Chapter 16
16.1

A sinkhole forms when the
roof of a cave weakens from
being dissolved by
groundwater and suddenly
collapses. One recorded
sinkhole swallowed a house,
several other buildings, five
cars, and a swimming pool!
You will learn how the
solution process occurs and
the factors that influence the
process.
1
16.1

Solution Formation
◦ What factors determine the rate at which a
substance dissolves?
16.1
 The compositions of the solvent and the solute
determine whether a substance will dissolve. The
factors that determine how fast a substance dissolves
are
1. stirring (agitation)
2. temperature
3. the surface area of the dissolving
particles
2
16.1
 A cube of sugar in cold tea dissolves slowly.
Little surface
area, cold
temperature and
no agitation.
16.1
 Granulated sugar dissolves in cold water more quickly
than a sugar cube, especially with stirring.
More surface area
and agitation.
3
16.1
 Granulated sugar dissolves very quickly in hot tea.
More surface
area and
increased
temperature.

Anything that increases solute to solvent
particle contact increases the rate of
dissolving.
◦ Stirring, heating, pulverizing
4
16.1
 A saturated solution contains the maximum amount of
solute for a given quantity of solvent at a given
temperature and pressure.
 An unsaturated solution contains less solute than a
saturated solution at a given temperature and
pressure.
16.1
 In a saturated solution, the rate of
dissolving equals the rate of
crystallization, so the total amount
of dissolved solute remains
constant.
 This is a dynamic equilibrium.
5
16.1
 The solubility of a substance is the amount of solute that
dissolves in a given quantity of a solvent at a specified
temperature and pressure to produce a saturated solution.
 Solubility is often expressed in grams of solute per 100 g
of solvent.
 Reference table G
6
16.1
 Some liquids combine in all proportions, while others
don’t mix at all.
 Two liquids are miscible if they dissolve in each other in
all proportions.
 Two liquids are immiscible if they are insoluble in each
other.
16.1
 Oil and water are immiscible.
7
16.1
 Vinegar and oil are immiscible.
16.1
 The mineral deposits
around hot springs result
from the cooling of the
hot, saturated solution of
minerals emerging from
the spring.
8
16.1

Factors Affecting Solubility
◦ What conditions determine the amount of solute
that will dissolve in a given solvent?
16.1
◦ Temperature affects the solubility of solid, liquid,
and gaseous solutes in a solvent; both temperature
and pressure affect the solubility of gaseous
solutes.
9
16.1
◦ Temperature
 The solubility of most solid substances increases as the
temperature of the solvent increases.
 The solubilities of most gases are greater in cold water
than in hot.
16.1
10
16.1
 A supersaturated solution contains more solute than it
can theoretically hold at a given temperature.
 The crystallization of a supersaturated solution can be
initiated if a very small crystal, called a seed crystal, of
the solute is added.
16.1
 A supersaturated solution is clear before a seed crystal
is added.
11
16.1
 Crystals begin to form in the solution immediately
after the addition of a seed crystal.
16.1
 Excess solute crystallizes rapidly.
12
16.1
◦ Pressure
 Changes in pressure have little effect on the solubility
of solids and liquids, but pressure strongly influences
the solubility of gases.
 Gas solubility increases as the partial pressure of the
gas above the solution increases.
16.2
◦ Water must be tested continually
to ensure that the concentrations
of contaminants do not exceed
established limits. These
contaminants include metals,
pesticides, bacteria, and even the
by-products of water treatment.
You will learn how solution
concentrations are calculated.
13
16.2
 The concentration of a solution is a measure of the
amount of solute that is dissolved in a given quantity
of solvent.
 A dilute solution is one that contains a small amount of
solute.
 A concentrated solution contains a large amount of solute.
16.2
 Molarity (M) is the number of moles of solute dissolved
in one liter of solution.
 To calculate the molarity of a solution, divide the
moles of solute by the volume of the solution.
 Reference Table T
14
16.2
 To make a 0.5-molar (0.5M) solution, first add 0.5 mol
of solute to a 1-L volumetric flask half filled with
distilled water.
16.2
 Swirl the flask carefully to dissolve the solute.
15
16.2
 Fill the flask with water exactly to the 1-L mark.
16.2
16
for Sample Problem 16.2
A solution has a volume of 2.0 L and contains 36.0 g of
glucose (C6H12O6). If the molar mass of glucose is 180
g/mol, what is the molarity of the solution?
16.3
17
for Sample
Problem
How many moles
of solute
are16.3
in 250 mL of 2.0 M
CaCl2? How many grams of CaCl2 is this?
16.2

Making Dilutions
◦ What effect does dilution have on the total moles of
solute in a solution?
◦ Diluting a solution reduces the number of moles of
solute per unit volume, but the total number of
moles of solute in solution does not change.
18
16.2
 The total number of moles of solute remains unchanged
upon dilution, so you can write this equation.
 M1 and V1 are the molarity and volume of the initial
solution, and M2 and V2 are the molarity and volume of
the diluted solution.
 Write this in your reference table T.
16.2
 Making a Dilute Solution
19
16.2
 To prepare 100 ml of 0.40M MgSO4 from a stock
solution of 2.0M MgSO4, a student first measures 20
mL of the stock solution with a 20-mL pipet.
16.2
 She then transfers the 20 mL to a 100-mL volumetric
flask.
20
16.2
 Finally she carefully adds water to the mark to make
100 mL of solution.
16.2
 Volume-Measuring Devices
21
16.4
for Sampleof
Problem
16.4 of 4.00M KI are needed to
How many milliliters
a solution
prepare 250.0 mL of 0.760M KI?
`
22
16.2

Percent Solutions
◦ What are two ways to express the percent
concentration of a solution?
1. as the ratio of the volume of the solute to the
volume of the solution or
2. as the ratio of the mass of the solute to the mass
of the solution.
16.2
◦ Concentration in Percent (Volume/Volume)
23
16.2
 Isopropyl alcohol (2-propanol) is sold as a 91%
solution. This solution consist of 91 mL of isopropyl
alcohol mixed with enough water to make 100 mL of
solution.
24

A bottle of antiseptic hydrogen peroxide
(H2O2) is labeled 3.0% (v/v). How many mL of
H2O2 are in a 400.0 mL bottle of this
solution?
16.2
◦ Concentration in Percent (Mass/Mass)
25
16.3

The wood frog is a remarkable creature because it
can survive being frozen. Scientists believe that a
substance in the cells of this frog acts as a natural
antifreeze, which prevents the cells from freezing.
You will discover how a solute can change the
freezing point of a solution.
16.3
 A property that depends only upon the number of
solute particles, and not upon their identity, is called a
colligative property.
26
16.3
◦ Three important colligative properties of solutions
are
1. vapor-pressure lowering
2. boiling-point elevation
3. freezing-point depression
16.3
 In a pure solvent, equilibrium is established between
the liquid and the vapor.
27
16.3
 In a solution, solute particles reduce the number of
free solvent particles able to escape the liquid.
Equilibrium is established at a lower vapor pressure.
16.3
◦ The decrease in a solution’s vapor pressure is
proportional to the number of particles the solute
makes in solution.
28
16.3
 Three moles of glucose dissolved in water produce 3
mol of particles because glucose does not dissociate.
16.3
 Three moles of sodium chloride dissolved in water
produce 6 mol of particles because each formula unit
of NaCl dissociates into two ions.
29
16.3
 Three moles of calcium chloride dissolved in water
produce 9 mol of particles because each formula unit
of CaCl2 dissociates into three ions.
16.3

Freezing-Point Depression
◦ The difference in temperature between the freezing
point of a solution and the freezing point of the
pure solvent is the freezing-point depression.
◦ What determines the amount by which a solution’s
freezing point differs from those properties of the
solvent?
30
16.3
◦ The magnitude of the freezing-point depression is
proportional to the number of solute particles dissolved
in the solvent and does not depend upon their identity.
◦ The freezing point of water decreases by 1.86°C for
every mole of particles that the solute forms when
dissolved in 1000 g of water.
◦ This is called the freezing point depression constant for
water.
16.3
 The freezing-point
depression of aqueous
solutions makes walks and
driveways safer when
people sprinkle salt on icy
surfaces to make ice melt.
The melted ice forms a
solution with a lower
freezing point than that of
pure water.
31
16.3

Boiling-Point Elevation
 The difference in temperature between the boiling
point of a solution and the boiling point of the pure
solvent is the boiling-point elevation.
 The same antifreeze added to automobile engines to
prevent freeze-ups in winter, protects the engine from
boiling over in summer.
16.3
 The magnitude of the boiling-point elevation is
proportional to the number of solute particles
dissolved in the solvent.
 The boiling point of water increases by 0.512°C for
every mole of particles that the solute forms when
dissolved in 1000 g of water.
 This is called the boiling point elevation constant for
water.
32
16.4


Cooking instructions often call for the
addition of a small amount of salt to the
cooking water. Dissolved salt elevates the
boiling point of water. You will learn how to
calculate the amount the boiling point of the
cooking water rises.
My favorite “pasta pot” is a 3 qt. stockpot. I
generally add about ½ tsp. salt to 2 qt. water. At
what T will the water boil at standard pressure? 1
tsp salt weighs 20 g. The molar mass of NaCl is
58.44 g/mol.
33
16.4
◦ What are two more ways of expressing the
concentration of a solution?
◦ The unit molality and mole fractions are two
additional ways in which chemists express the
concentration of a solution.
16.4
 The unit molality (m) is the number of moles of solute
dissolved in 1 kilogram (1000 g) of solvent. Molality is
also known as molal concentration.
34
16.4
 To make a 0.500m
solution of NaCl, use a
balance to measure 1.000
kg of water and add 0.500
mol (29.3 g) of NaCl.
16.4
 Ethlylene Glycol (EG) is added to water as antifreeze.
35
16.6
How many grams of sodium fluoride are needed to prepare a
0.400m NaF solution that contains 750 g of water?
36
16.4

Freezing-Point Depression and Boiling-Point
Elevation
 How are freezing-point depression and boiling-point
elevation related to molality?
 The magnitudes of the freezing-point depression (ΔTf)
and the boiling-point elevation (ΔTb) of a solution are
directly proportional to the molal concentration (m), when
the solute is molecular, not ionic.
16.4
 The constant, Kf, is the molal freezing-point
depression constant, which is equal to the change in
freezing point for a 1-molal solution of a nonvolatile
molecular solute.
37
 The constant, Kb, is the molal boiling-point elevation
constant, which is equal to the change in boiling point
for a 1-molal solution of a nonvolatile molecular
solute.
38
39
16.8
for Sample
Problem
16.8
What is the freezing
point
depression
of an aqueous solution
of 10.0 g of glucose (C6H6O12) in 50.0 g H2O?
40
16.9
for NaCl
Samplewould
Problem
16.9 to be dissolved in 1.000 kg
What mass of
have
of water to raise the boiling point by 2.00oC?
41