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…
16.1

Solution Formation
◦ What factors determine the rate at which a
substance dissolves?
 stirring (agitation)
 temperature
 the surface area of the dissolving particles
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.
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
16.1

Solubility
◦ How is solubility usually expressed?
 The solubility of a substance is the maximum amount
of solute that can dissolve in a given quantity of a
solvent at a specified temperature and pressure.
 Solubility is often expressed in grams of solute per 100 g
of solvent.
 Reference table G
16.1
 A saturated solution contains the maximum amount of
solute for a given quantity of solvent at a given
temperature and pressure.
 Points on the curve
 An unsaturated solution contains less solute than a
saturated solution at a given temperature and
pressure.
 Points below the curve
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.
16.1
 The mineral deposits
around hot springs result
from the cooling of the
hot, saturated solution of
minerals emerging from
the spring.
16.1
◦ What conditions determine the amount of solute
that will dissolve in a given solvent?
 Temperature affects the solubility of solids, liquids,
and gases
 Pressure also affects the solubility of gases (not liquids
or solids)
16.1
◦ Increasing temperature:
 Increases the solubility of most solids
 Decreases the solubility of gases
16.1
16.1
◦ Changing 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 vapor pressure of the gas
above the solution increases.
 Open bottles of soda go flat…
16.1
 A supersaturated solution is unstable and 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.
16.1
 Crystals begin to form in the solution immediately
after the addition of a seed crystal.
16.1
 Excess solute crystallizes rapidly.
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 byproducts of water
treatment.
16.2

Molarity
◦ How do you determine the concentration of a
solution?
 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 unit of concentration for
solutions.
 To calculate the molarity of a solution, divide the
moles of solute by the volume (in liters) of the
solution.
 Reference Table T
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.
16.2
 Fill the flask with water exactly to the 1-L mark.
16.2
16.2

Making Dilutions
◦ What effect does dilution have on the total moles of
solute in a solution?
 Nothing. 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.
16.2
 Making a Dilute Solution
16.2
 Volume-Measuring Devices
16.2
◦ Since the total number of moles of solute remains
unchanged upon dilution, you can use this equation:
M1V1 = M2V2
 Where 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 ref. table T
16.2
 How do you prepare 100 ml of 0.40M MgSO4 from a
stock solution of 2.0M MgSO4?
`
◦ Another way to express concentration is parts per
million (ppm).
 This is useful when very little solute is dissolved in a
very large volume of water; for example the amount of
dissolved oxygen in a lake.
 Ref. table T
ppm = (mass solute/mass solution) X 1,000,000
◦ A 1.0 L sample of water from a stream contains
0.008g of dissolved oxygen. What is the
concentration in parts per million?
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.
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.
16.3
◦ What are three colligative properties of solutions?
 vapor-pressure lowering
 boiling-point elevation
 freezing-point depression
16.3

Dissolving a solute in a solvent will
◦ Lower the vapor pressure
◦ Raise the boiling point and
◦ Lower the freezing point
The more you dissolve
the greater the change.
16.3
 In a pure solvent, equilibrium is established between
the liquid and the vapor.
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
◦ 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.
This solution will have a
lower vapor pressure than
the previous one.
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. This solution will
have the lowest vapor
pressure.

Which solution has the lowest vapor pressure
at STP?
◦ 1 M KCl
◦ 1 M CH3OH
◦ 1 M MgBr2
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.
16.3

Dissolving a solute in water will lower its
freezing point.
◦ The difference in temperature between the freezing
point of a solution and the freezing point of the
pure solvent is the freezing-point depression.

The more concentrated the solution the lower the freezing
point.

Which solution would have the lowest
freezing point?
◦ 1 mole of MgBr2 in 1 kg water
◦ 1 mole of CH3OH in 1 kg water
◦ 1 mole NaNO3 in 1 kg water
16.3

Dissolving a solute in water will raise its
boiling point.
◦ 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 more concentrated the solution the higher the boiling
point.

Which solution would have the highest boiling
point?
◦ 1 mole of NaCl in 1 kg water
◦ 1 mole of NaCl in 2 kg water
◦ 1 mole of NaCl in 3 kg water