Solution Properties
11.1
11.2
11.3
11.4
11.5
Solution Composition
The Energies of Solution Formation
Factors Affecting Solubility
The Vapor Pressures of Solutions
Boiling-Point Elevation and Freezing-Point
Depression
11.6 Osmotic Pressure
11.7 Colligative Properties of Electrolyte
Solutions
11.8 Colloids
Various Types of Solutions
Example
State of
Solution
State of
Solute
State of
Solvent
Air, natural gas
Gas
Gas
Gas
Vodka, antifreeze
Liquid
Liquid
Liquid
Brass
Solid
Solid
Solid
Carbonated water (soda)
Liquid
Gas
Liquid
Seawater, sugar solution
Liquid
Solid
Liquid
Hydrogen in platinum
Solid
Gas
Solid
Solution Composition
moles of solute
Molarity (M ) =
liters of solution
mass of solute
Mass (weight) percent =
 100%
mass of solution
moles A
Mole fraction (  A ) =
total moles of solution
moles of solute
Molality ( m ) =
kilogram of solvent
Molarity
moles of solute
Molarity (M ) =
liters of solution
Exercise #1
You have 1.00 mol of sugar in 125.0
mL of solution. Calculate the
concentration in units of molarity.
8.00 M
Exercise #2
You have a 10.0 M sugar solution.
What volume of this solution do you
need to have 2.00 mol of sugar?
0.200 L
Exercise #3
Consider separate solutions of NaOH and
KCl made by dissolving 100.0 g of each
solute in 250.0 mL of solution. Calculate the
concentration of each solution in units of
molarity.
10.0 M NaOH
5.37 M KCl
Mass Percent
mass of solute
Mass (weight) percent =
 100%
mass of solution
Exercise #4
What is the percent-by-mass concentration
of glucose in a solution made my dissolving
5.5 g of glucose in 78.2 g of water?
6.6%
Mole Fraction
moles A
Mole fraction ( A ) =
total moles of solution
Exercise #5
A solution of phosphoric acid was made by
dissolving 8.00 g of H3PO4 in 100.0 mL of
water. Calculate the mole fraction of H3PO4.
(Assume water has a density of 1.00 g/mL.)
0.0145
Molality
moles of solute
Molality (m) =
kilogram of solvent
Exercise #6
A solution of phosphoric acid was made by
dissolving 8.00 g of H3PO4 in 100.0 mL of
water. Calculate the molality of the solution.
(Assume water has a density of 1.00 g/mL.)
0.816 m
Formation of a Liquid Solution
1. Separating the solute into its individual
components (expanding the solute).
2. Overcoming intermolecular forces in the
solvent to make room for the solute
(expanding the solvent).
3. Allowing the solute and solvent to interact to
form the solution.
Steps in the Dissolving Process
Steps in the Dissolving Process
• Steps 1 and 2 require energy, since forces
must be overcome to expand the solute
and solvent.
• Step 3 usually releases energy.
• Steps 1 and 2 are endothermic, and step
3 is often exothermic.
Enthalpy (Heat) of Solution
• Enthalpy change associated with the
formation of the solution is the sum of the ΔH
values for the steps:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
• ΔHsoln may have a positive sign (energy
absorbed) or a negative sign (energy
released).
Enthalpy (Heat) of Solution
Concept Check
Explain why water and oil (a long
chain hydrocarbon) do not mix. In
your explanation, be sure to address
how ΔH plays a role.
The Energy Terms for Various Types of
Solutes and Solvents
H1
H2
H3
Hsoln
Outcome
Polar solute, polar
solvent
Large
Large
Large,
negative
Small
Solution
forms
Nonpolar solute,
polar solvent
Small
Large
Small
Large,
positive
No
solution
forms
Nonpolar solute,
nonpolar solvent
Small
Small
Small
Small
Solution
forms
Polar solute, nonpolar Large
solvent
Small
Small
Large,
positive
No
solution
forms
In General
• One factor that favors a process is an increase
in probability of the state when the solute and
solvent are mixed.
• Processes that require large amounts of energy
tend not to occur.
• Overall, remember that “like dissolves like”.
Factors Affecting Solubility
• Structural Effects:
 Polarity – “like dissolves like”
• Pressure Effects:
 Henry’s law – for solubility of gases
• Temperature Effects:
 Affecting aqueous solutions
Pressure Effects
• Henry’s law: c = kP
c = concentration of dissolved gas
k = constant
P = partial pressure of gas solute
above the solution
• Amount of gas dissolved in a solution is
directly proportional to the partial pressure of
gas above the solution.
A Gaseous Solute
Temperature Effects (for Aqueous Solutions)
• Although the solubility of most solids in water
increases with temperature, the solubilities of
some substances decrease with increasing
temperature.
• Predicting temperature dependence of
solubility is very difficult.
• Solubility of a gas in solvent typically
decreases with increasing temperature.
The Solubilities of Several Solids as a Function
of Temperature
The Solubilities of Several Gases in Water
Ideal Solution:
One that obeys Raoult’s Law
Ideal Solutions Consisting of Two
Volatile Liquids
• Two volatile Liquids form ideal solution if:
– they are structurally very similar, and
– molecular interactions between nonidentical molecules
were relatively similar to identical molecules.
• The vapor of each liquid obeys Raoult’s Law:
PA = XAPoA;
PB = XBPoB
PT = PA + PB = XAPoA + XBPoB
(X : mole fraction; Po : vapor pressure of pure liquid)
Summary of the Behavior of Various Types of
Solutions of Two Volatile Liquids
Interactive Forces
Between Solute (A) and
Solvent (B) Particles
Hsoln
T for
Solution
Formation
Deviation
from
Raoult’s
Law
Example
Benzenetoluene
A  A, B  B  A  B
Zero
Zero
None
(ideal
solution)
A  A, B  B < A  B
Negative
(exothermic)
Positive
Negative
Acetonewater
A  A, B  B > A  B
Positive
(endothermic)
Negative
Positive
Ethanolhexane
Vapor Pressure for a Solution of
Two Volatile Liquids
Laboratory Fractional Distillation Apparatus
Fractional Distillation Towers in Oil Refinaries
Refined Crude Oil Mixtures
Concept Check
For each of the following solutions, would
you expect it to be relatively ideal (with
respect to Raoult’s Law), show a positive
deviation, or show a negative deviation?
a) Hexane (C6H14) and chloroform (CHCl3)
b) Ethyl alcohol (C2H5OH) and water
c) Hexane (C6H14) and octane (C8H18)
Exercise #7
• A solution of benzene (C6H6) and toluene (C7H8)
contains 50.0% benzene by mass. The vapor
pressures of benzene and pure toluene at 25oC are
94.2 torr and 28.4 torr, respectively. Assuming ideal
behavior, calculate the following:
(a) The mole fractions of benzene and toluene;
(b) The vapor pressure of each component in the
mixture, and the total vapor pressure above the
solution.
(c) The composition of the vapor in mole percent.
Exercise #8
• A solution composed of 24.3 g acetone (CH3COCH3) and
39.5 g of carbon disilfide (CS2) has a measured vapor
pressure of 645 torr at 35oC.
(a) Is the solution ideal or nonideal?
(b) If not, does it deviate positively or negatively from
Raoult’s law?
(c) What can you say about the relative strength of carbon
disulfide-acetone interactions compared to the acetoneacetone and carbon disulfide-carbon disulfide interaction?
(Vapor pressures at 35oC of pure acetone and pure carbon
disulfide are 332 torr and 515 torr, respectively.)
An Aqueous Solution and Pure Water in a
Closed Environment
Liquid/Vapor Equilibrium
Vapor Pressure Lowering:
Addition of a Solute
Vapor Pressures of Solutions of Nonvolatile
Solutes
• Nonvolatile solute lowers the vapor
pressure of solvent.
• Raoult’s Law: Psoln = solvPsolv
Psoln = vapor pressure of solution
 solv = mole fraction of solvent
Psolv = vapor pressure of pure solvent
Colligative Properties




Lowering of solvent vapor pressure
Freezing-point depression
Boiling-point elevation
Osmotic pressure
 Colligative properties depend only on the
number, not on the identity, of the solute
particles in an ideal solution.
Lowering of Solvent Vapor Pressure
• The presence of nonvolatile solute particles lowers
the number of solvent molecules in the vapor that is
in equilibrium with the solution.
• The solvent vapor pressure is lowered;
• Assuming ideal behavior, the lowering of vapor
pressure is proportional to the mole fraction of solute:
P = Xsolute.Posolvent (for nonelectrolytes)
= iXsolute.Posolvent (for electrolytes)
(i is the van’t Hoff’s factor, which approximately relates to
the number of ions per formula unit of the compound)
Changes in Boiling Point and Freezing
Point of Water
Freezing-Point Depression
• When a solute is dissolved in a solvent, the
freezing point of the solution is lower than
that of the pure solvent.
• ΔT = Kfmsolute (for nonelectrolytes)
= iKfmsolute (for electrolytes)
ΔT = freezing-point depression
Kf = freezing-point depression constant
msolute = molality of solute
Freezing Point Depression:
Solid/Liquid Equilibrium
Freezing Point Depression:
Addition of a Solute
Freezing Point Depression:
Solid/Solution Equilibrium
Boiling-Point Elevation
• Nonvolatile solute elevates the boiling
point of the solvent.
• ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = boiling-point elevation constant
msolute = molality of solute
Boiling Point Elevation:
Liquid/Vapor Equilibrium
Boiling Point Elevation:
Addition of a Solute
Boiling Point Elevation:
Solution/Vapor Equilibrium
Exercise #9
• What mass of ethylene glycol (C2H6O2), in grams,
must be added to 1.50 kg of water to produce a solution
that boils at 105oC?
(Boiling point elevation constant for water is Kb = 0.512oC/m)
At what temperature will the solution freeze?
(Freezing point depression constant for water is Kf = 1.86oC/m)
Osmotic Pressure
• Osmosis – flow of solvent into the
solution through a semipermeable
membrane.
•  = MRT
 =
M =
R =
T =
osmotic pressure (atm)
molarity of the solution
gas law constant
temperature (Kelvin)
Osmosis
van’t Hoff Factor, i
• The relationship between the moles of
solute dissolved and the moles of
particles in solution is usually expressed
as:
moles of particles in solution
i =
moles of solute dissolved
Modified Equations for the Colligative
Properties of Electrolytes
T = imK
 = iMRT
Ion Pairing
• At a given instant a small percentage of the
sodium and chloride ions are paired and thus
count as a single particle.
Ion Pairing
• Ion pairing is most important in
concentrated solutions.
• As the solution becomes more dilute, the
ions are farther apart and less ion pairing
occurs.
• Ion pairing occurs to some extent in all
electrolyte solutions.
• Ion pairing is most important for highly
charged ions.
Exercise #10
A solution was prepared by dissolving 25.00 g
glucose in 200.0 g water. The molar mass of
glucose is 180.16 g/mol. What is the boiling
point of the resulting solution (in °C)?
Glucose is a molecular solid that is present as
individual molecules in solution.
100.35 °C
Exercise #11
You take 20.0 g of a sucrose (C12H22O11) and NaCl
mixture and dissolve it in 1.0 L of water. The
freezing point of this solution is found to be 0.426°C. Assuming ideal behavior, calculate the
mass percent composition of the original mixture,
and the mole fraction of sucrose in the original
mixture.
72.8% sucrose and 27.2% sodium chloride;
mole fraction of the sucrose is 0.313
Exercise #12
A plant cell has a natural concentration of
0.25 m. You immerse it in an aqueous solution
with a freezing point of – 0.246°C. Will the
cell explode/expand, shrivel, or do nothing?
Exercise #13
When 33.4 mg of a compound is dissolved
in 10.0 mL of water at 25°C, the solution has
an osmotic pressure of 558 torr. Calculate
the molar mass of this compound.
111 g/mol
Examples
• The expected value for i can be determined
for a salt by noting the number of ions per
formula unit (assuming complete dissociation
and that ion pairing does not occur).
 NaCl
 KNO3
 Na3PO4
i=2
i=2
i=4
Colloidal Mixtures
• A suspension of tiny particles in some
medium.
• Tyndall effect – scattering of light by
particles.
• Suspended particles are single large
molecules or aggregates of molecules or
ions ranging in size from 1 to 1000 nm.
Scattering of Light by Colloid Particles
Tyndall Effect of Colloidal Mixture
Tyndall Effect of Morning Mist
Types of Colloids
Micelle – A Colloidal Suspension
Micelle in Soap Bubbles
Coagulation
• Destruction of a colloid.
• Usually accomplished either by
heating or by adding an electrolyte.