1

CHAPTER 4

Solutions

CBy

Dr. Hisham Ezzat

2011- 2012

First year

1. Completely miscible liquids– Ideal solution– Non - ideal solution

2. Completely immiscible liquids H2O and aniline, H2O and chlorobenzene

3. Partially immiscible liquids, H2O and phenol, H2O and ether

Ideal Solution

1. The force of attraction between all molecules are identical i.e. the attraction force is not affected by addition of other components A - A = B-B = A - B.

2. No heat is evolved or absorbed during mixing i.e. H soln. = Zero

3. The volume of solution is the sum of volumes of the two liquids.

4. The solution obeys Raoult's law.

Figure (1): Vapor pressure of ideal solutions

Example 6:

Heptane (C7H16) and octane (C8H18) form ideal solutions What is the vapor pressure at 40°C of a solution that contains 3.0 mol of heptane and 5 mol of octane? At 40°C, the vapor pressure of heptane is 0.121 atm and the vapor pressure of octane is 0.041 atm.

Solution:

The total number of moles is 8.0. therefore X heptane = 3.0/8.0 = 0.375 X octane = 5.0/8.0 = 0.625Total = X heptane . Po heptane + X

octane. Po octane= 0.375 x 0.12 +0.625 x 0.04 = 0.045 atm + 0.026 atm. = 0.071 atm.

Example 7:

Assuming ideality, calculate the vapor pressure of 1.0 m solution of a non - volatile, on dissociating solute in water at 50°C. The vapor pressure of water 50°C is 0.122 atm.

Solution :

From example 2 the mole fraction of water in 1.0m solution is 0.982.

PH2O = XH2O PH2O = 0.982 x 0.122 = 0.120 atm.

Problem:

At 140°C, the V.P of C6H5CI is 939.4 torr and that of C6H5Br is 495.8 torr. Assuming that these two liquids from an ideal solution. Find the composition of a mixture of two liquids which boils at 140°C under 1 atm pressure?

Non- ideal solutions

Negative deviation Positive deviation1- The force of attraction

increase by mixing A - A, B-B < A-B

The force of attraction decrease by mixing A-A , B-B

> A-B2- The vapor pressure will be lower

than that given by Roault's lawThe vapor pressure will be higher than that given by

Raoult's law.3- H solution :- Ve (exothermic) H solution: + Ve

(endothermic)4- Temperature change when solution is formed: increase

Temperature change when solution is formed: decrease.

5- Example: Acetone-water Ethanol-hexane

Fig.2: Vapour pressure of non-ideal solution (-ve deviation)

Fig.3: Vapour pressure of non-ideal solution (+ve deviation)

Fractional Distillation of Binary Miscible liquids

• The separation of mixture of volatile liquids into their components is called fractional distillation,

• the distillate containing the more volatile component and the residue the less volatile one

a) Ideal solutions

• If a mixture of 2 liquids (A and B) form a completely miscible ideal solution and PA > PB result in B.P. of A < B.P of B thus on boiling:-– 1) The Liquid A boils at lower B.P than that of liquid B.– 2) The liquid A which is more volatile will be passed

from the fractionating column and the liquid B which is less volatile returned again to the distallating flask.

• A solution of intermediate B.p. between 2 pure liquid -called azeotropic solution

b) Non - ideal solutions (solutions that exhibit deviations from Raoults law)

If a solution having any other compositions is distilled, the azeotropic mixture will distill first and the excess of (A) or (B) will remains in the flask e.g 95 % ethanol and 5 % H2O.

Non - ideal solutions with minimum boiling point:

• If a solution having any other composition is distilled, the execs of acetone or CHCI3 will distill first leaving the azeotropic mixture in the flask.

2) Non - ideal solutions with maximum boiling point:

Example 8:

A solution is prepared by mixing 5.81 g acetone C3H6O, (M. wt = 58.1 g/mole) 11.9 g chloroform (CHCI3 M.wt 119.4 g/mole). At 35°C this solution has a total vapor pressure of 260 torr. Is this an ideal solution? Comment? The vapor pressure of pure acetone and pure CHCI3 at 35°C are 345 and 293 torr, respectively.

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Colligative Properties of Solutions

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Colligative Properties of Solutions There are four common types of colligative

properties:1. Vapor pressure lowering

2. Freezing point depression

3. Boiling point elevation

4. Osmotic pressure Vapor pressure lowering is the key to all

four of the colligative properties.

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Lowering of Vapor Pressure and Raoult’s Law Addition of a nonvolatile solute to a solution

lowers the vapor pressure of the solution. The effect is simply due to fewer solvent

molecules at the solution’s surface. The solute molecules occupy some of the spaces

that would normally be occupied by solvent. Raoult’s Law models this effect in ideal

solutions.

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Lowering of Vapor Pressure and Raoult’s Law Derivation of Raoult’s Law.

P P

where P vapor pressure of solvent

P vapor pressure of pure solvent

mole fraction of solvent

solvent solvent solvent0

solvent

solvent0

solvent

X

in solution

X in solution

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Lowering of Vapor Pressure and Raoult’s Law Lowering of vapor pressure, Psolvent, is defined as:

0solventsolvent

0solventsolvent

0solvent

solvent0solventsolvent

)P1(

)P)((- P

PP P

X

X

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Lowering of Vapor Pressure and Raoult’s Law Remember that the sum of the mole fractions

must equal 1. Thus Xsolvent + Xsolute = 1, which we can

substitute into our expression.

Law sRaoult' iswhich

P P

- 10solventsolutesolvent

solventsolute

X

XX

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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.

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The vapor pressure of water is 17.5 torr at 20°C. Imagine holding the temperature constant while adding glucose, C6H12O6, to the water so that the

resulting solution has XH2O = 0.80 and XGlu = 0.20.

What is , the vapor pressure of water over the solution 0

AAA PXP

torrXPXP AAA 5.1780.00

= 14 torr

Examples

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Glycerin, C3H8O3, is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr

The vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult's law is obeyed, what is the mole fraction of ethylene glycol in the solution? Answer: 0.290

P°H2O =1070 torrPH2O = 1 Atm = 760 torr

XH2O = ---------PH2O

P°H2O

= ---------760 torr1070 torr

= 0.71028XXH2OH2O + X + XEGEG = 1 = 10.7103 + XEG = 1

1- 0.7103 = XEG

XEG = 0.28972 = 0.290

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More ExamplesSucrose is a nonvolatile, nonionizing solute in water. Determine the vapor pressure lowering, at 27°C, of a solution of 75.0 grams of sucrose, C12H22O11, dissolved in 180. g of water. The vapor pressure of pure water at 27°C is 26.7 torr. Assume the solution is ideal.

molSucg

SucmolgSucnSuc 219.0

3.342

10.75

molWatyerg

WatermolgWaternWater 99.9

18

1180

978541.02191.0991.9

991.9

UcWater

waterWater nn

nX

13.2697854.07.260 XtorrXPP WaterWaterWater

Vapor Pressure Lowered = 26.7-26.1= 0.6

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solution is made by mixing 52.1 g of propyl chloride, C3H8Cl, and 38.4 g of propyl bromide, C3H8Br. What is the vapor pressure of propyl chloride in the solution at 25°C? The vapor pressure of pure propyl chloride is 347 torr at 25°C and that of pure propyl bromide is 133 torr at 25°C. Assume that the solution is an ideal solution.

6633.054.78

11.52

CPg

CPmolCPgnCP

312.099.122

14.38

CBg

CBmolCBgnCB

67996.03122.06633.0

6633.0

PBPC

PCrPC nn

nX

TorrXXPP PCPCPC 23695.235679964.03470

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. At 25°C a solution consists of 0.450 mole of pentane, C5H12, and 0.250 mole of cyclopentane, C5H10. What is the mole fraction of cyclopentane in the vapor that is in equilibrium with this solution? The vapor pressure of the pure liquids at 25°C are 451 torr for pentane and 321 torr for cyclopentane. Assume that the solution is an ideal solution.

95.202451450.0 XXPP PenPenPen

25.80321250.00 XXPP CPenCPenCPen

RT

VPn

RT

VPn

RT

PVn CPen

CPenPen

Pen ;;

PenCPen

CPen

PenCPen

CPen

PenCPen

CPenCPen PP

P

RT

VP

RT

VPRT

VP

nn

nX

283.095.20225.80

25.80

CPenP

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Boiling Point Elevation Addition of a nonvolatile solute to a solution

raises the boiling point of the solution above that of the pure solvent. This effect is because the solution’s vapor

pressure is lowered as described by Raoult’s law. The solution’s temperature must be raised to

make the solution’s vapor pressure equal to the atmospheric pressure.

The amount that the temperature is elevated is determined by the number of moles of solute dissolved in the solution.

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Boiling Point Elevation

Boiling point elevation relationship is:

solvent for the

constantelevation point boiling molal K

solution ofion concentrat molal

elevationpoint boiling T :where

KT

b

b

bb

m

m

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Boiling Point Elevation

Example 14-4: What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution?

C101.28=C28.1+C100.0 =solution theofPoint Boiling

C28.1T

)50.2)(C/ 512.0(T

K T

000

0b

0b

bb

mm

m

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Boiling-Point Elevation Boiling-Point Elevation The addition of a nonvolatile solute lowers the vapor pressure of the solution. At any given temperature, the vapor pressure of the solution is lower than that of the pure liquid

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The increase in boiling point relative to that of the pure solvent, Tb, is directly proportional to the number of solute particles per mole of solvent molecules. Molality expresses the number of moles of solute per 1000 g of solvent, which represents a fixed number of moles of solvent mKT bb

Solvent B.Point (°C) Kb (°C/m)

Freezing P. (°C)

Kf (°C/m)

Water, H2O 100.0 0.52 0.00 1.86Benzen, C6H6 80.1 2.53 5.5 5.12Ethanol, C2H6O 78.4 1.22 -114.0 1.99Carbon tetrachloride, CCl4 76.8 5.02 -22 29.8Chloroform, CHCl3

61.2 3.63 -63.5 4.68

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Automotive antifreeze consists of ethylene glycol, C2H6O2, a nonvolatile nonelectrolyte. Calculate the boiling point of a 25.0 mass percent solution of ethylene glycol in water.

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Freezing Point Depression Addition of a nonvolatile solute to a solution

lowers the freezing point of the solution relative to the pure solvent.

See table for a compilation of boiling point and freezing point elevation constants.

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Freezing Point Depression

Relationship for freezing point depression is:

T K

where: T freezing point depression of solvent

molal concentration of soltuion

K freezing point depression constant for solvent

f f

f

f

m

m

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Freezing Point Depression Notice the similarity of the two relationships

for freezing point depression and boiling point elevation.

Fundamentally, freezing point depression and boiling point elevation are the same phenomenon. The only differences are the size of the effect which is

reflected in the sizes of the constants, Kf & Kb.

This is easily seen on a phase diagram for a solution.

mm bbff K T vs.KT

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Freezing Point Depression

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Freezing Point Depression

Example 14-5: Calculate the freezing point of a 2.50 m aqueous glucose solution.

C4.65 - = C4.65 - C0.00=solution ofPoint Freezing

C65.4T

)50.2)(C/(1.86T

KT

000

0f

0f

ff

mm

m

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Freezing Point Depression

Example : Calculate the freezing point of a solution that contains 8.50 g of benzoic acid (C6H5COOH, MW = 122) in 75.0 g of benzene, C6H6.

You do it!You do it!

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Freezing Point Depression

C0.72=C4.76-C5.48 =F.P.

C76.4)929.0)(C/12.5(T

KT

solution. for this depression theCalculate .2

929.0COOHHC g 122

COOHHC mol 1

HC kg 0.0750

COOHHC g 50.8

HC kg

COOHHC mol ?

molality! Calculate .1

000

00f

ff

56

56

66

56

66

56

mm

m

m

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Determination of Molecular Weight by Freezing Point Depression The size of the freezing point depression

depends on two things:1. The size of the Kf for a given solvent, which are well

known.

2. And the molal concentration of the solution which depends on the number of moles of solute and the kg of solvent.

If Kf and kg of solvent are known, as is often the case in an experiment, then we can determine # of moles of solute and use it to determine the molecular weight.

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Determination of Molecular Weight by Freezing Point Depression Example : A 37.0 g sample of a new covalent

compound, a nonelectrolyte, was dissolved in 2.00 x 102 g of water. The resulting solution froze at -5.58oC. What is the molecular weight of the compound?

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Determination of Molecular Weight by Freezing Point Depression

g/mol 7.61mol 0.600

g 37 is massmolar theThus

compound mol 600.0

kg 0.200 3.00=OH kg 0.200in compound mol ?

water.of kg 0.200 mL 200

are thereproblem In this

00.3C1.86

C58.5

K

T

the thusKT

2

0

0

f

f

ff

m

mm

m

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Osmotic Pressure Osmosis is the net flow of a solvent

between two solutions separated by a semipermeable membrane.

The solvent passes from the lower concentration solution into the higher concentration solution.

Examples of semipermeable membranes include:

1. cellophane and saran wrap2. skin3. cell membranes

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Osmotic Pressure

H2O 2O

semipermeable membrane

H2O H2O

sugar dissolvedin water

H2O

H2O

H2O

H2O

net solvent flow

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Osmotic Pressure Osmosis is a rate controlled phenomenon.

The solvent is passing from the dilute solution into the concentrated solution at a faster rate than in opposite direction, i.e. establishing an equilibrium.

The osmotic pressure is the pressure exerted by a column of the solvent in an osmosis experiment.

M

M

RT

where: = osmotic pressure in atm

= molar concentration of solution

R = 0.0821L atmmol K

T = absolute temperature

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Osmotic Pressure

For very dilute aqueous solutions, molarity and molality are nearly equal. M m

m

for dilute aqueous solutions only

RT

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Osmotic Pressure Osmotic pressures can be very large.

For example, a 1 M sugar solution has an osmotic pressure of 22.4 atm or 330 p.s.i.

Since this is a large effect, the osmotic pressure measurements can be used to determine the molar masses of very large molecules such as:

1. Polymers

2. Biomolecules like proteins ribonucleotides

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Osmotic Pressure

Example : A 1.00 g sample of a biological material was dissolved in enough water to give 1.00 x 102 mL of solution. The osmotic pressure of the solution was 2.80 torr at 25oC. Calculate the molarity and approximate molecular weight of the material.

You do it!You do it!

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Osmotic Pressure

M M

M M

RT RT

atm = 2.80 torr1 atm

760 torr atm =

= atm

0.0821 KL atmmol K

? .

..

0 00368

0 00368298

150 10 4

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Osmotic Pressure

M M

M M

M

RT RT

atm = 2.80 torr1 atm

760 torr atm =

= atm

0.0821 K

g

mol

1.00 g

0.100 L

L

typical of small proteins

L atmmol K

gmol

? .

..

?

..

0 00368

0 00368

298150 10

1

150 106 67 10

4

44

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Application of Osmotic Pressure1. Water Purification by Reverse Osmosis If we apply enough external pressure to an

osmotic system to overcome the osmotic pressure, the semipermeable membrane becomes an efficient filter for salt and other dissolved solutes. Ft. Myers, FL gets it drinking water from the Gulf

of Mexico using reverse osmosis. Dialysis is another example of this phenomenon.

2) Isotonic solution: In the living cells, the osmotic pressure of solution is equal to the osmotic pressure of the cell.

e.g: NaCI (0.9%) has the same osmotic pressure as blood.

3) Hypertonic solution: A solution of higher osmotic pressure. In this solution red blood cells shrink. The cells are called plasmolysed.

4) Hypotonic solution: A solution of lower osmotic pressure. In this solution red blood cells swells up and burst. The cell is said to be haemolysed

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End of Chapter 2

Human Beings are solution chemistry in action!