1. Chapter 11
Properties of Solutions

2. 11.1 Solution composition
A solution is a homogenous mixture of 2 or more substances
The solute is(are) the substance(s) present in the smaller amount(s)
The solvent is the substance present in the larger amount

3. Various types of solutions

4. Concentration Units
The concentration of a solution is the amount of solute present in a given quantity of solvent or solution.
Percent by Mass
x 100%
mass of solute
mass of solute + mass of solvent
% by mass = =
x 100%
mass of solute
mass of solution
Mole Fraction (X)
XA =
moles of A
sum of moles of all components

5. volume of solution (liters)
Molarity (M)
moles of solute
M =
volume of solution (liters)
Molality (m)
m =
moles of solute
mass of solvent (kg)

7. Example
1.00 g C2H5OH is added to g of water to make 101 mL of solution. Find the molarity, mass %, mole fraction and molality of ethanol.

8. Molarity
Number of moles of solute per L or
solution

9. Mass Percent also called weight percent
percent by mass of the solute in the solution

10. Mole Fraction ratio of number of moles of a part of
solution to total number of moles of solution

11. Molality
Number of moles of solute per kg of solvent

12. Example
An aqueous antifreeze solution is 40% ethylene glycol (C2H6O2) by mass. The density of the solution is 1.05 g/cm3. Calculate the molality, molarity and mole fraction of ethylene glycol

13. Mass of water (solvent) = 100-40 = 60.0 g
where EG = ethylene glycol (C2H6O2)
# mol EG = mol EG
# mol water = 60.0/18.0 = 3.33 mol

14. What is the molality of a 5
What is the molality of a 5.86 M ethanol (C2H5OH) solution whose density is g/mL?
m =
moles of solute
mass of solvent (kg)
M =
moles of solute
liters of solution
Assume 1 L of solution:
Mass of solute = mass of 5.86 moles ethanol = 270 g ethanol
Mass of solvent = mass of 1 L solution=
(1000 mL x g/mL) = 927 g of solution
mass of solvent = mass of solution – mass of solute
= 927 g – 270 g = 657 g = kg
m =
moles of solute
mass of solvent (kg)
5.86 moles C2H5OH =
= 8.92 m
0.657 kg solvent

15. 11.2 Energies of Solution Formation
Dissolution of a solute in liquids
“like dissolves like”
Two substances with similar intermolecular forces are likely to be soluble in each other.
Non-polar molecules are soluble in non-polar solvents CCl4 in C6H6
Polar molecules are soluble in polar solvents C2H5OH in H2O
Ionic compounds are more soluble in polar solvents NaCl in H2O or NH3 (l)
Why this behavior occur?

16. Solubility Process The formation of a liquid solution takes place in
3 steps
1. Expand solute
molecules
2. Expand solvent
3. Mixing solute and
solvent
Endothermic
Endothermic
Often Exothermic

17. Energy of Solubility Process
Steps 1 and 2 require energy to overcome IMFs
endothermic
Step 3 usually releases energy
exothermic
enthalpy of solution
sum of ∆H values
can be – or +
∆Hsoln = ∆H1 + ∆H2 + ∆H3

18. Energy of Solubility Process

19. Case 1: oil and water Oil is nonpolar (LD forces)
Water is polar (H bonding)
∆H1 will be small for typical size
∆H2 will be large
∆H3 will be small since there won’t be much interaction between the two
∆Hsoln will be large and +ve because energy required by steps 1 and 2 is larger than the amount released by 3

20. Case 2: NaCl and water NaCl is ionic water is polar (H bonding)
∆H1 will be large
∆H2 will be large
∆H3 will be large and –ve because of the strong interaction between ions and water
∆Hsoln will be close to zero- small but +ve

21. Energy of solubility process
Enthalpy of hydration - ∆Hhyd
enthalpy change associated with dispersal
of gaseous solute in water
NaCl(s)  Na+(g) + Cl-(g)
∆H1=786 kJ/mol
H2O(l) + Na+(g) + Cl-(g)  Na+(aq) + Cl-(aq)
∆Hhyd=∆H2 + ∆H3=-783 kJ/mol
∆Hsoln=3 kJ/mol

22. DH3 DH2 DH1 DH3 Size of DH3 determines whether a solution will form
Solute and Solvent
DH3
Solution
DH2
DH3
Energy
Solvent
DH1
Reactants
Solution

23. Mixing solvent and solute
Two factors explains the solubility:
An increase in the probability of mixing favors the process
Processes that require large amounts of energy tend not to occur
If DHsoln is small and positive, a solution will still form because of entropy.
There are many more ways for them to become mixed than there is for them to stay separate.

24. Energy of Solubility Process

25. 1. Structure and Solubility
11.3 Factors Affecting Solubility
1. Structure and Solubility
Water soluble molecules must have dipole moments -polar bonds.
To be soluble in non polar solvents the molecules must be non polar.
Pentane C5H12 is miscible with hexane C6H14
and immiscible with water
Solubility of alcohols decreases with the molar
mass?
Cl3OH CH3(CH2)3OH CH3 (CH2)6OH
Soluble Insoluble
(Hydrophilic) (Hydrophobic)
Polarity decreases
OH is smaller portion
Hydrocarbon is larger 25

26. Structure effects: Vitamins and the body
hydrophobic: water- fearing: nonpolar
Insoluble in water
hydrophilic: water-loving: polar
Soluble in water
Hydrophilic, excreted by
the body and must be consumed regularly
Hydrophobic, accumulates
in the body. The body can
tolerate a diet deficient
in vitamin

27. 2. Pressure effects
Changing the pressure doesn’t effect the amount of solid or liquid that dissolves
They are incompressible.
It does effect gases.
Pressure effects the amount of gas that can dissolve in a liquid.
The dissolved gas is at equilibrium with the gas above the liquid. 27

28. The gas is at equilibrium with the dissolved gas in this solution.
The equilibrium is dynamic. 28

29. If the pressure is increased the gas molecules dissolve faster.
The equilibrium is disturbed. 29

30. The system reaches a new equilibrium with more gas dissolved.
Henry’s Law.
P= kC
Pressure = constant x Concentration of gas 30

31. Henry’s Law
Amount of gas dissolved is directly proportional to P of gas above solution
C  P C = kP
C is concentration and P is partial pressure of gaseous solute
The law is obeyed best by dilute solutions of gases that don’t dissociate or react with solvent

32. Example
A soft drink bottled at 25°C contains CO2 at pressure of 5.0 atm over liquid. Assume that PCO2 in atmosphere is 4.0 x 10-4 atm. Find the equilibrium concentration in soda before and after opening. k=32 L*atm/mol at 25°C

33. Example
before opening:
after opening:

34. Example
Solubility of pure N2 in blood at body temp, 37oC and 1 atm is 6.2X10-4 M. If a diver breaths air ( = 0.78) at a depth where the total pressure is 2.5 atm, calculate the concentration of N2 in his body.

36. 3. Temperature Effects (for aqueous solutions)
Increased temperature increases the rate at which a solid dissolves.
But it is not possible to predict whether it will increase the amount of solid that dissolves.
Solubility can be predicted only from a graph of experimental data. 36

37. 20 40 60 80
100 37

38. Temperature effects
dissolving a solid occurs faster at higher T
but the amount able to be dissolved does not change

39. Temperature effects
Solubility of gas in water decreases
with T

40. Temperature and Solubility
Solid solubility and temperature
solubility increases with increasing temperature
solubility decreases with increasing temperature
12.4

41. Solubility and environment
Thermal pollution
Water used as a coolant when pumped again into the source (lakes and rivers) floats on the cold water causing a decrease in solubility of O2 and consequently affecting the aquatic life.
CO2 dissolves in water that contains CO32- causing formation of HCO3- that is soluble in water. When temp increases CO2 will be driven off the water causing precipitation of CO32- again forming scales on the wools that blocks the pipes and reduce the heating efficiency
CO32- (aq) + CO2(aq) HCO3-

42. 11.4 The vapor pressures of solutions
A nonvolatile solvent lowers the vapor pressure of the solution.
For the molecules of the solvent to escape, they must overcome
the forces of both the other solvent molecules and the solute molecules.
Nonvolatile solute decreases
# of solvent molecules per unit
volume. Thus, # of solute
molecules escaping
will be lowered.

43. P1 Po1 How a nonvolatile solute affects a solvent?
Water has a higher vapor pressure than a solution P1
Po1
Aqueous Solution
Pure water

44. Water molecules evaporate faster from pure water than from the solution
<
Aqueous Solution
Pure water

45. The water molecules condense faster in the solution so it should all end up there.
empty
Aqueous Solution
Pure water

46. Raoult’s Law
The presence of a nonvolatile solute lowers the vapor pressure of the solvent.
Psolution = Observed Vapor pressure of the solution (vapor
pressure of solvent in solution)
solvent = Mole fraction of the solvent
P0solvent = Vapor pressure of the pure solvent
Applies only to an ideal solution where the solute doesn’t contribute to the vapor pressure (solute and solvent are alike: solute-solute, solvent-solvent and solute-solvent interactions
Are very similar ). Solute acts to dilute the solvent

47. Only one solute
Vapor pressure lowering

48. Deviations If solvent has a strong affinity for solute (H bonding).
Lowers solvents ability to escape.
Lower vapor pressure than expected.
Negative deviation from Raoult’s law.
If DHsoln is large and negative (exothermic): strong interaction exists between the solute and the solvent; thus a negtive deviation from Rault’s law because both components will have lower escaping tendency in the solution than in pure liquids.
If the two liquids mix endothermically (solute-solvent interactions are weaker than interactions among molecules in the pure liquids. More energy is required to expand liquids than is released when liquids are mixed). In this case molecules in the solution have higher tendency to escape than expected and positive deviations from Raul’s law are observed.

49. Liquid-liquid solutions in which both components are volatile
Nonideal solutions
Liquid-liquid solutions in which both components are volatile
Modified Raoult's Law:
P0 is the vapor pressure of the pure solvent
PA and PB are the partial pressures

50. Raoult’s Law – Ideal Solution
A solution of two liquids that obeys Raoult’s Law is called an ideal solution

51. Negative Deviations from Raoult’s Law
Strong solute-solvent interaction results in a vapor pressure lower than predicted
Exothermic mixing = Negative deviation

52. Positive Deviations from Raoult’s Law
Weak solute-solvent interaction results in a vapor pressure higher than predicted
Endothermic mixing = Positive deviation

53. An ______________ is one that obeys Raoult’s Law
PA = XA P A
PB = XB P B
Ideal Solution
PT = PA + PB
PT = XA P A
+ XB P B
An ______________ is one that obeys Raoult’s Law

54. < & > & PT is ______________ predicted by Raoults’s law
Force
A-B
A-A
B-B
<
&
Force
A-B
A-A
B-B
>
&

55. Vapor pressure lowering when ionic compounds are dissolved
VPL, when NaCl is dissolved in water, is twise as much as expected.
1 mol NaCl dissociates into 1 mol Na+ and 1 mol Cl-
# mols of solute = 2 X # mols NaCl
Thus, vapor pressure measurement can give information about the nature of the solute
When Na2SO4 is dissolved VPL is 3Xexpected

56. Example
Find the vapor pressure at 25°C for solution of g of sucrose (C12H22O11) in mL of water.
The density of water at 25°C is g/mL and the partial pressure of water vapor at 25°C is torr.

57. Example

58. Colligative Properties of solutions
11.5 Boiling point elevation and
freezing Point depression
Colligative Properties of solutions
Colligative properties depend only on the number of solute particles (molecules or ions) in a solution
They do not depend on the identity of particles
Colligative properties include:
Vapor pressure lowering
Boiling point elevation and freezing point depression
Osmotic pressure
Each of these properties is a consequence of a decrease in
the escaping tendency of solvent molecules brought by
the presence of solute particles.

59. Boiling Point Elevation
A nonvolatile solute lowers the vapor pressure
A higher T is required to reach the 1 atm of pressure which defines boiling point
A nonvolatile solute elevates the boiling point of the solvent
The amount of the elevation depends on concentration of the solute

60. Freezing Point Depression
Vapor pressure of solid and liquid are equal at freezing point
nonvolatile solute lowers the vapor pressure so a lower T is needed to decrease the vapor pressure to that of the solid
a nonvolatile solute depresses the freezing point of the solvent
the amount of the depression depends on concentration of the solute

61. 1 atm
Vapor Pressure of pure water
Vapor Pressure of solution

62. 1 atm
Freezing and boiling points of water

63. 1 atm
Freezing and boiling points of solution

64. 1 atm
DTb
DTf

65. Boiling-Point Elevation
DTb = Tb – T b
T b is the boiling point of
the pure solvent
T b is the boiling point of
the solution
Tb > T b
DTb > 0
DTb = Kb m
m is the molality of the solution
Kb is the molal boiling-point
elevation constant (0C/m)

66. Freezing-Point Depression
DTf = T f – Tf
T f is the freezing point of
the pure solvent
T f is the freezing point of
the solution
T f > Tf
DTf > 0
DTf = Kf m
m is the molality of the solution
Kf is the molal freezing-point
depression constant (0C/m)

67. Molal Boiling-Point Elevation and Freezing-Point Depression Constants of Several Common Liquids

68. Example 1
18.00 g of glucose are added to g of water. The boiling point of the solution is C. The boiling point constant is 0.51 C*kg/mol.
Find the molar mass of glucose.

69. Example 1

70. Example 1

71. Example 2
What mass of C2H6O2 (M=62.1 g/mol) needs to be added to 10.0 L H2O to make a solution that freezes at -23.3°C? density is 1.00 g/mL; boiling point constant is 1.86°C*kg/mol

73. DTf = Kf m Kf water = 1.86 0C/m DTf = Kf m
What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is g.
DTf = Kf m
Kf water = C/m =
3.202 kg solvent
478 g x
1 mol
62.01 g
m =
moles of solute
mass of solvent (kg)
= 2.41 m
DTf = Kf m
= C/m x 2.41 m = C
DTf = T f – Tf
Tf = T f – DTf
= C – C = C

74. Understanding osmoses: Experimental approach
Osmoses and Osmotic Pressure
Understanding osmoses: Experimental approach
Semipermeable membrane is a partition with pores that allow small solvent particles to pass through but not bigger solute particles. Thus, it separates a solution and a pure solvent
Final
Initial

75. Osmosis
The flow of solvent molecules from a pure solvent through a semi-permeable membrane into a solution
when the system has reached equilibrium, the water levels are different
Because the liquid levels are different, there is a greater hydrostatic pressure on the solution than on pure solvent

76. Osmotic pressure,  Osmotic pressure of a solution:
BEFORE
AFTER
Osmotic pressure of a solution:
The minimum pressure that stops the osmosis

77. Osmotic Pressure
The minimum pressure that stops the osmosis is equal to the osmotic pressure of the solution

78. Osmotic pressure, , and concentration of nonelectrolytes
  molarity of soultion
PV = nRT (Ideal gas equation)
Relation between  and M is the same: V = nRT
where
M is the molarity of the solution
R is the gas law constant
T is the temperature in Kelvin

79. Example
When 1.00x10-3 g of a protein is mixed with water to make 1.00 mL of solution, the osmotic pressure is 1.12 torr at 25.0°C.
Find the molar mass of this protein.

81. Osmosis and blood cells
>0.95% NaCl
>& 0.31 M glucose
isotonic
solution
Cell remains
stable
0.95% NaCl
& 0.31M
glucose
& 0.31 M glucose
<0.95% NaCl
<& 0.31 M glucose
0.95% NaCl
& 0.31M
glucose
hypotonic
solution
hypertonic
solution
Cell shrinks
Cell swells and burst
Crenation
Hemolyses

82. Dialysis
Separation of small species (molecules of solute and solvent and ions) from big species in a solution by means of a semi permeable membrane
In the artificial kidney, the blood is circulated from the patient through cellophane tubes immersed in solution containing all essential ions and small molecules in blood at the appropriate concentrations.
Only waste products dialyze from blood through the membrane.

83. Functioning of an artificial kidney

84. Reverse osmosis
Reverse osmosis is the process of reversing the normal net flow of solvent molecules through a semipermeable membrane by applying to the solution a pressure exceeding the osmotic pressure .

85. Reverse Osmosis A net flow of Solvent molecules
(blue) from solution to the solvent
Desalination (Water purification) is an application of reverse osmosis

87. Solutions of Electrolytes
The van’t Hoff factor (i) is used to modify the equations for colligative properties
FPD: DTf = –i Kfm BPE: DTb = i Kbm OP: p = i M RT
EOS
i is dependent on solution molality

88. 11.7 Collegative properties of electrolytes solutions
Since colligative properties only depend on the number of solute particles
Ionic compounds (salts) should have a bigger effect.
When they dissolve they dissociate.
Individual Na+ and Cl- ions fall apart.
1 mole of NaCl makes 2 moles of ions.
1mole Al(NO3)3 makes 4 moles ions.

89. Electrolytes have a bigger impact on melting and freezing points per mole because they make more species.
Relationship is expressed using the
van’t Hoff factor i
The expected value can be determined from the formula of the salt.

90. van’t Hoff Factor Observed i value is smaller than expected
Ion pairing most affects i value for highly charged ions
affects colligative properties

91. The actual value (effect on colligative properties) is usually less because
At any given instant some of the ions in solution will be paired.
Ion pairing tends to be higher for highly charged ions
Ion pairing increases with concentration.
i decreases with concentration.
Thus, van’t Hoff factor should be added to the equations of collegative properties:
DH = iKm

92. Colligative Properties of Electrolyte Solutions
0.1 m NaCl solution
0.1 m Na+ ions & 0.1 m Cl- ions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution
0.2 m ions in solution
van’t Hoff factor (i) =
actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
i should be
nonelectrolytes 1
NaCl 2
CaCl2 3

93. Dissociation Equations and the Determination of i
NaCl(s) 
Na+(aq) + Cl-(aq)
i = 2
AgNO3(s) 
Ag+(aq) + NO3-(aq)
i = 3
MgCl2(s) 
Mg2+(aq) + 2 Cl-(aq)
i = 3
Na2SO4(s) 
2 Na+(aq) + SO42-(aq)
AlCl3(s) 
Al3+(aq) + 3 Cl-(aq)
i = 4

94. Colligative Properties of Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iMRT
Vapor pressure lowering, VPL

95. Ideal vs. Real van’t Hoff Factor
The ideal van’t Hoff Factor is only achieved in VERY DILUTE solution.

97. iobs < iexp because of high ion pairing
Example 1
Osmotic pressure for 0.10 M solution of Fe(NH2)2(SO4)2 at 25°C was 10.8 atm. Compare the van’t Hoff Factor observed and expected.
iexp= 1+2+2=5
iobs < iexp because of high ion pairing

98. Ch. 11 Solutions
11.8 Colloids

99. Colloids Tyndall Effect Colloid
scattering of light particles used to determine whether something is a solution or suspension
Colloid
also called colloidal dispersion
suspension of tiny but visible particles in a medium

100. Types of Colloids

101. Colloids
particles in a colloid are nm in diameter and overall neutral
they have layer of negative ions on outside
so repel each other
Coagulation
when a colloid is destroyed through heating or addition of electrolyte