Chapter 11:
Solutions
11-1
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Questions for Consideration
1.
2.
3.
4.
5.
6.
What is a solution?
How do substances dissolve to make a solution?
What determines whether one substance will
dissolve in another?
How can we measure the concentration of a
solution?
In a chemical reaction, how can we relate the
amount of one reactant or product in solution to
the amount of another?
What properties of solutions depend only on the
concentration of dissolved particles, not on the
identity of the solute?
11-2
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Chapter 11 Topics:
1.
2.
3.
4.
5.
6.
The Composition of Solutions
The Solution Process
Factors That Affect Solubility
Measuring Concentrations of Solutions
Quantities for Reactions That Occur in
Aqueous Solution
Colligative Properties
11-3
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Solubility of Drugs
Figure 11.1
11-4
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Solutions are important in the medical
field:
Figure 11.3
11-5
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11.1 The Composition of Solutions




Solution – homogeneous mixture
Solute – substance being dissolved, usually
present in the smaller amount
Solvent – substance doing the dissolving,
usually present in the larger amount
Aqueous solution – a solution where the
solvent is water
11-6
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Composition of Solutions
Figure 11.4

Which is the solute? Which is the solvent?
NaCl
water
11-7
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Solubility

How readily or completely a solute dissolves
in a solvent




A soluble ionic compound dissociates into ions
when dissolved in water.
Insoluble compounds remain mostly undissociated
in their ionic, crystalline structure.
The solubilities for ionic compounds do not follow
a predictable pattern.
Solubility rules are listed in Table 11.1
11-8
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Mg(OH)2 is insoluble in water
Figure 11.6
Solubility rules are in Table 11.1
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11-9
11-10
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Electrolytes


A solution that conducts electricity
Two kinds of electrolytes:

Strong electrolyte


Weak electrolyte


A solute that dissociates completely into ions in
aqueous solution
A solute that dissociates partially into ions in
aqueous solution
Nonelectrolyte

A solute that does not dissociate into ions in
aqueous solution
11-11
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Electrolytes
Figure 11.7
11-12
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Electrolytes form ions in solution.
Strong electrolytes dissociate or ionize completely:
Figure 11.5
NaCl(s)
Na+(aq) + Cl(aq)
HCl(g)
H+(aq) + Cl(aq)
NaOH(s)
Na+(aq) + OH(aq)
11-13
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Dissolving NaCl
NaCl(s)
Na+(aq) + Cl(aq)
NaCl dissociates completely.
Figure 11.5
11-14
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Dissolving HCl
HCl(g)
H+(aq) + Cl(aq)
Or
HCl(g) + H2O(l)
H3O+(aq) + Cl(aq)
HCl ionizes completely.
Figure 11.5
11-15
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Dissolving NaOH
NaOH(s)
Na+(aq) + OH(aq)
NaOH dissociates completely
Figure 11.5
11-16
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Nonelectrolytes

Most molecular
compounds are
nonelectrolytes – they
retain their molecular
structure in aqueous
solution.

H2O2(l)
H2O2(aq)
Figure 11.8
11-17
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Activity: Composition of Solutions

What ions, atoms, or molecules are
present when the following substances
are mixed with water? Write a balanced
equation to represent the process of
dissolving each substance in water.



ZnCl2(s)
KNO3(s)
C2H5OH(l)
11-18
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Activity Solutions: Composition of
Solutions

ZnCl2(s)
ZnCl2(s) + H2O(l)

KNO3(s)
KNO3(s) + H2O(l)

Zn2+(aq) + 2 Cl–(aq)
C2H5OH(l)
C2H5OH(l)
K+(aq) + NO3–(aq)
C2H5OH(aq)
11-19
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11.2 The Solution Process

All substances that
dissolve, whether
they’re electrolytes or
nonelectrolytes, must
form new attractive
forces with the solvent
molecules.
Figure 11.8
11-20
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Dissolving NaCl in Water

What happens when NaCl dissolves in
water?
Figure 11.9
11-21
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Figure 11.9A
11-22
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When NaCl dissolves, NaCl ionic bonds break and
some H-bonds between water molecules are overcome.
Figure 11.9B
11-23
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New forces called ion-dipole forces form.
Figure 11.10
Figure 11.9C
11-24
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Heat of Solution


The difference between the energy required
for separation of solute and solvent and the
energy released upon formation of ion-dipole
interactions
When the solvation process provides more
energy than is needed to separate the pure
solvent and solvent particles, the heat of
solution is negative, and the overall dissolving
process is exothermic.
11-25
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
Some dissolving
processes are
endothermic (cold
pack).

Some dissolving
processes are
exothermic (hot pack).
Figure from p. 449
11-26
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Exothermic Dissolving Process
Figure 11.11A
11-27
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Endothermic Dissolving Process
Figure 11.11B
11-28
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Endothermic Dissolving Process


How and why do endothermic dissolving
processes occur?
Entropy, another driving force, overcomes
the unfavorable energy change.
11-29
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Entropy



A measure of the tendency for matter to
become disordered or random in its
distribution
Matter spontaneously changes from a
state of order to a state of randomness,
unless energy changes oppose it.
Solutions form because the makeup of
the solution naturally tends toward
disorganization.
11-30
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Driving Forces for Solution Formation


Decreasing energy is favorable.
Increasing entropy (disorder) is favorable.
Figure 11.13
Does
entropy
increase or
decrease in
the
formation
of a
solution?
11-31
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Activity: Dissolving of I2 in CCl4
1.
2.
3.
What forces are broken?
What forces are formed?
Does entropy increase or decrease?
Figure 11.12
11-32
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Activity Solution: The Solution Process



London forces must be broken to allow
I2, the solute, to dissolve. London forces
must also be broken to allow C6H14, the
solvent, to surround the I2 molecules.
London forces between I2 and C6H14
molecules are formed.
Entropy increases during formation of a
solution.
11-33
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Insoluble Substances

When a substance does not dissolve, the
process is too endothermic. As a result, the
entropy factor cannot overcome the large
difference in energy between the pure
substance and the solution.
11-34
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11.3 Factors that Affect Solubility



Structure
Temperature
Pressure (gas solutes only)
11-35
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Structure and Solubility


Why do some substances dissolve in one solvent
but not another?
Why is the fat-soluble vitamin A insoluble in
water, and vitamin C water-soluble?
Figure 11.15
11-36
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Structure and Solubility



A general rule “like-dissolves-like” allows us to predict
solubility.
Vitamin A has a large nonpolar section, so it dissolves
more easily in nonpolar solvents such as fat.
Vitamin C has many –OH groups (very polar) so it
dissolves readily in water.
Figure 11.15
11-37
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Fat is Mostly Nonpolar
Figure 11.16
11-38
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Why do oil and water not mix?
Figure 11.14
11-39
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Why are CCl4 and C6H14 soluble in one
another?
Figure 11.17
11-40
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Why do oil (hydrocarbons) and vinegar
not mix?
Figure 11.18
11-41
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Like-Dissolves-Like


When the bonding in the
solvent is similar to the
bonding in the solute, then
the solute will dissolve in
the solvent.
Which should be more
soluble in water?
CH3OH or I2?

Which should be more
soluble in benzene (C6H6)?
CH3OH or I2?
11-42
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Temperature

Affects the solubility of most substances
in water


Most ionic solids are more soluble in water
at higher temperatures
Gases are less soluble as temperature
increases
11-43
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Temperature
Figure 11.19
Solubility of Solids in Water
Solubility of Gases in Water
11-44
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Gas Solubility vs. Temperature
Figure 11.19

The lower O2 solubility
at higher temperatures
causes fish to die when
industries dump hot
water into lakes and
rivers.
11-45
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Gas Pressure



Affects gases but not liquids or solids
The solubility of a gas in a liquid is
directly proportional to the pressure
of the gas above the liquid.
An increase in pressure results in an
increase in solubility of a gas.
11-46
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Gas Pressure
Figure 11.20
11-47
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When pressure is released upon opening
the cap, what happens to the solubility of
CO2?
Figure 11.21
11-48
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Gas Solubility vs. Pressure



What happens to gas solubility in blood
when a scuba diver descends to lower
depths of the ocean?
What happens to gas solubility when the
scuba diver ascends?
The “bends” occurs when a scuba diver
ascends too quickly. How can the “bends”
be cured?
11-49
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11.4 Measuring Concentrations of
Solutions
Concentration – relative amounts of solute and
solvent.
 A saturated solution contains the maximum amount
of solute in a given amount of solvent.
 An unsaturated solution contains less solute than a
saturated solution.
 Solubility describes
the concentration of a
saturated solution.
(g solute/100 g solvent)

Figure 11.22
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11-50
Supersaturated Solution

Some solutions, when heated and treated
carefully, become supersaturated. They hold
more solute than a saturated solution. When
disturbed by adding a crystal or scratching the
container, they precipitate out the excess solute
and become saturated.
Figure 11.23
11-51
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Activity: Solubility

The solubility of NaCl is 38 g/100 grams of
water at 25ºC. Describe the resulting
solution after 45 g NaCl is added to 150
grams of water.
This solution contains 45 g NaCl/ 150 g H2O,
which equals 30 g NaCl/100 g H2O. Since
the solution contains less NaCl than is
soluble in 100 g of water, the solution is
unsaturated.
11-52
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Concentrations Expressed Quantitatively






Percent by Mass
Percent by Volume
Mass/volume percent
Parts per Million and Parts per Billion
Molarity (M)
Molality (m)
11-53
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Concentration Units
11-54
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Percent by Mass of Solute

Solution concentration is often expressed as the
mass percent of solute:
mass of solute
Percent Mass Solute =
100
total mass of solution

To find the mass of solution, we can add the mass
of the solute to the mass of the solvent
11-55
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Percent by Mass of Solute

Example:
 What is the percent-by-mass concentration of
acetic acid (CH3CO2H) in a vinegar solution that
contains 2.70 g of acetic acid and 122.8 g of water?
mass solute
Percent by mass =
 100%
mass solution
2.70 g
Percent by mass =
 100% = 2.15%
2.70 g + 122.8 g
11-56
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Activity: Percent by Mass of Solute

What is the mass percent of NaCl in a solution
that is prepared by adding 10.0 g NaCl to 50.0 g
water?
mass of solute
Percent Mass Solute =
100
total mass of solution
10.0 g
Percent by mass =
 100% = 16.7%
10.0 g + 50.0 g
11-57
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Percent by Volume
volume of solute
Percent by volume =
100
total volume of solution

The volume of the solution must be
measured. Volumes are not additive like
masses.
11-58
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Percent by Volume

Example:

To prepare a solution, we dissolve 205 mL of
ethanol in sufficient water to give a total volume of
235 mL. What is the percent by volume
concentration of ethanol?
volume of solute
Percent by volume =
100
total volume of solution
205 mL
Percent by volume =
 100% = 87.2%
235 mL
11-59
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Mass/Volume Percent
grams of solute
Mass/volume percent =
100
volume of solution

Example:

The concentration of NaCl in a saline bag is 0.9%.
Therefore, for every 100 mL of solution, the mass
of NaCl present is 0.9 g.
11-60
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Parts per Million and per Billion

Parts per million


Is the number of grams of solute in 1 million (106)
grams of solution.
mass of solute
Parts per million =
 106
mass of solution
Parts per billion

Is the number of grams of solute in 1 billion (109)
grams of solution.
mass of solute
9
Parts per billion =
 10
mass of solution
11-61
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Molarity

Molarity was first introduced in Chapter 4.
moles of solute
Molarity (M ) =
liters of solution
11-62
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Activity: Molarity

A solution contains 117 g of potassium
hydroxide (KOH) dissolved in sufficient water
to give a total volume of 2.00 L. The molar
mass of KOH is 56.11 g/mol. What is the
molarity of the aqueous potassium hydroxide
solution?
11-63
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Activity Solution: Molarity

A solution contains 117 g of potassium hydroxide
(KOH) dissolved in sufficient water to give a total
volume of 2.00 L. The molar mass of KOH is 56.11
g/mol. What is the molarity of the aqueous potassium
hydroxide solution?
moles of solute
Molarity (M ) =
liters of solution
1 mol KOH
moles KOH = 117 g KOH 
= 2.09 mol KOH
56.11 g KOH
2.09 mol KOH
Molarity (M ) =
= 1.04 M KOH
2.00 L
11-64
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Molality


If the temperature of a solution increases, the
molarity of the solution changes because the
volume changes.
Molality does not change with temperature
because it divides moles of solute with the
mass of solution.
moles of solute
molality (m) =
kilograms of solution
11-65
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Activity: Molality

A solution contains 56.7 g of potassium hydroxide
(KOH) dissolved in sufficient water to give a total
mass of 245.8 g. The molar mass of KOH is 56.11
g/mol. What is the molality of the aqueous
potassium hydroxide solution?
moles of solute
molality (m) =
kilograms of solution
1 mol KOH
moles of solute = 56.7 g KOH 
= 1.01 mol KOH
56.11 g KOH
1kg
kilograms of solvent = 245.8 g 
= 0.2458 kg
1000 g
1.01 mol KOH
molality =
= 4.11 m KOH
0.2458 kg solution
11-66
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11.5 Quantities for Reactions That Occur
in Aqueous Solution



Chapter 6 discussed procedures
for calculating amounts of
reactants and products in chemical
reactions.
Here we will apply those
procedures to reactions in aqueous
solutions, where concentration is
another variable.
Two types of reactions will be
considered:


Precipitation reactions
Acid-base neutralization reactions
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Figure 11.24
11-67
Precipitation Reactions

The cation from one reactant combines with
the anion from another reactant to form an
insoluble compound, called a precipitate.
11-68
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Activity: Precipitation Reactions

When a solution of lead(II) nitrate is mixed
with a solution of potassium chromate, a
yellow precipitate forms according to the
equation:
Pb(NO3)2(aq) + K2CrO4(aq)
2 KNO3(aq) + PbCrO4(s)
What volume of 0.105 M lead(II) nitrate is
required to react with 100.0 mL of 0.120 M
potassium chromate? What mass of PbCrO4
solid forms?
11-69
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Activity Solution: Precipitation Reactions
Pb(NO3)2(aq) + K2CrO4(aq) 2 KNO3(aq) + PbCrO4(s)
What volume of 0.105 M lead(II) nitrate is required
to react with 100.0 mL of 0.120 M potassium
chromate? What mass of PbCrO4 solid forms?
To solve for volume of Pb(NO3)2:
M × V = moles
100.0 mL K 2 CrO 4 
0.120 mol K 2CrO 4
= 0.0120 mol K 2CrO 4
1000 mL K 2CrO 4
0.0120 mol K 2CrO 4
1 mol Pb(NO3 ) 2

= 0.0120 mol Pb(NO3 ) 2
1 mol K 2 CrO 4
0.0120 mol Pb(NO3 ) 2
1 L Pb(NO3 ) 2

= 0.114 L Pb(NO3 ) 2
0.105 mol Pb(NO3 ) 2
11-70
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Activity Solution: Precipitation Reactions
Pb(NO3)2(aq) + K2CrO4(aq) 2 KNO3(aq) +
PbCrO4(s)
What volume of 0.105 M lead(II) nitrate is required
to react with 100.0 mL of 0.120 M potassium
chromate? What mass of PbCrO4 solid forms?
To solve for mass of PbCrO4:
0.0120 mol K 2CrO4
1 mol PbCrO4

= 0.0120 mol PbCrO4
1 mol K 2 CrO4
0.0120 mol PbCrO4
323.2 g PbCrO4

= 3.38 g PbCrO4
1 mol PbCrO4
11-71
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Acid-Base Titrations

Titration


Equivalence point


The process of determining the concentration of one
substance in solution by reacting it with a solution of
another substance that has a known concentration
The point in a titration at which the moles of the acid equal
the moles of the base
End point


The point in a titration where the indicator changes color
Slightly different than the equivalence point, but a good
estimation
11-72
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Acid-Base Titrations
Figure 11.25
11-73
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Activity: Acid-Base Titrations

We need 28.18 mL of 0.2437 M HCl solution to
completely titrate 25.00 mL of Ba(OH)2
solution of unknown concentration. What is
the molarity of the barium hydroxide solution?
11-74
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Activity Solution: Acid-Base
Titrations

We need 28.18 mL of 0.2437 M HCl solution to completely
titrate 25.00 mL of Ba(OH)2 solution of unknown
concentration. What is the molarity of the barium hydroxide
solution?
First, write the balanced chemical equation:
2 HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + 2 H2O(l)
Next, find the moles of HCl:
0.2437 moles HCl
28.17 mL HCl 
 6.865  10-3 moles HCl
1000 mL HCl
11-75
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Activity Solution: Acid-Base Titrations
HCl(aq) + Ba(OH)2(aq) → BaCl2(aq) + H2O(l)
Next, we need to find the volume (in L) of Ba(OH)2:
25.00 mL Ba(OH)2
1 L Ba(OH)2

1000 mL Ba(OH)2
 0.02500 L Ba(OH)2
Finally, divide the moles of Ba(OH)2 by the volume of
Ba(OH)2:
6.865  10-3 moles Ba(OH) 2
Molarity of Ba(OH) 2 
 0.2746 M Ba(OH) 2
0.02500 L Ba(OH) 2
11-76
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Acid-Base Titrations
Figure from p. 454
11-77
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