Solutions

1
CHAPTER 2

• Solutions
• By
• Dr. Hisham Ezzat

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les.asp?path./userdownloads/physical20chemistry
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2
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.

3
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

4
Ways of Expressing Concentration

• Qualitative Terms
• Dilute Solution A dilute solution has a
  relatively small concentration of solute.
• A concentrated solution has a relatively high
  concentration of solute.

5
Quantitative terms
Molarity
6
Molality

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

7
Molality and Mole Fraction

• Weight percent (wt )
• Volume percent or percent by volume (vol )

8
Molality and Mole Fraction

• Molarity

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

9
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

10
Molality and Mole Fraction

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

11
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!

12
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!

13
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.

14
Molality and Mole Fraction

• The mole fraction of component B - XB

15
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!

16
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)?

17
Molality and Mole Fraction

• Now we can calculate the mole fractions.

18
Chapter Goals

• 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

19
Chapter Goals

• Colligative Properties of Solutions
• Lowering of Vapor Pressure and Raoults 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

20
Chapter Goals

• Colloids
• The Tyndall Effect
• The Adsorption Phenomena
• Hydrophilic and Hydrophobic Colloids

21
Spontaneity of the Dissolution Process

• As an example of dissolution, lets assume that
  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 (lt 0) dissolution is
  favored.
• If ?Hsolution is endothermic (gt 0) dissolution is
  not favored.

22
Spontaneity of the Dissolution Process

• Change in disorder, or randomness, of the
  solution ?Smixing
• If ?Smixing increases (gt 0) dissolution is
  favored.
• If ?Smixing decreases (lt 0) dissolution is not
  favored.
• Thus the best conditions for dissolution are
• For the solution process to be exothermic.
• ?Hsolution lt 0
• For the solution to become more disordered.
• ?Smixing gt 0

23
Spontaneity of the Dissolution Process

• Disorder in mixing a solution is very common.
• ?Smixing is almost always gt 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.

24
Spontaneity of the Dissolution Process

• 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 lt 0.
• If solvation energy is less than the sum of the
  solute-solute and solvent-solvent attractions,
  the dissolution is endothermic, ?Hsolution gt 0.

25
Spontaneity of the Dissolution Process
26
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.

27
Dissolution of Solids in Liquids

• Dissolution is a competition between
• Solute -solute attractions
• crystal lattice energy for ionic solids
• Solvent-solvent attractions
• H-bonding for water
• Solute-solvent attractions
• Solvation or hydration energy

28
Dissolution of Solids in Liquids

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

29
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).
• Lets look at the dissolution of CaCl2.

30
Dissolution of Solids in Liquids
31
Dissolution of Solids in Liquids

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

32
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

33
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

34
Dissolution of Liquids in Liquids (Miscibility)

• Nonpolar molecules essentially slide in
  between each other.
• For example, carbon tetrachloride and benzene are
  very miscible.

35
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.

36
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.

37
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.

38
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.

39
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.

40
Rates of Dissolution and Saturation

• Supersaturated solutions have higher-than-saturate
  d concentrations of dissolved solutes.

41
Effect of Temperature on Solubility

• According to LeChateliers 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,
  LeChateliers principle applies to them.
• Possible stresses to chemical systems include
• Heating or cooling the system.
• Changing the pressure of the system.
• Changing the concentrations of reactants or
  products.

42
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.

43
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.

44
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.

45
Effect of Pressure on Solubility

• The effect of pressure on the solubility of gases
  in liquids is described by Henrys Law.

46
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.

47
Colligative Properties of Solutions

• There are four common types of colligative
  properties
• Vapor pressure lowering
• Freezing point depression
• Boiling point elevation
• Osmotic pressure
• Vapor pressure lowering is the key to all four of
  the colligative properties.

48
Lowering of Vapor Pressure and Raoults 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 solutions surface.
• The solute molecules occupy some of the spaces
  that would normally be occupied by solvent.
• Raoults Law models this effect in ideal
  solutions.

49
Lowering of Vapor Pressure and Raoults Law

• Derivation of Raoults Law.

50
Lowering of Vapor Pressure and Raoults Law

• Lowering of vapor pressure, ?Psolvent, is defined
  as

51
Lowering of Vapor Pressure and Raoults Law

• Remember that the sum of the mole fractions must
  equal 1.
• Thus Xsolvent Xsolute 1, which we can
  substitute into our expression.

52
Lowering of Vapor Pressure and Raoults Law

• This graph shows how the solutions vapor
  pressure is changed by the mole fraction of the
  solute, which is Raoults law.

53
Examples
The vapor pressure of water is 17.5 torr at 20C.
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
14 torr
54
Glycerin, C3H8O3, is a nonvolatile nonelectrolyte
with a density of 1.26 g/mL at 25C. Calculate
the vapor pressure at 25C of a solution made by
adding 50.0 mL of glycerin to 500.0 mL of water.
The vapor pressure of pure water at 25C is 23.8
torr
55
The vapor pressure of pure water at 110C is 1070
torr. A solution of ethylene glycol and water has
a vapor pressure of 1.00 atm at 110C. Assuming
that Raoult's law is obeyed, what is the mole
fraction of ethylene glycol in the solution?
Answer 0.290
PH2O 1070 torr PH2O 1 Atm 760 torr
PH2O PH2O
760 torr 1070 torr
XH2O ---------
---------

XH2O XEG 1
0.7103 XEG 1
1- 0.7103 XEG
0.290
XEG
56
Many solutions do not obey Raoult's law exactly
They are not ideal solutions. If the
intermolecular forces between solvent and solute
are weaker than those between solvent and solvent
and between solute and solute, then the solvent
vapor pressure tends to be greater than predicted
by Raoult's law. Conversely, when the
interactions between solute and solvent are
exceptionally strong, as might be the case when
hydrogen bonding exists, the solvent vapor
pressure is lower than Raoult's law predicts.
Although you should be aware that these
departures from ideal solution occur, we will
ignore them for the remainder of this chapter.
57
More Examples
Sucrose is a nonvolatile, nonionizing solute in
water. Determine the vapor pressure lowering, at
27C, of a solution of 75.0 grams of sucrose,
C12H22O11, dissolved in 180. g of water. The
vapor pressure of pure water at 27C is 26.7
torr. Assume the solution is ideal.
Vapor Pressure Lowered 26.7-26.1 0.6
58
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 25C? The vapor
pressure of pure propyl chloride is 347 torr at
25C and that of pure propyl bromide is 133 torr
at 25C. Assume that the solution is an ideal
solution.
59
. At 25C 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 25C are 451 torr for pentane and 321 torr for
cyclopentane. Assume that the solution is an
ideal solution.
60
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.

61
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 solutions vapor
  pressure is lowered as described by Raoults law.
• The solutions temperature must be raised to make
  the solutions 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.

62
Boiling Point Elevation

• Boiling point elevation relationship is

63
Boiling Point Elevation

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

64
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
65
The increase in boiling point relative to that of
the pure solvent, DTb, 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
66
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.
67
Calculate the boiling point of a solution of 2.0
molal of NaCl. Kb, water 0.52 C /mola.
Dt Kbm
NaCl(aq) ? Na Cl- 2.0 m 2.0 m
2.0 m
2.0 m 2.0 m 4.0m
Dt (0.52 C/molal)(4.0 molal) 2.08 C
BP NBP Dt 100.00C 2.08 C 102.08 C
68
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.

69
Freezing Point Depression

• Relationship for freezing point depression is

70
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.

71
Freezing Point Depression
72
Freezing Point Depression

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

73
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!

74
Freezing Point Depression
75
Determination of Molecular Weight by Freezing
Point Depression

• The size of the freezing point 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.

76
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?

77
Determination of Molecular Weight by Freezing
Point Depression
78
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

79
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.

80
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

81
Colligative Properties and Dissociation of
Electrolytes

• The vant 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.

82
Colligative Properties and Dissociation of
Electrolytes

• i has an ideal value of 2 for 11 electrolytes
  like NaCl, KI, LiBr, etc.

• i has an ideal value of 3 for 21 electrolytes
  like K2SO4, CaCl2, SrI2, etc.

83
Colligative Properties and Dissociation of
Electrolytes

• Example 14-8 The freezing point of 0.0100 m NaCl
  solution is -0.0360oC. Calculate the vant Hoff
  factor and apparent percent dissociation of NaCl
  in this aqueous solution.
• meffective total number of moles of solute
  particles/kg solvent
• First lets calculate the i factor.

84
Colligative Properties and Dissociation of
Electrolytes
85
Colligative Properties and Dissociation of
Electrolytes

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

86
Colligative Properties and Dissociation of
Electrolytes
87
Colligative Properties and Dissociation of
Electrolytes
88
Colligative Properties and Dissociation of
Electrolytes
89
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!

90
Colligative Properties and Dissociation of
Electrolytes
91
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
• cellophane and saran wrap
• skin
• cell membranes

92
Osmotic Pressure
semipermeable membrane
H2O
2O
H2O
H2O
sugar dissolved in water
H2O
H2O
net solvent flow
H2O
H2O
93
Osmotic Pressure
94
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.

95
Osmotic Pressure

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

96
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
• proteins
• ribonucleotides

97
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!

98
Osmotic Pressure
99
Osmotic Pressure
100
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.

101
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
• Fog
• Smoke
• Paint
• Milk
• Mayonnaise
• Shaving cream
• Clouds

102
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.

103
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.

104
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.

105
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.

106
Hydrophilic and Hydrophobic Colloids

• Sodium stearate

107
Hydrophilic and Hydrophobic Colloids
108
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.

109
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

110
Hydrophilic and Hydrophobic Colloids

• Linear alkylbenzenesulfonates are good detergents.

111
Synthesis Question

• The worlds 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 pilots blood if he had
  continued to fly higher?

112
Synthesis Question

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

113
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?

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End of Chapter 14

• Human Beings are solution chemistry in action!