Chapter 13
Notes
Solutions
13.1: The Nature of Solutions
1. Solution: A homogeneous mixture of 2 or
more substances in a single physical state
(visibly the same throughout.)
A. Properties of Solutions:
1. small particles
2. particles are evenly distributed
3. particles won’t separate when allowed to
stand
4. every substance has it’s own solubility
B. Vocabulary:
Solute: substance that is dissolved
(usually  50% of solution)
solvent
solute
Solvent: substance that does the dissolving
(usually  50% of the solution)
Soluble: means something can be dissolved in a
solvent. Ex) salt in H2O
Insoluble: means something cannot be dissolved
in a solvent. Ex) iodine in water
C.Types of Solutions:
1. Solid Solutions (solid solute)
Alloys: solid solution that contains two or more
metals.
Ex) steel – 99% iron & 1% carbon, gold jewelry, brass
Advantages of alloys over pure metals: stronger,
corrosion resistant, cheaper, attractive color
2. Gaseous Solutions: (gaseous solute) The
properties of gaseous solutions depend on
the properties of its compounds. Ex) air
3. Liquid Solutions: (liquid solute)
Miscible liquids: Two liquids that are able to
mix together in any proportion.
Ex) alcohol & water
Immiscible liquids: Two liquids that are
unable to mix together in any proportion.
Ex) oil & water (lava lamps)
Aqueous Solutions:
Solutions with H2O as the
solvent. Because water can
dissolve so many things, it
is called the universal
solvent.
Tincture: Solutions with
alcohol as the solvent.
Ex) iodine & alcohol
Electrolyte: ionic compounds dissolve in water
to form ions that can conduct an electric
current. Ex) salt & water
Nonelectrolyte
Electrolyte
Nonelectrolyte: molecular (covalent)
compounds dissolve in water to form
molecules that cannot conduct an electric
current. Ex) sugar & water
D. Separating Solutions: Done through a change
of state - evaporation or condensation.
(distillation)
Type of Solution
Gas in Gas
Air
Specific Solute IN Specific Solvent
O2
N2
CO2
H2O
hydrocarbons
Pt
H2O
air
lemon juice
water
mercury
silver
carbon
air
sugar
water
carbon
iron
Gas in Liquid
Soda
Gas in Solid
catalytic converter
Liquid in Gas
fog
Liquid in Liquid
Lemonade
Liquid in Solid
teeth fillings
Solid in Gas
smoke
Solid in Liquid
Kool-Aid
Solid in Solid
steel
13.2: Concentrations of Solutions
A. Terms Used for Concentration: The amount of
solute in a given amount of solvent or solution.
1. Concentrated: When a solution has a large amount of
solute compared to solvent.
2. Dilute: When a solution has a small amount of solute
compared to solvent.
Which solution is more
concentrated?
3. Saturated: A solution
is saturated if it
contains as much solute
as can possibly be
dissolved under existing
conditions of
temperature and
pressure. (gases only)
4. Unsaturated: Has less
than the maximum
amount that can be
dissolved.
5. Supersaturated: Has more than the
maximum amount that can be dissolved.
Solubility Graph for NaNO3
At 20oC, a saturated solution
contains how many grams of
NaNO3 in 100g of water?
What kind of solution is
formed when 90g
NaNO3 is dissolved in
100g water at 30oC?
unsaturated
What kind of solution is
formed when 120g
NaNO3 is dissolved in
100g water at 40oC?
supersaturated
Saturated sol’n
170
160
150
140
Supersaturated
solution
130
120
Solubility ( g/100 g water )
90 g
What is the solubility
at 70oC?
135 g/100 g water
180
110
100
90
80
70
Unsaturated solution
60
50
40
30
20
10
0
0
10
20
30
40
50
60
70
Temperature (deg C)
80
90
100
110
The Formation of Solutions:
B. How a Solution forms:
1. Process of dissolving takes place at the surface
of the solute.
Dissolving an Ionic Solute
Ionic solutes break up into
their ions in water
2. The interaction between the solute and solvent
to allow ions to separate is called solvation.
This interaction is called hydration when water
is the solvent.
3. Energy is absorbed when the bonds between
solute and solvent break. (endo)
4. Energy is released when bonds between solute
and solvent form. (exo)
C. Solubility
1. Solubility is the amount of solute that will
dissolve in a given amount of solvent at a
certain temperature (pressure.)
2. Factors that affect solubility:
A. Nature of Solute and Solvent:
General rule:
like dissolves like
polar
in
polar
nonpolar in
nonpolar
Example:
iodine in CCl4 (both nonpolar)
salt in water (both polar – ionic)
Solubility of Polar Substances
• Ethanol is soluble in water because of the polar OH bond
• “Like Dissolves Like”
Why is solid sugar soluble in water?
Some substances are Insoluble in
Water
• Nonpolar oil does not
interact with polar
water.
• Water-water hydrogen
bonds keep the water
from mixing with the
nonpolar molecules.
B. Temperature:
1. As temperature increases, the solubility of gases
decreases.
2. For an endothermic solid: If the temperature drops
when the solute and solvent are mixed, raising the
temperature will increase solubility.
3. For an exothermic solid: If the temperature rises
when the solute and solvent are mixed, raising the
temperature will decrease solubility.
Gases are less soluble at high
temperatures than at low temperatures.
C. Pressure:
1. As pressure increases over a gas, solubility
increases.
3. Factors that affect the
Rate of Dissolving:
A. surface area (crushing)
B. stirring
C. temperature
D. All three of these
factors affect the area
over which the solvent
can come into contact
with the solute.
Which solution is more
concentrated?
• In chemistry, the units used for concentration are
called molarity.
Which solution has a higher
concentration?
D. Molarity: (M)
molarity =
moles of solute
liters of solution
M =
n
V
units =
mol
L
Ex #1) What is the molarity of a solution formed
by mixing 10.0 g of sulfuric acid with enough
water to make 100.0 mL of solution?
10.0 g H 2SO4
 1 mol H SO
2
4

 98.1 g H 2SO4


 = 0.102 mol H 2SO4


0.102 mol
M =
 1.02 M
0.1000 L H 2 O
Ex #2) How many grams of bromine are needed to
prepare 0.500 L of a 0.0100 M solution of bromine in
water?
n
0.0100 M =
n = 0.00500 mol Br2
0.500 L
 159.8 g Br2 
0.00500 mol Br2 
 = 0.799 g Br2
 1 mol Br 2 
Ex #3) Describe how would you prepare 100.0 mL of
0.7500 M potassium nitrate?
n
0.7500 M KNO3 =
n = 0.07500 mol KNO3
0.1000 L
 101.1 g KNO 
3
0.07500 mol KNO3 
 = 7.583 g KNO3
 1mol KNO3 


Dissolve 7.583 g KNO3 in 100.0 mL of solution.
Ex #4) What volume of 0.600 M sodium hydroxide can
be prepared from 4.8 g of solute?
 1 mol NaOH
4.8 g NaOH 
 40.0 g NaOH


 = 0.12 mol NaOH


0.12 mol NaOH
0.600 M =
V
0.12 mol
V=
= 0.20 L
0.600 mol/L