Spencer L. Seager
Michael R. Slabaugh
www.cengage.com/chemistry/seager
Chapter 7:
Solutions and Colloids
Jennifer P. Harris
SOLUTIONS
• Solutions are homogeneous mixtures of two or more
substances in which the components are present as atoms,
molecules, or ions.
• Particles in liquid solutions are:
• too small to reflect light, thus solutions are transparent
(clear).
• in constant motion.
• not settled by the influence of gravity.
• Some solutions are colored.
SOLVENT & SOLUTE
• SOLVENT OF A SOLUTION
• The solvent of a solution is the substance present in the
largest amount in the solution
• SOLUTE OF A SOLUTION
• A solute of a solution is any substance present in an amount
less than that of the solvent. A solution may contain more
than one solute.
PHYSICAL STATES OF SOLUTIONS
• The physical state of a solution, solid, liquid, or gas, is usually
the same as the physical state of the solvent.
SOLUBILITY
• The solubility of a solute is the maximum amount of the solute
that can be dissolved in a specific amount of solvent under
specific conditions of temperature and pressure.
SOLUBLE vs. INSOLUABLE
• SOLUBLE SUBSTANCE
• This is a term used to describe a substance that dissolves to
a significant extent in a solvent without stating how much
actually will dissolve.
• INSOLUBLE SUBSTANCE
• This is a term used to describe a substance that does not
dissolve to a significant extent in a solvent.
Sugar has limited solubility in water
IMMISCIBLE
• This is a term used to describe liquids that are not soluble in
each other. The liquids don't mix together to form a solution.
DEGREES OF SATURATION
• A saturated solution is a solution that contains the
maximum amount possible of dissolved solute in a stable
situation under the prevailing conditions of temperature and
pressure.
• A supersaturated solution is an unstable solution that
contains an amount of solute greater than the solute
solubility under the prevailing conditions of temperature and
pressure.
• An unsaturated solution is a solution that contains an
amount of solute less than the amount required to form a
saturated solution under the prevailing conditions of
temperature and pressure.
Crystallization
1. A seed crystal is added to a supersaturated solution
2. Crystallization is initiated (crystal growth)
3. Crystallization is completed and purified compound can be separated
from solvent.
EXAMPLES OF SOLUTE SOLUBILITES
AT 0°C
EXAMPLES OF SOLUTE SOLUBILITES
AT 0°C (continued)
EFFECT OF TEMPERATURE ON
SOLUBILITY
THE SOLUTION PROCESS
• The solution process involves interactions between solvent
molecules (often water) and the particles of solute.
• An example of the solution process for an ionic solute in water:
THE SOLUTION PROCESS (continued)
• An example of the solution process for a polar solute in water:
THE SOLUTION PROCESS (continued)
• A solute will not dissolve in a solvent if:
• the forces between solute particles are too strong to be
overcome by interactions with solvent particles.
• the solvent particles are more strongly attracted to each
other than to solute particles.
• A good rule of thumb for solubility is “like dissolves like.”
• Polar solvents dissolve polar or ionic solutes.
• Nonpolar solvents dissolve nonpolar or nonionic solutes.
GENERAL SOLUBILITIES OF IONIC
COMPOUNDS IN WATER
INCREASING THE RATE OF DISSOLVING
• Crush or grind the solute.
• Small particles provide more surface area for solvent attack
and dissolve more rapidly than larger particles.
• Heat the solvent.
• Solvent molecules move faster and have more frequent
collisions with solute at higher temperatures.
• Stir or agitate the solution.
• Stirring removes locally saturated solution
from the vicinity of the solute and allows
unsaturated solvent to take its place.
HEAT AND SOLUTION FORMATION
• Endothermic: Solute
• Exothermic:
+ solvent + heat→ solution
Solute + solvent → solution + heat
SOLUTION CONCENTRATIONS
• Solution concentrations express a quantitative relationship
about the amount of solute contained in a specific amount of
solution.
• Concentration units discussed include molarity and
percentage.
MOLARITY
• The molarity of a solution expresses the number of moles of
solute contained in one liter of solution.
• The mathematical calculation of the molarity of a solution
involves the use of the following equation:
moles of solute
M
liters of solution
• In this equation, the number of moles of solute in a sample of
solution is divided by the volume in liters of the same sample of
solution.
MOLARITY CALCULATION EXAMPLES
• Example 1: A 250-mL sample of solution contains 0.134
moles of solute. Calculate the molarity of the solution.
• Solution: The solution sample volume in liters is 0.250 L. The
number of moles of solute in the sample is 0.134 mol. These
two quantities are substituted into the equation given earlier:
moles of solute 0.134 mole
mole
M

 0.536
liters of solution
0.250 L
liter
MOLARITY CALCULATION EXAMPLES
(continued)
• Example 2: 9.45 g of methyl alcohol, CH3OH, was dissolved in
enough pure water to give 500 mL of solution. What was the
molarity of the solution?
• Solution: The solution volume was 500 mL or 0.500L. The
amount of solute was given in grams, but needs to be expressed
in moles. The molecular weight of methyl alcohol is 32.0 u, so
1 mol = 32.0 g. This fact can be used to convert the mass of
methyl alcohol into moles:
1mole alcohol
9.45 g alcohol 
 0.295 mole alcohol
32.0 g alcohol
• The number of moles of alcohol and the number of liters of
solution can now be substituted into the equation for molarity:
moles of solute 0.295 mole
mole
M

 0.590
liters of solution
0.500 L
liter
PERCENT CONCENTRATIONS
• Percent concentrations express the amount of solute
contained in 100 parts of solution. The parts of solution may
be expressed in different units.
part
%
 100
total
WEIGHT/WEIGHT PERCENT
• Weight/weight percent, abbreviated %(w/w), is a
concentration that expresses the mass of solute contained in
100 mass units of solution. Any mass units may be used, but
the mass of solute and solution must be in the same units.
• The equation used for mathematical calculations is:
solute mass
%( w / w) 
 100
solution mass
WEIGHT/WEIGHT PERCENT EXAMPLE
• Calculation example: Calculate the %(w/w) of a solution
prepared by dissolving 15.0 grams of table sugar in 100 mL of
water. The density of the water is 1.00 g/mL.
• Solution: The mass of the water used is 100 grams because
according to the density each mL has a mass of 1.00 grams.
The mass of solution is equal to the mass of water plus the mass
of the sugar solute or 100 + 15.0 = 115 grams. The calculation
is:
15.0 g
%( w / w) 
 100  13.0%
115 g
WEIGHT/VOLUME PERCENT
• Weight/volume percent, abbreviated %(w/v), is a concentration
that expresses the number of grams of solute contained in 100
mL of solution. The solute mass must be expressed in grams,
and the amount of solution must be expressed in mL.
• This percent concentration is normally used when the
solute is a solid and the solvent and resulting solutions are
liquids.
• The equation used for mathematical calculations is:
grams of solute
%( w / v) 
 100
mL of solution
WEIGHT/VOLUME PERCENT EXAMPLE
• Calculation example: Calculate the %(w/v) of a solution
prepared by dissolving 8.95 grams of sodium chloride in enough
water to give 50.0 mL of solution.
• Solution: The data are given in units that may be put directly into
the equation. The calculation is:
8.95 g
%( w / v) 
 100  17.9%
50.0 mL
VOLUME/VOLUME PERCENT
• Volume/volume percent, abbreviated %(v/v), is a concentration
that expresses the volume of liquid solute contained in 100
volumes of solution. Any volume units may be used, but they
must be the same for both the solute and the solution.
• The equation used for mathematical calculations is:
solute volume
%( v / v) 
 100
solution volume
VOLUME/VOLUME PERCENT EXAMPLE
• Calculation example: A solution is made by dissolving 250 mL of
glycerin in enough water to give 1.50 L of solution. Calculate the
%(v/v) of the resulting solution.
• Solution: The data give the solute volume and solution volume in
different units. They must be the same. Either quantity could be
changed, but the solute volume will be expressed in liters as
0.250 L. The calculation is:
0.250 L
%( v / v) 
 100  16.7%
1.50 L
SOLUTION PREPARATION
• Solutions of known concentration are usually prepared in one
of two ways.
• In one method, the necessary quantity of pure solute is
measured using a balance or volumetric equipment. The solute
is put into a container and solvent, usually water, is added until
the desired volume of solution is obtained.
SOLUTION PREPARATION EXAMPLE
• Calculation example: Describe how to prepare 500 mL of
0.250 M NaCl solution.
• Solution: The mass of NaCl needed must first be
determined. The volume and concentration of the desired
solution are known, so the equation for molarity is rearranged
to solve for the number of moles of solute needed. The result
is:
moles of solute = M x liters of solution
= 0.250 M x 0.500 L = 0.125 mole
• Thus, 0.125 moles of NaCl is needed. NaCl has a formula
weight of 58.4 u, so 0.125 moles has a mass of 0.125 x 58.4g
or 7.30 grams. The solution is prepared by weighing a
sample of NaCl with a mass of 7.30 grams. The sample is put
into a 500 mL volumetric flask and pure water is added up to
the mark on the flask.
SOLUTION PREPARATION (continued)
• In a second method, a quantity of solution with a
concentration greater than the desired concentration is
diluted with an appropriate amount of solvent to give a
solution with a lower concentration. This type of problem
is made simpler by using the following equation:
(Cc)(Vc) = (Cd)(Vd)
• In this equation, Cc is the concentration of the
concentrated solution that is to be diluted, Vc is the
volume of concentrated solution that is needed, Cd is the
concentration of the dilute solution, and Vd is the volume
of dilute solution.
SOLUTION PREPARATION EXAMPLE
• Calculation example: Describe how to prepare 250 mL of 0.500
M HCl solution from a 1.50 M HCl solution.
• Solution: According to the definitions given above, Cc = 1.50 M,
Cd = 0.500 M, and Vd = 250 mL. The equation given above can
be solved for Vc, the volume of concentrated solution needed:
Vc

C d Vd  0.500 M250 mL 


 83.3 mL
Cc 
1.50 M
• The solution is prepared by measuring 83.3 mL of 1.50 M HCl and
pouring it into a 250 mL volumetric flask. Pure water is then added
up to the mark on the flask to give 250 mL of 0.500 M solution.
SOLUTION STOICHIOMETRY
• As shown earlier, the number of moles of solute in a volume of
solution of known molarity can be obtained by multiplying
together the known molarity and the solution volume in liters.
• Molarity is a ratio of moles of solute to liters of solution. This
ratio can be written as two conversion factors:
moles solute
liters solution
or
liters solution
moles solute
• The conversion factor on the left is used to multiply by the
molarity. It is selected to cancel the units of liters of solution
and obtain the units of moles of solute.
• The conversion factor on the right is used to divide by the
molarity. It is selected to cancel the units of moles of solute
and obtain the units of liters of solution.
SOLUTION STOICHIOMETRY EXAMPLE
• Calculation example: Consider the balanced equation
HCl(aq) + NaOH(aq)
NaCl(aq) + H2O(l)
How many mL of 0.100 M HCl solution would exactly react
with 25.00 mL of 0.125 M NaOH solution?
• Solution: The number of moles of NaOH in 25.00 mL of 0.125
M solution is calculated by multiplying the molarity and the
solution volume in liters:
moles of NaOH = 0.125 M x 0.02500 L
= 0.00313 moles of NaOH
• According to the reaction, 1 mole of NaOH reacts with 1 mole
of HCl, so 0.00313 moles of HCl will be needed.
SOLUTION STOICHIOMETRY EXAMPLE
(continued)
• The volume of 0.100 M HCl solution that contains 0.00313 can
be calculated by dividing the needed number of moles by the
solution molarity:
moles of solute
liters solution 
M
0.00313 moles
liters solution 
 0.0313 liters
0.100 M
• Thus, it is seen that 0.0313 liters or 31.3 mL of 0.100 M HCl
solution will be required to react with the 25.00 mL of 0.125 M
NaOH solution.
SOLUTION PROPERTIES
• Absolutely pure water conducts electricity very poorly.
• Some solutes called electrolytes produce water solutions that
conduct electricity well.
• Some solutes called nonelectrolytes produce water solutions
that do not conduct electricity.
A solution of a
strong electrolyte
conducts electricity
well.
A solution of a weak
electrolyte conducts
electricity poorly.
A solution of a
nonelectrolyte does not
conduct electricity.
ELECTROLYTES
• STRONG ELECTROLYTES
• Strong electrolytes form solutions that conduct
electricity because they dissociate completely into
charged ions when they dissolve.
• WEAK ELECTROLYTES
• Weak electrolytes form weakly conducting solutions
because they dissociate into ions only slightly when
they dissolve.
• NONELECTROLYTES
• Nonelectrolytes form nonconducting solutions because
they do not dissociate into ions at all when they
dissolve.
COLLIGATIVE PROPERTIES OF SOLUTIONS
• Colligative solution properties are properties that depend
only on the concentration of solute particles in the solution.
Three colligative properties are boiling point, freezing point,
and osmotic pressure.
• Experiments demonstrate that the vapor pressure of water
(solvent) above a solution is lower than the vapor pressure of
pure water.
SOLUTION BOILING POINT
• The boiling point of a solution is always higher than the
boiling point of the pure solvent of the solution.
• The difference between the boiling point of the pure solvent
and the solution is represented by the symbol ∆tb.
• The difference in boiling point between pure solvent and
solution depends on the concentration of solute particles,
and is calculated using the following equation:
∆tb = nKbM
• In this equation, ∆tb is the difference between the boiling point
of the solution and the boiling point of the pure solvent.
SOLUTION BOILING POINT (continued)
• The n in the equation is the number of moles of solute particles
put into the solution when 1 mole of solute dissolves. For
solutes that do not dissociate, n = 1. For solutes that are
strong electrolytes, n varies depending on the number of ions
formed when the solute dissolves.
• For example, the dissociation of calcium chloride is represented
as:
CaCl2
Ca2+ + 2 Cl• Thus, when 1 mole of CaCl2 dissolves, 3 moles of particles (ions)
are put into the solution and n = 3.
• Kb in the equation is a constant called the boiling point
constant. Kb is characteristic of the solvent used to make the
solution.
• M is the molarity concentration of solute in the solution.
SOLUTION FREEZING POINT
• The freezing point of a solution is always lower than the
freezing point of the pure solvent of the solution.
• The difference in freezing point of the solution and pure
solvent is ∆tf, which is calculated by subtracting the freezing
point of the solution from the freezing point of the pure
solvent.
• The value of ∆tf is calculated by using the following equation:
∆tf = nKfM
• The symbols in this equation have meaning similar to those
given for the boiling point elevation equation. Kf is a constant
characteristic of the solvent used to form the solution.
OSMOTIC PRESSURE OF SOLUTIONS
• When solutions having different concentrations of solute
are separated by a semipermeable membrane, solvent
tends to flow through the membrane from the less
concentrated solution into the more concentrated solution
in a process called osmosis.
• When the more concentrated solution involved in
osmosis is put under sufficient pressure, the net osmotic
flow of solvent into the solution can be stopped.
• The pressure necessary to prevent the osmotic flow of
solvent into a solution is called the osmotic pressure of
the solution and can be calculated by using the following
equation, which is similar to the ideal gas law given earlier:
π = nMRT
OSMOTIC PRESSURE OF SOLUTIONS
(continued)
• In this equation, π is the osmotic pressure, n is the number of
moles of solute particles put into solution when 1 mole of
solute dissolves, M is the molarity of the solution, R is the
universal gas constant written as 62.4 L torr/K mol, and T is the
solution temperature in Kelvin.
• The product of n and M is called the osmolarity of the solution.
COLLOIDS
• Colloids are homogeneous mixtures of two or more
components called the dispersing medium and the
dispersed phase. The dispersed phase substances in a
colloid are in the form of particles larger than those found in
solutions.
• DISPERSING MEDIUM OF A COLLOID
• The dispersing medium of a colloid is the substance
present in the largest amount. It is analogous to the
solvent of a solution.
• DISPERSED PHASE OF A COLLOID
• The dispersed phase of a colloid is the substance
present in a smaller amount than the dispersing
medium. It is analogous to the solute of a solution.
COLLOID PROPERTIES
• In colloids, the dispersed phase particles cannot be seen and
do not settle under the influence of gravity.
• Colloids appear to be cloudy because the larger particles in the
dispersed phase scatter light.
• Colloids demonstrate the Tyndall effect in which the path of the
light through a colloid is visible because the light is scattered.
TYPES OF COLLOIDS
STABILIZING COLLOIDS
• Ions in a dispersing medium are attracted to colloid
particles and stick on their surfaces.
• All colloid particles within a particular system will attract ions
of only one charge or the other.
• The colloid particles all acquire the same charge and repel
each other.
• The repulsion helps prevent the particles from coalescing into
aggregates large enough to settle out.
• Substances known as emulsifying agents or stabilizing
agents are used to prevent some colloids from coalescing
(e.g. egg yolk in oil and water to form mayonnaise, soap/
detergent ions forming a charged layer around nonpolar oils
and greases).
STABILIZING COLLOIDS (continued)
DESTRUCTION OF COLLOIDS
• Colloid solids are removed from gaseous smoke stack
wastes before they are released into the atmosphere in the
Cottrel precipitator.
• The precipitator contains a number of highly charged plates or
electrodes.
• As smoke passes over the charged surfaces, the colloid
particles lose their charges.
• The particles then coalesce into larger
particles that settle out and are
collected for disposal.
• Dialysis can be used to separate
small particles from colloids
(e.g. cleaning the blood of people
suffering from kidney malfunction).
DIALYSIS
• A dialyzing membrane is a semipermeable membrane with
larger pores than osmotic membranes that allow solvent
molecules, other small molecules, and hydrated ions to pass
through.
• Dialysis is a process in which solvent molecules, other small
molecules, and hydrated ions pass from a solution through a
membrane.