Chapter 13 – Properties of Solutions
Many chemical reactions occur when the
reactants are in the aqueous phase.
Therefore, we need a way to quantify the
amount of reactants in an aqueous
solution. We quantify products and
reactants in an aqueous solution by
measuring the concentration.
Properties of Solutions
•
Factors that affect the solubility of a compound.
•
The process of a solute dissolving in a solvent is
called solvation.
Solutions
•
Factors that affect the solubility of a compound.
•
Agitation – When a solute is agitated, or stirred,
more water molecules can come in contact with the
solute, thus speeding up solvation.
Solutions
•
Factors that affect the solubility of a compound.
•
Heat – An increase in the temperature causes the
water molecules to move faster. So there will be
more contacts between
the water molecules and
the solute, increasing the
process of solvation.
Solutions
•
Factors that affect the solubility of a compound.
•
Increasing the surface area of the solute – breaking
up or crushing the solute increase the surface area
of the solute. As a result, more water molecules can
contact the solute making is solvate faster.
Solutions
•
Heats of Solution (ΔHsoln).
•
Solvation is a three part process –
1.
2.
3.
Breaking up of the solute.
Breaking up the solvent.
Formation of new intermolecular (or ion-dipole) forces
between te solute and the solvent.
Solutions
•
Heats of Solution (ΔHsoln).
•
Solvation is a three part process –
Solutions
•
Heats of Solution (ΔHsoln).
•
Solvation is a three part process –
1.
If the sum of the three parts is positive (+ΔHsoln), then
the process is endothermic.
Solutions
•
Heats of Solution (ΔHsoln).
•
Solvation is a three part process –
1.
If the sum of the three parts is negative (-ΔHsoln), then
the process is exothermic.
Solutions
•
Calculating Solution Concentration
•
Expressing Concentrations
•
•
•
•
Molarity (M)
Molality (m)
% by volume or mass
Mole fraction
Solutions
• Calculating Solution Concentration
• Molarity (M): Molarity is defined as
the number of moles of solute per
liter of solution.
Molarity (M) = moles of solute
1.0 L of H2O
Solutions
• What is the molarity of a solution that
was made by dissolving 5.0 grams of
NaCl in 500. mL of water.
Solutions
• How many moles of calcium chloride
would be contained in 30.0 mL of a
1.5 M calcium chloride solution?
Solutions
• Calculating Solution Concentration
• Molality (m): Molality (m) is defined
as the number of moles of solute per
kilogram of solvent.
Molality (m) = moles of solute
kg of solvent
Solutions
• Calculating Solution Concentration
• Calculate the molality of solution that
was made by dissolving 3.0 grams of
ammonium chloride in 100.0 mL of
water.
Solutions
• Calculating Solution Concentration
• How many grams of water must 60.0
grams of sodium sulfate be dissolved
in to make a 0.80 m sodium sulfate
solution?
Solutions
• Calculating Solution Concentration
• % by volume: It is what the name says
it is, the percentage of the whole
solution that is the solute.
% by volume = volume of solute
volume of solution
Solutions
• Calculating Solution Concentration
• % by mass: The percentage of the
mass of the solution that is the solute.
% by mass = mass of solute
mass of solution
Solutions
• Calculating Solution Concentration
• Vinegar is made by adding 3.0 grams
of concentrated acetic acid to
97.0
grams of water. What is the % by
mass of acetic acid in vinegar?
Solutions
• Calculating Solution Concentration
• Most red wines are 12.0 % ethyl
alcohol by volume. How many
milliters of ethyl alcohol are contained
in a 1.0 L bottle of red wine?
Solutions
• Calculating Solution Concentration
•
Mole Fractions – Is defined as the number
of moles of solute per moles of solvent
plus the number of moles of solute.
Mole fraction =
moles solute
moles solvent + moles solute
Solutions
• Calculating Solution Concentration
•
Calculate the mole fraction of a solution
that is made by dissolving 15.0 grams of
sucrose, C12H22O11, in 600.0 mL of water.
Solutions
• Colligative Properties
•
A physical property of a substance that
varies depending on the number of solute
particles dissolved in the solution.
Solutions
• Colligative Properties
•
Solute particles affect;
•
The vapor pressure of a solution
•
The boiling point of a solution
•
The freezing point of a solution
•
The rate of diffusion of water particles
(osmosis)
Solutions
• Colligative Properties
•
Vapor pressure – Defined as the pressure
of the vapor of the solvent when dynamic
equilibrium is reached between the rate of
evaporation and the rate of condensation.
Solutions
• Colligative Properties
•
Vapor pressure of water at 25°C is 3.17
kPa.
•
If we let pure water reach dynamic
equilibrium with its vapor, it will exert a
pressure of 3.17 kPa on its container.
Solutions
• Colligative Properties
Solutions
• Colligative Properties
•
Vapor Pressure
•
If solute particles are added to pure water, they will
interfere with the evaporation of the water molecules.
•
The interference of the water particles by the solute
particles causes the rate of evaporation to slow down.
•
Solute particles
cause the vapor
pressure to
decrease.
Solutions
• Colligative Properties
•
Calculating Vapor Pressure
PA = XA(P°A)
The vapor pressure of a solution is equal to the
product of the mole fraction of the solvent and
vapor pressue of the pure solvent (at a specified
temperature).
Solutions
• Colligative Properties
•
Calculating Vapor Pressure
PA = XA(P°A)
Calculate the vapor pressure of a 1.50 M MgCl2
solution.
Solutions
• Colligative Properties
•
Boiling Point Temperature
•
The formal definition of ‘boiling point’ is the
temperature at which the vapor pressure of a
liquid becomes equal
to the atmospheric
pressure around
the liquid.
Solutions
• Colligative Properties
•
Boiling Point Temperature
•
If the vapor pressure is lowered by the addition
of solute particles, then it must take more energy
for the water molecules
to evaporate. As a result,
the aqueous solution will
boil at a higher
temperature.
Solutions
• Colligative Properties
•
Boiling Point Elevation
•
The change in the boiling temperature (ΔTb) is
equal to the boiling point constant for water (kb)
times the molality of the
solution (m) times the
Van Hoff Factor (i)*.
ΔTb = kb . m . i
*
i = the # of particles the
Solid produces when it
dissolves.
Solutions
• Colligative Properties
•
Calculate the boiling point of a solution that is
made by dissolving 40.0 grams of NaCl in
500.0 mL of water.
ΔTb = kb . m . I
(kb = 0.512 °C/m)
Solutions
• Colligative Properties
•
Freezing Point Depression
•
Solute particles can interfere with the
arrangement of water molecules when they
freeze. As a result, a
colder temperature
must be reached before
water molecules can
arrange themselves to
form the solid (ice).
Solutions
• Colligative Properties
•
Freezing Point Depression
Solutions
• Colligative Properties
•
Freezing Point Depression
• The change in the freezing point of
water (ΔTf) is equal equal to the
freezing point constant of water(kf)
times the molality of the solution (m)
times the Van Hoff Factor (i).
ΔTf = kf . m . i
(kf = 1.86 °C/m)
Solutions
• Colligative Properties
•
Freezing Point Depression
•
Calculate the freezing point of an aqueous
solution consisting of 10.0 grams of CaCl2
dissolved in 1.0 L of water.
ΔTf = kf . m . i
(kf = 1.86 °C/m)
Solutions
• Colligative Properties
•
Osmotic Pressure
•
•
Osmosis is the movement of water molecules through a
semi-permeable membrane. Osmotic pressure is a
measure of the pressure that the water molecules expert
on the semi-permeable membrane.
Solute particles can cause
water molecules to
migrate to area
where the water
molecules are less
concentrated.
Solutions
• Colligative Properties
•
Osmotic Pressure
• Now you know why salt and snails don’t mix.
Solutions
• Colligative Properties
•
Calculating Osmotic Pressure
Π = MRT
Osmotic pressure (Π) is equal to the product of
the molar mass (M), the gas law constant (R), and
the temperature.
Solutions
• Colligative Properties
•
Calculating Osmotic Pressure
Π = MRT
What is the osmotic pressure at 20.0°C of a
0.0020 M solution of sucrose (C12H22O11)?
• Colloids
•
•
Solutions
Solutions that have small particles suspended in
solvent-like medium.
The particles are so small (5 – 1000 nm)that they
won’t settle out of the solution under the influence
of gravity.
• Colloids
•
•
Solutions
Some colloids consist of particles so small that they
are microscopic.
But the particles can scatter and reflect light. This is
called the Tyndall Effect.
• Colloids
•
Solutions
There are many forms of colloids;
•
•
•
•
•
•
•
•
Gases dispersed in liquids
Gases dispersed in solids
Liquids dispersed in liquids
Liquids dispersed in gases
Liquids dispersed in solids
Solids dispersed in gases
Solids dispersed in liquids
Solids dispersed in solids
• Colloids
•
Solutions
Biochemistry
o
o
The blood is a colloid consisting of many suspended
particles.
Proteins stay suspended from interaction between
the polar water molecules and the polar ends on the
surface of a protein.
• Colloids
•
Solutions
Biochemistry
o
Hydrophobic colloids, ‘water fearing’, can stay
suspended in the blood by cations coating the surface
of the suspended particle.
Solutions
• Solubility of Gases and Liquids
• Solubility of Gases in Water – Generally, as the
temperature of water increases, the solubility of
a gas decreases.
• Fast moving water
molecules are no
longer in contact
with dissolved gases
long enough to keep
it dissolved.
Solutions
• Solubility of Gases and Liquids
• Solubility of Gases in water
• Consequences -
Solutions
• Solubility of Gases and Liquids
• Solubility of Gases in Water
• Henry’s Law – The solubility of a gas
increases as the pressure of the gas
increases.
S1 = S2
P1 P2
Solutions
• Solubility of Gases and Liquids
• Solubility of Gases in Water
• Henry’s Law Problem – At standard pressure,
the solubiligy of oxygen gas is 0.041 g/L. If you
wanted to increase the solubility of O2 to 0.
100 g/L, to which pressure would you have to
pressurize the O2?
S1 = S2
P1 P2
Solutions
• Solubility of Gases and Liquids
• Solubility of Solids in Water • Generally, the solubility
of a solid increases as
the temperature of the
water increases.
• Some solids become
less soluble as the
temperature of the
water increases.
Solutions
• Solubility of Gases and Liquids
• Solubility of Solids in Water • The solubility graph
represents the
maximum amount
of solute that can
dissolve in 1 gram of
water as a function of
the temperature.
• Saturation occurs
when a solution can
no longer hold any
more solute.
Solutions
• Solubility of Gases and Liquids
• Solubility of Solids in Water • An unsaturated solution
does not have the
maximum amount of
solute dissolved.
• A supersaturated
solution has more
than the maximum
allowable dissolved
solute.
Solutions
• Solubility of Gases and Liquids
• Solubility of Solids in Water • What mass of KNO3
would precipitate from
a saturated solution
that was cooled from
23°C to 10°C?