Chapter 13
Solutions
Cameroon: location of Lake Nyos
50 miles underneath lake Nyos
CO2 is produced by molten
volcanic rock (magma) and
held in solution by the pressure
of the water above it.
CHAPTER OUTLINE







Type of Solutions
Solubility & Saturation
Soluble & Insoluble Salts
Concentration Units
Dilution
Osmolarity
Tonicity of Solutions
2
Tragedy in Cameroon
• Lake Nyos
– lake in Cameroon, West Africa
– on August 22, 1986, > 1700
people & 3000 cattle died
• Burped Carbon Dioxide
Cloud Appeared From Nyos
– CO2 seeps in from underground
and dissolves in lake water to
levels above normal saturation
– though not toxic, CO2 is heavier
than air – the people died from
asphyxiation
Possible Resolution
• scientists have studied Lake
Nyos and similar lakes in the
region to try and keep such
tragedies from reoccurring
• currently, they are trying to
keep the CO2 levels in the
lake water from reaching very
high supersaturation levels by
pumping air into the water to
agitate it
By understanding solutions we are able to divert natural tragedies
TYPE OF
SOLUTIONS
 A solution is a homogeneous mixture of two
substances:
Solute:
substance being dissolved
present in smaller amount
Solvent:
substance doing the dissolving
present in larger amount
Solutes and solvents may be of any form of matter:
solid, liquid or gas.
5
Common Types of Solution
Solution Phase
gaseous
solutions
Solute Solvent
Phase Phase Example
gas
gas
liquid solutions
gas
liquid
solid
liquid
liquid
liquid
soda (CO2 in H2O)
vodka (C2H5OH in H2O)
seawater (NaCl in H2O)
solid solutions
solid
solid
brass (Zn in Cu)
air (mostly N2 & O2)
e.g. Brass (varies in composition)
Type
Color
% Cu
% Zn
Density
g/cm3
MP
°C
Tensile
Strength
psi
Uses
Gilding
reddish
95
5
8.86
106
6
50K
pre-83 pennies,
munitions, plaques
Commercial bronze
90
10
8.80
104
3
61K
door knobs,
grillwork
Jewelry
bronze
87.5
12.5
8.78
103
5
66K
costume jewelry
Red
golden
85
15
8.75
102
7
70K
electrical sockets,
fasteners & eyelets
Low
deep
yellow
80
20
8.67
999
74K
musical
instruments,
clock dials
Cartridge
yellow
70
30
8.47
954
76K
car radiator cores
Common
yellow
67
33
8.42
940
70K
lamp fixtures,
bead chain
Muntz
metal
yellow
60
40
8.39
904
70K
7
nuts & bolts,
brazing rods
SOLUBILITY
 Solutions form between solute and solvent
molecules because of similarities between them.
Like dissolves Like
Ionic solids dissolve in water because the charged ions
(polar) are attracted to the polar water molecules.
Non-polar molecules such as oil and grease dissolve in
non-polar solvents such as kerosene.
8
SOLUBILITY
• solutions that contain metal solutes and a metal solvent
are called alloys
• when one substance (solute) dissolves in another
(solvent) it is said to be soluble
– salt is soluble in water,
– bromine is soluble in methylene chloride
chlorocarbon that is not miscible with water,
but will dissolve in most organic solvents
• when one substance does not dissolve in another it is
said to be insoluble
– oil is insoluble in water
Salt Dissolving in Water
partial + surround the anion
partial – surround the cation
Solvation is the process of attraction and association of molecules of a
solvent with molecules or ions of a solute. As ions dissolve in a solvent
they spread out and become surrounded by solvent molecules.
SOLUBILITY
• there is usually a limit to the solubility of one
substance in another
– gases are always soluble in each other
– two liquids that are mutually soluble are
said to be miscible
• alcohol and water are miscible
• oil and water are immiscible
Descriptions of Solubility
• saturated solutions have the maximum
amount of solute that will dissolve in that
solvent at that temperature
• unsaturated solutions can dissolve
more solute
• supersaturated solutions are holding
more solute than they should be able to
at that temperature
– unstable
Adding Solute to various Solutions
unsaturated
saturated
supersaturated
Supersaturated Solution
A supersaturated solution has more dissolved solute than
the solvent can hold. When disturbed, all the solute above
the saturation level comes out of solution.
Electrolytes
• electrolytes are substances
whose aqueous solution is a
conductor of electricity
• strong electrolytes, all the
electrolyte molecules are
dissociated into ions, SALTS
• nonelectrolytes, none of the
molecules are dissociated into
ions, SUGARS
• weak electrolytes, a small
percentage of the molecules are
dissociated into ions
SOLUBILITY
 Solubility refers to the maximum amount of solute
that can be dissolved in a given amount of solvent.
 Many factors affect the solubility of a solute in a
solution.
Type of solute
Type of solvent
Temperature
Solubility is measured in grams of solute per 100 grams
of solvent at a given temperature.
16
SOLUBILITY
 Solubility of most solids
in water increases as
temperature increases.
 Using a solubility chart,
the solubility of a solute
at a given temperature
can be determined.
 For example, KNO3 has
a solubility of 80 g/100 g
H2O (80%) at 40 C.
17
SOLUBILITY
OF GASES
 Solubility of gases in water decreases as temperature
increases.
 At higher temperatures more gas molecules have the
energy to escape from solution.
 Henry’s law states that the solubility of a gas is
directly proportional to the pressure above the
liquid.
 For example, a can of soda is carbonated at high
pressures in order to increase the solubility of CO2.
Once the can is opened, the pressure is reduced and
the excess gas escapes from the solution.
18
Solubility and Pressure
• the solubility of gases in water depends on
the pressure of the gas
• higher pressure = higher solubility
Solubility and Pressure
When soda pop is sealed, the CO2 is under pressure.
Opening the container lowers the pressure, which decreases
the solubility of CO2 and causes bubbles to form.
Solution Concentration
• dilute solutions have low solute
concentrations
• concentrated solutions have high
solute concentrations
CONCENTRATION
UNITS
 The amount of solute dissolved in a certain amount
of solution (occasionally amount of solvent) is called
concentration.
Concentration =
amount of solute
amount of solution
 Three types of concentration units will be studied in
this class:
Mass Percent: (m/m) and (m/v)
Molarity
22
MASS PERCENT
 Mass percent (% m/m) is defined as the mass of
solute divided by the mass of solution.
m
a
s
s
o
f
s
o
l
u
t
e
M
a
s
s
%
(
m
/
m
)
=
x
1
0
0
m
a
s
s
o
f
s
o
l
u
t
i
o
n
mass of solute +
mass of solvent
23
MASS/VOLUME
PERCENT
 Mass/Volume percent (% m/v) is defined as the mass
of solute divided by the volume of solution.
m
a
s
s
o
f
s
o
l
u
t
e
M
a
s
s
%
(
m
/
v
)
=
x
1
0
0
v
o
l
u
m
e
o
f
s
o
l
u
t
i
o
n
24
Example 1:
What is the mass % (m/m) of a NaOH solution that
is made by dissolving 30.0 g of NaOH in 120.0 g of
water?
Mass of solution = 30.0 g + 120.0 g = 150.0 g
3
0
.
0
g
M
a
s
s
%
(
m
/
m
)
=
x
1
0
0
=
2
0
.
0
%
1
5
0
.
0
g
25
Example 2:
What is the mass % (m/v) of a solution prepared by
dissolving 5.0 g of KI to give a final volume of 250 mL?
5
.
0
g
M
a
s
s
%
(
m
/
v
)
=
x
1
0
0
=
2
.
0
%
2
5
0
m
L
26
USING PERCENT
CONCENTRATION

 Some
In theexamples
preparation
of percent
of solutions,
compositions,
one oftentheir
needs to
meanings,
calculate the
andamount
possible
ofconversion
solute or solution.
factors are
in thethis,
table
below:composition can be used as
 shown
To achieve
percent
a conversion factor.
27
Example 1:
A topical antibiotic solution is 1.0% (m/v) Clindamycin.
How many grams of Clindamycin are in 65 mL of this
solution?
1.0 g Clindamycin
65 mL solution x
100 mL solution = 0.65 g
28
Example 2:
How many grams of solute are needed to prepare 150 mL
of a 40.0% (m/v) solution of LiNO3?
40.0 g LiNO3
150 mL solution x
100 mL solution = 60. g LiNO3
29
MOLARITY
 The most common unit of concentration used
in the laboratory is molarity (M).
 Molarity is defined as:
Molarity =
moles of solute
Liter of solution
30
Example 1:
What is the molarity of a solution containing 1.4 mol
of acetic acid in 250 mL of solution?
1
L
5
0
m
L
x
=
0
.
2
5
L
Vol. of solution = 2
1
0
0
0
m
L
Molarity =
1
.4
m
o
la
c
e
t
i
c
a
c
i
d
= 5.6 M
0
.2
5
L
31
Example 2:
What is the molarity of a solution that contains 75 g
of KNO3 in 350 mL of solution?
Mol of solute =
Vol of solvent =
1m
o
l
7
5gx
=
1
0
1
.1g
0.74 mol
1
L
3
5
0
m
L
x
=
0
.
3
5
L
1
0
0
0
m
L
0
.
7
4
m
o
l
M
o
l
a
r
i
t
y
=
=
2
.
1
M
0
.
3
5
0
L
32
Preparing a 1.00 M NaCl
Solution
Weigh out
1 mole (58.45 g)
of NaCl and add
it to a 1.00 L
volumetric flask.
Step 1
Add water to
dissolve the
NaCl, then
add water to
the mark.
Step 2
Swirl to Mix
Step 3
USING
MOLARITY
 Molarity relationship can be used to calculate:
m
o
l
e
s
s
o
l
u
t
e
M
o
l
a
r
i
t
y
=
v
o
l
u
m
e
o
f
s
o
l
u
t
i
o
n
Amount of solute:
Moles solute = Molarity x volume
Volume of solution:
m
o
l
e
s
s
o
l
u
t
e
V
o
l
u
m
e
o
f
s
o
l
u
t
i
o
n
=
M
o
l
a
r
i
t
y
34
Example 1:
How many moles of nitric acid are in 325 mL of 16 M
HNO3 solution?
1
L
2
5
m
L
x
=
0
.
3
2
5
L
Vol. of solution = 3
1
0
0
0
m
L
16
.
3
2
5
L
x
mol of solute = 0
1
m
o
l
= 5.2 mol
L
35
Example 2:
How many grams of NaHCO3 are in 325 mL of 4.50 M
solution of NaHCO3?
Vol. of solution =
1
L
3
2
5
m
L
x
=
0
.
3
2
5
L
1
0
0
0
m
L
4.50
.
3
2
5
L
x
mol of solute = 0
1
mass of solute =
m
o
l
= 1.46 mol
L
8
4
.0
g
=
1
.4
6
m
o
lx
1
m
o
l
123 g
36
Example 3:
What volume (mL) of 2.0 M NaOH solution contains
20.0 g of NaOH?
mol of solute =
1m
o
l
2
0
.0gx
=
4
0
.0g
1
.
5
0
0
m
o
lx
Vol. In L = 0
2.0
Vol. In mL =
0.500 mol
L
= 0.25 L
m
o
l
1
0
0
0
m
L
0
.2
5
0
L
x
=
1
L
250 mL
37
Example 4:
How many mL of a 0.300 M glucose (C6H12O6) IV solution
is needed to deliver 10.0 g of glucose to the patient?
mol of solute =
1m
o
l
1
0
.0gx
=
1
8
0
.1g
1
.
0
5
5
5
m
o
l
x
Vol. In L = 0
0.300
Vol. In mL =
0.0555 mol
L
= 0.185 L
m
o
l
1
0
0
0
m
L
0
.1
8
5
L
x
=
1
L
185 mL
38
DILUTION
Amount of
solute

Solutions
When
arewater
oftenisprepared
added
tofrom
a solution,
more
Volume
andmore
concentration
are inversely
proportional
remains
concentrated
ones
by adding water. This
Volume
constant
process is called
dilution.
increases
Concentration
decreases
Frozen
juice
Water
Diluted
juice
39
DILUTION
 The amount of solute depends on the
concentration and the volume of the solution.
Therefore,
M1 x V1 = M2 x V2
Concentrated
solution
Dilute
solution
40
Example 1:
What is the molarity of the final solution when 75 mL of
Concentration
6.0 M KCl solution is diluted to 150 mL?
M1 = 6.0 M
V1 = 75 mL
decreases
Volume
increases
M1 x V1 = M2 x V2
M
V
(
6
.0
M
)
(
7
5
m
L
)
1
1
M=
=
2
V
1
5
0
m
L
2
M2 = ???
V2 = 150 mL
M2 = 3.0 M
41
Example 2:
What volume (mL) of 0.20 M HCl solution can be
Volume
prepared by diluting 50.0
mL
of
1.0
M
HCl?
Concentration
decreases
M1 = 1.0 M
V1 = 50.0 mL
M2 = 0.20 M
V2 = ???
increases
M1 x V1 = M2 x V2
M1 V1 (
1
.
0
M
)
(
5
0
.
0
m
L
)
V=
=
2
M2
0
.
2
0
M
V2 = 250 mL
42
Making a Solution by Dilution
M1 x V1 = M2 x V2
M1 = 12.0 M V1 = ? L
M2 = 1.50 M V2 = 5.00 L
M 1  V1  M 2  V 2
V1 
V1 
1.50
M 2  V2
M1
M   5 . 00 L 
12.0
M
 0 . 625 L
dilute 0.625 L of 12.0 M solution to 5.00 L
Solution Stoichiometry
• we know that the balanced chemical equation tells
us the relationship between moles of reactants
and products in a reaction
– 2 H2(g) + O2(g) → 2 H2O(l) implies for every 2 moles of
H2 used, you need 1 mole of O2 and to make 2 moles of
H2O
• molarity is the relationship between moles of
solute and liters of solution, thus we can measure
the moles of a material in a reaction within a
solution by knowing its molarity and volume
Example 1:
•How much 0.115 M KI solution, in liters, is required to
completely precipitate all the Pb2+ in 0.104 L of 0.225 M
Pb(NO3)2?
2 KI(aq) + Pb(NO3)2(aq) → PbI2(s) + 2 KNO3(aq)
Identify what the question is looking for:
volume of KI solution, L
0.115 M KI  0.115 mol KI  1 L solution
0.225 M Pb(NO3)2  0.225 mol Pb(NO3)2  1 L solution
Chem. Eq’n  2 mol KI  1 mol Pb(NO3)2
0.104 L Pb(NO
3 ) 2 sol' n 
0.225 mol Pb(NO
1 L sol' n
= 0.40696 L
= 0.407 L
3 )2

2 mol KI
1 mol Pb(NO

3 )2
1 L KI sol' n
0.115 mol KI
OSMOLARITY
 Many
Recall important
that when properties
ionic substances
of solutions
(strongdepend
on the number
electrolytes)
dissolve
of particles
in water
formed
they form
in solution.
several
particles for each formula unit.
 For example:
NaCl (s)
1 formula
unit
Na+ (aq) + Cl (aq)
2 particles
47
OSMOLARITY
CaCl2 (s)
1 formula
unit
Ca2+ (aq) + 2 Cl (aq)
3 particles
48
OSMOLARITY
 When covalent substances (non- or weak
electrolytes) dissolve in water they form only
one particle for each formula unit.
 For example:
C12H22O11 (s)
1 formula
unit
C12H22O11 (aq)
1 particle
49
OSMOLARITY
 Osmolarity of a solution is its molarity
multiplied by the number of particles formed
in solution.
Osmolarity = i x Molarity
Number of
particles in
solution
50
Examples:
0.10 M NaCl =
1 particle
2 x 0.10 M
= 0.20inosmol
solution
0.10 M CaCl2 = 3 x 0.10 M = 0.30 osmol
0.10 M C12H22O112 =
1
x
0.10
M
=
0.10
osmol
particles
3 particles
in solution
in solution
Same molarities
but different osmolarities
51
TONICITY OF
SOLUTIONS
 Because the cell membranes in biological systems
are semipermeable, particles of solute in solutions
can travel in and out of the membranes. This
process is called osmosis.
 The direction of the flow of solutions in or out of the
cell membranes is determined by the relative
osmolarity of the cell and the solution.
 The comparison of osmolarity of a solution with
those in body fluids determines the tonicity of a
solution.
52
ISOTONIC
SOLUTIONS
 Solutions with the same osmolarity as the cells
(0.30) are called isotonic.
 These solutions are called physiological
solutions and allow red blood cells to
retain their normal volume.
53
HYPOTONIC
SOLUTIONS
 Solutions with lower osmolarity than the cells are
called hypotonic.
 In these solutions, water flows into a
red blood cell, causing it to swell and
burst (hemolysis).
54
HYPERTONIC
SOLUTIONS
 Solutions with greater osmolarity than the cells
are called hypertonic.
 In these solutions, water leaves the
red blood cells causing it to shrink
(crenation).
55
Examples:
0.10 M NaCl = 0.20 osmol
hypotonic
0.10 M CaCl2 = 0.30 osmol
isotonic
0.10 M C12H22O11 = 0.10 osmol
hypotonic
56
THE END
57