Solutions
IMRAN TARIQ
ASSISTANT PROFESSOR
UNIVERSITY COLLEGE OF PHARMACY,
UNIVERSITY OF THE PUNJAB
TYPES OF MIXTURES
HOMOGENOUS e.g. True solutions
HETEROGENOUS e.g. coarse and colloidal
dispersions
Homogeneity: something evenly distributed
Heterogeneity: something not distributed evenly in
space; a clump or cluster (not a single phase as in case
of colloidal and coarse dispersion).
Definitions
A solution is a homogeneous mixture of two or more
substances
OR
A solution is a homogenous mixture of solute and
solvent.
OR
 A solution is a homogenous mixture of two
substances but consisting of one phase.
A solute is dissolved in a solvent.
– solute is the substance being dissolved
– solvent is the liquid in which the solute is dissolved
– an aqueous solution has water as solvent
Binary solution: A homogenous mixture consisting
of one phase and containing only two components
i.e. one solute and one solvent e.g. Solution of Nacl
in water.
Dilute Solutions: A solution containing relatively
small quantity of solute as compared with the amount
of solvent.
Concentrated Solution: A solution containing large
amount of solute in the solution than that in dilute
solution.
Un-saturated solution: a solution in which more
solute can be dissolved at a given temperature is
called as an unsaturated solution.
A saturated solution is one where the concentration
is at a maximum - no more solute is able to dissolve
at a given temperature.
– A saturated solution represents an equilibrium: the rate of
dissolving is equal to the rate of crystallization. The salt
continues to dissolve, but crystallizes at the same rate or
under these condition the number of molecules leaving the
solute is equal to the number of molecule returning to the
solid phase i.e. solute.
Super Saturated Solution:
A solution that contains relatively larger amount of
solute than that required for saturation it is prepared
by heating and adding more and more solute.
Types of Solutions: Based on
physical states of solute and solvent:
Gas
Gas Solid Smoke in air
Gas
Gas
liquid Water vapors in air
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 (∆H) changes with each interaction broken or
formed.
Ionic solid dissolving in water
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.
Dissolution vs reaction
Dissolution is a physical change—you can get back the
original solute by evaporating the solvent.
If you can’t, the substance didn’t dissolve, it reacted.
Degree of saturation
Saturated solution
Solvent holds as much
solute as is possible at
that temperature.
Undissolved solid
remains in flask.
Dissolved solute is in
dynamic equilibrium
with solid solute
particles.
Degree of saturation
Unsaturated Solution
Less than the
maximum amount of
solute for that
temperature is
dissolved in the
solvent.
No solid remains in
flask.
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.
3 Stages of Solution Process
Separation of Solute
– must overcome IMF or ion-ion attractions in solute
– requires energy, ENDOTHERMIC ( + ∆H)
Separation of Solvent
– must overcome IMF of solvent particles
– requires energy, ENDOTHERMIC (+ ∆H)
Interaction of Solute & Solvent
– attractive bonds form between solute particles and solvent
particles
– “Solvation” or “Hydration” (where water = solvent)
– releases energy, EXOTHERMIC (- ∆H)
Dissolution at the molecular level?
Consider the dissolution of NaOH in H2O
TYPES OF HETEROGENEOUS
MIXTURES
SUSPENSIONS
A suspension is a mixture in
which particles of a material are
dispersed throughout a liquid or a
gas but are large enough that they
settle out.
The particles in a suspension are
large enough to scatter or block
light. A suspension can be
separated by passing it through a
filter.
SUSPENSIONS
Different components are in different phase, such as
solids in liquids or liquids in gases
It is necessary to shake the substance before using it
COLLOIDS
Is a mixture in which the particles are
spread throughout but are not large
enough to settle out.
The particles are not as small as
those of a solution, however are
smaller than those of a suspension
COLLOIDS
Particles in a colloid are large enough to scatter light.
A colloid cannot be separated by passing it through a
filter.
COLLOIDS
Consists of two separated
phases: Disperse phase (or
internal phase) and a continuos
phase (or dispersion medium).
May be solid, liquid or gas
Some are translucent because of
the Tyndall Effect (which is the
scattering of light)
Solutions Colloids Suspensions
Homogeneous Heterogeneous Heterogeneous
Solution in which dispersoid
or consists of small
molecules or ions.
(Evenly distributed)
Solution in which dispersoid
consists of single large
molecule or group of small
molecule.
Solution in which dispersoid
consists of group of large
molecules. ( not evenly
distributed at all)
Particle size: 0.01-1 nm;
can be atoms, ions,
molecules
Particle size: 1-1000 nm,
dispersed; can be combined
or large molecules
Particle size: over 1000 nm,
suspended; can be large
particles or combined
particles
Do not separate on standing Do not separate on standing Particles settle out
Do not scatter light Scatter light (Tyndall effect) May scatter light, but are
not transparent
They may pass through
ordinary as well as ultra
filters.
They may pass through
ordinary filter but not usually
through ultra filters
Thy cant pass through
ordinary as well as ultra
filters.
Examples:
1)Solution of Nacl in water.
2)Solution of glucose in
Examples:
1)Solution of satrch
2)Milk
Examples:
1)Pharmaceutical
suspension and emulsions
Ultra filtration is a separation process using
membranes with a pore sizes in the range of 0.1 to
0.001 micron. Typically, ultrafiltration will remove high
molecular weight substances, colloidal materials and
organic polymeric molecules. Low molecular weight
organics and ions such as sodium, calcium,
magnesium and sulfates are not removed.
HETEROGENEOUS MIXTURES
You can distinguish the two or more phases.
Gases in Solution
In general, the solubility of
gases in water increases
with increasing mass.
Larger molecules have
stronger dispersion
forces.
Gases in Solution
The solubility of
liquids and solids
does not change
appreciably with
pressure.
The solubility of a gas
in a liquid is directly
proportional to its
pressure.
Henry’s Law
Sg = kPg
where
Sg is the solubility of the
gas;
k is the Henry’s law
constant for that gas in that
solvent;
Pg is the partial pressure of
the gas above the liquid.
At a certain temperature, the Henry’s
law constant for N2 is 6.0 × 10−4
M/atm.
If N2 is present at 3.0 atm, what is the
solubility of N2?
1. 6.0 × 10−4
M
2. 1.8 × 10−3
M
3. 2.0 × 10−4
M
4. 5.0 × 10−5
M
Correct Answer:
Henry’s law, Sg
= kPg
Sg = (6.0 × 10−4
M/atm)(3.0 atm)
S = 1.8 × 10−3
M
1. 6.0 × 10−4
M
2. 1.8 × 10−3
M
3. 2.0 × 10−4
M
4. 5.0 × 10−5
M
Calculate the concentration of CO2 in a soft
drink that is bottled with a partial pressure
of CO2 of 4.0 atm over the liquid at 25°C.
The Henry’s law constant for CO2 in water
at this temperature is 3.1 × 10–2
mol/L-atm.
Solve:
Temperature
Generally, the
solubility of solid
solutes in liquid
solvents
increases with
increasing
temperature.
Temperature
The opposite is
true of gases:
Carbonated soft
drinks are more
“bubbly” if stored
in the refrigerator.
Warm lakes have
less O2 dissolved
in them than cool
lakes.
CONCENTRATION
EXPRESSION
5 ways of expressing concentration5 ways of expressing concentration
–Mass percent: (mass solute / mass of solution) * 100
–Molarity(M): moles solute / Liter solution
–Molality* (m) - moles solute / Kg solvent
–Normality (N)- gram equivalent of solute/ liter solution
–Mole Fraction(χ
A) - moles solute / total moles solution
* Note that molality is the only concentration unit in which
denominator contains only solvent information rather than
solution.
Mass Percentage or Percentage
Expression
Mass % of A =
mass of A in solution
total mass of solution
× 100
% expression is an expression of parts of solute per 100 parts
Of the solution
Determine the mass percentage of
hexane in a solution containing
11 g of butane in 110 g of hexane.
1.9.0 %
2.10. %
3.90.%
4.91 %
Correct Answer:
Thus,
110 g
(110 g + 11 g)
100
solutionofmasstotal
solutionincomponentofmass
componentof%mass ×=
× 100 = 91%
1. 9.0 %
2. 10. %
3. 90.%
4. 91 %
% (w/w) =
% (w/v) =
% (v/v) =
% Concentration
100x
solutionmass
solutemass
100x
solutionvolume
solutemass
100x
solutionvolume
solutevolume
% w/w:
It expresses the no. of grams of the solute per 100
gram of the solution.
e.g. a 10 % w/w aqueous glycerine solution means 10 g
of glycerine dissolved in sufficient water to make
overall 100 gram of the solution.
% v/v:
It expresses the no. of milliliters of the solute per 100
milliliters of the solution.
e.g. a 10 % v/v aqueous ethanolic solution means 10 ml
of ethanol dissolved in sufficient water to make
overall 100 mls of the solution.
% w/v
It expresses the no. of grams of the solute per 100
mls of the solution.
e.g. a 10 % w/v aqueous Nacl solution means 10 g of
Nacl dissolved in sufficient water to make overall 100
mls of the solution.
% v/w
It expresses the no. of mls of the solute per 100 gram
of the solution.
e.g. a 10 % v/w aqueous glycerine solution means 10
ml of glycerine dissolved in sufficient water to make
overall 100 gram of the solution.
This is another way to expressing concentration,
particularly those of very dilute solutions. e.g. to
express the impurities of substances in water.
ppm denotes the amount of given substance in a total
amount of 1,000,000 of solution e.g. one milligram
per kilogram. 1 part in 10 6
ppb denotes the amount of given substance in a total
amount of 1,000,000,000 of solution e.g. 0.001
milligram per kilogram. 1 part in 10 9
Parts per Million and
Parts per Billion
Parts per Million and
Parts per Billion
ppm =
mass of A in solution
total mass of solution
× 106
Parts per Million (ppm)
Parts per Billion (ppb)
ppb =
mass of A in solution
total mass of solution
× 109
If 3.6 mg of Na+
is detected in a 200. g
sample of water from Lake Erie, what is
its concentration in ppm?
1. 7.2 ppm
2. 1.8 ppm
3. 18 ppm
4. 72 ppm
Correct Answer:
6
10
solutionofmasstotal
solutionincomponentofmass
componentofppm ×=
ppm1810
g200.
g0.0036
g200.
mg3.6 6
=×=
1. 7.2 ppm
2. 1.8 ppm
3. 18 ppm
4. 72 ppm
mole of solute
L of solution
M =
Molarity (M)
molarity= Given weight / molecular weight substance x
1/ volume of solution in liters
e.g.1 mole of Nacl= 58.5 gm of Nacl
No. of moles= Given weight / molecular weight substance
Because volume is temperature dependent, molarity can
change with temperature.
It expresses the no. of moles of solute
dissolved per liter of the solution.
Concentration: Molarity Example
If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of
solution, what is the molarity of KMnO4?
Now that the number of moles of substance is known, this can be
combined with the volume of solution — which must be in
liters — to give the molarity. Because 250. mL is equivalent to
0.250 L .
As is almost always the
case, the first step is to
convert the mass of
material to moles.
0.435 g KMnO4• 1 mol KMnO4 = 0.00275 mol KMnO4
158.0 g KMnO4
Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M
0.250 L solution
moles of solute
kg of solvent
m =
Molality (m)
Because both moles and mass do not
change with temperature, molality
(unlike molarity) is not temperature
dependent.
It expresses the no. of moles of solute dissolved
per kg of the solvent.
Changing Molarity to Molality
If we know the
density of the
solution, we can
calculate the
molality from the
molarity, and
vice versa.
A solution is made by dissolving 4.35 g glucose (C6H12O6)
in 25.0 mL of water at 25°C. Calculate the molality of
glucose in the solution.
Solution molar mass of glucose, 180.2 g/mol
water has a density of 1.00 g/mL, so the mass of the solvent is
Normality (N)
It expresses the no. of gram equivalent of solute
dissolved per liter of the solution.
Normality= Given weight / equivalent weight x 1/
volume of solution in liters
No. gram equivalent of solute
L of solution
N =
EQUIVALENT WEIGHT
The no. of parts by weight of that substance that will
combine with one part by weight of replaceable H+ or
OH- or number of positive charge on elements
Solution
moles of A
total moles in solution
XA =
Mole Fraction (X)
The sum of the mole fractions of all the components
is always equal to unity i.e. 1
The ratio of the number of the moles of that
component to the total number of moles of all the
components of the solution
Colligative Properties
Colligative Properties are properties of a liquid that
change when a solute is added.
The magnitude of the change depends on the numbernumber
of solute particles in the solution, NOT on the naturenature
and sizeand size of the solute particles.
The colligative properties arise from the attractive
forces that are exerted by the solute on the solvent.
e.g. such attractive forces reduces the tendency of the
solvent molecules to escape from the liquids as
vapours and vapours pressure of the solvent is
therefore reduced by the presence of the solvent.
Following are the colligative properties of the solution.
 Lowering of the vapour pressure
 Elevation of the boiling point
 Depression of the freezing point
 Osmotic pressure
Lowering of the vapour pressure
If any non-volatile and non- electrolyte solute is
dissolved in the solvent, the escaping tendency of the
solvent molecule is reduced and in turn vapour
pressure is also reduced. This is because solvent
molecules are hindered by the solute molecules and
so escaping tendency of the solvent molecules is
reduced. Therefore a less no. of solvent molecules
will be able to escape and so less no. of the vapours
will be formed and ultimately the vapour pressure of
the solvent is reduced. Greater the no. of the solute
particles, greater will be the hindrance thus lower is
the vapour pressure.
In case, when any non-volatile and electrolyte is
dissolved in the solvent, the lowering in the vapour
pressure will be abnormally high than that of the non-
electrolyte. This is because, the electrolyte dissociate
into ions in the solution, so no. of the particle will
increased and as a result, lowering in vapour
pressure will also high.
Elevation in boiling point
Greater the vapour pressure of a liquid, lower will be
its boiling point. Similarly, lower the vapour pressure
of the liquid, higher will be its boiling point. So If any
non-volatile and non- electrolyte solute is dissolved in
the solvent, the escaping tendency of the solvent
molecule is reduced and in turn vapour pressure is
also reduced and ultimately its boiling point will be
elevated.
In case, when any non-volatile and electrolyte is
dissolved in the solvent, the lowering in the vapour
pressure will be abnormally high than that of the non-
electrolyte. This is because, the electrolyte dissociate
into ions in the solution, so no. of the particle will
increased and as a result, lowering in vapour
pressure will also high. There fore elevation in boiling
point is also greater than that in case of non
electrolyte.
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 BoilingExpress by Boiling
point Elevationpoint Elevation
equationequation
Depression in freezing point
The normal freezing point of solvent (pure liquid) is
defined as the temperature at which solid and liquids
forms of the solvent co- exist in equilibrium at a fixed
external pressure commonly 1 atm. ( i.e. 760 mm Hg)
Pure water has freezing point at 0˚C.
“At this equilibrium state the solid and liquid forms of the
solvent must have the same vapour pressure. If it is
not so, the form having the higher vapour pressure
would change into that form having lower vapour
pressure.”
( lowering in vapour pressure of the liquid after addition
of solute has been discussed earlier)
Solution
So the freezing point of the solution is the
temperature at which solid forms of the pure solvent
co- exist in equilibrium with the solution at a fixed
external pressure commonly 1 atm. ( i.e. 760 mm Hg)
As the vapour pressure of the solution is lower than
that of pure solvent so it is obvious that the solid and
solution cant be co exist in equilibrium ( this is
because at equilibrium both states must have same
vapour pressure). They can now co – exist a t some
lower temperature ( new freezing point), where the
solid and solution have the same vapour pressure.
That’s why freezing point of the solvent will be lowered
upon addition of solute.
Greater no. of particles= lower of V.P= Depression in F.P
Osmotic pressure
Osmotic pressure
Diffusion:
Is the movement of particles along the concentration gradient i.e the
movement of particles from higher solute concentration to lower solute
concentration.
Osmosis is the spontaneous movement of water across a semi-
permeable membrane from an area of low solute concentration to an
area of high solute concentration
or
osmosis is the movement of solvent molecule from dilute solution to
concentrated solution through semi permeable membrane.
( semi permeable membrane is a type of membrane which permits only
solvent molecules to pass through it)
Osmotic Pressure - The hydro static pressure that builds up on the
semi permeable membrane which just stops the osmosis of pure
solvent into the solution through semi permeable membrane.
Solution
Demonstration
Tie a cellophane paper (semi permeable) at the end
of the thistle funnel and immersed it in the bath of
pure water. Then add sugar solution in the funnel up
to the mark “X” and kept the apparatus for few hours.
After the level of the sugar solution will rise from
mark “X” to mark “Y”. This is due to the osmosis of
water from the water bath to the sugar solution in the
thistle funnel due to the movement of solvent
molecules. After reaching at mark “Y”, the movement
of the molecules will be stopped further that was due
to the hydrostatic pressure exerted by the column of the
solution on the semipermeable membrane which
prevents the entry of the pure solvent to the solution.
So, if the concentration of the solution will high, more
solvent molecules will move the solution and the level of
the column of the solution will rise higher and greater
will be the pressure on the membrane and ultimately
greater will be the osmotic pressure and vice versa.
Osmotic pressure α concentration of the solution
Osmosis and Blood Cells
(a) A cell placed in an isotonic solution. The net movement of
water in and out of the cell is zero because the concentration of
solutes inside and outside the cell is the same.
(b) In a hypertonic solution, the concentration of solutes outside
the cell is greater than that inside. There is a net flow of water
out of the cell, causing the cell to dehydrate, shrink, and
perhaps die.
(c) In a hypotonic solution, the concentration of solutes outside of
the cell is less than that inside. There is a net flow of water into
the cell, causing the cell to swell and perhaps to burst.
IDEAL SOLUTIONS
Cohesive forces: forces that exist between the two
similar types of the molecules i.e. between solute-
solute or between solvent-solvent.
Adhesive forces: forces that exist between different
types of molecules i.e. between solute and solvent.
“ Solutions in which the adhesive and cohesive forces
are same (equal) are known as ideal solutions”
Or
“Solutions that obey Raoult's law”
Raoult`s Law:
At a definite temperature, the partial pressure (PA) of
component (A) in a liquid mixture is equal to the
vapour pressure of that component in the pure state
(P°A ) multiplied by the mole fraction (χ A) of that
component in the solution.
PA = χA P°A
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 = PA + PB
The vapor pressure from each component follows
Raoult’s Law:
PT = χAP°A + χBP°B
Also, χA + χB = 1 (since there are 2 components)
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, χBz + χTol = 1
• Benzene: when PBz = P°Bz = 384 torr & χBz = 0.5
• Toluene: when PTol = P°Tol = 133 torr & χTol = 0.5
PT = χBz P°BZ + χTol P°TOL
384 torr
133 torr
X Benzene
X Toluene
0 1
1 0
P (Total)
P (Benzene)
P (Toluene)
133 torr
384 torr
Characteristics of an ideal solution:
Ideal behavior is expected to be exhibited by the
systems which comprises of the chemical similar
compounds, because it is only in such systems that
the conditions of equal intermolecular forces between
components are likely to be satisfied.
Examples: solutions of ethyl alcohol- methyl alcohol,
chloroform-bromoform, benzene- toluene.
Ideal solutions have zero enthalpy change i.e. heat is
neither absorbed nor evolved during solution
formation.
The volume of the solution is exactly equal to the
sum of the individual volumes of the components.
Non- Ideal or Real Solutions
Solutions in which cohesive and adhesive forces are
not equal are known as non-ideal or real solutions
Solutions in which solute-solute, solvent-solvent and
solute-solvent attractive forces are not equal.
Solutions which don’t obey Raoults`s law
Deviations from Raoult`s Law
Negative deviation from Raoult`s Law:
Real solution which showed negative deviation from
Raoult`s Law are those solution in which adhesive
forces ( i.e. solute-solvent) are stronger than the
cohesive forces (i.e. solute-solute or solvent-solvent)
and vapour pressure of the solution is less than
expected from Raoult`s law.
i.e. A-B > A-A, B-B
Reasoning of lowering of vapour
pressure
It attractive forces between solute and solvent (A-B)
are stronger than those exerted between solute –
solute (A-A) and solvent- solvent (B-B) molecules, then
this strong mutual affinity between solute and solvent
molecules results in the formation of complex or
compound which results in strong holding of solvent
molecules and results in lowering of escaping
tendency of solvent molecules and ultimately lowering
of vapour pressure. When this occur, there may be
decrease in solution volume occur than the sum of
volume of the components.
E.g. Chloroform- Ethanol, Benzene - Ethanol
Positive deviation from Raoult`s Law:
Real solution which showed positive deviation from
Raoult`s Law are those solution in which cohesive
forces (i.e. solute-solute or solvent-solvent) are
stronger than the adhesive forces ( i.e. solute-
solvent)and vapour pressure of the solution is greater
than expected from Raoult`s law.
i.e. A-B < A-A, B-B
Reasoning of elevation of vapour
pressure
It attractive forces between solute and solvent (A-B)
are less than those exerted between solute – solute
(A-A) and solvent- solvent (B-B) molecules, then the
presence of “A” reduces the (B-B) attraction and
similarly presence of “B” molecules reduces (A-A)
attraction. This results in greater escaping tendency of
A and B and ultimately partial vapour pressure of the
components are greater than expected from Raoult`s
law showed positive deviation. When this occur, there
may be increase in solution volume occur than the sum
of volume of the components.
E.g. Chloroform- Acetone, Pyridine- Acetic Acid

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Solution

  • 1. Solutions IMRAN TARIQ ASSISTANT PROFESSOR UNIVERSITY COLLEGE OF PHARMACY, UNIVERSITY OF THE PUNJAB
  • 2. TYPES OF MIXTURES HOMOGENOUS e.g. True solutions HETEROGENOUS e.g. coarse and colloidal dispersions Homogeneity: something evenly distributed Heterogeneity: something not distributed evenly in space; a clump or cluster (not a single phase as in case of colloidal and coarse dispersion).
  • 3. Definitions A solution is a homogeneous mixture of two or more substances OR A solution is a homogenous mixture of solute and solvent. OR  A solution is a homogenous mixture of two substances but consisting of one phase.
  • 4. A solute is dissolved in a solvent. – solute is the substance being dissolved – solvent is the liquid in which the solute is dissolved – an aqueous solution has water as solvent Binary solution: A homogenous mixture consisting of one phase and containing only two components i.e. one solute and one solvent e.g. Solution of Nacl in water.
  • 5. Dilute Solutions: A solution containing relatively small quantity of solute as compared with the amount of solvent. Concentrated Solution: A solution containing large amount of solute in the solution than that in dilute solution. Un-saturated solution: a solution in which more solute can be dissolved at a given temperature is called as an unsaturated solution.
  • 6. A saturated solution is one where the concentration is at a maximum - no more solute is able to dissolve at a given temperature. – A saturated solution represents an equilibrium: the rate of dissolving is equal to the rate of crystallization. The salt continues to dissolve, but crystallizes at the same rate or under these condition the number of molecules leaving the solute is equal to the number of molecule returning to the solid phase i.e. solute.
  • 7. Super Saturated Solution: A solution that contains relatively larger amount of solute than that required for saturation it is prepared by heating and adding more and more solute.
  • 8. Types of Solutions: Based on physical states of solute and solvent: Gas Gas Solid Smoke in air Gas Gas liquid Water vapors in air
  • 9. Solutions How does a solid dissolve into a liquid? What ‘drives’ the dissolution process? What are the energetics of dissolution?
  • 10. How Does a Solution Form? 1. Solvent molecules attracted to surface ions. 2. Each ion is surrounded by solvent molecules. 3. Enthalpy (∆H) changes with each interaction broken or formed. Ionic solid dissolving in water
  • 11. 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.
  • 12. Dissolution vs reaction Dissolution is a physical change—you can get back the original solute by evaporating the solvent. If you can’t, the substance didn’t dissolve, it reacted.
  • 13. Degree of saturation Saturated solution Solvent holds as much solute as is possible at that temperature. Undissolved solid remains in flask. Dissolved solute is in dynamic equilibrium with solid solute particles.
  • 14. Degree of saturation Unsaturated Solution Less than the maximum amount of solute for that temperature is dissolved in the solvent. No solid remains in flask.
  • 15. 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.
  • 16. 3 Stages of Solution Process Separation of Solute – must overcome IMF or ion-ion attractions in solute – requires energy, ENDOTHERMIC ( + ∆H) Separation of Solvent – must overcome IMF of solvent particles – requires energy, ENDOTHERMIC (+ ∆H) Interaction of Solute & Solvent – attractive bonds form between solute particles and solvent particles – “Solvation” or “Hydration” (where water = solvent) – releases energy, EXOTHERMIC (- ∆H)
  • 17. Dissolution at the molecular level? Consider the dissolution of NaOH in H2O
  • 18. TYPES OF HETEROGENEOUS MIXTURES SUSPENSIONS A suspension is a mixture in which particles of a material are dispersed throughout a liquid or a gas but are large enough that they settle out. The particles in a suspension are large enough to scatter or block light. A suspension can be separated by passing it through a filter.
  • 19. SUSPENSIONS Different components are in different phase, such as solids in liquids or liquids in gases It is necessary to shake the substance before using it
  • 20. COLLOIDS Is a mixture in which the particles are spread throughout but are not large enough to settle out. The particles are not as small as those of a solution, however are smaller than those of a suspension
  • 21. COLLOIDS Particles in a colloid are large enough to scatter light. A colloid cannot be separated by passing it through a filter.
  • 22. COLLOIDS Consists of two separated phases: Disperse phase (or internal phase) and a continuos phase (or dispersion medium). May be solid, liquid or gas Some are translucent because of the Tyndall Effect (which is the scattering of light)
  • 23. Solutions Colloids Suspensions Homogeneous Heterogeneous Heterogeneous Solution in which dispersoid or consists of small molecules or ions. (Evenly distributed) Solution in which dispersoid consists of single large molecule or group of small molecule. Solution in which dispersoid consists of group of large molecules. ( not evenly distributed at all) Particle size: 0.01-1 nm; can be atoms, ions, molecules Particle size: 1-1000 nm, dispersed; can be combined or large molecules Particle size: over 1000 nm, suspended; can be large particles or combined particles Do not separate on standing Do not separate on standing Particles settle out Do not scatter light Scatter light (Tyndall effect) May scatter light, but are not transparent They may pass through ordinary as well as ultra filters. They may pass through ordinary filter but not usually through ultra filters Thy cant pass through ordinary as well as ultra filters. Examples: 1)Solution of Nacl in water. 2)Solution of glucose in Examples: 1)Solution of satrch 2)Milk Examples: 1)Pharmaceutical suspension and emulsions
  • 24. Ultra filtration is a separation process using membranes with a pore sizes in the range of 0.1 to 0.001 micron. Typically, ultrafiltration will remove high molecular weight substances, colloidal materials and organic polymeric molecules. Low molecular weight organics and ions such as sodium, calcium, magnesium and sulfates are not removed.
  • 25. HETEROGENEOUS MIXTURES You can distinguish the two or more phases.
  • 26. Gases in Solution In general, the solubility of gases in water increases with increasing mass. Larger molecules have stronger dispersion forces.
  • 27. Gases in Solution The solubility of liquids and solids does not change appreciably with pressure. The solubility of a gas in a liquid is directly proportional to its pressure.
  • 28. Henry’s Law Sg = kPg where Sg is the solubility of the gas; k is the Henry’s law constant for that gas in that solvent; Pg is the partial pressure of the gas above the liquid.
  • 29. At a certain temperature, the Henry’s law constant for N2 is 6.0 × 10−4 M/atm. If N2 is present at 3.0 atm, what is the solubility of N2? 1. 6.0 × 10−4 M 2. 1.8 × 10−3 M 3. 2.0 × 10−4 M 4. 5.0 × 10−5 M
  • 30. Correct Answer: Henry’s law, Sg = kPg Sg = (6.0 × 10−4 M/atm)(3.0 atm) S = 1.8 × 10−3 M 1. 6.0 × 10−4 M 2. 1.8 × 10−3 M 3. 2.0 × 10−4 M 4. 5.0 × 10−5 M
  • 31. Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this temperature is 3.1 × 10–2 mol/L-atm. Solve:
  • 32. Temperature Generally, the solubility of solid solutes in liquid solvents increases with increasing temperature.
  • 33. Temperature The opposite is true of gases: Carbonated soft drinks are more “bubbly” if stored in the refrigerator. Warm lakes have less O2 dissolved in them than cool lakes.
  • 35. 5 ways of expressing concentration5 ways of expressing concentration –Mass percent: (mass solute / mass of solution) * 100 –Molarity(M): moles solute / Liter solution –Molality* (m) - moles solute / Kg solvent –Normality (N)- gram equivalent of solute/ liter solution –Mole Fraction(χ A) - moles solute / total moles solution * Note that molality is the only concentration unit in which denominator contains only solvent information rather than solution.
  • 36. Mass Percentage or Percentage Expression Mass % of A = mass of A in solution total mass of solution × 100 % expression is an expression of parts of solute per 100 parts Of the solution
  • 37. Determine the mass percentage of hexane in a solution containing 11 g of butane in 110 g of hexane. 1.9.0 % 2.10. % 3.90.% 4.91 %
  • 38. Correct Answer: Thus, 110 g (110 g + 11 g) 100 solutionofmasstotal solutionincomponentofmass componentof%mass ×= × 100 = 91% 1. 9.0 % 2. 10. % 3. 90.% 4. 91 %
  • 39. % (w/w) = % (w/v) = % (v/v) = % Concentration 100x solutionmass solutemass 100x solutionvolume solutemass 100x solutionvolume solutevolume
  • 40. % w/w: It expresses the no. of grams of the solute per 100 gram of the solution. e.g. a 10 % w/w aqueous glycerine solution means 10 g of glycerine dissolved in sufficient water to make overall 100 gram of the solution.
  • 41. % v/v: It expresses the no. of milliliters of the solute per 100 milliliters of the solution. e.g. a 10 % v/v aqueous ethanolic solution means 10 ml of ethanol dissolved in sufficient water to make overall 100 mls of the solution.
  • 42. % w/v It expresses the no. of grams of the solute per 100 mls of the solution. e.g. a 10 % w/v aqueous Nacl solution means 10 g of Nacl dissolved in sufficient water to make overall 100 mls of the solution.
  • 43. % v/w It expresses the no. of mls of the solute per 100 gram of the solution. e.g. a 10 % v/w aqueous glycerine solution means 10 ml of glycerine dissolved in sufficient water to make overall 100 gram of the solution.
  • 44. This is another way to expressing concentration, particularly those of very dilute solutions. e.g. to express the impurities of substances in water. ppm denotes the amount of given substance in a total amount of 1,000,000 of solution e.g. one milligram per kilogram. 1 part in 10 6 ppb denotes the amount of given substance in a total amount of 1,000,000,000 of solution e.g. 0.001 milligram per kilogram. 1 part in 10 9 Parts per Million and Parts per Billion
  • 45. Parts per Million and Parts per Billion ppm = mass of A in solution total mass of solution × 106 Parts per Million (ppm) Parts per Billion (ppb) ppb = mass of A in solution total mass of solution × 109
  • 46. If 3.6 mg of Na+ is detected in a 200. g sample of water from Lake Erie, what is its concentration in ppm? 1. 7.2 ppm 2. 1.8 ppm 3. 18 ppm 4. 72 ppm
  • 48. mole of solute L of solution M = Molarity (M) molarity= Given weight / molecular weight substance x 1/ volume of solution in liters e.g.1 mole of Nacl= 58.5 gm of Nacl No. of moles= Given weight / molecular weight substance Because volume is temperature dependent, molarity can change with temperature. It expresses the no. of moles of solute dissolved per liter of the solution.
  • 49. Concentration: Molarity Example If 0.435 g of KMnO4 is dissolved in enough water to give 250. mL of solution, what is the molarity of KMnO4? Now that the number of moles of substance is known, this can be combined with the volume of solution — which must be in liters — to give the molarity. Because 250. mL is equivalent to 0.250 L . As is almost always the case, the first step is to convert the mass of material to moles. 0.435 g KMnO4• 1 mol KMnO4 = 0.00275 mol KMnO4 158.0 g KMnO4 Molarity KMnO4 = 0.00275 mol KMnO4 = 0.0110 M 0.250 L solution
  • 50. moles of solute kg of solvent m = Molality (m) Because both moles and mass do not change with temperature, molality (unlike molarity) is not temperature dependent. It expresses the no. of moles of solute dissolved per kg of the solvent.
  • 51. Changing Molarity to Molality If we know the density of the solution, we can calculate the molality from the molarity, and vice versa.
  • 52. A solution is made by dissolving 4.35 g glucose (C6H12O6) in 25.0 mL of water at 25°C. Calculate the molality of glucose in the solution. Solution molar mass of glucose, 180.2 g/mol water has a density of 1.00 g/mL, so the mass of the solvent is
  • 53. Normality (N) It expresses the no. of gram equivalent of solute dissolved per liter of the solution. Normality= Given weight / equivalent weight x 1/ volume of solution in liters No. gram equivalent of solute L of solution N =
  • 54. EQUIVALENT WEIGHT The no. of parts by weight of that substance that will combine with one part by weight of replaceable H+ or OH- or number of positive charge on elements
  • 56. moles of A total moles in solution XA = Mole Fraction (X) The sum of the mole fractions of all the components is always equal to unity i.e. 1 The ratio of the number of the moles of that component to the total number of moles of all the components of the solution
  • 57. Colligative Properties Colligative Properties are properties of a liquid that change when a solute is added. The magnitude of the change depends on the numbernumber of solute particles in the solution, NOT on the naturenature and sizeand size of the solute particles. The colligative properties arise from the attractive forces that are exerted by the solute on the solvent. e.g. such attractive forces reduces the tendency of the solvent molecules to escape from the liquids as vapours and vapours pressure of the solvent is therefore reduced by the presence of the solvent.
  • 58. Following are the colligative properties of the solution.  Lowering of the vapour pressure  Elevation of the boiling point  Depression of the freezing point  Osmotic pressure
  • 59. Lowering of the vapour pressure
  • 60. If any non-volatile and non- electrolyte solute is dissolved in the solvent, the escaping tendency of the solvent molecule is reduced and in turn vapour pressure is also reduced. This is because solvent molecules are hindered by the solute molecules and so escaping tendency of the solvent molecules is reduced. Therefore a less no. of solvent molecules will be able to escape and so less no. of the vapours will be formed and ultimately the vapour pressure of the solvent is reduced. Greater the no. of the solute particles, greater will be the hindrance thus lower is the vapour pressure.
  • 61. In case, when any non-volatile and electrolyte is dissolved in the solvent, the lowering in the vapour pressure will be abnormally high than that of the non- electrolyte. This is because, the electrolyte dissociate into ions in the solution, so no. of the particle will increased and as a result, lowering in vapour pressure will also high.
  • 63. Greater the vapour pressure of a liquid, lower will be its boiling point. Similarly, lower the vapour pressure of the liquid, higher will be its boiling point. So If any non-volatile and non- electrolyte solute is dissolved in the solvent, the escaping tendency of the solvent molecule is reduced and in turn vapour pressure is also reduced and ultimately its boiling point will be elevated. In case, when any non-volatile and electrolyte is dissolved in the solvent, the lowering in the vapour pressure will be abnormally high than that of the non- electrolyte. This is because, the electrolyte dissociate into ions in the solution, so no. of the particle will increased and as a result, lowering in vapour pressure will also high. There fore elevation in boiling point is also greater than that in case of non electrolyte.
  • 64. 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 BoilingExpress by Boiling point Elevationpoint Elevation equationequation
  • 66. The normal freezing point of solvent (pure liquid) is defined as the temperature at which solid and liquids forms of the solvent co- exist in equilibrium at a fixed external pressure commonly 1 atm. ( i.e. 760 mm Hg) Pure water has freezing point at 0˚C. “At this equilibrium state the solid and liquid forms of the solvent must have the same vapour pressure. If it is not so, the form having the higher vapour pressure would change into that form having lower vapour pressure.” ( lowering in vapour pressure of the liquid after addition of solute has been discussed earlier)
  • 68. So the freezing point of the solution is the temperature at which solid forms of the pure solvent co- exist in equilibrium with the solution at a fixed external pressure commonly 1 atm. ( i.e. 760 mm Hg) As the vapour pressure of the solution is lower than that of pure solvent so it is obvious that the solid and solution cant be co exist in equilibrium ( this is because at equilibrium both states must have same vapour pressure). They can now co – exist a t some lower temperature ( new freezing point), where the solid and solution have the same vapour pressure. That’s why freezing point of the solvent will be lowered upon addition of solute. Greater no. of particles= lower of V.P= Depression in F.P
  • 70. Osmotic pressure Diffusion: Is the movement of particles along the concentration gradient i.e the movement of particles from higher solute concentration to lower solute concentration. Osmosis is the spontaneous movement of water across a semi- permeable membrane from an area of low solute concentration to an area of high solute concentration or osmosis is the movement of solvent molecule from dilute solution to concentrated solution through semi permeable membrane. ( semi permeable membrane is a type of membrane which permits only solvent molecules to pass through it) Osmotic Pressure - The hydro static pressure that builds up on the semi permeable membrane which just stops the osmosis of pure solvent into the solution through semi permeable membrane.
  • 72. Demonstration Tie a cellophane paper (semi permeable) at the end of the thistle funnel and immersed it in the bath of pure water. Then add sugar solution in the funnel up to the mark “X” and kept the apparatus for few hours. After the level of the sugar solution will rise from mark “X” to mark “Y”. This is due to the osmosis of water from the water bath to the sugar solution in the thistle funnel due to the movement of solvent molecules. After reaching at mark “Y”, the movement of the molecules will be stopped further that was due
  • 73. to the hydrostatic pressure exerted by the column of the solution on the semipermeable membrane which prevents the entry of the pure solvent to the solution. So, if the concentration of the solution will high, more solvent molecules will move the solution and the level of the column of the solution will rise higher and greater will be the pressure on the membrane and ultimately greater will be the osmotic pressure and vice versa. Osmotic pressure α concentration of the solution
  • 74. Osmosis and Blood Cells (a) A cell placed in an isotonic solution. The net movement of water in and out of the cell is zero because the concentration of solutes inside and outside the cell is the same. (b) In a hypertonic solution, the concentration of solutes outside the cell is greater than that inside. There is a net flow of water out of the cell, causing the cell to dehydrate, shrink, and perhaps die. (c) In a hypotonic solution, the concentration of solutes outside of the cell is less than that inside. There is a net flow of water into the cell, causing the cell to swell and perhaps to burst.
  • 76. Cohesive forces: forces that exist between the two similar types of the molecules i.e. between solute- solute or between solvent-solvent. Adhesive forces: forces that exist between different types of molecules i.e. between solute and solvent. “ Solutions in which the adhesive and cohesive forces are same (equal) are known as ideal solutions” Or “Solutions that obey Raoult's law”
  • 77. Raoult`s Law: At a definite temperature, the partial pressure (PA) of component (A) in a liquid mixture is equal to the vapour pressure of that component in the pure state (P°A ) multiplied by the mole fraction (χ A) of that component in the solution. PA = χA P°A
  • 78. Mixtures of Volatile Liquids Both liquids evaporate & contribute to the vapor pressure
  • 79. 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 = PA + PB The vapor pressure from each component follows Raoult’s Law: PT = χAP°A + χBP°B Also, χA + χB = 1 (since there are 2 components)
  • 80. 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, χBz + χTol = 1 • Benzene: when PBz = P°Bz = 384 torr & χBz = 0.5 • Toluene: when PTol = P°Tol = 133 torr & χTol = 0.5 PT = χBz P°BZ + χTol P°TOL
  • 81. 384 torr 133 torr X Benzene X Toluene 0 1 1 0 P (Total) P (Benzene) P (Toluene) 133 torr 384 torr
  • 82. Characteristics of an ideal solution: Ideal behavior is expected to be exhibited by the systems which comprises of the chemical similar compounds, because it is only in such systems that the conditions of equal intermolecular forces between components are likely to be satisfied. Examples: solutions of ethyl alcohol- methyl alcohol, chloroform-bromoform, benzene- toluene. Ideal solutions have zero enthalpy change i.e. heat is neither absorbed nor evolved during solution formation. The volume of the solution is exactly equal to the sum of the individual volumes of the components.
  • 83. Non- Ideal or Real Solutions
  • 84. Solutions in which cohesive and adhesive forces are not equal are known as non-ideal or real solutions Solutions in which solute-solute, solvent-solvent and solute-solvent attractive forces are not equal. Solutions which don’t obey Raoults`s law
  • 86. Negative deviation from Raoult`s Law: Real solution which showed negative deviation from Raoult`s Law are those solution in which adhesive forces ( i.e. solute-solvent) are stronger than the cohesive forces (i.e. solute-solute or solvent-solvent) and vapour pressure of the solution is less than expected from Raoult`s law. i.e. A-B > A-A, B-B
  • 87. Reasoning of lowering of vapour pressure It attractive forces between solute and solvent (A-B) are stronger than those exerted between solute – solute (A-A) and solvent- solvent (B-B) molecules, then this strong mutual affinity between solute and solvent molecules results in the formation of complex or compound which results in strong holding of solvent molecules and results in lowering of escaping tendency of solvent molecules and ultimately lowering of vapour pressure. When this occur, there may be decrease in solution volume occur than the sum of volume of the components. E.g. Chloroform- Ethanol, Benzene - Ethanol
  • 88. Positive deviation from Raoult`s Law: Real solution which showed positive deviation from Raoult`s Law are those solution in which cohesive forces (i.e. solute-solute or solvent-solvent) are stronger than the adhesive forces ( i.e. solute- solvent)and vapour pressure of the solution is greater than expected from Raoult`s law. i.e. A-B < A-A, B-B
  • 89. Reasoning of elevation of vapour pressure It attractive forces between solute and solvent (A-B) are less than those exerted between solute – solute (A-A) and solvent- solvent (B-B) molecules, then the presence of “A” reduces the (B-B) attraction and similarly presence of “B” molecules reduces (A-A) attraction. This results in greater escaping tendency of A and B and ultimately partial vapour pressure of the components are greater than expected from Raoult`s law showed positive deviation. When this occur, there may be increase in solution volume occur than the sum of volume of the components. E.g. Chloroform- Acetone, Pyridine- Acetic Acid