1. Chapter 11 Properties of Solutions
DE Chemistry
Dr. Walker

2. Solutions
In a solution, the excess material is the solvent, the smaller amount is the solute

3. Measuring Solution Composition
Molarity is most common, covered previously in Chapter 4

4. Measuring Solution Composition
Mass Percent

5. Mass Percent Example
What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol (C2H5OH) in 85.0 mL of water?

6. Mass Percent Example
What is the weight percent of ethanol in a solution made by dissolving 5.3 g of ethanol (C2H5OH) in 85.0 mL of water?
Density of water = 1.0 g/mL, therefore 85 mL of water = 85 g
Mass of solute/mass of solution x 100 = mass %
5.3 g ethanol/(5.3 g ethanol g water) = 5.9%

7. Measuring Solution Composition
Mole Fraction

8. Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

9. Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Calculate moles of ethanol:
5.30 g C2H5OH
1 mole C2H5OH
= moles C2H5OH
46.08 g C2H5OH

10. Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Calculate moles of water:
85.0 g H2O
1 mole H2O
= moles H2O
18.02 g H2O

11. Mole Fraction Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Mole Fraction of ethanol
Mole Fraction of Water
0.115 moles C2H5OH =
moles C2H5OH moles H2O
4.72 moles H2O =
0.115 moles C2H5OH moles H2O

12. Measuring Solution Composition
Molality

13. Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g

14. Example
What is the weight percent of ethanol in a solution made by dissolving 5.30 g of ethanol (C2H5OH) in 85.0 mL of water?
Note: If the density of water is 1.0 g/mL, 85 mL = 85 g
Molality = moles solute/kg solvent
5.30 g C2H5OH
1 mole C2H5OH
= moles C2H5OH
46.08 g C2H5OH

15. Measuring Solution Composition
Normality
Equivalents of acids and bases
Mass that donates or accepts a mole of protons (usually a subscript of H+ or OH-)
Equivalents of oxidizing and reducing agents
Mass that provides or accepts a mole of electrons (usually a subscript)

16. Normality Find molarity first
Normality = Molarity x number of acidic protons
1.5 M HCl = 1.5 N HCl (1 acidic proton)
1.5 M H2SO4 = 3.0 N H2SO4 (2 acidic protons)
1.5 M H3PO4 = 4.5 N H3PO4 (3 acidic protons)

17. Energy of Solution Formation
“Like Dissolves Like”
Polar molecules and ionic compounds tend to dissolve in polar solvents
Nonpolar molecules dissolve in nonpolar compounds
Water - polar
Butane - nonpolar

18. Heat of Solution
The Heat of Solution is the amount of heat energy absorbed (endothermic) or released (exothermic) when a specific amount of solute dissolves in a solvent.
Substance
Heat of Solution
(kJ/mol)
NaOH
-44.51
NH4NO3
+25.69
KNO3
+34.89
HCl
-74.84

19. Steps In Solution Formation

20. Steps in Solution Formation
H1 Step 1 - Expanding the solute
Separating the solute into individual components – usually endothermic

21. Steps in Solution Formation
H2 Step 2 - Expanding the solvent
Overcoming intermolecular forces of the solvent molecules – usually endothermic

22. Steps in Solution Formation
H3 Step 3 - Interaction of solute and solvent to form the solution – usually exothermic

23. Enthalpy of Solution May be positive or negative
Negative is favorable (exothermic)
Positive values are endothermic, does not dissolve spontaneously
Solution formation increases entropy (favorable)

24. Factors Affecting Solubility
Structure Effects
Polar (hydrophilic) dissolves in polar
Water soluble vitamins
Nonpolar (hydrophobic) in nonpolar
Fat soluble vitamins

25. Pressure Effects – Henry’s Law
The concentration of a dissolved gas in a solution is directly proportional to the pressure of the gas above the solution
Applies most accurately for dilute solutions of gases that do not dissociate or react with the solvent
Yes  CO2, N2, O2
No  HCl, HI  remember, these dissociate in solution!

26. Henry’s Law Example

27. Henry’s Law Example Unopened bottle:
C = kP C = (3.1 x 10-2 mol/L . Atm)(5.0 atm) =
0.16 mol/L
Opened bottle:
C = kP C = (3.1 x 10-2 mol/L . Atm)(4.0 x atm) =
1.2 x 10-5 mol/L

28. Solubility – Temperature Effects
Solids
Increases in temperature always cause dissolving to occur more rapidly
Increases in temperature usually increases solubility (the amount that can be dissolved)
Gases
Solubility of gases always decreases with increasing temperatures

29. Vapor Pressure – Nonvolatile solutes
Nonvolatile electrolytes lower the vapor pressure of a solute
Nonvolatile molecules do not enter the vapor phase
Fewer molecules are available to enter the vapor phase
Dissociation of ionic compounds has nearly two, three or more times the vapor pressure lowering of nonionic (nonelectrolyte) solutes.

31. Raoult’s Law Psolution = Observed Vapor pressure of
The presence of a nonvolatile solute lowers the vapor pressure of the solvent.
Psolution = Observed Vapor pressure of
the solution
solvent = Mole fraction of the solvent
P0solvent = Vapor pressure of the pure solvent
This should make sense, since you’ve learned previously that dissolved
Solutes raise boiling points. Lower vapor pressure = higher boiling point

32. Raoult’s Law Example
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25o C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of Kool-Aid?

33. Raoult’s Law Example Mole Fraction of water (remember 1 L = 1000 g)
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25o C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of Kool-Aid?
Usually, you must solve for the mole fraction of the solvent first:
Mole Fraction of water (remember 1 L = 1000 g)
2000 g H2O
1 mole H2O
= moles H2O
18.02 g H2O
moles H2O =
1.5 moles kool aid moles H2O

34. Raoult’s Law Example
1.5 moles of cherry Kool-Aid are added to a pitcher containing 2 liters of water on a nice day at 25o C. The vapor pressure of water alone is 23.8 mm Hg at 25o C. What is the new vapor pressure of Kool-Aid?
Using the mole fraction from the previous page
PKool-Aid = cH2O Ppure H2O = (.987)(23.8 mm Hg) = 23.5 mm Hg

36. Liquid-liquid solutions in which both components are volatile (non-ideal solutions)
Modified Raoult's Law:
P0 is the vapor pressure of the pure solvent
PA and PB are the partial pressures
This is a variation on Dalton’s Law of Partial Pressures that accounts for Raoult’s Law

37. Modified Raoult’s Law Example
What is the vapor pressure
of the solution?

38. Modified Raoult’s Law Example

39. Modified Raoult’s Law Example
What is the vapor pressure
of the solution?

40. Ideal Solutions Liquid-liquid solution that obeys Raoult’s Law
Like gases, none are ideal, but some are close
Negative Deviations
Lower than predicted vapor pressure
Solute and solvent are similar, strong forces of attraction
In a solution of acetone and
ethanol, there is a negative
deviation due to the hydrogen
bonding interactions shown

41. Ideal Solutions Positive Deviations
Higher than predicted vapor pressure
Particles easily escape attractions in solution to enter the vapor phase
Ethanol and hexane are not attracted to each other due to differences
In polarity. As a result, its solution is a positive deviation (higher DH)
From Raoult’s Law.

42. Colligative Properties
Properties dependent on the number of solute particles but not on their identity
Boiling-Point elevation
Freezing-Point depression
Osmotic Pressure

43. Boiling Point Elevation
Each mole of solute particles raises the boiling point of 1 kilogram of water by 0.51 degrees Celsius.
Kb = 0.51 C  kilogram/mol
m = molality of the solution
i = van’t Hoff factor

44. Freezing Point Depression
Each mole of solute particles lowers the freezing point of 1 kilogram of water by 1.86 degrees Celsius.
Kf = 1.86 C  kilogram/mol
m = molality of the solution
i = van’t Hoff factor

45. Boiling Point Elevation
A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C2H6O2, d=1.11 g/mL) and water is used, what will be the new boiling point?

46. Boiling Point Elevation
A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C2H6O2, d=1.11 g/mL) and water is used, what will be the new boiling point?
Van’t Hoff factor = 1 (not an ionic salt)
Kb = kg/mol
Molality = moles/kg
3000 mL C2H6O2
1.11 g C2H6O2
1 mole C2H6O2
= 53.15
moles
1 mL C2H6O2
62.08 g C2H6O2

47. Boiling Point Elevation
A radiator in a car has a capacity of 6 L. If a 50/50 (v/v) solution of ethylene glycol (C2H6O2, d=1.11 g/mL) and water is used, what will be the new boiling point?
Van’t Hoff factor = 1 (not an ionic salt)
Kb = kg/mol
Molality = moles/kg
3 L water = 3 kg water
m = moles solute/kg solvent
m = moles C2H6O2 /3 kg water = m
DT = i . Kb. m= 1 x x = 9.15 C = degrees C

48. Freezing Point Depression
Give the freezing point depression from adding 10 g ethylene glycol to 100 g water.
DT = (1) (1.86) (m)
10 g C2H6O2 / 620.8g g/mol = 0.16 moles
m = 0.16 moles/0.1 kg = 1.6 m
DT = (1)(1.86)(1.6) = 2.88 degrees

49. Determination of Molar Mass by Freezing Point Depression

50. Determination of Molar Mass by Freezing Point Depression
Rearrange equation
Msolute = DT/Kf
Msolute = C / 5.12 C . kg/mol = 4.69 x 10-2 mol/kg

51. Determination of Molar Mass by Freezing Point Depression
Rearrange equation
Msolute = DT/Kf
Msolute = C / 5.12 C . kg/mol = 4.69 x 10-2 mol/kg
Moles hormone
4.69 x 10-2 mol/kg =
0.015 kg benzene

52. Determination of Molar Mass by Freezing Point Depression
Molality
Moles hormone =
4.69 x 10-2 mol/kg
0.015 kg benzene
Moles hormone = 7.04 x 10-4 moles
Determine molar mass

53. Van’t Hoff Factor, i
For ionic compounds, the expected value of i is an integer greater than 1
NaCl, i = 2
BaCl2, i = 3
Al2(SO4)3, i = 5
Total number of ions in solution

54. 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

55. Freezing Point Depression and Boiling Point Elevation Constants, C/m
Solvent Kf Kb
Acetic acid
3.90
3.07
Benzene
5.12
2.53
Nitrobenzene
8.1
5.24
Phenol
7.27
3.56
Water
1.86
0.512

56. Osmotic Pressure Semipermeable Membrane
Membrane which allows solvent but not solute molecules to pass through.
As time passes…
One side of the membrane is mostly solvent
The other side (which didn’t pass through) is more concentrated since the solute can’t go through the membrane
Osmosis
The flow of solvent into the solution through the semipermeable membrane

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

58. Osmotic Pressure Calculations
i = van’t Hoff Factor
M = Molarity of the solution
R = Gas Constant = Latm/molK
See page 509 in text for example

59. Example

60. Dialysis
Transfer of solvent molecules as well as small solute molecules and ions
Remember in osmosis, only solute molecules are transferred
This description fits the “dialysis” used to filter the blood when the kidneys do not work properly.

61. Kidney Dialysis
Dialyser contains ions and small molecules in blood, but none of the waste products.

62. Osmotic Pressure and Living Cells
Crenation
Cells placed in a hypertonic (higher osmotic pressure) solution lose water to the solution, and shrink
Hemolysis
Cells placed in a hypotonic (lower osmotic pressure) solution gain water from the solution and swell, possibly bursting

63. Reverse Osmosis
External pressure applied to a solution can cause water to leave the solution
Concentrates impurities (such as salt) in the remaining solution
Pure solvent (such as water) is recovered on the other side of the semipermeable membrane
Applicable to desalination plants which can make drinking water from ocean water.

64. Colloids Tiny particles suspended in some medium
Particles range in size from 1 to 1000 nm
Noticeable by shining light through the medium
Particles are large enough that they scatter light

65. Examples of Colloids

66. Tyndall Effect Scattering of light by particles
Light passes through a solution
Light is scattered in a colloid