Definitions

 Heterogeneous and homogeneous systems

 Systems with one phase are homogeneous (solutions)
 Systems with more than one phase are heterogeneous
(suspensions or colloids)
 Heterogeneous Mixtures
 Suspensions - mixture with large particles that settle out if left
undisturbed
 Have very large particles
 Can be filtered
 Ex. Muddy water
some medicines (antibiotics)

Colloids - mixture with medium size particles… larger than

solution particles, but smaller than suspension particles
 Colloids
 Have medium sized particles
 Can not be filtered.
 Many colloids are cloudy or milky in appearance.
 Show Tyndall effect (the path of light is visible)
 Colloids in Biological Systems

Some molecules have a

polar, hydrophilic
(water-loving) end and a
nonpolar, hydrophobic
(water-hating) end.
Colloids in Biological Systems

Sodium stearate is
one example of
such a molecule.
 Colloids in Biological Systems
These molecules can
aid in the
emulsification of fats
and oils in aqueous
solutions.
Colloids
 Definitions

 Solution - homogeneous mixture

Solute - substance being

dissolved

Solvent - present in
greater amount
 Solutions

How does a solid dissolve

into a liquid?

What ‘drives’ the dissolution

process?

What are the energetics of

dissolution?
 How Does a Solution Form?
1. Solvent molecules attracted to surface ions.
2. Each ion is surrounded by solvent molecules.
3. Enthalpy (DH) changes with each interaction broken or
formed.

Ionic solid dissolving in water

 Solvation or hydratation

 Solvation – the process of dissolving

solute particles are surrounded by

solvent particles

solute particles are separated and

pulled into solution
 Hydration: process by which water
molecules remove and surround individual
ions from the solid.
How Does a Solution Form

The ions are solvated

(surrounded by
solvent).
If the solvent is water,
the ions are hydrated.
The intermolecular force
here is ion-dipole.
Solvatation: polar water molecules interact with the positive
and negative ions of a salt.
Aqueous Solutions of Molecular
Compounds

Similar process but molecular

compounds do not break down into
ions, they just separate into individual
molecules
 Factors affecting solubility
 “Like dissolves like”
 Rate of stirring
 Particle size
 Temperature
 “Like dissolves like”

 Polar solutes only dissolve into polar solvents

 Water and ethanol
 Same relationship as non-polar solutes and solvents
 Examples in human body:
 Fat digestion
 Lipids are non-polar
 Water is the main solvent of the body
 Lipids coalesce together because they can’t dissolve
 Bile used to separate them to increase surface area
 Allows digestion
 Solvation

“Like Dissolves Like”

NONPOLAR

POLAR
NONPOLAR POLAR
 Solvation

 Molecular
Solvation
 molecules stay
intact

C6H12O6(s)  C6H12O6(aq)
 Factors Affecting Solubility

The stronger the

intermolecular
attractions between
solute and solvent, the
more likely the solute
Example: ethanol in water will dissolve.
Ethanol = CH3CH2OH
Intermolecular forces = H-bonds; dipole-dipole; dispersion

Ions in water also have ion-dipole forces.

An oil layer
floating on
water.
The association of amphipathic molecules in aqueous solutions
Amphipathic compounds
in aqueous solution

Polar
head group
Hydrophillic

Non-polar
Hydrophobic
tail
 Hydrophobic Interactions
• A nonpolar solute
"organizes" water
• The H-bond network of
water reorganizes to
accommodate the
nonpolar solute
• There is an decrease in
"order" of water
• This is an increase in
ENTROPY
 Factors that Increase the Rate of Solvation

Stirring the solution

- moves dissolved particles away from crystal and allows fresh
solvent to get to undissolved solute.

Smaller particle size

- increases surface area so there are more places the solvent
can get to the solute & less bonds for solvent to break

Heating the solvent

- speeds up the solvent particles so there are more frequent
collisions between solvent and solute
 Particle size and Temperature
 Particle size:
 Smaller particles more surface area
 More space for particles to come into contact with
the solvent
 Temperature:
 Higher temperature, higher kinetic energy
 More frequent and successful collisions
 Higher rate of dissolution
 Dissolution at the molecular level?
 Consider the total energy for a reaction to be spontaneous
 Enthalpy Changes in Solution

The enthalpy change

of the overall process
depends on H for
each of these steps.

Start

End
Start End
Temperature

Generally, the
solubility of solid
solutes in liquid
solvents increases with
increasing
temperature.
 Rate of stirring
 Stirring exposes more of the solute
 Makes more of the solute come
in contact with the solvent
 Increasing the rate of
dissolution.

 Put in another way, stirring brings

"fresh parts" of the solute into
contact with the solvent.
 Solubility
 Amount of solute that can dissolve in
a given amount of solvent
 Unsaturated solution:
 Less than maximum amount of
solute
 Saturated solution:
 Contains the maximum amount
of solute
 Dissolved solute will be in
equilibrium with solid solute
left at the bottom of the flask
 Supersaturated solution:
 Contains more solute than is
possible at given temperature
 Unstable solution ->
crystallization
Solubility
 Solubility:
 Amount of a solute that
can dissolve in a given
amount of solvent
 Unsaturated Solution:
 Contain less than the
maximum amount of
solute.
 Can dissolve more solute.
 Saturated Solution:
 Contain the maximum
amount of solute that can
dissolve.
 Have undissolved solute at
the bottom of the
container.
 Degree of saturation

 Supersaturated
 Solvent holds more solute than is normally possible at
that temperature.
 These solutions are unstable; crystallization can often
be stimulated by adding a “seed crystal” or scratching
the side of the flask.
 The Dissolution Process
Seven Homogeneous Possibilities
Solute Solvent Example
 Solid Liquid salt water
 Liquid Liquid mixed drinks
 Gas Liquid oxygen in water (fishes)
 Liquid Solid dental amalgams
 Solid Solid alloys
 Gas Solid metal pipes
 Gas Gas air

Two Heterogeneous Possibilities (colloids)

 Solid Gas dust in air
 Liquid Gas clouds, fog
 Gases in Solution

 The solubility of
Increasing liquids and solids
pressure
does not change
above
solution appreciably with
forces more pressure.
gas to  But, the solubility of a
dissolve. gas in a liquid is
directly proportional
to its pressure.
 Henry’s Law
The effect of partial pressure on solubility of gases

At pressure of few atmosphere or less, solubility of gas solute follows

Henry Law which states that the amount of solute gas dissolved in solution
is directly proportional to the amount of pressure above the solution.

c=kP

c = solubility of the gas (M)

k = Henry’s Law Constant
P = partial pressure of gas

Henry’s Law Constants (25°C), k

N2 8.42 •10-7 M/mmHg
O2 1.66 •10-6 M/mmHg
CO2 4.48•10-5 M/mmHg
Henry’s Law & Soft Drinks

 Soft drinks contain “carbonated

water” – water with dissolved carbon
dioxide gas.
 The drinks are bottled with a CO2
pressure greater than 1 atm.
 When the bottle is opened, the
pressure of CO2 decreases and the
solubility of CO2 also decreases,
according to Henry’s Law.
 Therefore, bubbles of CO2 escape
from solution.
 CO2 + H2O  H2CO3  H + + CO3 2-
Henry law is applied to all gases that do not
interact with the solvent
 Temperature
 The opposite is true of
gases. Higher
temperature drives gases
out of solution.

 Carbonated soft drinks are

more “bubbly” if stored in
the refrigerator.
 Warm lakes have less O2
dissolved in them than
cool lakes.
 Gases in Solution

 In general, the solubility

of gases in water
increases with increasing
mass.
Why?
 Larger molecules have
stronger dispersion
forces.
Increase of temperature can cause
so much oxygen to come out of
solution that it can no longer
support life!!
Ways of Expressing
Concentrations of
Solutions
 Concentration of Solutions
 Concentration is the amount of solute
dissolved in a given amount of solvent.
 Qualitative expressions of concentration
 Concentrated – higher ratio of solute to
solvent
 Dilute - smaller ratio of solute to solvent
Comparison of a Concentrated and Dilute Solution
 % Concentration
mass solute
 % (w/w) = x 100
mass solution
mass solute
 % (w/v) = x 100
volume solution
volume solute
 % (v/v) = x 100
volume solution
 Parts per Million and

Parts per Million (ppm)

mass of A in solution
ppm =  106
total mass of solution

ppm= mass of A in solution(mg)/total mass of solution (kg)

 Mole Fraction (X)

moles of A
cA =
total moles in solution

 In some applications, one needs the mole fraction of

solvent, not solute—make sure you find the quantity you
need! In a solution composed by A(solvent) and B +C
(solutes)
 cA+ cB + cC=1
 Molarity (M)

mol of solute
M=
L of solution

 Because volume is temperature dependent,

molarity can change with temperature.
 Molality (m)

mol of solute
m=
kg of solvent

Because neither moles nor mass change

with temperature, molality (unlike molarity)
is not temperature dependent.
 Colligative Properties
 Colligative properties depend only on the
number of solute particles present, not on
the identity of the solute particles.
 Among colligative properties are
Vapor pressure lowering
Boiling point elevation
Melting point depression
Osmotic pressure
Vapor pressure lowering is the key to all
four of the colligative properties.
 Vapor Pressure

As solute molecules are

added to a solution, the
solvent become less
volatile (=decreased
vapor pressure).
Solute-solvent interactions
contribute to this effect.
 Vapor Pressure

Therefore, the vapor

pressure of a solution is
lower than that of the
pure solvent.
 Vapor Pressure Lowering

 The presence of a non-volatile solute means that fewer

solvent particles are at the solution’s surface, so less
solvent evaporates!
In this picture we can see the
effect of a NON VOLATILE
SOLUTE on the Sature
Vapour Tension at a
determinate Temperature.
The presence of solute’s
molecules at the interface
between liquid fase and
vapour fase lowers the
chance of break out that
may accour with solvent’s
molecules.
 Raoult’s Law
only the solvent (A) contributes to
the vapour pressure of the solution

Psolution = csolventPA (pure solvent)

where
cA is the mole fraction of compound A
PA is the normal vapor pressure of A at
that temperature
Psolution is the total vapour pressure of the
solution
 Lowering of Vapor Pressure and Raoult’s Law

 Lowering of vapor pressure, Psolvent, is defined as:

Psolvent  Psolvent
0
 Psolvent
 Psolvent
0
- ( c solvent )( Psolvent
0
)
 (1  c solvent )Psolvent
0 Remember that the sum of the mole
fractions must equal 1.
Thus csolvent + csolute = 1, which we
can substitute into our expression.
c solute  1 - c solvent
Psolvent  c solute P 0
solvent

which is Raoult' s Law

 Lowering of Vapor Pressure and Raoult’s
Law
 This graph shows how the solution’s vapor pressure is
changed by the mole fraction of the solute, which is
Raoult’s law.
Non-Volatile Solutes
The draws on the side illustrate
how the vapor pressure of
water is affected by the
addition of the non-volatile
solute, NaCl.
Note that:
 there are fewer water
molecules in the vapor (i.e.,
lower vapor pressure) above Pure water - microscopic view

the NaCl solution than in the

vapor above pure water, and
there are no sodium ions or
chloride ions in the vapor
above the NaCl solution.

1.0 M NaCl solution - microscopic view.

Note that the ionic solid, NaCl, produces Na+ ions (blue)

and Cl- ions (green) when dissolved in water.

Mixtures of Volatile Liquids
Both liquids evaporate & contribute to the vapor pressure
Raoult’s Law: Mixing Two Volatile Liquids

 Since BOTH liquids are volatile and contribute to

the vapour, the total vapor pressure can be
represented using Dalton’s Law:
PT = P A + P B

The vapor pressure from each component follows Raoult’s

Law:
PT = cAP°A + cBP°B

Also, cA + cB = 1 (since there are 2 components)

384 torr 384 torr

P (Total)
P (Benzene)
133 torr 133 torr

P (Toluene)

0 X Benzene 1
1 X Toluene 0
 Benzene and Toluene
Consider a two solvent (volatile) system
 The vapor pressure from each component follows Raoult's
Law.
 Benzene - Toluene mixture:
 Recall that with only two components, cBz + cTol = 1

 Benzene: when cBz = 1, PBz = P°Bz = 384 torr & when

cBz = 0 , PBz = 0

 Toluene: when cTol = 1, PTol = P°Tol = 133 torr & when

cTol = 0, PBz = 0
 Colligative Properties
Boiling-Point Elevation
• At 1 atm (normal boiling point of pure liquid) there is a
lower vapor pressure of the solution. Therefore, a higher
temperature is required to teach a vapor pressure of 1 atm
for the solution (Tb).
 THE BOILING POINT
 The Boiling Point is the temperature at
which internal vapor pressure of the
liquid is equal to the pressure exerted
by its surroundings
 If the liquid is open to the atmosphere,
the boiling point is the temperature at
which the internal vapor pressure of the
liquid becomes equal to atmospheric
pressure (~760 mm Hg).
 The internal vapor pressure of a pure
liquid rises steadily as the temperature
is increased until the boiling point is
reached.
Boiling-Point Elevation
Tb = Tb – T b 0
T b0 is the boiling point of
the pure solvent
T b is the boiling point of
the solution

Tb > T b 0 Tb > 0

Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)

12.6
 Vapor Pressure Lowering for a Solution
 The diagram below shows how a phase diagram is affected by
dissolving a solute in a solvent.
 The black curve represents the pure liquid and the blue curve
represents the solution.
 Notice the changes in the freezing & boiling points.
 Normal Boiling Process
Extension of vapor pressure concept:
Normal Boiling Point: BP of Substance @ 1atm
When solute is added, BP > Normal BP
Boiling point is elevated when solute inhibits solvent from escaping.

Elevation of B. pt.

Express by Boiling
point Elevation
equation
 Freezing Point Depression
 Normal Freezing Point: FP of Substance @ 1atm
 When solute is added, FP < Normal FP
 FP is depressed when solute inhibits solvent from crystallizing.

When solution freezes the solid form is almost

always pure.

Solute particles does not fit into the crystal lattice

of the solvent because of the differences in size.
The solute essentially remains in solution and
blocks other solvent from fitting into the crystal
lattice during the freezing process.
Freezing-Point Depression
Tf = T f0 – Tf
0
T fis the freezing point of
the pure solvent
T f is the freezing point of
the solution

T f0 > Tf Tf > 0

Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)

12.6
 Colligative Properties
Freezing Point Depression
• The solution freezes at a lower temperature (Tf) than the
pure solvent.
• Decrease in freezing point (Tf) is directly proportional
to molality (Kf is the molal freezing-point-depression
constant):
T f  K f m
 Change in Freezing Point
Common Applications
of Freezing Point
Depression

Ethylene
glycol –
deadly to
Propylene glycol small animals
 Colligative Properties

Boiling-Point Elevation
• Molal boiling-point-elevation constant, Kb, expresses
how much Tb changes with molality, mS:

Tb  Kb  mS
• Decrease in freezing point (Tf) is directly proportional
to molality (Kf is the molal freezing-point-depression
constant):
 T f  K f  mS
Boiling Point Elevation and Freezing Point Depression
In both equations, T does
not depend on what the
Tb = Kb  m
solute is, but only on how
many particles are
dissolved.
Tf = Kf  m
 Antartic fishes

Fishes that live in the

Antartic Ocean, have an
high concentrations of
Etilenic Glycol.
EG allowes them to live at
very low temperatures
as their freezing point is
lower then common
fishes.
The etilenic glycol can be
found in the “Paraflu” that
is used as antifreezing
fluid.
An antifreezing fluid is a
chemical additive which
lowers the freezing point
of a water-based liquid.
An “antifreeze mixture” is
used to achieve freezing-
point depression in very
cold environments .
 Salts to ice’s melt

Salts (in particoular NaCl) are

used to melt ice on streets and
roads.
It occurs because adding salts
on iced surfaces, we increase
solute’s concentration. So at the
same time the freezing point
decreases.
 Osmosis

 Semipermeable membranes allow some

particles to pass through while blocking
others.
 In biological systems, most semipermeable
membranes (such as cell walls) allow water
to pass through, but block solutes.
Semipermeable Membranes

 Ions and hydrophilic

molecules cannot easily
pass thru the hydrophobic
membrane
 Small and hydrophobic
molecules can
 Colligative Properties
Osmosis
• movement of a solvent from low solute concentration to
high solute concentration across a semipermeable
membrane.
 Osmosis
at the Particulate Level
Osmotic Pressure (p)
Osmosis is the selective passage of solvent molecules through a porous membrane
from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but blocks the
passage of solute molecules.
Osmotic pressure (p) is the pressure required to stop osmosis.

more
dilute
concentrated
Since the water molecules are more concentrated in right
compartment than that in the left compartment the velocity
with which they pass from right to left is higher than the
velocity witch which thay pass from left to right.

This phenomenon increase the solution’ volume on the

right side thus increasing the idrostatic pressure that, in
turn, increase the velocity of water passage from left to
right until the two velocities have the same value
(equilibrium reached).
 Osmosis
In osmosis, there is net
movement of solvent
from the area of higher
solvent concentration
(lower solute
concentration) to the are
of lower solvent
concentration (higher
solute concentration).

Water tries to equalize the concentration on both

sides until pressure is too high.
 Osmotic Pressure

 The pressure required to stop osmosis,

known as osmotic pressure, p, is

n
p=( ) RT = MRT
V
where M is the molarity of the solution

If the osmotic pressure is the same on both sides of a

membrane (i.e., the concentrations are the same), the
solutions are isotonic.
Chemistry In Action:
Desalination
 Iso/Hypo/Hypertonic Solutions
 Isotonic: same osmotic pressure inside and out
 Example: physiological solutions
0.90% (m/v) NaCl and 5.0% (m/v) glucose

 Hypotonic: lower osmotic pressure outside

 Hypertonic: higher osmotic pressure outside

 Osmotic Solutions – Tonicity
(tonos = tension)
 Isotonic – equal solute on each side of the membrane
 Hypotonic – less solute outside cell, water rushes into cell
and cell bursts
 Hypertonic – more solute outside cell, water rushes out of
cell and cell shrivels
 Osmotic Swelling
 Animal cells maintain normal cell structure with Na+-K+
pump (moves out Na+ and prevents Cl- from moving in)
 Plants have cell walls – turgor pressure is the effect of
osmosis and active transport of ions into the cell – keeps
leaves and stems upright
 Protozoans have special water collecting vacuoles to remove
excess water
Iso/Hypo/Hypertonic Solutions
Dialysis
 Oncotic Pressure
OP is a kind of osmotic pressure exerted by proteins in
blood vessels' plasma that usually tends to pull water into
the circulatory system.
It is the force that opposes to the hydrostatic pressure.

The most of Oncotic Pressure in capillaries is generated by

the presence of an high quantity of macromolecules,with
special regard to proteins;such as albumin, which
constitutes approximately 80% of the total oncotic pressure
exerted by blood plasma on interstitial fluids.
 Water, water, everwhere…
 Seawater is water from a sea or ocean. On average,
seawater in the world's oceans has a salinity of ~3.5%.
This means that for every 1 liter of seawater there
are 35 grams of salts (mostly, but not entirely, sodium
chloride) dissolved in it.
 A person who drinks undiluted sea water will actually
become more dehydrated & may salt in the intestine
may cause diarrhea. To could potentially extend your
drinking supply though; it can be diluted with potable
water by a factor of 4 or greater to bring it below a
concentration of 0.9% solute, rendering it safer for
consumption.
 Glycogen instead of single
sugar monomers: a simple
method to reduce osmotic
pressure.
BIOLOGICAL
REFERENCES
The hydrophilic phosphate heads are in the outside layer and exposed
to the water content outside and within the cell. The hydrophobic
tails are the layer hidden in the inside of the membrane. The
phospholipidic bilayer is more permeable to small, uncharged
solutes. Protein channels float through the Phospholipids. This model
is known as the fluid mosaic model.
 Erythrocytes

Red blood cells, or erythrocytes, are the most common type

of blood cell and the vertebrate organism's principal means of
delivering oxygen (O2) to the body tissues via the blood flow
through the circulatory system.They take up oxygen in
the lungs or gills and release it while squeezing through the
body's capillaries.
These cells' cytoplasm is rich in haemoglobin, an iron-
containing biomolecule that can bind oxygen and is
responsible for the blood's red color.
In humans, mature red blood cells are oval and flexible
biconcave disks. They lack a cell nucleus and
most organelles to accommodate maximum space for
haemoglobin. 2.4 million new erythrocytes are produced per
second .
 An important application of osmosis phenomena in
medicine
Osmotic Solutions – Tonicity
Isotonic: equal solute on each side of the
membrane, so there is the same osmotic pressure
inside and outside the cell.
Hypotonic: less solute outside cell, water rushes
into cell and cell bursts, hence there is lower
osmotic pressure outside.
Hypertonic: more solute outside cell, water rushes
out of cell and cell shrivels, so there is a higher
osmotic pressure outside.
Isotonic Solution or Isotonicity

When the osmotic pressure outside the red blood cells is the same as the pressure inside the cells, the
Solution is isotonic with respect to the cytoplasm. This is the usual condition of red blood cells in plasma. The
cells are normal.

Hypertonic Solution or Hypertonicicty

When the osmotic pressure of the solution outside the blood cells is higher than the osmotic pressure inside
the red blood cells, the solution is hypertonic. The water inside the blood cells exits the cells in an attempt to
equalize the osmotic pressure, causing the cells to shrink.

Hypotonic Solution or Hypotonicity

When the solution outside of the red blood cells has a lower osmotic pressure than the cytoplasm of the red
blood cells, the solution is hypotonic with respect to the cells. The cells take in water in an attempt to equalize
the osmotic pressure, causing them to swell and potentially burst.
Two drops of blood are shown
with a bright red oxygenated
drop on the left and a
deoxygenated drop on the right.

Human erythrocytes (red blood cells)

viewed by phase contrast light
microscopy. Three conditions are
shown: hypertonic conditions (where
the erythrocytes contract and appear
"spiky"), isotonic conditions (where the
erythrocytes appear normal) and
hypotonic conditions (where the
etrythrocytes expand and become more
round).
Medical Aspects
 Saline Solution:

Saline solution(SS) is
the commonly-used
phrase for a solution of
0.9% w/v of NaCl.
 Saline solution’s uses:
1) For medical uses, saline is often
used to flush wounds and skin
abrasions, because it has got the
same concentration of blood cells.
Normal saline will not burn or sting
when applied.

2) Saline is also often used for nasal

washes to relieve some of the
symptoms of the common cold. The
solution exerts a softening and
loosening influence on the mucus to
make it easier to wash out and
clear the nasal passages for both
babies and adults
3) Saline is also used in I.V.
therapy, intravenously
supplying extra water to a
dehydrated patient or
supplying the daily water
and salt needs
("maintenance" needs) of a
patient who is unable to take
them by mouth.

4) used frequently
in intravenous drips (IVs)
for patients who cannot take
fluids orally and have
developed or are in danger
of developing dehydration
or hypovolemia (state of
decreased blood volume)
 5) Saline solution has an extremely important role as
a carrier of drugs that need to be transported through
the organism.
 Cerebral Edema

Accumulation of fluid in the brain tissues. Causes

include infection, tumor, trauma, or exposure to
certain toxins. Because the skull cannot expand to
accommodate the fluid pressure, brain tissues are
compressed. Early symptoms are changes in level of
consciousness: sluggishness, then dilatation of one or
both pupils, and a gradual loss of consciousness.
Cerebral edema can be fatal.
 Treatment is aimed at enhancing gas
exchange, reducing fluid overload, and
strengthening and slowing the heart
beat. To accomplish these goals the
patient is often given oxygen by mask
or through mechanically assisted
ventilation.

Treatment approaches can

include osmotherapy using
Mannitol or Glicerol, diuretics and
surgical decompression.

Treatment with Mannitol or

Glicerol allowes the water diffusion from
the cerebral tissue to the plasma hence
reducing cerebral volume .

Cerebral edema
 Low Sodium Diet

A low sodium diet (NaCl) is a diet that

includes no more than 1,500 to
2,400 mg of sodium per day. The
human minimum requirement for
sodium in the diet is about 500 mg
per day.
With a lesser amount of sodium the
osmotic pressure decreases.Thus
less H2O throgh blood,reducing the
volume of the ematic flow.
 Method to Distinguish Types of Electrolytes

nonelectrolyte weak electrolyte strong electrolyte

Copyright McGraw-Hill 2009 114
 Electrolytes and Nonelectrolytes
 Electrolyte: substance that dissolved in water
produces a solution that conducts electricity
 Contains ions

 Nonelectrolyte: substance that dissolved in water

produces a solution that does not conduct
electricity
 Does not contain ions
 Strong and weak electrolytes
 Strong Electrolyte: 100% dissociation
 All water soluble ionic compounds, strong
acids and strong bases
 Weak electrolytes
 Partially ionized in solution
 Exist mostly as the molecular form in
solution
 Weak acids and weak bases
 Strong Electrolytes Are…
 Strong acids
 Strong bases
 Soluble ionic salts
 Aqueous Solutions
Some compounds dissolve in water
but do not conduct electricity.
They are called nonelectrolytes.

Examples include:
sugar
ethanol
ethylene glycol
 Colligative Properties of
Electrolytes
Because these properties depend on the number of particles
dissolved, solutions of electrolytes (which dissociate in
solution) show greater changes than those of nonelectrolytes.
e.g. NaCl dissociates to form 2 ion particles; its limiting van’t
Hoff factor is 2.
 Electrolytes
In a solution,an electrolyte is a specific chemical specie that forms ions.
We may classify electrolytes in:

-STRONG ELECTROLYTES, which are completely dissociated such as salts

(NaCl, Barium Sulfate ecc.), strong bases (NaOH) and strong acids (HCl).

-WEAK ELECTROLYTES, These, conversly, are only partially dissociated.

(For example weak bases and acids).

Also we might find a particoular class of substances that are not dissociable.
We call this class NON-ELECTROLYTES.
 Colligative Properties of
Electrolytes
However, a 1 M solution of NaCl does not show twice
the change in freezing point that a 1 M solution of
methanol does.
It doesn’t act like there are really 2 particles.
van’t Hoff Factor

One mole of NaCl in

water does not really
give rise to two moles
of ions.
If we want to know the number of particles
(fragments) generated by an electrolyte, we could, of
course, go easly to an answer considering strong
electrolytes as these dissociate completely.
We cannot do the same thing if the case concernes
weak electrolytes.
 Van’t Hoff Factor

The van 't Hoff factor is the ratio between the actual
concentration of particles produced when the
substance is dissociate, and the concentration of a
substance as calculated from its mass.

n[1+α(ν-1)]
Where: - α is the dissociation degree which is the
fraction of the number of the original solute
moles and dissociated ones.
- ν is the number of fragment generated by
the dissociation
- With α=0 we have no dissociation trought molecules (so we have
non- electrolytes)
- With α=1 we have complete dissociation so we have strong
electrolytes.
- All values we find out to be in between 0 and 1 indicate weak
electrolytes

Let’s go to understand better!!!

V is number of molecules/ions. So if we have:

-NaCl  Na+ + Clˉ ( v= 2)
-CaCl2  Ca²+ +2Cl ˉ (V=3)

α= n° moles dissociates/ n° moles initial

Solve the exercises!

1) The blood plasma is isotonic compared to a solution of NaCl

(0,9 % w/w). Calculate the blood’s osmotic pressure (p) at
37 °C.

p= CRT[1+α(ν-1)]

n° mol NaCl= Gr/Mw = 0,9/58,5 = 0,0153

M= n/V = 0,0153/ 0,1L = 0,153
p= 0,153 x 0,0821 x 310 K [1+α(ν-1)]
NaCl  Na+ + Clˉ ( v= 2) (α=1)
[1+α(ν-1)]= [1+1(2-1)]=2

p = 0,153 x 0,0821 x 310 K x 2= 7,79 atm

 The van’t Hoff Factor

We modify the
previous equations by
multiplying by the
van’t Hoff factor, i

Tf = Kf  m  i

i = 1 for non-elecrtolytes
 The Concept of Equilibrium

Tube at 0° contains N2O4.

As temperature increases
reaction starts to arrive at
an equilibrium.

N2O4(g) 2NO2(g)

Chemical equilibrium occurs when a reaction

and its reverse reaction proceed at the same
rate.
 Chemical Equilibrium
• Consider colorless N2O4. At room temperature, it
decomposes to brown NO2:
N2O4(g)  2NO2(g).
• At some time, the color stops changing and we have a
mixture of N2O4 and NO2. We now write it like this…
N
N22OO44(g)
(g) 2NO
2NO22(g)
(g)
• Each is constantly being formed at the same rate that it is
being consumed. It is therefore a “dynamic equilibrium”.
• Chemical equilibrium is the point at which the
concentrations of all species are constant.
 A System at Equilibrium

Once equilibrium is
achieved, the amount
of each reactant and
product remains
constant.
 Depicting Equilibrium

In a system at equilibrium, both the forward

and reverse reactions are being carried out;
as a result, we write its equation with a
double arrow

N2O4 (g) 2 NO2 (g)

The Equilibrium Constant
 La legge che esprime la dipendenza della velocità di reazione dalla T e dalla concentrazione
dei reagenti, è la seguente equazione cinetica differenziale:
 Velocità di reazione = Kcinetica ∙[H2] ∙ [I2]
 (le parentesi quadre indicano le concentrazioni molari, le quali vengono elevate ai loro
coefficienti stechiometrici, se la cinetica della reazione è ad un solo stadio...)
 equazione di Arrhenius: Kcinetica = A ∙ e -Ea/RT
 ove Kcinetica è una costante, la quale aumenta all'aumentare della T
 e diminuisce all'aumentare della energia di attivazione Ea secondo la
 equazione di Arrhenius: Kcinetica = A ∙ e -Ea/RT
 (A è il fattore di frequenza degli urti, Ea l'energia di attivazione, R la costante dei gas, T la
temperatura assoluta in °K).

 Quindi se aumentiamo la T, aumenta la Kcinetica e quindi aumenta la Velocità di reazione.
 Se abbassiamo la Ea, ad esempio adoperando un adatto catalizzatore,
 aumenta la Kcinetica ed aumenta anche la Velocità di reazione.
 Se aumentiamo la concentrazione [ ] dei reagenti, aumenta la Velocità di reazione.
 E tutto ciò perché, se aumentiamo T e concentrazione dei reagenti, aumenta il numero dei
famosi “urti efficaci”...
 The Equilibrium Constant
 Forward reaction:
N2O4 (g)  2 NO2 (g)

 Rate law:
Rate = kf [N2O4]
 The Equilibrium Constant
 Reverse reaction:
2 NO2 (g)  N2O4 (g)

 Rate law:
Rate = kr [NO2]2
The Concept of Equilibrium
 As a system approaches
equilibrium, both the
forward and reverse
reactions are occurring.
 At equilibrium, the
forward and reverse
reactions are proceeding at
the same rate.
 The Equilibrium Constant
 Therefore, at equilibrium

Ratef = Rater

kf [N2O4] = kr [NO2]2

 Rewriting this, it becomes

kf [NO2]2
=
kr [N2O4]
 The Equilibrium Constant

The ratio of the rate constants is a constant

at that temperature, and the expression
becomes
kf [NO2]2
Keq = =
kr [N2O4]
 The Equilibrium Constant
• Consider the following reaction:
aA + bB cC + dD
• There are two reactions going on, forward and reverse. The rate of
each reaction can be expressed separately:
Ratef = kf[A]a[B]b
Rater = kr[C]c[D]d
• At equilibrium, Rf = Rr or…kf[A]a[B]b = kr[C]c[D]d
• We can rearrange this equation and combine the rate constants into
a single constant. We end up with this…

K eq 
Cc Dd
A a Bb
where Keq is called the equilibrium constant.
 The Equilibrium Constant

 To generalize this expression, consider the

reaction
aA + bB cC + dD
• The equilibrium expression for this reaction
would be
[C]c[D]d
Kc =
[A]a[B]b
 IL CATALIZZATORE NON INFLUENZA L'EQUILIBRIO:


 ad es. se non introduciamo per nulla il catalizzatore, nel recipiente di

reazione, possiamo cmq avere un'alta resa nei prodotti, seppur in
tempi lunghissimi.
 Infatti il catalizzatore fa aumentare sia la velocità della reazione
diretta, che la velocità della reazione inversa; quindi
il catalizzatore non fa aumentare la resa della reazione,
 non fa spostare l'equilibrio a destra( come potremmo erroneamente
pensare ),
 non fa aumentare la Keq, ma fa solo aumentare la velocità con cui si
ottiene sempre quella stessa quantità di prodotto.
 Equilibrium Can Be Reached
from Either Direction

As you can see, the ratio of [NO2]2 to [N2O4] remains

constant at this temperature no matter what the initial
concentrations of NO2 and N2O4 are.
 Equilibrium Can Be Reached
from Either Direction

It does not matter whether we start with N2 and H2

or whether we start with NH3. We will have the
same proportions of all three substances at
equilibrium.
 Facts About Keq
• Keq does not depend on initial concentrations of the products or
reactants since it’s only the equilibrium concentrations that we are
concerned with!

• The larger Keq , the more products are present at equilibrium. If

Keq>> 1, then products dominate at equilibrium and equilibrium lies
to the right.
• Conversely, the smaller Keq , the more reactants are present at
equilibrium. If Keq << 1, then reactants dominate at equilibrium
and the equilibrium lies to the left.
The rate of the forward reaction diminishes with time, while that of the
backward reaction increases, until they are equal.
A large K means the reaction lies far to the right at equilibrium.
The Magnitude of Equilibrium Constants
 The Equilibrium Constant

Because pressure is proportional to

concentration for gases in a closed system,
the equilibrium expression can also be
written
(PC)c (PD)d
Kp =
(PA)a (PB)b
 Relationship between Kc and Kp

 From the ideal gas law we know that

PV = nRT
• Rearranging it, we get

n
P= RT
V
 Relationship between Kc and Kp

Plugging this into the expression for Kp for

each substance, the relationship between Kc
and Kp becomes

Kp = Kc (RT)n
Where
n = (moles of gaseous product) − (moles of gaseous reactant)
 Manipulating Equilibrium Constants

The equilibrium constant of a reaction in the

reverse reaction is the reciprocal of the
equilibrium constant of the forward
reaction.
[NO2]2
N2O4 (g) 2 NO2 (g) Kc = [N O ] = 0.212 at 100C
2 4
[N2O4] 1
2 NO2 (g) N2O4 (g) Kc = [NO ]2 =
2 0.212
= 4.72 at 100C
 Manipulating Equilibrium Constants
The equilibrium constant of a reaction that has been
multiplied by a number is the equilibrium constant
raised to a power that is equal to that number.

[NO2]2
N2O4 (g) 2 NO2 (g) Kc = [N O ] = 0.212 at 100C
2 4

[NO2]4
2 N2O4 (g) 4 NO2 (g) Kc = [N O ]2 = (0.212)2 at 100C
2 4
 LE CHÂTELIER’S PRINCIPLE
When a chemical system at equilibrium is disturbed, it
reattains equilibrium by undergoing a net reaction that
reduces the effect of the disturbance.

Thus, when a disturbance occurs, we say that the

equilibrium position shifts, which means that
concentrations (or pressures) change in a way that reduces
the disturbance, and the system attains a new equilibrium
position (Q = K again).
We’re going to examine the three types of disturbances
that can occur in a reaction:

• Change in Concentration

• Change in Pressure (Volume)

• Change in Temperature
Let’s use the following reaction as example, the
reversible gaseous reaction between phosphorus
trichloride and chlorine to produce phosphorus
pentachloride:

PCl3(g) + Cl2(g)  PCl5(g)

 THE EFFECT OF A CHANGE
IN CONCENTRATION
When a system at equilibrium is disturbed by a change in
concentration of one
of the components, the system reacts in the direction that
reduces the change:

• If the concentration increases, the system reacts to

consume some of it.

• If the concentration decreases, the system reacts to

produce some of it.
Chemical Equilibrium
Le Châtelier’s Principal

1.) What Happens When a System at Equilibrium is Perturbed?

 Change concentration, temperature, pressure or add other chemicals

 Equilibrium is re-established
- Reaction accommodates the change in products, reactants, temperature, pressure, etc.
- Rates of forward and reverse reactions re-equilibrate
 Changes in Equilibrium
• Le Chatelier’s Principle: If a stress is applied to a system that is
already at equilibrium, the equilibrium will shift to reduce the effect
of the stress.

• We will now look at changing various things on a system at

equilibrium and discuss how the equilibrium will shift to “relieve the
stress.”

(1) Changing Concentrations: A+B↔C+D

• If more [A] is added, the rate of the forward reaction increases
to relieve the stress. As this occurs [C] and [D] increase, the rate
of the reverse reaction increases and quickly equilibrium is re-
established. At equilibrium –Rf = Rr. The concentrations of A, B,
C, D have changed but:
[C][D] remains constant at a given temperature.
[A][B]
 Changes in Equilibrium
Changing Concentrations Continued.. A+B↔C+D
• If [C] is removed for the system, the rate of the forward reaction
will increase to replace the [C] that was removed. As this occurs
[D] increases, and [A] and [B] decreases the rate of the reverse
reaction increases and quickly equilibrium is re-established. Once
again, Keq is the same even though the concentrations are
different.
• The equilibrium position shifts to the right if a reactant is
added or a product is removed: [reactant] increases or
[product] decreases.

• The equilibrium position shifts to the left if a reactant is

removed or a product is added: [reactant] decreases or
[product] increases.
 Chemical Equilibrium
Le Châtelier’s Principal

2.) Example:
Consider this reaction:

[ Br - ][Cr2O72- ][ H  ]8
K  1  10 11 at 25 o C
[ BrO3- ][Cr 3 ]2

At one equilibrium state:

[H  ]  5.0 M [Cr2O72- ]  0.10 M [Cr 3 ]  0.0030 M

[Br  ]  1.0 M [BrO 3- ]  0.043 M
 Chemical Equilibrium
Le Châtelier’s Principal

2.) Example:
What happens when:
[Cr2O72- ] increased from 0.10 M to 0.20 M
According to Le Châtelier’s Principal, reaction should go back to
left to off-set dichormate on right:

Use reaction quotient (Q), Same form of equilibrium equation,

but not at equilibrium:
[ Br - ][Cr2O72- ][ H  ]8 1.0 0.20 5.0 8
Q   2  10 11  K
[ BrO3- ][Cr 3 ]2 0.0430.0030 2
 Chemical Equilibrium
Le Châtelier’s Principal

2.) Example:
Because Q > K, the reaction must go to the left to decrease
numerator and increase denominator.

Continues until Q = K:

1. If the reaction is at equilibrium and products are added (or reactants

removed), the reaction goes to the left

2. If the reaction is at equilibrium and reactants are added ( or products

removed), the reaction goes to the right
 THE EFFECT OF A CHANGE
IN PRESSURE (VOLUME)
It only affects chemical reaction in which gases are
involved. Pressure changes can occur in three ways:

• Changing the concentration of a gaseous component

• Adding an inert gas (one that does not take part in the
reaction)

• Changing the volume of the reaction vessel

 Changes in Equilibrium

(2) Pressure: ONLY AFFECTS CHEMICAL REACTIONS WHICH INVOLVE GASES

• As the pressure on the gaseous system increases, the gaseous

substances are compressed, their concentrations and # of
molecules/liter increase. The reaction, (forward or reverse),
which favors the reduction of the number of molecules per liter
will be favored. (Remember, if V↓, then this would cause P↑.)
 Changes in Equilibrium

Changing pressure continued…

Example: 2H2(g) + O2(g) ↔ 2H2O(g)
• If pressure is increased by decreasing the volume of the gases in the
container at equilibrium, then the forward reaction is favored. Why?
- This reduces the # of gas particles from 3 to 2.
- Note: If the # of gas particles on both sides of the equation is the
same, then changing pressure has NO EFFECT on the equilibrium.
• If an inert gas is added, it WILL NOT change the equilibrium.
If we halve the volume the gas pressure immediately
doubles ( the Volume of a given mass of gas is inversely
proportional to its Pressure)

To reduce this increase in gas pressure, the system

responds by reducing the number
of gas molecules.

PCl3(g) + Cl2(g)  PCl5(g)

2 mol gas  1 mol gas

Notice that a change in volume results in a change in
concentration: a decrease in container volume raises the
concentration, and an increase in volume lowers the
concentration.
 THE EFFECT OF A CHANGE
IN TEMPERATURE
Of the three types of disturbances only Temperature
changes alter the value of Keq.

PCl3(g) + Cl2(g)  PCl5(g) + heat (exothermic)

 Changes in Equilibrium

(3) Temperature: All chemical reactions either give off heat

(exothermic) or take in heat (endothermic). An increase in
temperature favors the endothermic reaction; a decrease in
temperature favors the exothermic reaction.
• The temperature of a reaction will change the value of Keq, but for
now, let’s just focus on how the equilibrium will shift.
 Changes in Equilibrium
Changes in temperature continued…
• Example: H2 + I2 ↔ 2HI ∆H = - 25 kJ/mol (exothermic)
Another way of looking at the reaction…
H2 + I2 ↔ 2HI + heat
Lowering the temperature favors HI formation. (You can
think of it as though we are removing the “heat” product from
the equation.) Raising the temperature favors the reverse
reaction.
• van Hoff’s Law: In a system at equilibrium, an increase in heat
energy is displaced so that heat is absorbed.
(4) Catalyst: A catalyst increases the rate of both the forward and the
reverse reaction by decreasing Ea. It has no effect on Keq but does
cause equilibrium to be reached more quickly.
If we consider heat as a component of the equilibrium
system, a rise in temperature occurs when heat is
“added” to the system and a drop in temperature occurs
when heat is “removed” from the system. As with a
change in any other component, the system shifts to
reduce the effect of the change. Therefore, a temperature
increase (adding heat) favors the endothermic (heat-
absorbing) direction, and a temperature decrease
(removing heat) favors the exothermic (heat-releasing)
direction.
Heterogeneous Equilibrium
 The Concentrations of Solids and
Liquids Are Essentially Constant

Both can be obtained by dividing the density of the

substance by its molar mass—and both of these are
constants at constant temperature.
[ ]= n/V but n= Weight/PM so
[ ]= Weight/(V x PM) and then
[ ]= density/PM
The Concentrations of Solids and
Liquids Are Essentially Constant

Therefore, the concentrations of solids and

liquids do not appear in the equilibrium
expression

PbCl2 (s) Pb2+ (aq) + 2 Cl−(aq)

Kc = [Pb2+] [Cl−]2
 More Facts About Keq

•The equilibrium expression only contains the concentrations of gases

or aqueous substances and NEVER solids or pure liquids. Why?
- Consider the decomposition of calcium carbonate:
CaCO3(s)  CaO(s) + CO2(g)
- The concentrations of solids and pure liquids are constant.
The amount of CO2 formed will not depend greatly on the
amounts of CaO and CaCO3 present.
Kp = [PCO2]
CaCO3 (s) CO2 (g) + CaO(s)
As long as some CaCO3 or CaO remain in the
system, the amount of CO2 above the solid will
remain the same.
Note: Although the concentrations of these species are not included in
the equilibrium expression, they do participate in the reaction and must
be present for an equilibrium to be established!
 Multiple equilibria
CO2 + H2O  H2CO3  H + + CO3 2-

Remember Henry’ law.

Experiments: lemon slice in Coke

Equilibrium Calculations
 Equilibrium Calculations

A closed system initially containing

1.000 x 10−3 M H2 and 2.000 x 10−3 M I2
At 448C and V= 1 L reaction is allowed to reach
equilibrium. Analysis of the equilibrium mixture shows
that the concentration of HI is 1.87 x 10−3 M. Calculate
Kc at 448C for the reaction taking place, which is

H2 (g) + I2 (g) 2 HI (g)

 What Do We Know?

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change

At 1.87 x 10-3
equilibrium

H2 (g) + I2 (g) 2 HI (g)

 [HI] Increases by 1.87 x 10-3 M

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change +1.87 x 10-3

At 1.87 x 10-3
equilibrium

H2 (g) + I2 (g) 2 HI (g)

 Stoichiometry tells us [H2] and [I2]
decrease by half as much

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 1.87 x 10-3
equilibrium

H2 (g) + I2 (g) 2 HI (g)

 We can now calculate the equilibrium
concentrations of all three compounds…

[H2], M [I2], M [HI], M

Initially 1.000 x 10-3 2.000 x 10-3 0

Change -9.35 x 10-4 -9.35 x 10-4 +1.87 x 10-3

At 6.5 x 10-5 1.065 x 10-3 1.87 x 10-3

equilibrium

H2 (g) + I2 (g) 2 HI (g)

…and, therefore, the equilibrium constant

[HI]2
Kc =
[H2] [I2]
(1.87 x 10-3)2
=
(6.5 x 10-5)(1.065 x 10-3)
= 51
BE CAREFULL: do not mix up concentrations with moles!!!