Colligative Properties Kausar Ahmad Kulliyyah of Pharmacy 1PHM1153 Physical Pharmacy 1 2010/11

Colligative Properties

Kausar AhmadKulliyyah of Pharmacy

http://staff.iiu.edu.my/akausar

1PHM1153 Physical Pharmacy 1

2010/11

http://staff.iiu.edu.my/akausar

Contents

• Effect of solute on vapour pressure• Boiling point elevation• Freezing point depression• Osmotic pressure

Lecture 1

• Raoult’s law• Deviation from Raoult’s law

Lecture 2

• Distillation• Azeotropic mixtures• Clausius-Clapeyron equation

Lecture 3

PHM1153 Physical Pharmacy 1 2010/11

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Introduction

Solutions have different properties than either the pure solvent or the solute.

• a solution of sugar in water is neither crystalline like sugar nor tasteless like water.

Some of the properties unique to solutions depend only on the number of dissolved particles and not their identity.


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Definition

• Properties of a liquid that may be altered by the presence of a solute.

• Depend only on the amount of solute in a solution • Does not depend on the identity of the solute

Colligative properties

• Depend on the identity of the dissolved solute and the solvent

Non-colligative properties


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Colligative vs Non-colligative

5

Compare 1.0 M aqueous sugar solution to a 0.5 M solution of salt (NaCl) in water.

both solutions have the same number of dissolved particles

any difference in the properties of those two solutions is due to a non-colligative property.

Both have the same freezing point, boiling point, vapor pressure, and osmotic pressure


Non-Colligative Properties

Sugar solution is sweet and salt solution is salty.

Therefore, the taste of the solution is not a colligative property.

Another non-colligative property is the color of a solution. • A 0.5 M solution of CuSO4 is bright blue in contrast to the colorless

salt and sugar solutions. Other non-colligative properties include viscosity, surface tension, and solubility.


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Examples of Colligative Properties

vapour pressurefreezing point

depression (melting)

boiling point elevation osmotic pressure

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All of these properties relate to the vapor pressure.


Effect of Solute on Vapour Pressure

When a nonvolatile solute is dissolved in a solvent, the vapor pressure of the resulting solution is lower than that of the pure solvent.

The amount of the vapor pressure lowering is proportional to the amount of solute and not its identity.

Therefore, vapor pressure lowering is a colligative property.


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Vapour pressure:Solution < pure solvent


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Boiling point

Boiling point elevation is a colligative property related to vapor pressure lowering.

The boiling point is defined as the temperature at which the vapor pressure of a liquid equals the atmospheric pressure.

Due to vapor pressure lowering, a solution will require a higher temperature to reach its boiling point than the pure solvent.


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Freezing Point

Every liquid has a freezing point - the temperature at which a liquid undergoes a phase change from liquid to solid.

When solutes are added to a liquid, forming a solution, the solute molecules disrupt the formation of crystals of the solvent.

That disruption in the freezing process results in a depression of the freezing point for the solution relative to the pure solvent.


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Osmotic Pressure

When a solution is separated from a volume of pure solvent by a semi-permeable membrane that allows only the passage of solvent molecules, the height of the solution begins to rise.

The value of the height difference between the two compartments reflects a property called the osmotic pressure of a solution.


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Phase Diagrams and the effect of solutes on freezing and boiling points


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end of lecture 1/3

Raoult's Law

Raoult's law states that the

vapor pressure of a solution, P1,

equals the

mole fraction of the solvent, X1 ,

multiplied by the

vapor pressure of the pure solvent, Po.

P1 = Po x X1


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A Visual Demonstration of Raoult's Law

• Description: The intensity of color of bromine vapor is reduced (P↓) when a colorless volatile liquid is added (x↓).

• Source: Journal of Chemical Education - Vol. 67

• Year : 1990 page: 598


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Total Vapour Pressure

In an ideal solution made up of two volatile components:

• P = p1 +p2

In an ideal solution made up of one volatile component:

• P = p1


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Ideal & Non-Ideal/Real Solutions

Solutions that obey Raoult's law are called ideal solutions because they behave exactly as we would predict.

Solutions that show a deviation from Raoult's law are called non-ideal solutions or real solutions because they deviate from the expected behavior.


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Deviations from Raoult's law• the "law" is approximately obeyed by most solutions, some

show deviations from the expected behavior.

• Deviations from Raoult's law can either be positive or negative.

– A positive deviation means that there is a higher than expected vapor pressure above the solution:

P1 > P1o x X1

– A negative deviation means that we find a lower than expected vapor pressure for the solution:

P1 < P1o x X1


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Reason for the deviation• In vapor pressure lowering we assumed that the solute did

not interact with the solvent at all.

• If the solute is strongly held by the solvent, the solution will show a negative deviation from Raoult's law, because the solvent will find it more difficult to escape from solution.

• If the solute and solvent are not as tightly bound to each other as they are to themselves, then the solution will show a positive deviation from Raoult's law because the solvent molecules will find it easier to escape from solution into the gas phase.


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Deviation from Raoult’s Law


Vapor Pressure of a Mixture: Raoult's Law

• The measurement of pressure exerted by a vapour is demonstrated using barometers.

• Vapor pressure varies with the strength of the intermolecular forces in the liquid.

• We can calculate the vapor pressure of a mixture using Raoult's law.


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Source: http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Manometer/Manometer.html

http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Manometer/Manometer.html

http://www.upscale.utoronto.ca/GeneralInterest/Harrison/Manometer/Manometer.html

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Example:

Consider a solution that contains 0.6 mole fraction of decane and 0.4 mole fraction of diethyl ether.

We can calculate the vapor pressure of a mixture using Raoult's law:

P = P1

o

x X1 + P2

o

x X2

= (5 x 0.6) + (460 x 0.4) = 187


Factors that affect the magnitude of the changes in melting point and boiling point.

Concentration Effect of ionic compounds


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Effect of ionic compounds

• 0.1 mole of an ionic compound such as NaCl,, has an effect on the melting and boiling points that is almost twice what we would observe for 0.1 mole sugar.

– salt ionizes into Na+ and Cl- ions in water – these ions act as independent particles on the vapor pressure

of water. – Thus NaCl, NaNO3, and CaCO3 would have multipliers of 2,

while Na2SO4, and CaCl2 would have multipliers of 3.


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Van’t Hoff factor• In dilute solutions, ionic compounds have simple multiple

effects, but as the solution concentration increases the multiplier effect diminishes.

• This phenomenon was first discovered by van't Hoff and is generally called the van't Hoff factor.

• As concentration increases, some of the ions floating in solution find one another and form ion pairs, in which two oppositely charged ions briefly stick together and act as a single particle.

• the higher the concentration, the more likely it is that two ions will find one another.


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Change in Boiling Point• One consequence of Raoult's law is that the boiling point of a

solution made of a liquid solvent with a nonvolatile solute is greater than the boiling point of the pure solvent.

• For a solution, the vapor pressure of the solvent is lower at any given temperature.

• Therefore, a higher temperature is required to boil the solution than the pure solvent.

• The change in boiling point is

ΔTb = iKbm


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m is molality

• because molality is temperature independent.

Kb is a boiling point elevation constant

• that depends on the particular solvent used.

i is the van't Hoff factor • number of dissociated moles of particles per mole of solute. • i=1 for all non-electrolyte solutes and equals the total number of ions

released for electrolytes. • Na2SO4 is 3 because that salt releases three moles of ions per mole

of the salt. 29PHM1153 Physical Pharmacy 1

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ΔTb = iKbm

Change in Freezing Point• In order for a liquid to freeze it must achieve a very ordered

state that results in the formation of a crystal.

• If there are impurities in the liquid, i.e. solutes, the liquid is inherently less ordered.

• Therefore, a solution is more difficult to freeze than the pure solvent so a lower temperature is required to freeze the liquid.

• The change in freezing point is ΔTf = -iKfm

– Note that the sign of the change in freezing point is negative because the freezing point of the solution is less than that of the pure solvent.


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Molal Boiling Point Elevation and Freezing Point Depression Constants at 1 Atm pressure

Compound Normal bp Kb (oC/m) Normal fp Kf (oC/m)

Water, H2O 100.0 0.52 0.0 1.86

Benzene, C6H6 80.1 2.53 5.5 5.12

Ethanol, C2H5OH 78.4 1.22 -114.6 1.99

Carbon tetrachloride, CCl4 76.8 5.02 -22.3 29.8

Chloroform, CHCl3 61.2 3.63 -63.5 4.68

Thus the boiling point of water would increase 0.52 oC for a one molal solution, while the freezing point of this solution would decrease by 1.86 oC.


Osmotic pressureSolvent molecules flow into solution

Volume of the solution rise

Net flow through the membrane ceases due to the extra pressure exerted by the excess height of the solution chamber.

Converting the difference in height into pressure gives the osmotic pressure

P = ρghPHM1153 Physical Pharmacy 1

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Effect of concentration on osmotic pressure

PV = inRT

• Since n / V gives the concentration of the solute in units of molarity, M

P = iMRT

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End of lecture 2/3PHM1153 Physical Pharmacy 1

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Application of Raoult’s Law

Distillation


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DISTILLATION

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Distillation is the process of boiling &

condensing the vapour back to

liquid

A simple distillation:• To change from

liquid to vapour by boiling

• To change from vapour to liquid by condensing


Distillation of Binary MixtureIdeal Solution

• When a liquid and its vapour are in equilibrium, vapour is richer in the more volatile component compared to liquid mixture

• At equilibrium, liquid and vapour phase can be separated and analysed


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Binary Mixtures Obeying Raoult’s Law Ideal Solution

• Attractive forces between molecules of different component equal those of the same component.A-B = A-A = B-B– Without a maximum or minimum at intermediate compositions.

• No change in properties of components (except dilution to form solution).

• Vapour pressure, refractive index, surface tension and viscosity of solution are averages of properties of pure individual constituents.

• No heat evolved or absorbed during mixing process.• No change in solution temperature.• No shrinkage or expansion. • Final properties of solution are additive properties of individual

constituents.• E.g. methyl alcohol-water, benzene-toluene, methanol-ethanol.


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Definition - Azeotropic Mixture• Describes a mixture of miscible liquids which

boils at a constant composition and thus the composition cannot be changed by simple distillation

• Composition of vapour similar to that of liquid• The composition as well as the boiling point of

an azeotropic mixture changes with pressure• If a liquid mixture represented by a composition

X1 is distilled, the vapour has composition X2 and condenses to form a liquid of that composition.

• Distillation starting at composition X1 produces an azeotropic mixture Z as the distillate, and the residue tends towards pure B

• Similarly, a mixture of composition Y1, yields an azeotropic mixture as distillate, and pure A as a residue


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Positive Deviation from Raoult’s Law

Vapour pressure greater than expected from Raoult’s law.

• Escaping tendency higher

The two components differ in properties e.g. polarity

Attractive forces: A-B < A-A or B-B

A max is found in vapour pressure-composition curve

Presence of azeotropic mixtures lead to min boiling point

Examples: 1. ethyl, isopropyl, n-propyl alcohol - water,2. volatile drug (methyl amphetamine) chloroform - ethanol


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Positively deviated solution mixturewith a minimum boiling point

• The mixture shows positive deviation from Raoult’s law

• A minimum boiling point is obtained

• The solution has an azeotropic mixture whose vapour pressure is the highest and boiling point is the lowest

• example of a positive azeotrope is 95.6% ethanol and 4.4% water (by weight). Ethanol boils at 78.4°C, water boils at 100°C, but the azeotrope boils at 78.1°C


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http://en.wikipedia.org/wiki/Ethanol

Distillation of positive azeotrope 95.6% ethanol and 4.4% water

• distillation of any mixture will result in the distillate being closer in composition to the azeotrope than the starting mixture.

• E.g. if a 50/50 mixture of ethanol and water is distilled once, the distillate will be 80% ethanol and 20% water i.e. closer to the azeotropic mixture than the original.

• Distilling the 80/20% mixture produces a distillate that is 87% ethanol and 13% water.

• Further repeated distillations will produce mixtures that are progressively closer to the azeotropic ratio of 95.5/4.5%.

• increasing distillations will not give distillate that exceeds the azeotropic ratio.


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http://en.wikipedia.org/wiki/Ethanol

http://en.wikipedia.org/wiki/Distillate

Negative Deviation from Raoult’s Law

Vapour pressure of solution less than that expected from Raoult’s law

• Escaping tendency lower

Attractive forces: A-B > A-A or B-B

A min is found in vapour pressure – composition curve

Presence of azeotropic mixtures lead to max boiling point

E.g. pyridine-acetic acid, chloroform-acetone (formation of loose compounds from hydrogen bonding), formation of hydrates, nitric acid-water, sulfuric acid-water.


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Negatively deviated solutionwith maximum boiling point

• Mixtures showing negative deviation from Raoult’s law

• A maximum boiling point is obtained

• The solution has an azeotropic mixture whose vapour pressure is the lowest and the boiling point is the highest

• example of a negative azeotrope is 20.2% hydrogen chloride and 79.8% water (by weight). Hydrogen chloride boils at –84°C and water at 100°C, but the azeotrope boils at 110°C


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http://en.wikipedia.org/wiki/Hydrogen_chloride

Distillation of negative azeotrope 20.2% hydrogen chloride and 79.8% water

• distillation of any mixture of those constituents will result in the residue being closer in composition to the azeotrope than the original mixture.

• E.g. hydrochloric acid solution contains less than 20.2% hydrogen chloride,

• boiling the mixture will give a solution that is richer in hydrogen chloride than the original.

• If the solution initially contains more than 20.2% hydrogen chloride, boiling will give a solution that is poorer in hydrogen chloride than the original.

• Boiling of any hydrochloric acid solution long enough will cause the solution left behind to approach the azeotropic ratio.


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http://en.wikipedia.org/wiki/Residue

http://en.wikipedia.org/wiki/Hydrochloric_acid


Application of Raoult’s Law in Aerosol Formulation

Vapour pressure of propellant control the spray

• As pressure increases, spray rate increases -> fine spray, propellant gas vapourises fast

• As pressure decreases, spray rate decreases -> coarse spray, propellant gas vapourises slowly -> wet

Thus, improvise by using appropriate mixture of propellant

• Examples of components:• HFA-125: pentafluoroethane CHCF5

• HFC-134a: tetrafluoroethane CHCHF4

• HFA-152: difluoroethane CH3CHF2

• HFA-227: heptafluoropropane CHFCF3CF3 45PHM1153 Physical Pharmacy 1

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Exercise: Application of Raoult’s Law in Aerosol Formulation

The vapour pressure of pure propellant 11 (MW 137.4) at 21oC is P11

0 = 13.4 psi

The vapour pressure of pure propellant 12 (MW 120.9) at 21oC is P12

0 = 84.9 psi

A total of 100 g propellants consisting of 50:50 mixture by gram weight was used.

What is the total vapour pressure?


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Variation of Vapour Pressure with Temperature

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As T ↑, energy of molecules

↑

More escape into

vapour phase

increase in vapour

pressure

Clausius-Clapeyron Equation


Clausius-Clapeyron Equation

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lnor

RT

H

dT

Pd

RT

HP

dT

dP vapvap

• The variation of vapour pressure with temperature in terms of molar enthalpy of the liquid, Hvap

• V is the difference in molar volume of the two phases

• Since molar volume of vapour is very much greater than the molar volume of liquid, V approaches volume of vapour, Vv

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• Assuming that the vapour obeys ideal gas behaviour,

PV = RT

V = RT/P

Thus, Vv = RT/P

The equation becomes

Assuming Hvap to be constant:

VT

H

dT

dP vap

constant303.2

log

RT

HP

vap


Application of Clausius-Clapeyron Equation

• To estimate vapour pressure at any temperature.

• To calculate enthalpy of vaporisation based on the slope of plot

• To study phase transition – important to determine the extent of weight loss during processing or testing.

End lecture 3/3

constant303.2

log

RT

HP

vap


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References

The required texts you shall seek,The recommended ones you may flirt,Revise! Revise! You’d better be quick!When you get your grades you won’t be hurt!


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Documents

Colligative Properties Kausar Ahmad Kulliyyah of Pharmacy 1PHM1153 Physical Pharmacy 1 2010/11