Unit: 11: Solution Chemistry
Solutions
Mixture

A combination of 2 or more kinds of matter, each of
which retains its own composition and properties

To distinguish the components of a mixture one would
have to inspect the mixture molecule by molecule
Mixtures differ according to the size of the particles in
the mixture

Mixtures:

Are classified as either suspensions, colloids or
solutions based on size of particles

Solutions contains smallest particles and
suspension contains the largest particles
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.
Soluble: means it can be dissolved



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.
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
3 Factors that influence the rate of
dissolving
1.
2.
3.
Increase the surface area of the solute
Agitate the solution (stir or shake)
Heat the solvent
Solubility
Solubility is
• the maximum amount of solute that dissolves in a
specific amount of solvent.
• expressed as grams of solute in 100 grams of solvent
water.
g of solute
100 g water
Learning Check
At 40C, the solubility of KBr is 80 g/100 g H2O.
Identify the following solutions as either
1) saturated or (2) unsaturated. Explain.
A. 60 g KBr added to 100 g of water at 40C.
B. 200 g KBr added to 200 g of water at 40C.
C. 25 g KBr added to 50 g of water at 40C.
Solution
A. 2
Amount of 60 g KBr/100 g water is less than the
solubility of 80 g KBr/100 g water.
B. 1
In 100 g of water, 100 g KBr exceeds the
solubility of 80 g KBr water at 40C.
C. 2
This is the same as 50 g KBr in 100 g of water,
which is less than the solubility of 80 g KBr/100
g water at 40C.
Solubility and the Solution Process

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
Molecules in gas phase are far apart from each other and not
interacting strongly with each other in solution.
Mixing of Gas Molecules
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
Solute
Solution
Solute – solute
Solute – solvent
+
Solvent – solvent

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.
Molecular Solutions

Molecular compounds with similar chemical structures and
polarities tend to be miscible.(not soluble in each other)

Homologous alcohol series have polar and non-polar ends.
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.
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.
Effect of Temperature on Solubility
Solubility
• Depends on
temperature.
• Of most solids
increases as
temperature
increases.
• Of gases decreases
as temperature
increases.
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.

Learning Check
A. Why do fish die in water that is too warm?
Solution
Because O2 gas is less soluble in warm water, fish
cannot obtain the amount of O2 required for their
survival.
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.
Gas solubilities in liquids always increases with increase
pressure
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.
Solubility and Pressure
Henry’s Law states
• the solubility of a gas
in a liquid is directly
related to the
pressure of that gas
above the liquid.
• at higher pressures,
more gas molecules
dissolve in the liquid.
• (increase pressure
= increase
solubility of gas)
Example Problem Henry’s Law

The solubility of methane, the major component of natural gas, in
water at 20C and 1.00atm pressure is 0.026g/L. If the
temperature remains constant, what will the solubility of this gas
be at the following pressure: .6 atm

Solution:
Given 1 L = 0.26 g
K = equilibrium constant
[CH4] = KP[CH4]
.026g/L =K (1atm)
K = .026g/atm
solubility = (.026g atm)(.6 atm)= .0156g




Example Henry’s Law

To increase the solubility of a gas at constant
temperature from 0.85g/ml (1 atm) to 5.1 g/ml, the
pressure would have to be increased to what?

Solution:
 Given .85g = 1 ml
 [gas] = KP[gas]
 .85g/L =K (1atm)
 K = .85g/atm
solubility:

5.1g/ml = (.85g/atm) (x)
x= 5.96 atm
Example Problem Henry’s Law

The solubility of methane, the major component of
natural gas, in water at 20C and 1.00atm pressure is
0.026g/L. If the temperature remains constant, what
will the solubility of this gas be at the following
pressure: 1.8 atm
Solution Henry’s Law

Solution:
 Given 1 L = 0.26 g
K = equilibrium constant
 [CH4] = KP[CH4]
 .026g/L =K (1atm)
 K = .026g/atm
solubility = (.026g atm)(1.8 atm)=
.0468g
Why could a bottle of carbonated drink possibly burst
(explode) when it is left out in the hot sun ?
A. The pressure in a bottle increases as the gas leaves
solution as it becomes less soluble at high
temperatures. As pressure increases, the bottle could
burst.
Units of Concentration


Physical properties of solutions
moles of solute
are often related to the
Molarity 
liter of solution
concentration of the solute in the
mol of A
solution.Molarity
X
mol A  mol B
Mole fraction: The same
mass of solute
quantity we have used in
wt% 
x10 2
mass of solution
fractional abundances as well as
with gases (Dalton’s law). A
unitless number.
Weight (mass) Percent (wt%)
– similar to mole fraction except
use mass of each.
Percent by mass

The number of grams of solute dissolved in 100g of solvent
wt% 
mass of solute
x10 2
mass of solution
Percent by mass

A solution of sodium chloride is prepared by dissolving 5 g of slat
in 550 g of water. What is the concentration of this solution given
as percent by mass?
mass of solute
wt% 
x10 2
mass of solution
wt % 
5g of NaCl
x10 2
5g NaCl  550g H2O
Percent by mass example
Determine mass % of solution made from dissolving
30.0 g H2O2 with 70.0 g H2O.
30.0 g H2O2
.
30.0 g H2O2 +70.0 g H2O.
Answer: 30%
x100
Percent by mass example

A 7.5% by mass aqueous solution of sodium chloride
has a mass of 650 g. What mass of sodium chloride is
contained in the solution? What mass of water is
contained in the solution?

Solution:

650g solution x 7.5g NaCl/100g solution = 48.8g NaCl

650 g solution x 92.5gH2O/100g solution = 601 g H2O
Percent by mass example

A 60% by mass solution of H2SO4 in water is
prepared using 225g of H2SO4. What mass of
solution is produced? What mass of water is
required?

Solution:

225g H2SO4 x 100g solution/60g H2SO4 = 375 g solution

225 g H2SO4 x 40gH2O/60g H2SO4 = 150 gH2O
Units of Concentration

Other units: parts per million
(ppm) and parts per billion (ppb)
for small concentrations.
moles of solute
Molarity 
liter of solution
mass of solute
ppm 
x10 6
mass of solution
wt% 
mass of solute
x10 2
mass of solution
mass of solute
ppb 
x10 9
mass of solution
Units of Concentration: Molality

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)
Molality example

A solution contains 17.1 g sucrose ( C12H22 O11)
dissolved in 125G H2O . Find the molal concentration of
the solution

Solution

Mass solute/mass solvent

17.1g C12H22 O11 /125g H2O x 1 mol C12H22 O11 / 342g C12H22 O11 x 1000g/1kg =

x 1/molar mass solute x 1000 g/1kg
.4 molality
Molality example

What is the molality of a solution in which 3 moles of
NaCl is dissolved in 1.5Kg of water?

Solution
mol solute
Molality (m) 
mass of solvent (kg)

3moles/1.5Kg = 2m
Molality example

What is the molality of solution in which 25g of NaCl is
dissolved in 2.0 Kg of water?
Molality (m) 
mol solute
mass of solvent (kg)

Solution

Convert grams NaCl to Moles Nacl

25g NaCl x 1 molNaCl/58g NaCl = .4274 moles NaCl

.4274NaCl / 2Kg = .2137m
Molarity

The Molarity of a solution is the number of moles of
solute per liter of solution.
mol solute
Molarity (m) 
liters of solvent
Molarity example Problem

What mass of K3PO4 is required to prepare 4.00 L of
1.5 M solution?

Solution:


1.5M = X/4L solve for x
X= 6 moles

Convert moles to grams

6 moles K3PO4 x 212gK3PO4 /1 molK3PO4 = 1272 g
mol solute
Molarity (m) 
liters of solvent
Molarity example Problem

What volume of .075M solution can be prepared using
90g of NH4Cl
Molarity (m) 
mol solute
liters of solvent

Solution:

Convert grams NH4Cl to moles
90g x 1/153g(wt PT) = 1.698 mole NH4Cl



.75M = 16.98moesl/x solve for x
X =2.26L
Molarity example Problem

What is the molarity of a solution that contains 210g of
Al2(SO4)3 in 2.75 liters of solution

Solution:


Convert grams Al2(SO4)3 to moles
210g x 1mole/ 369g (wt PT) = .0569 moles Al2(SO4)3

.0569/2.7 L = .2069M
mol solute
Molarity (m) 
liters of solvent
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)

Raoult’s law:




states that the partial vapor pressure of each
component of an ideal mixture of liquids is equal to the
vapor pressure of the pure component multiplied by its
mole fraction in the mixture.
Thus the total vapor pressure of the ideal solution
depends only on the vapor pressure of each chemical
component (as a pure liquid) and the mole fraction of
the component present in the solution
Is used to determine the vapor pressure of a solution
when a solute has been added to it

Raoult's law is based on the assumption that
intermolecular forces between unlike molecules are
equal to those between similar molecules: the
conditions of an ideal solution. This is analogous to the
ideal gas law
Raoult's law: Example

25 grams of cyclohexane (Po = 80.5 torr, MM =
84.16g/mol) and 30 grams of ethanol (Po = 52.3 torr ,
MM = 92.14) are both volatile components present in a
solution. What is the partial pressure of ethanol?
Raoult's law
 Solution:

Moles cyclohexane: 25g x 1mol/84.16 = .297
moles
 Moles ethanol: 30 g x 1mol/92.14 = .326
moles
 X ethanol: .326/(.326)+(.297) =.523

PxPo = (.523) (52.3 torr) =27.4 torr
Raoult's law: Example

A solution contains 15 g of mannitol C6H14O6,
dissolved in 500g of water at 40C. The vapor
pressure of water at 40C is 55.3 mm Hg.
Calculate the vapor pressure of solution (
assume mannitol is nonvolatile)
Solution:









Molecular wt water = 18g
Convert grams of water to moles
500g x 1mol/18g =27.78 mole water
Mass mannitol = 182 g
Convert grams mannitol to moles
15g x 1mol/182g = .0824 moles
Total moles = 27.78 + .0824 =27.86
Mole fraction water = 27.78 (water)/27.86 (total
moles) = .997
Solution vapor pressure = .997 x 55.3 mmHg = 55.13
mmHg
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:
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.
Tf  K f  m
E.g. Determine the BP elevation for the sucrose solution in the previous
example. Kb = 0.521 C/m.
 A property
that depends on the
number of solute particles but is
independent of their nature is
called colligative property
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.
 Colligative properties are directly proportional to
the molal conc. Of a molecular solvent.
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.
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.