File

CHAPTER 4

First year

 Solutions

C By Dr. Hisham Ezzat 2011- 2012

1

Completely miscible liquids

1. – –

2.

3.

Ideal solution Non - ideal solution

Completely immiscible liquids H2O and aniline, H2O and chlorobenzene Partially immiscible liquids, H2O and phenol, H2O and ether

Ideal Solution 1.

2.

3.

4.

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.

No heat is evolved or absorbed during mixing i.e.  H soln. = Zero The volume of solution is the sum of volumes of the two liquids. 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 = 0.625 Total = 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 deviation 1- The force of attraction The force of attraction increase by mixing decrease by mixing A-A , B-B A - A, B-B < A-B > A-B 2- The vapor pressure will be lower The vapor pressure will be than that given by Roault's law higher than that given by Raoult's law. 3- H solution :- Ve (exothermic)  H solution: + Ve (endothermic) 4- Temperature change when Temperature change when solution is formed: increase solution is formed: decrease. 5- Example: Acetone-water

Ethanol-hexane

Fig.2: Vapour pressure of non-ideal Fig.3: Vapour pressure of non-ideal solution (-ve deviation) 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)

Non - ideal solutions with minimum boiling point: 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.

2) Non - ideal solutions with maximum 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.

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.



Colligative Properties of Solutions

20

Colligative Properties of Solutions There are four common types of colligative properties:



1. 2. 3. 4.



Vapor pressure lowering Freezing point depression Boiling point elevation Osmotic pressure

Vapor pressure lowering is the key to all four of the colligative properties.

21

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.

22

Lowering of Vapor Pressure and Raoult’s Law 

Derivation of Raoult’s Law. Psolvent  X solvent Psolvent 0

where Psolvent  vapor pressure of solvent in solution Psolvent  vapor pressure of pure solvent 0

X solvent  mole fraction of solvent in solution

23

Lowering of Vapor Pressure and Raoult’s Law 

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

Psolvent  P

0 solvent

 Psolvent

 Psolvent - ( X solvent)( Psolvent) 0

0

 (1  X solvent)P

0 solvent

24

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. X solute  1 - X solvent Psolvent  X solute Psolvent 0

which is Raoult' s Law

25

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.

26

Examples 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

PA  X A PA

PA  X A P  0 . 80 X 17 . 5 torr 0 A

= 14 torr 27

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

28

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 torr PH2O = 1 Atm = 760 torr PH2O torr XH2O = --------- = 760 --------P°H2O 1070 torr = XH2O + XEG = 1 0.7103 + XEG = 1

0.71028

1- 0.7103 = XEG XEG =

0.28972

= 0.290

More Examples

Sucrose 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. n Suc  75 . 0 gSuc nW ater

1 mol

Suc

 0 . 219 mol

342 . 3 g Suc 1 mol Water  180 gWater  9 . 99 mol 18 g Watyer

X W ater 

n water n W ater  n Uc



9 . 991 9 . 991  0 . 2191

PW ater  PW ater X W ater  26 . 7 torr 0

 0 . 978541

X 0 . 97854  26 . 13

Vapor Pressure Lowered = 26.7-26.1= 0.6

30

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. n CP  52 . 1 g n CB  38 . 4 g

X

PC



CP CB

n PCr n PC  n PB

1

mol

CP

78 . 54 g

CP

1

CB

mol

122 . 99 g



 0 . 6633  0 . 312

CB

0 . 6633 0 . 6633  0 . 3122

 0 . 67996

PPC  PPC X PC  347 X 0 . 679964  235 . 95  236 Torr 0

31

. 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. PPen  PPen X 0

Pen

 . 450 X 451  202 . 95

PCPen  PCPen X CPen  0 . 250 X 321  80 . 25 0

n 

PV

; n Pen 

RT

X CPen 

PCPen 

PPen V RT

n CPen n CPen  n Pen

; n CPen 

PCPen V

RT PCPen V



80 . 25 80 . 25  202 . 95

PCPen RT  PCPen V P V PCPen  PPen  Pen RT RT  0 . 283 32

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. 33

Boiling Point Elevation 

Boiling point elevation relationship is: Tb  K b m where : Tb  boiling point elevation m  molal concentrat ion of solution K b  molal boiling point elevation constant for the solvent

34

Boiling Point Elevation 

Example 14-4: What is the normal boiling point of a 2.50 m glucose, C6H12O6, solution? Tb  K b m Tb  (0.512 C/m)( 2.50m) 0

Tb  1.28 C 0

0

0

0

Boiling Point of the solution = 100.0 C + 1.28 C = 101.28 C

35

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

36

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  Tb  K b m B.Point (°C)

Solvent Water, H2O Benzen, C6H6 Ethanol, C2H6O Carbon tetrachloride,

CCl4

Chloroform, CHCl3

100.0 80.1 78.4 76.8 61.2

Kb (°C/m)

0.52 2.53 1.22 5.02 3.63

Freezing P. (°C)

Kf (°C/m)

0.00 5.5 -114.0 -22 -63.5

1.86 5.12 1.99 29.8 4.68 37

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.

38

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.

39

Freezing Point Depression 

Relationship for freezing point depression is: Tf  K f m where: Tf  freezing point depression of solvent m  molal concentration of soltuion K f  freezing point depression constant for solvent

40

Freezing Point Depression 

Notice the similarity of the two relationships for freezing point depression and boiling point elevation.

Tf  K f m vs.  Tb  K b m 

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.

41

Freezing Point Depression

42

Freezing Point Depression 

Example 14-5: Calculate the freezing point of a 2.50 m aqueous glucose solution. Tf  K f m Tf  (1.86 C/m)( 2.50m) 0

Tf  4.65 C 0

0

0

0

Freezing Point of solution = 0.00 C - 4.65 C = - 4.65 C

43

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!

44

Freezing Point Depression 1. Calculate molality! ? mol C 6 H 5 COOH kg C 6 H 6



8.50 g C 6 H 5 COOH



0.0750 kg C 6 H 6

1 mol C 6 H 5 COOH

 0.929m

122 g C 6 H 5 COOH 2. Calculate the depression for this solution. Tf  K f m Tf  (5.12 C/m)( 0.929 m)  4.76 C 0

0

0

0

0

F.P. = 5.48 C - 4.76 C = 0.72 C 45

Determination of Molecular Weight by Freezing Point Depression The size of the freezing point depression depends on two things:



1.

2.



The size of the Kf for a given solvent, which are well known. 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.

46

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?

47

Determination of Molecular Weight by Freezing Point Depression Tf  K f m thus the m

Tf Kf

0



5.58 C 0

 3.00m

1.86 C

In this problem there are 200 mL  0.200 kg of water. ? mol compound in 0.200 kg H 2 O = 3.00 m  0.200 kg  0.600 mol compound Thus the molar mass is

37 g

 61.7 g/mol

0.600 mol 48

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. 2.

3.

cellophane and saran wrap skin cell membranes

49

Osmotic Pressure semipermeable membrane

sugar dissolved H2O in water

H2O

H2O

H2O

H2O

H2O

H2O

net2solvent flow O

50

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.   MRT

where:  = osmotic pressure in atm M = molar concentration of solution R = 0.0821

L atm mol K

T = absolute temperature 51

Osmotic Pressure  For very dilute aqueous solutions, molarity and molality are nearly equal.  Mm

  mRT for dilute aqueous solutions only

52

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:



Polymers Biomolecules like

1. 2.  

proteins ribonucleotides 53

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!

54

Osmotic Pressure   MRT  M  ? atm = 2.80 torr  M =



RT 1 atm

 0.00368 atm = 

760 torr 0.00368 atm



L atm 0.0821 mol K

298 K

 150 .  10

4

M

55

Osmotic Pressure   MRT  M 

 RT

? atm = 2.80 torr 

1 atm

 0.00368 atm = 

760 torr M = ? g mol



0.00368 atm L atm 298 K  0.0821 mol K 

1.00 g 0.100 L



 150 .  10

1L 150 .  10

4

4

 6.67  10 M

M

4 g mol

typical of small proteins 56

Application of Osmotic Pressure 1. 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.

57









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

58

End of Chapter 2 

Human Beings are solution chemistry in action!

59