝐇𝐦𝐢𝐱

ΔHsoln can either be positive or negative
depending on the intermolecular forces
present
Polar solute, polar
solvent
Nonpolar solute, polar
solvent
Nonpolar solute,
nonpolar solvent
Polar solute, nonpolar
solvent



If ΔHsoln is small and positive, a solution will still form because of
entropy
This is because there are many more ways for them to become
mixed than there is for them to stay separate
Processes that require large amounts of energy tend not to
occur
As a solid solute begins to dissolve in a
solvent, the concentration of solute
particles in solution increases
 Eventually, no more solute can dissolve in
solution and thus, a dynamic equilibrium is
established between two opposing
processes:

› Dissolution
› Crystallization (precipitation)

This equilibrium is also known as saturated
solution
•
•
No ions are
initially present
•
As the solid dissolves,
the concentration of
ions increase until
equilibrium
is
established
•
The solution is then
saturated – no more
solid forms
The rate of solid
dissolution is equal to
solid
formation
(precipitation)
–
EQUILIBRIUM
(saturation)!

The maximum amount of solute needed to form a
saturated solution in a given quantity of solvent at
a specified temperature is known as the solubility

“Soluble” is often defined as 3 grams of solid dissolving in
100 mL
If we dissolve less solute than the amount needed
to form a saturated solution, the solution is
unsaturated
 Under suitable conditions, it is possible to form very
unstable solutions that contain a greater amount of
solute needed to form a saturated solution
› Called supersaturated solutions

Solubility is favored if solute and solvent
have similar molecular structures and
thus, IMFs
 The stronger the attractions between
solute and solvent molecules, the
greater the solubility of the solute in that
solvent


Increased temperature
increases RATE at which
a solid dissolves NOT
amount of solid
› Note that while most
substances become
more soluble in water
with increasing
temperature, cerium
sulfate becomes less
soluble

The solubility of gases in
water decreases with
increasing temperature
Solubility
(g/100ml water)
300
SO2
KCl
glycine
NaBr
KNO3
sucrose
200
100
0
0
20
40
60
80
Temperature (oC)
100

Using the graph on the previous slide, how
does the solubility of KCl at 80°C compare with
that of NaBr at the same temperature?

Using the graph on the NEXT slide, how much
solute will crystallize out of solution if a warm
solution containing 150 g KNO3 in 100 g of
water is cooled to 50°C?

Using the graph on the NEXT slide, how much
more solid can be dissolved in solution if 50 g of
KNO3 in 100 g of water is warmed to a
temperature of 90°C?
100 g will dissolve at 50oC,
so 150-100 = 50g
50g of KNO3 crystallizes out
Solubility
(g/100ml water)
300
200
100
KNO3
0
0
20
40
60
80
Temperature (oC)
100
210g will dissolve at 90oC,
so 210-50 = 160g,
160 more g of KNO3 dissolves
Solubility
(g/100ml water)
300
200
100
KNO3
0
0
20
40
60
80
Temperature (oC)
100


Little effect on solids and
liquids due to
incompressibility
Greatly increase the
solubility of a gas
› Gas enters the solution at
a higher rate than it
leaves

Henry’s law gives the
relationship between gas
pressure and
concentration of dissolved
gas

At constant temperature, the solubility
of a gas is directly proportional to the
pressure of the gas above the solution
This law is accurate to
within 1-3% for slightly
soluble gases and
pressures up to one
atmosphere
Solubility
(g/100g water)
Sg = kPg
0.010
O2
0.005
N2
He
0.000
0
1
Pressure (atm)
2
Substance
Temperature
oC
Solubility
g/100 mL water
NaCl (s)
100
39.12
PbCl2 (s)
100
3.34
AgCl (s)
100
0.0021
CH3CH2OH (l)
0 -100
Infinity
CH3CH2OCH2CH3 (l)
15
8.43
O2 (g)
60
0.0023
CO2 (g)
40
0.097
SO2 (g)
40
5.41