Chapter 14

1

CHAPTER 14 Solutions

2

Chapter GoalsThe Dissolution Process

1. Spontaneity of the Dissolution Process2. Dissolution of Solids in Liquids3. Dissolution of Liquids in Liquids (Miscibility)4. Dissolution of Gases in Liquids5. Rates of Dissolution and Saturation6. Effect of Temperature on Solubility 7. Effect of Pressure on Solubility8. Molality and Mole Fraction

3

Chapter GoalsColligative Properties of Solutions

9. Lowering of Vapor Pressure and Raoult’s Law

10. Fractional Distillation

11. Boiling Point Elevation

12. Freezing Point Depression

13. Determination of Molecular Weight by Freezing Point Depression or Boiling Point Elevation

14. Colligative Properties and Dissociation of Electrolytes

15. Osmotic Pressure

4

Chapter Goals

Colloids

16. The Tyndall Effect

17. The Adsorption Phenomena

18. Hydrophilic and Hydrophobic Colloids

5

The Dissolution Process

Solutions are homogeneous mixtures of two or more substances. Dissolving medium is called the solventsolvent. Dissolved species are called the solutesolute.

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.

6

The Dissolution ProcessSeven Homogeneous Possibilities

Solute Solvent Example Solid Liquid salt water Liquid Liquid mixed drinks Gas Liquid carbonated

beverages Liquid Solid dental amalgams Solid Solid alloys Gas Solid metal pipes Gas Gas air

Two Heterogeneous Possibilities Solid Gas dust in air Liquid Gas clouds, fog

7

Spontaneity of the Dissolution Process As an example of dissolution, let’s assume that

the solvent is a liquid. Two major factors affect dissolution of solutes

1. 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.

8

Spontaneity of the Dissolution Process2. Change in disorder, or randomness, of the

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

9

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

Smixing is almost always > 0.

What factors affect Hsolution? 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.

10

Spontaneity of the Dissolution Process Solvent-solvent attractions such as hydrogen

bonding in water. This also requires an absorption of E.

Solvent-solute attractions, solvationsolvation, 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.

11

Spontaneity of the Dissolution Process

12

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

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 energycrystal lattice energy.

+ - + -(g) (g) (s)M X M X crystal lattice energy

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

13

Dissolution of Solids in Liquids Dissolution is a competition between:

1. Solute -solute attractions crystal lattice energy for ionic solids

2. Solvent-solvent attractions H-bonding for water

Solute-solvent attractions Solvation or hydration energy

14

Dissolution of Solids in Liquids Solvation is directed by the water to ion

attractions as shown in these electrostatic potentials.

15

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.

16

Dissolution of Solids in Liquids

8or 7ely approximat is where

OHCl2)Ca(OHCaCl 2-2

62OH

(s)22

x

x

Ca

OH2

OH2

OH2

OH2

H2O

H2O

2+

Cl-

OH H

H OH

HO H

H H O

17

Dissolution of Solids in Liquids The energy absorbed when one mole of

formula units becomes hydrated is the molar molar energy of hydrationenergy of hydration.

-y

n22-y

+nn22

+n

Xfor Ehydration O)X(HOH )(X

Mfor Ehydration )M(OHOH +(g)M

y

x

ng

x

18

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

19

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

20

Dissolution of Liquids in Liquids (Miscibility) Nonpolar molecules essentially “slide” in

between each other. For example, carbon tetrachloride and benzene

are very miscible.

C Cl

Cl

Cl

Cl

C

CC

C

CC

H

H

H

H

H

H

21

Dissolution of Gases in Liquids Polar gases are more soluble in water than

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.

acid weak

HSO OHSOH

SOH OH+SO

acid strong

Br OH OH+HBr

-3

+3

OH

)3(2

3222

-+32

2

aqaqaq

aq

aqaq

22

Dissolution of Gases in Liquids A few nonpolar gases are soluble in water because

they react with water.

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

acidk very wea

HCOOHCOHOHCO 33

OH

3222

2

aqaqg

23

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.

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

NaCl

Breaks

upmany smaller crystals

24

Rates of Dissolution and Saturation Saturated solutionsSaturated solutions have established an

equilbrium between dissolved and undissolved solutes Examples of saturated solutions include:

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

25

Rates of Dissolution and Saturation Symbolically this equilibrium is written as:

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

-aqaqs XM MX

26

Rates of Dissolution and Saturation SupersaturatedSupersaturated solutions have higher-than-

saturated concentrations of dissolved solutes.

27

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. Heating or cooling the system.

2. Changing the pressure of the system.

3. Changing the concentrations of reactants or products.

28

Effect of Temperature on Solubility What will be the effect of heating or cooling the 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.

Warming the water will decrease solubility and cooling the water will increase the solubility.

Predict the effect on an endothermic dissolution like this one.

LiBr Li Br kJ / molsH O + -2 aq aq 48 8.

KMnO kJ / mol K MnO4 sH O +

4-2 436. aq aq

29

Effect of Temperature on Solubility For ionic solids that dissolve endothermicallyendothermically

dissolution is enhanced by heatingheating. For ionic solids that dissolve exothermically exothermically

dissolution is enhanced by coolingcooling. Be sure you understand these trends.

30

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”.

31

Effect of Pressure on Solubility The effect of pressure on the solubility of gases in

liquids is described by Henry’s Law.

gas ofion concentratmolar where

k P

gas

gasgas

M

M

gas of pressure partial = P

ncombinatio liquid-gas

each for number unique constant, Law sHenry' =k

gas ofion concentratmolar where

k P

gas

gas

gasgas

M

M

32

Molality and Mole Fraction

In Chapter 3 we introduced two important concentration units.

1. % by mass of solute

%100solution of mass

solute of mass = w/w%

33


2. Molarity

solution of Liters

solute of moles = M

We must introduce two new concentration units in this chapter.

34


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

m moles of solute

kg of solvent

in dilute aqueous solutions molarity and

molality are nearly equal

35

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 H O

kg H O

g C H O

90.0 g H O 6 12 6

2

6 12 6

2

10 0.

OH kg 1

OH g 1000

OH g 90.0

OHC g 0.10

OH kg

OHC mol ?

2

2

2

6126

2

6126

.molalityin ion concentrat theis This

OHC 617.0OHC g 180

OHC mol 1

OH kg 1

OH g 1000

OH g 90.0

OHC g 0.10

OH kg

OHC mol ?

61266126

6126

2

2

2

6126

2

6126

m

36

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!

61266126

6126

6126

2

6126

OHC 578.0OHC g 180

OHC mol 1

L 1

mL 1000

nsol' mL

nsol' g 04.1

n sol' g 100.0

OHC g 0.10

OH L

OHC mol ?

M

37


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!You do it!

? mol C H COOH

kg C H

g C H COOH

200.0 mL C H

mL C H

0.879 g C H

g C H

1 kg C H

mol C H COOH

122 g C H COOH C H COOH

6 5

6 6

6 5

6 6

6 6

6 6

6 6

6 6

6 5

6 56 5

7 25 1

1000 10 338

.

. m

38

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.

B of moles ofnumber +A of moles ofnumber

A of moles ofnumber AX

39


The mole fraction of component B - XB

1.00. equalmust fractions mole theall of sum The

1 that NoteB of moles ofnumber +A of moles ofnumber

B of moles ofnumber

A

B

B

XX

X

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)?


61266126

6126

61266126

2

OHC mol 0556.0OHC g 180

OHC mol 1

OHC g 0.10OHC mol ?

water.of g 90.0 and glucose of g 10.0

are heresolution t thisof g 101.00In

41


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

OH mol 00.5OH g 18

OH mol 1OH g 0.90OH mol ? 2

2

222

42


Now we can calculate the mole fractions.

X

X

H O2

2 6 12 6

C H O6 12 6

2 6 12 6

2

6 12 6

mol H O

mol H O + 0.0556 mol C H O

mol C H O

mol H O + 0.0556 mol C H O

5 00

5 00

0 989

0 0556

5 00

0 011

100 0 989 0 011

.

.

.

.

.

.

. . .

43

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.

44

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.

45

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.

46

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

47

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

48

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

49

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.

50

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.

51

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.

52

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

53

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

54

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.

55

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

56

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

57


58


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

59


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.


60


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

61

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.

62

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?

63

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

64

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

65

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.

66

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

67

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.

lytenonelectro iff

actualf

T

T

i

68

Colligative Properties and Dissociation of Electrolytes

i has an ideal value of 2 for 1:1 electrolytes like NaCl, KI, LiBr, etc.

i has an ideal value of 3 for 2:1 electrolytes like K2SO4, CaCl2, SrI2, etc.

unit formulaions 2 ClNaClNa -

aq+aq

OH-+ 2

unit formulaions 3 Cl 2CaClCa -

aq+2aq

OH-2

+2 2

69

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.

i

m

m

m

m

T

T

K

Kf actual

f if nonelectrolyte

f effective

f stated

effective

stated

70


im

m

m

m

m mm

m m im

m

m

m

T

T

K

K

T KT

K

C

1.86 C

f actual

f if nonelectrolyte

f effective

f stated

effective

stated

f actual f effective effectivef actual

f0

effectiveeffective

stated

0 0360

0 01940 0194

0 0100194

0.

..

..

71

Colligative Properties and Dissociation of Electrolytes Next, we will calculate the apparent percent

dissociation. Let x = mNaCl that is apparently dissociated.

72


mxmxmx )0100.0(

Cl + Na NaCl -+OH2

73


mx

mmx

mxxxm

mxmxmx

0094.0

0194.0 0100.0

0100.0

)0100.0(

Cl + Na NaCl

effective

-+OH2

74


%94

%1000100.0

0094.0

%100 =on dissociati %apparent stated

diss app

m

m

m

m

75

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.


76


unionized 98.0% and ionized %0.2

%100 0500.0

0010.0%100 = ionized %

0010.0

0510.0 0500.0

0510.0C1.86

C0948.0

K

T

KT

0500.0 0500.0

0500.0

COOCH + H COOHCH

original

ionized

eff

0

0

f

feff

effff

eff

-3

+3

m

m

m

m

mx

mmxm

mm

m

m

mxmxxxm

mxmxmx

77

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

78

Osmotic Pressure

H2O 2O

semipermeable membrane

H2O H2O

sugar dissolvedin water

H2O

H2O

H2O

H2O

net solvent flow

79

Osmotic Pressure

80

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

81

Osmotic Pressure

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

m

for dilute aqueous solutions only

RT

82

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

83

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.


84

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

85

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

86

Osmotic PressureWater 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.

87

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. Fog2. Smoke3. Paint4. Milk5. Mayonnaise6. Shaving cream7. Clouds

88

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.

89

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 3 3

Fe Cl H O Fe O H O + 6 H + Cl

A colloidal particle contains many Fe O H O units with Fe ions

bound to its surface. The + charged particles repel each other and

keep the colloid from precipitating.

2+ -2 2 3 2

+ -

2 3 23+

y y

y

90

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

Examples include many biological proteins like blood plasma. Hydrophobic colloids dislike water and are water insoluble.

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.

91

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.

92

Hydrophilic and Hydrophobic Colloids Sodium stearate

CH2

CH2CH2

CH2CH2

CH2CH3

CH2

CH2CH2

CH2CH2

CH2CH2

CH2CH2

CH2C

O

O

-Na+nonpolar tailhydrophobic portion

polar (ionic) headhydrophilic portion

93

Hydrophilic and Hydrophobic Colloids

94

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 soap anion Ca(soap anion)

insoluble scum

22(s)

95

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

96

Hydrophilic and Hydrophobic Colloids Linear alkylbenzenesulfonates are good

detergents.

CH2CH2CH2

CH2CH3CH2 CH2CH2

CH2CH2CH2CH2

S

O

O

O-Na+

97

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?

98

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.

99

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?

100

End of Chapter 14

Human Beings are solution chemistry in action!

Chapter 14

Documents

example of dissolution

dissolution disorder

dissolution change

dissolution of solutes

dissolution process

liquids dissolution

solutesolute attractions

liquids rates of dissolution