14 Solutions - Mr. Coe's Science Class Website @ DHS - Homepcoe.weebly.com/uploads/2/3/1/8/23185096/apchapter1… · · 2014-03-07Change of energy content or enthalpy of solution,

1

14 Solutions

2

Chapter Goals The Dissolution Process

1. Spontaneity of the Dissolution Process 2. Dissolution of Solids in Liquids 3. Dissolution of Liquids in Liquids (Miscibility) 4. Dissolution of Gases in Liquids 5. Rates of Dissolution and Saturation 6. Effect of Temperature on Solubility 7. Effect of Pressure on Solubility 8. Molality and Mole Fraction

3

Chapter Goals

Colligative 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 solvent. – Dissolved species are called the 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.

6

The Dissolution Process Seven 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


2. 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


• 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


• Solvent-solvent attractions such as hydrogen bonding in water. – This also requires an absorption of E.

• Solvent-solute attractions, 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.

11


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 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

3. 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


• 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


CaCl2(s)H2O → Ca(OH2)6[ ]2+

+ 2Cl- H2O( )x

where x is approximately 7 or 8

Ca

OH2

OH2

OH2

OH2

H2O

H2O

2+

Cl-

O H H

H O H

H O H

H H O

17


• The energy absorbed when one mole of formula units becomes hydrated is the molar energy of hydration.

[ ][ ] -y

n22-y

+nn22

+n

Xfor Ehydration O)X(HOH )(X

Mfor Ehydration )M(OHOH +(g)M

+→+

+→−

+

yx

ng

x

18


• 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.

22

Dissolution of Gases in Liquids

23

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

32222

← + →→+ −+aqaqg

24

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

up many smaller crystals

25

Rates of Dissolution and Saturation

• Saturated 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.

26

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

27

Rates of Dissolution and Saturation • Supersaturated solutions have higher-

than-saturated concentrations of dissolved solutes.

28

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.

29

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

30

Effect of Temperature on Solubility • For ionic solids that dissolve

endothermically dissolution is enhanced by heating.

• For ionic solids that dissolve exothermically dissolution is enhanced by cooling.

• Be sure you understand these trends.

31

Effect of Temperature on Solubility

32

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”.

33


• 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

=

=

MM

gas of pressure partial = Pncombinatio liquid-gas

each for number unique constant, Law sHenry' =k

gas ofion concentratmolar where

k P

gas

gas

gasgas

=

=

MM

34


35

Molality and Mole Fraction

• In Chapter 3 we introduced two important concentration units.

1. % by mass of solute

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

36


2. Molarity

solution of Literssolute of moles = M

• We must introduce two new concentration units in this chapter.

37


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

m =moles of solutekg of solvent

in dilute aqueous solutions molarity and molality are nearly equal

38


• 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 0.10OH kg

OHC mol ?

2

2

2

6126

2

6126

.molalityin ion concentrat theis This

OHC 617.0OHC g 180OHC mol 1

OH kg 1OH g 1000

OH g 90.0 OHC g 0.10

OH kgOHC mol ?

61266126

6126

2

2

2

6126

2

6126

m=

××=

39


• 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 180OHC mol 1

L 1mL 1000

nsol' mLnsol' g 04.1

n sol' g 100.0 OHC g 0.10

OH LOHC mol ?

M=×

××=

40


• 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 H COOH

kg C Hg 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 1 0 338

.

. m

41

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

42


• The mole fraction of component B - XB

1.00. equalmust fractions mole theall of sum The1 that Note

B of moles ofnumber +A of moles ofnumber B of moles ofnumber

A =+

=

B

B

XX

X

43


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

You do it!

61266126

6126

61266126

2

OHC mol 0556.0OHC g 180OHC 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

=

×=

×

44


• 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 18OH mol 1OH g 0.90OH mol ? 2

2

222 =×=

45


• 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

=

=

=

== +

5 005 000 989

0 05565 000 011

100 0 989 0 011

...

...

. . .

46


• Let’s combine all of these solution calculations into one problem.

• The sulfuric acid in a car battery has a density of 1.225 g/cm3 and is 3.75 M. What is the molality, mole fraction and percentage of sulfuric acid by mass in this solution? – Molarity has the solute, solvent, and solution

intertwined. To do this problem you must get each one separated. As a reminder, in this sulfuric acid solution:

• The solute is _____? • The solvent is ____? • The solution is _____?

47


• Problem solution pathway: 1. Use definition of molarity and density to separate the

components of the solution. 2. Use the appropriate components and recombine them to get

the other concentration units. 3. You are allowed to assume at the start of the calculation that

you have any amount of the solution necessary to help make your calculation easier. • You may want to use 1.0 L for molarity, 1.0 kg for molality, 1.0 mole

for mole fraction, or 100. g for % w/w.

48


solute of mass SOH of g 5.367SOH mol 1SOH g 0.98SOH moles 3.75

.SOH of moles 3.75 have that weus tells 3.75solution. thisof L 1.00 haveyou that Assume

42

42

4242

42

==

M

49


solution of g 1225cm

g 225.1L 1cm 1000L 1.00 solution of mass



3

3

42

42

4242

42

=

=

==

M

50


PROBLEM! THE OF REST THE FOR NECESSARY .COMPONENTS ITS OF ALL OF MASSES THE INTO DIVIDED SOLUTION THIS HAVENOW WENOTICE,

waterof g 857.5 solute of g 367.5 -solution of g 1225 solvent of mass

solution of g 1225cm

g 225.1L 1cm 1000L 1.00 solution of mass



3

3

4242

4242

42

==

=

=

==

M

51


kg 8575.0g 1000

kg 1OH g 857.5

kg. tog 857.5 eConvert thsolvent. the water,of g 857.5 are e that therknow wenscalculatio previous theFrom

solute. the,SOH of moles 3.75 have that weknow we, 3.75 theFrom

.solvent of kg

solute of moles ismolality Remember,

2

42

=

M

52


mm

M

37.4kg 0.8575

mol 75.3solvent of kg

solute of moles

molality. thecalculate Now

kg 8575.0g 1000

kg 1OH g 857.5

kg. tog 857.5 eConvert thsolvent. the water,of g 857.5 are e that therknow wenscalculatio previous theFrom


.solvent of kg

solute of moles ismolality Remember,

2

42

===

=

53


water.of moles water toof g 857.5 eConvert thsolvent. the water,of g 857.5 are e that therknow wenscalculatio previous theFrom


.solvent of moles solute of moles

solute of moles isfraction mole Remember,

42M+

54


0730.047.6 3.75

mol 75.3solvent of molessolute of moles

solute of moles

fraction. mole thecalculate Now

OH of mol 6.47g 18.0

mole 1OH g 857.5

water.of moles water toof g 857.5 eConvert thsolvent. the water,of g 857.5 are e that therknow wenscalculatio previous theFrom


.solvent of moles solute of moles

solute of moles isfraction mole Remember,

42SOH

22

42

=+

=+

=

=

+

X

M

55


w/w%0.30%100solution of g 1225

SOH of g 367.5 w/w%

w/w.% thecalculate Nowsolution. of g 1225 are e that therknow wenscalculatio previous theFrom

g. 367.5 of mass a have SOH of moles 3.75 that theknow wenscalculatio previous theFrom

100%.x solution of masssolute of mass is massby % Remember,

42

42

=×=

56

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.

57

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.

58

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.

59

Lowering of Vapor Pressure and Raoult’s Law

60

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

P Pwhere P vapor pressure of solvent

P vapor pressure of pure solventmole fraction of solvent

solvent solvent solvent0

solvent

solvent0

solvent

==

==

Xin solution

X in solution

61

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

0solventsolvent

0solventsolvent

0solvent

solvent0solventsolvent

)P1(

)P)((- P

PP P

XX

−=

=

−=∆


62

• 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

XXX

=∆

=


63

• This graph shows how the solution’s vapor pressure is changed by the mole fraction of the solute, which is Raoult’s law.


64

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.

65

Fractional Distillation

66

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.

67

Boiling Point Elevation

• Boiling point elevation relationship is:

solvent for the constantelevation point boiling molal K

solute ofion concentrat molal elevationpoint boiling T :where

KT

b

b

bb

==

=∆=∆

m

m

68

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

=∆

=∆

=∆

mmm

69

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.

70

Freezing Point Depression

• Relationship for freezing point depression is:

solventfor constant depressionpoint freezing Ksolute ofion concentrat molal

solvent of depressionpoint freezingT :whereKT

f

f

ff

==

=∆=∆

m

m

71


• 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 =∆=∆

72


73


• 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

=∆

=∆

=∆

mmm

74


• 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!

75


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

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

KTsolution. for this depression theCalculate .2

929.0COOHHC g 122

COOHHC mol 1HC kg 0.0750

COOHHC g 50.8HC kgCOOHHC mol ?

molality! Calculate .1

000

00f

ff

56

56

66

56

66

56

==∆

=∆

=

×=

mmm

m

76

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.

77


• 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?

78


g/mol 7.61mol 0.600g 37 is massmolar theThus

compound mol 600.0kg 0.200 3.00=OH kg 0.200in compound mol ?

water.of kg 0.200 mL 200 are thereproblem In this

00.3C1.86C58.5

KT

the thusKT

2

0

0

f

f

ff

=

=×

=

==∆

=

=∆

m

mm

m

79

Cooling Curves of Pure Water and Three 0.1 m Solutions

-5

-4

-3

-2

-1

0

1

2

3

4

5

0 100 200 300 400 500 600 700 800 900

Time (sec)

Tem

pera

ture

(o C

)

80

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

81

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.

82

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

83

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

TT

∆∆

=i

84

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 +→

85

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

mmm

= = =∆

∆

TT

KK

f actual

f if nonelectrolyte

f effective

f stated

effective

stated

86


( )

( )

( )( )

i mm

mm

m mm

m m i mm

mm

= = =

= ∴ = =

= ∴ = = =

∆

∆

∆∆

TT

KK

T KT

KC

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 0194 0 01940 0100

194

0.

. ..

.

87

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

percent dissociation. • Let x = mNaCl that is apparently

dissociated.

88


mxmxmx )0100.0(Cl + Na NaCl -+OH2

−→

89


[ ][ ]

mxmmx

mxxxmmxmxmx

0094.0 0194.0 0100.0

0100.0 )0100.0(

Cl + Na NaCl

effective

-+OH2

==+=++−=

−→

90


%94

%1000100.00094.0

%100 =on dissociati %apparent stated

diss app

=

×=

×

mm

mm

91

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!

92


( )( )[ ] ( )

( )

unionized 98.0% and ionized %0.2

%100 0500.0 0010.0%100 = ionized %

0010.0 0510.0 0500.0

0510.0C1.86

C0948.0KT

KT 0500.0 0500.0

0500.0COOCH + H COOHCH

original

ionized

eff

0

0

f

feff

effff

eff

-3

+3

=

×=×

==+=

==∆

=

×=∆+=++−=

−←→

mm

mm

mxmmxm

mm

m

mmxmxxxm

mxmxmx

93

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 wrap 2. skin 3. cell membranes

94

Osmotic Pressure

95

Osmotic Pressure

96

Osmotic Pressure

H2O 2O

semipermeable membrane

H2O H2O

sugar dissolved in water

H2O

H2O

H2O

H2O

net solvent flow

97

Osmotic Pressure

98

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.

ππ

= M

M

RTwhere: = osmotic pressure in atm

= molar concentration of solution

R = 0.0821 L atmmol K

T = absolute temperature

99

Osmotic Pressure

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

π = mfor dilute aqueous solutions only

RT

100

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

101

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!

102

Osmotic Pressure

( )( )

π π

π

= ∴ =

× =

= × −

M M

M M

RT RT

atm = 2.80 torr 1 atm760 torr

atm =

= atm0.0821 KL atm

mol K

? .

. .

0 00368

0 00368298

150 10 4

103

Osmotic Pressure

( )( )

π π

π

= ∴ =

× =

= ×

= ××

= ×

−

−

M M

M M

M

RT RT

atm = 2.80 torr 1 atm760 torr

atm =

= atm0.0821 K

gmol

1.00 g0.100 L

L

typical of small proteins

L atmmol K

gmol

? .

. .

?.

.

0 00368

0 00368298

150 10

1150 10

6 67 10

4

44

104

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.

105

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. Fog 2. Smoke 3. Paint 4. Milk 5. Mayonnaise 6. Shaving cream 7. Clouds

106

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.

107

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 ionsbound 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

108

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.

109

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.

110

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

111


112


113


• 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)

+ + →

114

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

115

Hydrophilic and Hydrophobic Colloids • Linear alkylbenzenesulfonates are good

detergents.

CH2CH2CH2

CH2CH3CH2 CH2CH2

CH2CH2CH2

CH2

S

O

O

O-Na+

116

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?

117

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.

118

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?

119

14 Solutions

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14 Solutions - Mr. Coe's Science Class Website @ DHS - Homepcoe.weebly.com/uploads/2/3/1/8/23185096/apchapter1… · · 2014-03-07Change of energy content or enthalpy of solution,