Solution
Chapter 12-1
Overview
• Solution formation
– Types of solution
– Solubility and the solution process
– Effect of temperature and pressure on solubility
• Colligative properties
–
–
–
–
–
Ways of expressing concentration
Vapor pressure of a solution
Boiling-point elevation and freezing point depression.
Osmosis
Colligative properties of ionic solutions
• Colloid Formation
– Colloids
Chapter 12-2
Types of Solution
• Solution – homogeneous mixture of two or more
substances of ions or molecules. E.g. NaCl (aq)
– Solvent = component which is the component in
greater amount.
– Solute = component which is present in the smaller
amount.
– Gaseous = gases are completely miscible in each other.
– Liquid = gas, liquid or solid solute dissolved in solute.
– Solid = mixture of two solids that are miscible in each
other to form a single phase.
• Colloid – appears to be a homogeneous mixture,
but particles are much bigger, but not filterable. E.g.
Fog, smoke, whipped cream, mayonnaise, etc.
• Suspension: larger particle sizes, filterable. E.g.
mud, freshly squeezed orange juice.
Chapter 12-3
Solubility and the Solution
Process
• The solid dissolves rapidly at first but
as the solution approaches
saturation the net rate of dissolution
decreases since the process is in
dynamic equilibrium.
• When the solution has reached
equilibrium the amount of solute
does not change with time;
• At equilibrium: the rate of
dissolution = rate of solution
Fig. 12.2 Solubility Equilibrium
Solubility and the Solution Process II
• Saturated solution: maximum amount of solute is
dissolved in solvent. Trying to dissolve more results
in undissolved solute in container.
• Solubility: Amount of solute that dissolves in a
solvent to produce a saturated solution. (Solubility
often expressed in g/100 mL.)
E.g. 0.30 g of I2 dissolved in 1000 g of H2O.
• Unsaturated solution: less than max. amount of
solute is dissolved in solvent.
E.g. 0.20 g of I2 dissolved in 1000 g of H2O.
• Supersaturation = more solute in solution than
normally allowed; we call this a supersaturated
solution.
Factors Affecting Solubility
• “like dissolves like” = substances with similar molecular structure
are usually soluble in each other.
• Gases = generally completely soluble in each other because of
entropy (Ch. 19, tendency towards maximum randomness).
• Molecules in gas phase are far apart from each other and not
interacting strongly with each other in solution.
Fig. 12.5 Mixing of Gas Molecules
Chapter 12-6
Energy Changes and the Solution
Process
• Intermolecular forces are also important in determining the solubility
of a substance.
– “like” intermolecular forces for solute and solvent will make the solute
soluble in the solvent.
• Hsoln is sometimes negative and sometimes positive.
– Solvent – solvent interactions: energy required to break weak bonds
between solvent molecules.
– Solute – solute interactions: energy required to break intermolecular
bonds between the solute molecules.
– Solute – solvent interactions: H is negative since bonds are formed
between them.
Solvent
Solute
Solution
Solute – solute
Solute – solvent
+
Solvent – solvent
Chapter 12-7
Molecular Solutions
• Molecular compounds with similar chemical structures and
polarities tend to be miscible.
• Homologous alcohol series have polar and non-polar ends.
Chapter 12-8
Ionic Solutions
• Solubility affected by:
– Energy of attraction (due Ion-dipole force) affects the solubility. Also called
hydration energy,
– Lattice energy (energy holding the ions together in the lattice. Related
• to the charge on ions; larger charge means higher lattice energy.
• Inversely proportional to the size of the ion; large ions mean smaller
lattice energy.
• Solubility increases with increasing ion size, due to decreasing lattice
energy; Mg(OH)2(least soluble), Ca(OH)2, Sr(OH)2, Ba(OH)2(most
soluble) (lattice energy changes dominant).
• Energy of hydration increases with for smaller ions than bigger ones;
thus ion size. MgSO4(most soluble),... BaSO4 (least soluble.) Hydration
energy dominant.
Chapter 12-9
Solubility: Temperature
Dependence
•
•
•
•
All solubilities are temperature dependent; must report temperatures with
solubilities.
Most solids are more soluble at higher temperatures. Exceptions exist.
All gases are less soluble at higher temperatures.
Temperature related to sign of Hsoln;
– negative means less soluble at high temperatures
– positive means more soluble (Le Chatelier’s principle).
•
E.g. Predict the temperature dependence of the solubility of Li2SO4, Na2SO4
and K2SO4 if their Hsoln are 29.8 kJ/mol, 2.4 kJ/mol and +23.8 kJ/mol,
respectively.
Chapter 12-10
Solubility: Pressure Dependence
• Pressure has little effect on the solubility
of a liquid or solid, but has dramatic effect
on gas solubility in a liquid.
• Henry’s law S = kHP. Allows us to predict
the solubility of a gas at any pressure.
E.g. At 25C P(O2 in air) = 0.21 atm. Its
solubility in water is 3.2x104M. Determine
its solubility when pressure of O2 = 1.00
atm.
Chapter 12-11
Units of Concentration
• Physical properties of solutions are often
related to the concentration of the solute
moles of solute
in the solution.Molarity
Molarity 
liter of solution
• Mole fraction: The same quantity we have
used in fractional abundances as well as
mol of A
X
with gases (Dalton’s law). A unitless
mol A  mol B
number.
Weight (mass) Percent (wt%) – similar to
mass of solute
mole fraction except use mass of each.
wt% 
x10 2
mass of solution
• E.g. determine the wt% of a solution
prepared by dissolving 1.44 g of NaCl in
100.0 mL of water. Assume that the
density of water is 1.00 g/mL
• Other units: parts per million (ppm) and
mass of solute
parts per billion (ppb) for small
ppm 
x10 6
concentrations.
mass of solution
ppb 
mass of solute
x10 9
mass of solution
Chapter 12-12
Units of Concentration2
• Molality(m): defined as the mol of solute per kg of
solvent. Unlike Molarity this unit is temperature
independent.
mol solute
Molality (m) 
mass of solvent (kg)
E.g. determine the molality of a solution prepared by
dissolving 1.44 g NaCl into exactly 100.0 mL of water.
Assume the density of water is 1.00 g/mL.
E.g.2 Determine mass % of solution made from dissolving 30.0
g H2O2 with 70.0 g H2O.
E.g.3 Determine molality of 30% H2O2(aq)
E.g.4 Determine the mole fraction of the compound in E.g. 3
E.g.5 Concentrated ammonia is 14.8 M and has a density of
0.900 g/mL. What is the molar volume and the molality?
Chapter 12-13
Physical Behavior of Solutions: Colligative Properties
• Compared with the pure solvent the
solution’s:
– Vapor pressure is lower
– Boiling point is elevated
– Freezing point is lower
– Osmosis occurs from solvent to solution when
separated by a membrane.
Chapter 12-14
Vapor-Pressure Lowering of Solutions: Raoult’s Law
• Raoult’s Law: Psoln = PsolvxXsolv
• Non–volatile solute: vapor pressure decreases upon addition of
solute.
• Linear for dilute solutions
• Vapor pressure lowering : P = Po  P = Po(1Xsolv)
E.g. Determine vapor pressure lowering when 5.00 g of sucrose added
to 100.0 g of H2O. FM(sucrose) = 342.3 g/mol. The vapor pressure of
water at 25°C is 23.8 mmHg.
E.g. 2 Determine the mass of sucrose dissolved in 100.0 g of water if
the vapor pressure was 20.0 mmHg.
Chapter 12-15
Vapor-Pressure Lowering of Solutions: Volatile Solute
•
•
•
In ideal solutions each substance in the solution obeys
Raoult’s law. Thus,
PA = XAP°o and Ptotal = PA + PB.
Combining gives: .
Ptotal  X A PAo  XBPBo  X A PAo  (1  X A )PBo
•
•
 PBo  X A (PAo  PBo )
Notice the equation predicts the linear variation in
total vapor pressure with variations in the mole
fraction of one of the components in the liquid phase.
E.g. Predict VP of ethylene (C2H4Br2) and propylene
bromides (C3H6Br2) above solution that is 60.0 mass %
in C2H4Br2. Also, predict X of each in gas phase. Given
pressure pure C2H4Br2: P = 173 torr and of pure
C3H6Br2 is P = 127 torr.
Determine:
–
–
–
•
moles and then X of each (solution).
VP of each above solution.
Mole fraction in gas from ideal gas law.
Fractional Distillation: differences between gas and
liquid phases leads to separation of complex mixtures.
Chapter 12-16
BP Elevation and FP Depression of Solutions
The magnitude of the change in FP and BP is directly proportional to the concentration of
the solute (molality) – expressed in terms of the total number of particles in the
solution.
• BP Elevation
The magnitude of the BP increase is given by the equation:
Tb  Kb  m
•
where Kb has units of °Ckg/mol or °C/m
FP Depression: linear variation with composition and given by:
Tf  K f  m
where the units for this constant are the same as for Kb
E.g. Determine freezing point depression when 5.00 g of sucrose is added to 100.0 g
of H2O. FM(sucrose) = 342.3 g/mol. Kf = 1.86°C/m.
E.g. Determine the BP elevation for the sucrose solution in the previous example. Kb
= 0.521 C/m.
Chapter 12-17
Osmosis and Osmotic Pressure
Osmosis: the passage of solvent through a membrane from the less concentrated side to the
more concentrated side.
• Osmotic pressure: the amount of pressure necessary to stop Osmosis.
• Small molecules such as water can move through certain types of materials (membranes).
• The tendency for this to occur is related to the molarity of the solution, is also a function
of the temperature and is measured with a device called a Thistle tube.
where M = is molarity of solute particles
  MRT
E.g. Determine osmotic pressure of a solution containing 0.100 g of hemoglobin
(molecular mass = 6.41x104 amu) in 0.0100 L at 1.00C.
E.g. Osmotic pressure of a solution containing 50.0 mg of a compound in 10.0 mL of water
was 4.80 torr at 5.00C. Determine FM of the compound.
Chapter 12-18
Reverse Osmosis
• Application of a pressure to
the solution (that is equal to
or greater than the Osmotic
pressure) and the solvent
flows from the more
concentrated side to the
other one.
• This process is used to obtain
pure water from salt water.
Chapter 12-19
Colligative Properties of Ionic
Solutions
• Colligative properties of solutions depends upon the total
concentration of particles.
• Each equation describing colligative properties must be modified to
account for this with ionic solutions since each ionic compound gives
more than one mole of ions for every mole of compound.
– BP elevation:
– Freezing point depression:
– Osmotic pressure:
Where i = van’t Hoff factor.
Tb  Kb  m  i
Tf  K f  m  i
  i MR  T
• van’t Hoff factor can be something other than integer under certain
circumstances, but for completely ionic solutions is equal to the
number of ions/ionic compound to be found in solution:
E.g. NaCl: i = 2; Na2SO4: i = 3;
Chapter 12-20