Chapter 14

CHAPTER 14 

Solutions

1

Chapter Goals 2. 3. 4. 5. 6. 7. 8. 9.

The Dissolution Process Spontaneity of the Dissolution Process Dissolution of Solids in Liquids Dissolution of Liquids in Liquids (Miscibility) Dissolution of Gases in Liquids Rates of Dissolution and Saturation Effect of Temperature on Solubility Effect of Pressure on Solubility Molality and Mole Fraction 2

Chapter Goals 2. 3. 4. 5. 6.

7.

8.

Colligative Properties of Solutions Lowering of Vapor Pressure and Raoult’s Law Fractional Distillation Boiling Point Elevation Freezing Point Depression Determination of Molecular Weight by Freezing Point Depression or Boiling Point Elevation Colligative Properties and Dissociation of Electrolytes Osmotic Pressure 3

Chapter Goals

2. 3. 4.

Colloids The Tyndall Effect The Adsorption Phenomena Hydrophilic and Hydrophobic Colloids

4

The Dissolution Process 

Solutions are homogeneous mixtures of two or more substances.  



Dissolving medium is called the solvent. solvent Dissolved species are called the solute. solute

There are three states of matter (solid, liquid, and gas) which when mixed two at a time gives nine different kinds of mixtures.  

Seven of the possibilities can be homogeneous. Two of the possibilities must be heterogeneous. 5

The Dissolution Process       

 

Solute Solid Liquid Gas Liquid Solid Gas Gas Solid Liquid

Seven Homogeneous Possibilities Solvent Example Liquid salt water Liquid mixed drinks Liquid carbonated beverages Solid dental amalgams Solid alloys Solid metal pipes Gas air Two Heterogeneous Possibilities Gas dust in air Gas clouds, fog

6

Spontaneity of the Dissolution  As an example of dissolution, let’s assume that Process the solvent is a liquid. Two major factors affect dissolution of solutes Change of energy content or enthalpy of solution, ∆Hsolution

 

 

If ∆Hsolution is exothermic (< 0) dissolution is favored. If ∆Hsolution is endothermic (> 0) dissolution is not favored.

7

Spontaneity of the Dissolution  Change in disorder, or randomness, of the Process solution ∆Smixing 

If ∆Smixing increases (> 0) dissolution is favored.



If ∆Smixing decreases (< 0) dissolution is not favored.

Thus the best conditions for dissolution are:



For the solution process to be exothermic.

 

∆Hsolution < 0

For the solution to become more disordered.

 

∆Smixing > 0

8

Spontaneity of the Dissolution  Disorder in mixing a solution is very common. Process 



What factors affect ∆Hsolution? 



∆Smixing is almost always > 0.

There is a competition between several different attractions.

Solute-solute attractions such as ion-ion attraction, dipole-dipole, etc. 

Breaking the solute-solute attraction requires an absorption of E.

9

Spontaneity of the Dissolution  Solvent-solvent attractions such as hydrogen Process bonding in water. 



This also requires an absorption of E.

Solvent-solute attractions, solvation, solvation releases energy. 

If solvation energy is greater than the sum of the solute-solute and solvent-solvent attractions, the dissolution is exothermic, ∆Hsolution < 0.



If solvation energy is less than the sum of the solute-solute and solvent-solvent attractions, the dissolution is endothermic, ∆Hsolution > 0. 10

Spontaneity of the Dissolution Process

11

Dissolution of Solids in  The energy released (exothermic) when a Liquids

mole of formula units of a solid is formed from its constituent ions (molecules or atoms for nonionic solids) in the gas phase is called the crystal lattice energy. energy

M

+

 X (g)  M X (s)  crystal lattice energy -

(g)

+

-

•The crystal lattice energy is a measure of the attractive forces in a solid. •The crystal lattice energy increases as the charge density increases. 12

Dissolution of Solids in  Dissolution is a competition between: Liquids Solute -solute attractions

1. 

crystal lattice energy for ionic solids

Solvent-solvent attractions

2. 

H-bonding for water

Solute-solvent attractions

 

Solvation or hydration energy

13

Dissolution of Solids in Liquids 

Solvation is directed by the water to ion attractions as shown in these electrostatic potentials.

14

Dissolution of Solids in Liquids 

In an exothermic dissolution, energy is released when solute particles are dissolved. 



This energy is called the energy of solvation or the hydration energy (if solvent is water).

Let’s look at the dissolution of CaCl2.

15

Dissolution of Solids in Liquids 2+ H O CaCl 2 (s) →  [ Ca(OH 2 ) 6 ] + 2Cl ( H 2 O ) x 2

where x is approximately 7 or 8 OH2

2+ H

OH2

H2O Ca H2O

O

OH2 OH2

H

H

O

Cl-

O H

H

H H

H O 16

Dissolution of Solids in Liquids 

The energy absorbed when one mole of formula units becomes hydrated is the molar energy of hydration. hydration M n + (g) + x H 2 O → [ M(OH 2 ) x ] X y- ( g ) + n H 2 O → [ X(H 2 O) n ]

n+

+ hydration E for M n +

y−

+ hydration E for X y-

17

Dissolution of Solids in Liquids Hydration energy increases with increasing charge density Ion Radius(Å) Charge/radius Heat of Hydration K+ 1.33 0.75 -351 kJ/mol Ca2+ 0.99 2.02 -1650 kJ/mol Cu2+ 0.72 2.78 -2160 kJ/mol Al3+ 0.50 6.00 -4750 kJ/mol 

18

Dissolution of Liquids in Liquids (Miscibility) 



Most polar liquids are miscible in other polar liquids. In general, liquids obey the “like dissolves like” rule.  



Polar molecules are soluble in polar solvents. Nonpolar molecules are soluble in nonpolar solvents.

For example, methanol, CH3OH, is very soluble in water 19

Dissolution of Liquids in Liquids (Miscibility) 

Nonpolar molecules essentially “slide” in between each other. 

For example, carbon tetrachloride and benzene are very miscible. H Cl Cl C Cl

H Cl H

C C

C C

C C

H H

H 20

Dissolution of Gases in  Polar gases are more soluble in water than Liquids nonpolar gases. 

 

This is the “like dissolves like” rule in action.

Polar gases can hydrogen bond with water Some polar gases enhance their solubility by reacting with water. HBr + H 2 O → H 3O

+

( aq )

+ Br

-

( aq )

strong acid SO 2 + H 2 O → H 2SO 3( aq ) H 2SO 3( aq )

H 2O  →  H O + ( aq ) + HSO - ( aq ) ←  3 3

weak acid

21

Dissolution of Gases in Liquids 

A few nonpolar gases are soluble in water because they react with water.

CO2 ( g )

− +    → + H 2O → H 2CO3 ←  H 3O ( aq ) + HCO3 ( aq ) H 2O

very weak acid 

Because gases have very weak solute-solute interactions, gases dissolve in liquids in exothermic processes. 22

Rates of Dissolution and Saturation 

Finely divided solids dissolve more rapidly than large crystals. 



Compare the dissolution of granulated sugar and sugar cubes in cold water.

The reason is simple, look at a single cube of NaCl.

Breaks NaCl 

up

many smaller crystals

The enormous increase in surface area helps the solid to dissolve faster.

23

Rates of Dissolution and  Saturated solutions have established an Saturation

equilbrium between dissolved and undissolved solutes 

Examples of saturated solutions include:  

Air that has 100% humidity. Some solids dissolved in liquids.

24

Rates of Dissolution and Saturation 

Symbolically this equilibrium is written as:

+ → MX ( s ) ← M ( aq ) + X ( aq )



In an equilibrium reaction, the forward rate of reaction is equal to the reverse rate of reaction.

25

Rates of Dissolution and Saturation 

Supersaturated solutions have higher-thansaturated concentrations of dissolved solutes.

26

Effect of Temperature on Solubility  According to LeChatelier’s Principle when stress is applied to a system at equilibrium, the system responds in a way that best relieves the stress. 

Since saturated solutions are at equilibrium, LeChatelier’s principle applies to them.

Possible stresses to chemical systems include:

 1. 2. 3.

Heating or cooling the system. Changing the pressure of the system. Changing the concentrations of reactants or products.

27

Effect of Temperature on  What will be the effect of heating or cooling the Solubility water in which we wish to dissolve a solid? 



It depends on whether the dissolution is exo- or endothermic.

For an exothermic dissolution, heat can be considered as a product. H 2O

+

-

LiBr( s)  → Li ( aq ) + Br ( aq ) + 48.8kJ / mol 



Warming the water will decrease solubility and cooling the water will increase the solubility. Predict the effect on an endothermic dissolution like this one. H 2O

+

-

KMnO 4 ( s ) + 43.6 kJ / mol  → K ( aq ) + MnO 4 ( aq ) 28

Effect of Temperature on Solubility 





For ionic solids that dissolve endothermically dissolution is enhanced by heating. heating For ionic solids that dissolve exothermically dissolution is enhanced by cooling. cooling Be sure you understand these trends.

29

Effect of Pressure on Solubility 

Pressure changes have little or no effect on solubility of liquids and solids in liquids. 



Liquids and solids are not compressible.

Pressure changes have large effects on the solubility of gases in liquids. 



Sudden pressure change is why carbonated drinks fizz when opened. It is also the cause of several scuba diving related problems including the “bends”. 30

Effect of Pressure on Solubility 

The effect of pressure on the solubility of gases in liquids is described by Henry’s Law.

Pgas gasgas gas ==kkMM where M M gas = molar concentration of gas where gas = molar concentration of gas k = Henry's Law constant, unique number for each gas - liquid combination Pgas = partial pressure of gas

31

Molality and Mole Fraction 

2.

In Chapter 3 we introduced two important concentration units. % by mass of solute

mass of solute % w/w = ×100% mass of solution

32

Molality and Mole Fraction 1.

Molarity

moles of solute M = Liters of solution 

We must introduce two new concentration units in this chapter.

33

Molality and Mole Fraction 

Molality is a concentration unit based on the number of moles of solute per kilogram of solvent.

moles of solute m= kg of solvent in dilute aqueous solutions molarity and molality are nearly equal 34

Molality and Mole Fraction 

Example 14-1: Calculate the molarity and the molality of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g

? mol C?6mol H12 O H 2 Og H 2 O C66 H1210 O.60 g C 106 H .012gOC66 H121000 O 6 g 1000 × =.0 g C 6 H 12 O×6 × × ? mol C 6 H 12=O 6 10 = ×1 kg H12 O kg H 2 Okg H 2 O 90.0 gH 90.0 kg H 2 O 2O g H 2O kg H 2 O

90.0 g H 2 O

1 mol C 6 H12 O 6 = 0.617m C 6 H12 O 6 180 g C 6 H12 O 6

This is the concentration in molality. 35

Molality and Mole Fraction 

Example 14-1: Calculate the molality and the molarity of an aqueous solution that is 10.0% glucose, C6H12O6. The density of the solution is 1.04 g/mL. 10.0% glucose solution has several medical uses. 1 mol C6H12O6 = 180 g You calculate the molarity!

? mol C 6 H12 O 6 10.0 g C 6 H12 O 6 1.04 g sol' n = × × L H 2O 100.0 g sol' n mL sol' n 1000 mL 1 mol C 6 H12 O 6 × = 0.578 M C 6 H12 O 6 1L 180 g C 6 H12 O 6 36

Molality and Mole Fraction 

Example 14-2: Calculate the molality of a solution that contains 7.25 g of benzoic acid C6H5COOH, in 2.00 x 102 mL of benzene, C6H6. The density of benzene is 0.879 g/mL. 1 mol C6H5COOH = 122 g You do it!

? mol C 6 H 5COOH 7.25g C 6 H 5COOH 1 mL C 6 H 6 = × kg C 6 H 6 200.0 mL C 6 H 6 0.879 g C 6 H 6 1000 g C 6 H 6 1 mol C 6 H 5COOH × = 0.338 m C 6 H 5COOH 1 kg C 6 H 6 122 g C 6 H 5COOH 37

Molality and Mole Fraction 

Mole fraction is the number of moles of one component divided by the moles of all the components of the solution 



Mole fraction is literally a fraction using moles of one component as the numerator and moles of all the components as the denominator.

In a two component solution, the mole fraction of one component, A, has the symbol XA.

number of moles of A XA = number of moles of A + number of moles of B 38

Molality and Mole Fraction 

The mole fraction of component B - XB

number of moles of B XB = number of moles of A + number of moles of B Note that X A + X B = 1 The sum of all the mole fractions must equal 1.00.

39

Molality and Mole Fraction 

Example 14-3: What are the mole fractions of glucose and water in a 10.0% glucose solution (Example 14-1)? You do it! In 1.00 ×10 g of this solution there are 10.0 g of glucose and 90.0 g of water. 2

? mol C 6 H12 O 6 = 10.0 g C 6 H12 O 6 × 1 mol C 6 H12 O 6 = 0.0556 mol C 6 H12 O 6 180 g C 6 H12 O 6 40

Molality and Mole Fraction 

Example 14-3: What are the mole fractions of glucose and water in a 10.0% glucose solution (Example 14-1)?

1 mol H 2 O ? mol H 2 O = 90.0 g H 2 O × = 5.00 mol H 2 O 18 g H 2 O

41

Molality and Mole Fraction 

Now we can calculate the mole fractions. 5.00 mol H 2 O X H 2O = 5.00 mol H 2 O + 0.0556 mol C 6 H 12 O 6 = 0.989 0.0556 mol C 6 H 12 O 6 X C 6 H12 O 6 = 5.00 mol H 2 O + 0.0556 mol C 6 H 12 O 6 = 0.011 100 . = 0.989 + 0.011

42

Colligative Properties of Solutions 

Colligative properties are properties of solutions that depend solely on the number of particles dissolved in the solution. 



Colligative properties do not depend on the kinds of particles dissolved.

Colligative properties are a physical property of solutions.

43

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.

44

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.

45

Lowering of Vapor Pressure and Raoult’s Law 

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

where Psolvent = vapor pressure of solvent in solution 0 Psolvent = vapor pressure of pure solvent X solvent = mole fraction of solvent in solution

46

Lowering of Vapor Pressure and Raoult’s Law 

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

∆Psolvent = P

0 solvent

=P

0 solvent

− Psolvent 0 solvent

- ( X solvent )(P

)

= (1 − X solvent )P

0 solvent

47

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 P

0 solvent

which is Raoult' s Law 48

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.

49

Fractional Distillation 



Distillation is a technique used to separate solutions that have two or more volatile components with differing boiling points. A simple distillation has a single distilling column. 



Simple distillations give reasonable separations.

A fractional distillation gives increased separations because of the increased surface area. 

Commonly, glass beads or steel wool are inserted into the distilling column.

50

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. 51

Boiling Point Elevation 

Boiling point elevation relationship is: ∆Tb = K b m

where : ∆Tb = boiling point elevation m = molal concentration of solution K b = molal boiling point elevation constant for the solvent

52

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 0 C/m)(2.50m) ∆Tb = 1.280 C

Boiling Point of the solution = 100.0 0 C + 1.280 C = 101.280 C

53

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 14-2 for a compilation of boiling point and freezing point elevation constants.

54

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

55

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. 56

Freezing Point Depression

57

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 = 4.650 C Freezing Point of solution = 0.00 0 C - 4.650 C = - 4.650 C

58

Freezing Point Depression 

Example 14-6: 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!

59

Freezing Point Depression 1. Calculate molality! ? mol C 6 H 5 COOH 8.50 g C 6 H 5 COOH = × kg C 6 H 6 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 0 C/m)(0.929m) = 4.76 0 C F.P. = 5.480 C - 4.76 0 C = 0.72 0 C 60

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







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.

61

Determination of Molecular Weight by Freezing Point  Depression Example 14-7: 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?

62

Determination of Molecular Weight by Freezing Point ∆T = K m thus the Depression f

f

∆Tf 5.580 C m= = = 3.00m 0 K f 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 37 g Thus the molar mass is = 61.7 g/mol 0.600 mol 63

Colligative Properties and Dissociation of Electrolytes 

Electrolytes have larger effects on boiling point elevation and freezing point depression than nonelectrolytes. 





This is because the number of particles released in solution is greater for electrolytes

One mole of sugar dissolves in water to produce one mole of aqueous sugar molecules. One mole of NaCl dissolves in water to produce two moles of aqueous ions: 

1 mole of Na+ and 1 mole of Cl- ions 64

Colligative Properties and Dissociation of Electrolytes 

Remember colligative properties depend on the number of dissolved particles. 



Since NaCl has twice the number of particles we can expect twice the effect for NaCl than for sugar.

The table of observed freezing point depressions in the lecture outline shows this effect.

65

Colligative Properties and Dissociation of Electrolytes 

Ion pairing or association of ions prevents the effect from being exactly equal to the number of dissociated ions

66

Colligative Properties and Dissociation of Electrolytes 



The van’t Hoff factor, symbol i, is used to introduce this effect into the calculations. i is a measure of the extent of ionization or dissociation of the electrolyte in the solution.

i=

∆Tf ( actual ) ∆Tf ( if nonelectrolyte) 67

Colligative Properties and Dissociation of Electrolytes 

i has an ideal value of 2 for 1:1 electrolytes like NaCl, KI, LiBr, etc. 2O Na + Cl - H  → Na (+aq ) + Cl (-aq ) 2 ions



formula unit

i has an ideal value of 3 for 2:1 electrolytes like K2SO4, CaCl2, SrI2, etc. 2O Ca 2+ Cl -2 H  → Ca (2aq+ ) + 2 Cl (-aq )

3 ions

formula unit 68

Colligative Properties and Dissociation of Electrolytes 





Example 14-8: The freezing point of 0.0100 m NaCl solution is -0.0360oC. Calculate the van’t Hoff factor and apparent percent dissociation of NaCl in this aqueous solution. meffective = total number of moles of solute particles/kg solvent First let’s calculate the i factor.

∆ Tf ( actual )

K f meffective meffective i= = = ∆Tf ( if nonelectrolyte ) K f mstated mstated 69

Colligative Properties and Dissociation of Electrolytes ∆Tf ( actual )

K f meffective meffective i= = = ∆Tf ( if nonelectrolyte ) K f mstated mstated ∆Tf ( actual ) 0.0360 0 C ∆Tf ( actual ) = K f meffective ∴ meffective = = 0 Kf 1.86 C m meffective

meffective 0.0194 m = 0.0194 m ∴i = = = 194 . mstated 0.0100 m

70

Colligative Properties and Dissociation of Electrolytes 



Next, we will calculate the apparent percent dissociation. Let x = mNaCl that is apparently dissociated.

71

Colligative Properties and Dissociation of Electrolytes NaCl  → Na + Cl (0.0100 − x)m xm xm H 2O

+

-

72

Colligative Properties and Dissociation of Electrolytes NaCl  → Na + Cl (0.0100 − x)m xm xm = [ 0.0100 − x + x + x ] m H 2O

meffective

+

-

= [ 0.0100 + x ] m = 0.0194 m x = 0.0094 m

73

Colligative Properties and Dissociation of Electrolytes apparent % dissociation =

mapp diss mstated

×100%

0.0094m = ×100% 0.0100m = 94%

74

Colligative Properties and Dissociation of Electrolytes 

Example 14-9: A 0.0500 m acetic acid solution freezes at -0.0948oC. Calculate the percent ionization of CH3COOH in this solution. You do it!

75

Colligative Properties and Dissociation of Electrolytes →H + + CH COO CH 3COOH ← 3

( 0.0500 − x )m x m xm meff = [( 0.0500 − x ) + x + x ] m = ( 0.0500 + x ) m ∆Tf = K f ×meff meff

∆Tf 0.09480 C = = = 0.0510 m 0 Kf 1.86 C m

meff = ( 0.0500 + x ) m = 0.0510 m x = 0.0010 m

mionized 0.0010 m % ionized = ×100% = ×100% moriginal 0.0500 m = 2.0% ionized and 98.0% unionized 76

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

77

Osmotic Pressure semipermeable membrane

sugar dissolved H2O in water

H2O

H2O

H2O

H2O

H2O

H2O

net2O solvent flow

78

Osmotic Pressure

79

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. π = MRT where: π = osmotic pressure in atm

M = molar concentration of solution L atm R = 0.0821 mol K T = absolute temperature 80

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

π = mRT for dilute aqueous solutions only

81

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 82

Osmotic Pressure 

Example 14-18: 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!

83

Osmotic Pressure π π = MRT ∴ M = RT 1 atm ? atm = 2.80 torr × = 0.00368 atm = π 760 torr 0.00368 atm −4 M = = 150 . × 10 M L atm ( 0.0821 mol K ) ( 298 K)

84

Osmotic Pressure π π = MRT ∴ M = RT 1 atm ? atm = 2.80 torr × = 0.00368 atm = π 760 torr 0.00368 atm −4 M = = 150 . × 10 M L atm 298 K) ( 0.0821 mol K)( ?g 1.00 g 1L 4g = × = 6.67 × 10 mol −4 mol 0.100 L 150 . × 10 M typical of small proteins 85

Osmotic Pressure 

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. US Navy submarines do as well. Dialysis is another example of this phenomenon.

86

Colloids Colloids are an intermediate type of mixture that has a particle size between those of true solutions and suspensions.





The particles do not settle out of the solution but they make the solution cloudy or opaque.

Examples of colloids include:

 1. 2. 3. 4. 5. 6. 7.

Fog Smoke Paint Milk Mayonnaise Shaving cream Clouds 87

The Tyndall Effect 

Colloids scatter light when it is shined upon them.  

Why they appear cloudy or opaque. This is also why we use low beams on cars when driving in fog. 

See Figure 14-18 in Textbook.

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The Adsorption Phenomenon Colloids have very large surface areas.





They interact strongly with substances near their surfaces.

One of the reasons why rivers can carry so much suspended silt in the water.



[

]

[

2 × Fe 2+ + 3 Cl - + ( y + 3) H 2 O → Fe 2 O 3 ⋅ y H 2 O + 6 H + + Cl -

]

A colloidal particle contains many Fe 2 O 3 ⋅ y H 2 O units with Fe3+ ions bound to its surface. The + charged particles repel each other and keep the colloid from precipitating.

89

Hydrophilic and Hydrophobic Colloids Hydrophilic colloids like water and are water soluble. 





Hydrophobic colloids dislike water and are water insoluble. 



Examples include many biological proteins like blood plasma. Hydrophobic colloids require emulsifying agents to stabilize in water.

Homogenized milk is a hydrophobic colloid. 



Milk is an emulsion of butterfat and protein particles dispersed in water The protein casein is the emulsifying agent.

90

Hydrophilic and Hydrophobic Colloids 

Mayonnaise is also a hydrophobic colloid. 





Mayonnaise is vegetable oil and eggs in a colloidal suspension with water. The protein lecithin from egg yolk is the emulsifying agent.

Soaps and detergents are excellent emulsifying agents.  

Soaps are the Na or K salts of long chain fatty acids. Sodium stearate is an example of a typical soap. 91

Hydrophilic and Hydrophobic Colloids Sodium stearate O

CH2 C O-Na+ CH2 CH2 CH2 CH2 H3C CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 CH2 polar (ionic) head hydrophilic portion CH2 CH2 nonpolar tail hydrophobic portion

92

Hydrophilic and Hydrophobic Colloids

93

Hydrophilic and Hydrophobic Colloids So called “hard water” contains Fe3+, Ca2+, and/or Mg2+ ions 



These ions come primarily from minerals that are dissolved in the water.

These metal ions react with soap anions and precipitate forming bathtub scum and ring around the collar.

Ca

2+

+ soap anion → Ca(soap anion)2(s) insoluble scum 94

Hydrophilic and Hydrophobic Colloids Synthetic detergents were designed as soap substitutes that do not precipitate in hard water.  

Detergents are good emulsifying agents. Chemically, we can replace COO- on soaps with sulfonate or sulfate groups

95

Hydrophilic and Hydrophobic Colloids 

Linear alkylbenzenesulfonates are good detergents.

CH2 CH2 CH2 CH2 CH2 CH2 CH3 CH2 CH2 CH2 CH2 CH2

O S O

O-Na+

96

Synthesis Question 

The world’s record for altitude in flying gliders was 60,000 feet for many years. It was set by a pilot in Texas who flew into an updraft in front of an approaching storm. The pilot had to fly out of the updraft and head home not because he was out of air, there was still plenty in the bottle of compressed air on board, but because he did not have a pressurized suit on. What would have happened to this pilot’s blood if he had continued to fly higher? 97

Synthesis Question 

As the pilot flew higher, the atmospheric pressure became less and less. With the lower atmospheric pressure, eventually the blood in the pilot’s veins would have begun to boil. This is a deadly phenomenon which the pilot wisely recognized.

98

Group Question 

Medicines that are injected into humans, intravenous fluids and/or shots, must be at the same concentration as the existing chemical compounds in blood. For example, if the medicine contains potassium ions, they must be at the same concentration as the potassium ions in our blood. Such solutions are called isotonic. Why must medicines be formulated in this fashion? 99

End of Chapter 14 

Human Beings are solution chemistry in action!

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