1. Physical Properties of Solutions
Chemistry
Second Edition
Julia Burdge
Lecture Powerpoints
Jason A. Kautz
University of Nebraska-Lincoln 13
Physical Properties of Solutions
Copyright (c) The McGraw-Hill Companies, Inc. Permission required for reproduction or display.

2. Physical Properties of Solutions13
13.1 Types of Solutions
13.2 A Molecular View of the Solution Process
The Importance of Intermolecular Forces
Energy and Entropy in Solution Formation
13.3 Concentration Units
Molality
Percent by Mass
Comparison of Concentration Units
13.4 Factors that Affect Solubility
Temperature
Pressure
13.5 Colligative Properties
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
Electrolyte Solutions
13.6 Calculations Using Colligative Properties
13.7 Colloids

3. Types of Solutions
13.1
A solution is a homogeneous mixture of two or more substances.
A solution consists of a solvent and one or more solutes.

4. Types of Solutions
Solutions can be classified by the amount of solute dissolved.
An unsaturated solution is one that contains less solute than the solvent has the capacity to dissolve at a specific temperature.

5. Types of Solutions
Solutions can be classified by the amount of solute dissolved.
A saturated solution is one that contains the maximum amount of solute that will dissolve in a solvent at a specific temperature.

6. Types of Solutions
Supersaturated solutions are generally unstable.

7. The Solution Process
13.2
Solvation occurs when solute molecules are separated from one another and surrounded by solvent molecules.
Solvation depends on three types of interactions:
1) Solute-solute interactions
2) Solvent-solvent interactions
3) Solute-solvent interactions

8. The Solution Process ΔHsoln = ΔH1 + ΔH2 + ΔH3 Separated solute
Separated solvent
Step 3
ΔH3 < 0
Energy
Step 2
ΔH2 > 0
Separated solute
Solvent
Solution
Step 1
ΔH1 > 0
ΔHsoln > 0
Solute
Solvent

9. Both non-polar liquids, solution forms when mixed
The Solution Process
“Like dissolves like”
Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other.
Toluene, C7H8
Octane, C8H18
Both non-polar liquids, solution forms when mixed
Two liquids are said to be miscible if they are completely soluble in each other in all proportions.

10. Polar and non-polar liquids, solution does not form when mixed
The Solution Process
“Like dissolves like”
Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other.
Water, H2O
Octane, C8H18
Polar and non-polar liquids, solution does not form when mixed

11. solution forms when mixed
The Solution Process
“Like dissolves like”
Two substances with similar type and magnitude of intermolecular forces are likely to be soluble in each other.
Water, H2O
Ethanol, C2H6O
Both polar liquids,
solution forms when mixed

12. Concentration Units
13.3
The amount of solute relative to the volume of a solution or to the amount of solvent in a solution is called concentration.
Molarity:
Mole fraction:

13. Concentration Units
Molality (m) is the number of moles of solute dissolved in 1 kg (1000 g) solvent:
Percent by Mass:

14. Concentration Units
An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C.
Solution:
Step 1: Assume 100 grams of solution, and use mass percent to determine the grams and moles of the solute.

15. Concentration Units
An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C.
Solution:
Step 2: Assume 100 g solution and determine the volume of the solution using the density of the solution.

16. Concentration Units
An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C.
Solution:
Step 3: Use the equation below to determine the molarity.

17. Concentration Units
An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C.
Solution:
Step 4: Determine the kg of water (solvent).
grams solution = grams solute + grams solvent
100 g solution  16 g solute = 84 g solvent = kg solvent

18. Concentration Units
An aqueous solution that is 16 percent sulfuric acid (H2SO4) by mass has a density of g/mL at 25°C. Determine (a) the molarity and (b) the molality of the solution at 25°C.
Solution:
Step 5: Use the equation below to determine the molality.

19. Factors That Affect Solubility
13.4
Temperature affects the solubility of most substances.

20. Factors That Affect Solubility
Pressure greatly influences the solubility of a gas.
Henry’s law states that the solubility of a gas in a liquid is proportional to the pressure of the gas over the solution.
c molar concentration
(mol/L)
P pressure (atm)
k proportionality
constant called
Henry’s law constant.
c = kP

21. Factors That Affect Solubility
Calculate the concentration of CO2 in water at 25°C when the pressure of CO2 over the solution is 4.0 atm.
Solution:
Step 1: Use the equation below to calculate concentration. The Henry’s law constant for CO2 at 25°C is 3.1 x 10‒3 mol/L atm.
c = (3.1 x 10‒3 mol/L atm)(4.0 atm)
c = mol/L
c = kP

22. Colligative Properties
13.5
Colligative properties are properties that depend on the number of solute particles in solution.
Colligative properties do not depend on the nature of the solute particles.
The colligative properties are:
vapor-pressure lowering
boiling-point elevation
freezing-point depression
osmotic pressure

23. Colligative Properties
Raoult’s law states that the partial pressure of a solvent over a solution is given by the vapor pressure of the pure solvent times the mole fraction of the solvent in the solution.
P1 partial pressure of
solvent over solution
P° vapor pressure of
pure solvent
χ1 mole fraction of solvent
ΔP vapor pressure lowering
χ2 mole fraction of solute

24. Colligative Properties
Calculate the vapor pressure of a solution made by dissolving 114 g of urea (molar mass = g/mol) in 485 g of water at 25°C. (At 25°C the vapor pressure of water is 23.8 mmHg)
Solution:
Step 1: Determine the mole fraction of water in the solution.

25. Colligative Properties
Calculate the vapor pressure of a solution made by dissolving 114 g of urea (molar mass = g/mol) in 485 g of water at 25°C. (At 25°C the vapor pressure of water is 23.8 mmHg)
Solution:
Step 2: Determine the vapor pressure of the solution using the equation below.
Pwater = (23.8 mmHg)
Pwater = 22.2 mmHg

26. Colligative Properties
If both components of a solution are volatile, the vapor pressure of the solution is the sum of the individual partial pressures.
Benzene
Toluene

27. Colligative Properties
Benzene
Toluene
An ideal solution obeys Raoult’s law.

28. Colligative Properties
Solutions boil at a higher temperature than the pure solvent.
ΔTb boiling point elevation
Kb boiling point elevation constant (°C/m)
m molality

29. Colligative Properties
Solutions freeze at a lower temperature than the pure solvent.
ΔTf freezing point
depression
Kf freezing point
depression constant (°C/m)
m molality

30. Colligative Properties

31. Colligative Properties
Calculate the freezing point of a solution containing 268 g of ethylene glycol (C2H6O2) and 1015 g of water. Kf for water is 1.86C/m.
Solution:
Step 1: Determine the moles of ethylene glycol.

32. Colligative Properties
Calculate the freezing point of a solution containing 268 g of ethylene glycol (C2H6O2) and 1015 g of water. Kf for water is 1.86C/m.
Solution:
Step 2: Use the equation below to determine the molality of the solution.

33. Colligative Properties
Calculate the freezing point of a solution containing 268 g of ethylene glycol (C2H6O2) and 1015 g of water. Kf for water is 1.86C/m.
Solution:
Step 3: Use the equation below to determine Tf for the solution and the freezing point of the solution.

34. Colligative Properties
Osmosis is the selective passage of solvent molecules through a porous membrane from a more dilute solution to a more concentrated one.

35. Colligative Properties
Osmotic pressure () of a solution is the pressure required to stop osmosis.
 Osmotic pressure (atm)
M molarity (moles/L)
R gas constant ( L atm/mol K)
T absolute temperature (Kelvin)

36. Colligative Properties
Electrolytes undergo dissociation when dissolved in water.
The van’t Hoff factor (i) accounts for this effect.

37. Colligative Properties
The van’t Hoff factor (i) is 1 for all nonelectrolytes:
For strong electrolytes i should be equal to the number of ions:
C12H22O11(s)
C12H22O11(aq)
H2O
1 particle dissolved, i = 1
NaCl(s)
Na+(aq) + Cl–(aq)
H2O
2 particles dissolved, i = 2
Na2SO4(s)
2Na+(aq) + SO42–(aq)
H2O
3 particle dissolved, i = 3

38. Colligative Properties
The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.
An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.
ion pair

39. Colligative Properties
The van’t Hoff factor (i) is usually smaller than predicted due to the formation of ion pairs.
An ion pair is made up of one or more cations and one or more anions held together by electrostatic forces.
Table 13.3

40. Colligative Properties
Concentration has an effect on experimentally measured van’t Hoff factors (i).
Table 13.4

41. Colligative Properties
The osmotic pressure of a M NaCl solution at 25C is atm. Determine the experimental van’t Hoff factor for NaCl at this concentration.
Solution:
Step 1: Solve for the van’t Hoff factor (i) using the equation below.

42. Calculations Using Colligative Properties
13.6
Calculate the molar mass of naphthalene, the organic compound in “mothballs,” if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.00C below that of pure benzene. Kf for benzene is 5.12C/m.
Solution:
Step 1: Use the freezing point depression equation below to calculate the molality of the solution.

43. Calculations Using Colligative Properties
Calculate the molar mass of naphthalene, the organic compound in “mothballs,” if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.00C below that of pure benzene. Kf for benzene is 5.12C/m.
Solution:
Step 2: Using the equation for molality, determine the moles of naphthalene.

44. Calculations Using Colligative Properties
Calculate the molar mass of naphthalene, the organic compound in “mothballs,” if a solution prepared by dissolving 5.00 g of naphthalene in exactly 100 g of benzene has a freezing point 2.00C below that of pure benzene. Kf for benzene is 5.12C/m.
Solution:
Step 3: Calculate the molar mass of naphthalene.

45. Calculations Using Colligative Properties
Percent dissociation is the percentage of dissolved molecules (or formula units, in the case of an ionic compound) that separate into ions in a solution.
Strong electrolytes should have complete, or 100%, dissociation, however, experimentally determined van’t Hoff factors indicate that this is not the case.
Percent dissociation of a strong electrolyte is more complete at lower concentration.
Percent ionization of weak electrolytes is also dependent on concentration.

46. Colloids
13.7
A colloid is a dispersion of particles of one substance throughout another substance. Colloid particles are much larger than the normal solute molecules.
Categories of colloids:
aerosols
foams
emulsions
sols
gels

47. Colloids
Examples of colloids

48. Colloids
Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing).
Hydrophilic groups on the
surface of a large molecule
stabilize the molecule in
water.

49. Colloids
Colloids with water as the dispersing medium can be categorized as hydrophilic (water loving) or hydrophobic (water fearing).
Negative ions are adsorbed onto the surface of hydrophobic colloids.
The repulsion between like charges prevents aggregation of the articles.

50. Colloids
Hydrophobic colloids can be stabilized by the presence of hydrophilic groups on their surface.

51. Colloids
Emulsification is the process of stabilizing a colloid that would otherwise not stay dispersed.

52. 13 Key Concepts Types of Solutions
A Molecular View of the Solution Process
The Importance of Intermolecular Forces
Energy and Entropy in Solution Formation
Concentration Units
Molality
Percent by Mass
Comparison of Concentration Units
Factors that Affect Solubility
Temperature
Pressure
Colligative Properties
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
Electrolyte Solutions
Calculations Using Colligative Properties
Colloids