1. Chapter 14 Solutions and Their Behavior

2. Solutions
A solution is a homogeneous mixture of two or more substances in a single phase.
By convention, the component present in largest amount is identified as the solvent and the other component(s) as the solute(s)

3. Solutions Solutions can be classified as saturated or unsaturated.
A saturated solution contains the maximum quantity of solute that dissolves at that temperature.
See Chapter 14 Video Presentation Slide 5

4. Solutions
See Chapter 14 Video Presentation Slide 6
An unsaturated solution can still take on more solute at a given temperature.
SUPERSATURATED SOLUTIONS contain more than is possible and are unstable.

5. Concentration Units
Molarity (M)
moles solute M =
L of solution

6. Concentration Units Weight (mass) % mass solute %  100=
Total mass of solution
x 100%
mass of solute
mass of solute + mass of solvent
% by mass =

7. Concentration Units Mole Fraction (X) moles solute (i) Xi =
Total moles in solution

8. Concentration Units Mole Fraction (X)
Where “i” is the moles of one component of the solute.
Total moles are all species:
mols solute (i) + mols solvent.
The sum of all of the mole fractions for each component are 1 exactly.

9. Concentration Units Molarity (M) moles of solute M =
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)

10. Concentration Units Parts Per Million (ppm) mg solute m = kg solution
Parts per million (ppm): grams of solute/grams of solution (then multiplied by 106 or 1 million)

11. Solution Concentration
Amount of solute
Most concentration units are expressed as:
Amount of solvent or solution
Molarity: moles of solute/liter of solution
Percent by mass: grams of solute/grams of solution (then multiplied by 100%)
Percent by volume: milliliters of solute/milliliters of solution (then multiplied by 100%)
Mass/volume percent: grams of solute/milliliters of solution (then multiplied by 100%) 11

12. Calculating Concentrations
Problem:
62.1 g (1.00 mol) of ethylene glycol is dissolved in g of water.
Calculate mol fraction, molality, and weight % of the solution.

13. Calculating Concentrations
Problem:
62.1 g (1.00 mol) of ethylene glycol is dissolved in g of water.
Mole Fraction:

14. Calculating Concentrations
Problem:
62.1 g (1.00 mol) of ethylene glycol is dissolved in g of water.
Mole Fraction:

15. Calculating Concentrations
Problem:
62.1 g (1.00 mol) of ethylene glycol is dissolved in g of water.
Molality:

16. Calculating Concentrations
Problem:
62.1 g (1.00 mol) of ethylene glycol is dissolved in g of water.
Wt. %:

17. 5.86 moles ethanol = 270 g ethanol
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:
5.86 moles ethanol = 270 g ethanol
927 g of solution (1000 mL x g/mL)
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
0.657 kg solvent
= 8.92 m 17

18. The Solution Process
Solutes dissolve in solvents by a process called solvation.
Polar solvent dissolve polar solutes, non- polar solvents dissolve non-polar solutes. (aka: “like dissolves like”.
If two liquids mix to an appreciable extent to form a solution, they are said to be miscible.
In contrast, immiscible liquids do not mix to form a solution; they exist in contact with each other as separate layers.

19. The Solution Process

20. Solvation of Ions +
When a cation exists in solution, it is surrounded by the negative dipole ends of water molecules.
When as anion exists in solution, it is surrounded by the positive dipole ends of water molecules.

21. Energetics of the Solution Process: solutionH
Energy must be supplied to separate the ions from their attractive forces. (an endothermic process)
latticeH

22. Energetics of the Solution Process: solutionH
Energy must be supplied to separate the ions from their attractive forces. (an endothermic process)
latticeH
Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation.

23. Energetics of the Solution Process: solutionH
Energy must be supplied to separate the ions from their attractive forces. (an endothermic process)
latticeH
Energy is evolved when the individual ions dissolve in water where each ion is stabilized by solvation.
This process, referred to as the Energy of Hydration when water is the solvent, is strongly exothermic.

24. Energetics of the Solution Process: solutionH
We can therefore represent the process of dissolving KF in terms of these chemical equations:
Step 1: KF(s) → K+(g) + F(g) = latticeH
Step 2: K+(g) + F (g) → K+(aq) +F(aq) =hydrationH

25. Energetics of the Solution Process: solutionH
The overall reaction is the sum of these two steps. The enthalpy of the overall reaction, called the enthalpy of solution (solnH), is the sum of the two enthalpies.
Overall:
KF(s)→ K+(aq) + F(aq)
solnH = latticeH + hydrationH

26. Energetics of the Solution Process: solutionH

27. Energetics of the Solution Process
If the enthalpy of formation of the solution is more negative than that of the solvent and solute, the enthalpy of solution is negative.
The solution process is exothermic!
See Chapter 14 Video Presentation Slide 7

28. Supersaturated Sodium Acetate
One application of a supersaturated solution is the sodium acetate “heat pack.”
The enthalpy of solution for sodium acetate is ENDOthermic.
See Chapter 14 Video Presentation Slide 8

29. Supersaturated Sodium Acetate
Sodium acetate has an ENDOthermic
heat of solution.
NaCH3CO2 (s) + heat f Na+(aq) + CH3CO2-(aq)
Therefore, formation of solid sodium acetate from its ions is EXOTHERMIC.
Na+(aq) + CH3CO2-(aq) f NaCH3CO2 (s) + heat

30. Temperature and Solubility
Solid solubility and temperature
solubility increases with increasing temperature
solubility decreases with increasing temperature 30

31. Factors Affecting: Solubility Pressure & Temperature— “Henry’s Law”
See Chapter 14 Video Presentation Slide 9
Gas solubility (mol/L)
Sg =KHPg

32. Temperature and Solubility
O2 gas solubility and temperature
solubility usually decreases with increasing temperature
Do you like your coke hot or cold? Why? 32

33. Chemistry In Action: The Killer Lake
8/21/86
CO2 Cloud Released
1700 Casualties
Trigger?
earthquake
landslide
strong Winds
Lake Nyos, West Africa 33

34. Colligative Properties
Relative to a pure solvent, a solution has:
Lower vapor pressure
Higher boiling point
Lower freezing point
A higher osmotic pressure
Example:
Pure water: b.p.=100°C f.p.= 0°C
1.00 m NaCl (aq) b.p.= 101°C f.p. = -3.7 °C

35. Colligative properties
Upon adding a solute to a solvent, the properties of the solvent are affected:
Vapor pressure decreases
Melting point decreases
Boiling point increases
Osmosis is possible (osmotic pressure)
Collectively these changes are called COLLIGATIVE PROPERTIES.
They depend only on the NUMBER of solute particles relative to solvent particles, not on the KIND of solute particles.

36. Understanding Colligative Properties
To understand colligative properties, one must consider the LIQUID-VAPOR EQUILIBRIUM for a solution.

37. Understanding Colligative Properties
See Chapter 14 Video Presentation Slides 10 & 11
LIQUID-VAPOR EQUILIBRIUM

38. Changes in Vapor Pressure: Raoult’s Law
Vapor pressure: pressure exerted by the vapor when a liquid in a closed container is at equilibrium with the vapor.
Vapor pressure of solution is decreased by the presence of a solute.
More solute particles, lower vapor pressure of solution.

39. Psolution = Xsolvent  Po
Raoult’s Law
Psolution = Xsolvent  Po
Psolution = the vapor pressure of a mixture of solute and solvent
Po = the vapor pressure of the pure solvent
Xsolvent = the mole fraction of the solvent.
The expression can also be written in the form:
P is the change to the vapor pressure of the pure solvent.

40. Special case of 2 Liquids
PA = XA P A
Ideal Solution
PB = XB P B
PT = PA + PB
PT = XA P A
+ XB P B
Special case of 2 Liquids 40

41. < & > & PT is greater than predicted by Raoults’s law
PT is less than
predicted by Raoults’s law
Force
A-B
A-A
B-B
<
&
Force
A-B
A-A
B-B
>
& 41

42. Vapor Pressure Lowering of Benzene + non volatile solute

43. Raoult’s Law
Problem:
Pure iodine (105 g) is dissolved in 325 g of CCl4 at
65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

44. Raoult’s Law
Problem:
Pure iodine (105 g) is dissolved in 325 g of CCl4 at
65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

45. Raoult’s Law
Problem:
Pure iodine (105 g) is dissolved in 325 g of CCl4 at
65 °C. Given that the vapor pressure of CCl4 at this temperature is 531 mm Hg, what is the vapor pressure of the CCl4–I2 solution at 65 °C? (Assume that I2 does not contribute to the vapor pressure.)

46. Boiling Point Elevation & Freezing Point Depression
Molecular Compounds
The temperature of the normal boiling point of a solution is increased by:
The temperature of the normal freezing point of a solution is decreased by:
Where the K’s are the respective boiling and freezing point constants and msolute is the molality of the solution.

47. Boiling Point Elevation & Freezing Point Depression
Molecular Compounds
For boiling point elevation: Kb > 0 (positive)
For freezing point depression: Kf < 0 (negative)

48. Boiling Point Elevation & Freezing Point Depression

49. van’t Hoff Factor Boiling and Freezing point effects involving ions:
When solutions containing ions are involved, the total concentration of solute particles must be considered.
The change in b.p. or f.p. is given by the equation:
Where m is the calculated molaity based on formula wt.
“i” = the number of ions (van’t Hoff factor)
compound Type i
CH3OH molecular
NaCl strong electrolyte
Ba(NO3)2 strong electrolyte
HNO2 weak electrolyte

50. The Boiling Point of a Solution is Higher Than That of a Pure Solvent
See Chapter 14 Video Presentation Slide 13

51. Change in Freezing Point
Pure water
Water solution
See Chapter 14 Video Presentation Slides 13 & 14
The freezing point of a solution is LOWER than that of the pure solvent.

52. Lowering the Freezing Point
Water with and without antifreeze
When a solution freezes, the solid phase is pure water. The solution becomes more concentrated.

53. Freezing Point Depression
Problem:
If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

54. Freezing Point Depression
Problem:
If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

55. Freezing Point Depression
Problem:
If 52.5 g of LiF is dissolved in 306 g of water, what is the expected freezing point of the solution? (Assume the van’t Hoff factor, i, for LiF is 2.)

56. Molar Mass By Boiling Point Elevation
Problem:
Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

57. Molar Mass By Boiling Point Elevation
Problem:
Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?
Solution: The molality of the solution can be found from the freezing point depression.
Knowing molality and mass of solvent, one can find the moles of solute.
Knowing mass and moles of solute, the molar mass of the compound can be determined.

58. Molar Mass By Boiling Point Elevation
Problem:
Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

59. Molar Mass By Boiling Point Elevation
Problem:
Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

60. Molar Mass By Boiling Point Elevation
Problem:
Benzyl acetate is one of the active components of oil of jasmine. If g of the compound is added to 25.0 g of chloroform (CHCl3), the boiling point of the solution is °C. What is the molar mass of benzyl acetate?

61. Osmosis Dissolving the shell in vinegar Egg in pure water
Egg in corn syrup
See Chapter 14 Video Presentation Slide 4

62. Osmosis
A semipermeable membrane allows only the movement of solvent molecules.
Solvent molecules move from pure solvent to solution in an attempt to make both have the same concentration of solute.
See Chapter 14 Video Presentation Slides 15 & 16

63. Osmosis
Osmosis of solvent from one solution to another will occur as they try to equalize one another’s concentration.
At the point where they have equal osmotic pressures, they are said to be Isotonic.

64. Osmosis at the Particulate Level

65. Process of Osmosis

66. Osmotic Pressure, 
Equilibrium is reached when the internal pressure of the apparatus equals the external pressure of the open tube.
Since the internal pressure, Pint is greater then the atmospheric pressure, a column of liquid rises:
Patm +Pcol = Pint
Recall that:
Pcol = g  d h

67.  = cRT Osmotic Pressure,  c = concentration of solute (mols/L)
Pcol is referred to as the Osmotic Pressure, ∏.
Osmotic pressure
 = cRT
c = concentration of solute (mols/L)
T = the absolute temperature (K)
 is in units of atm.

68. Osmosis & Living Cells

69. Reverse Osmosis: Water Desalination
Water desalination plant

70. Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
Vapor-Pressure Lowering
P1 = X1 P 1
Boiling-Point Elevation
DTb = Kb m
Freezing-Point Depression
DTf = Kf m
Osmotic Pressure (p)
p = CRT
Electrolyte Solutions
Boiling-Point Elevation
DTb = i Kb m
Freezing-Point Depression
DTf = i Kf m
Osmotic Pressure (p)
p = iCRT 70

71. Colloids
A colloid is a dispersion of particles of one substance throughout a dispersing medium of another substance.
In a solution, dispersed particles are molecules, atoms, or ions (roughly 0.1 nm in size). Solute particles do not “settle out” of solution.
In a suspension (e.g., sand in water) the dispersed particles are relatively large, and will settle from suspension.
In a colloid, the dispersed particles are on the order of 1–1000 nm in size.
Although they are larger than molecules/atoms/ions, colloidal particles are small enough to remain dispersed indefinitely.

72. Colloids
Colloidal dispersions scatter light, a phenomenon known as the “The Tyndall effect”

73. Hydrophobic Colloids
A hydrophobic colloid is stabilized by positive ions absorbed onto each particle and a secondary layer of negative ions.
Because the particles bear similar charges, they repel one another, and precipitation is prevented.

74. Soap
Soap molecules interact with water through the charged, hydrophilic end of the molecule.
The long, hydrophobic end of the molecule binds through dispersion forces with nonpolar hydrocarbons and other non polar substances.

75. Emulsifiers & Surfactants
An emulsifier (also known as an emulgent) is a substance which stabilizes an emulsion by increasing its kinetic stability.
One class of emulsifiers is known as surface active substances, or surfactants.
Examples of food emulsifiers are egg yolk (where the main emulsifying agent is lecithin) and honey.
In some cases, particles can stabilize emulsions as well through a mechanism called Pickering stabilization.
Both mayonnaise and Hollandaise sauce are oil-in- water emulsions that are stabilized with egg yolk lecithin or other types of food additives such as Sodium stearoyl lactylate.