1. Chapter 13
The Properties of Mixtures:
Solutions and Colloids
13-1

2. SOLUTIONS I
Unit 13
• Concentrations of solutions
• Chapter 11.1-11.4 McM
• Chapter 13.5 Silberberg
13-2

3. The Properties of Mixtures: Solutions and Colloids
13.1 Types of Solutions: Intermolecular Forces and Solubility
13.3 Why Substances Dissolve: Understanding the Solution Process
13.5 Concentration Terms
13-3

4. Goals & Objectives
• See the following Learning
Objectives on pages 512.
• Understand these Concepts:
• 13.14
• Master these Skills:
• 13.3-4.
13-4

5. Solutions and Colloids
A solution is a homogeneous mixture and exists as a
single phase.
The particles in a solution are individual atoms, ions, or small molecules.
A colloid is a heterogeneous mixture and exists as two or
more phases, which may be visibly distinct.
The particles in a colloid are typically macromolecules or aggregations of
small molecules.
13-5

6. Solutions and Solubility
A solute dissolves in a solvent to form a solution.
Usually, the solvent is the most abundant component.
The solubility (S) of a solute is the maximum amount
that dissolves in a fixed quantity of solvent at a given
temperature.
Substances that exhibit similar types of intermolecular
force dissolve in each other.
This is often expressed by saying “like dissolves in like.”
13-6

7. A Sodium Chloride Solution
• Sodium chloride is the solute
• water is the solvent (understood)
13-7

8. Figure 13.1
Ion-dipole
(40-600)
H bond
(10-40)
Dipole-dipole
(5-25)
13-8
Types of intermolecular forces in solutions.
Ion-induced dipole
(3-15)
Dipole-induced dipole
(2-10)
Dispersion
(0.05-40)

9. Concentration
• A measure of the amount of solute
dissolved per given amount of
solvent or solution.
• Typical units
– Percent by mass (% w/w)
– Molarity (M, mol/L)
– Molality (m)
– Proof
13-9

10. Proof
• Approximately twice the percent by
volume of alcohol
•
•
•
•
13-10
Proof
26
80
190
%Alcohol by Volume
13%
40%
95%

11. Percent by Mass of Solute
• %mass of solute = g solute x 100
•
g solution
• A 25.9% solution of glucose contains:
• 35.0g of glucose /35.g glucose + 100
g of water. The density of water is
1.00 g/ml
13-11

12. 13-12

13. Calculate the concentration in %(w/w) of the following
solutions. Assume that water has a density of 1.00 g/mL.
A 3.50 g sample of morphine hydrochloride is dissolved
in 750 mL of water.
13-13

14. 13-14

15. Calculate the concentration in %(w/w) of the following
solutions. Assume that water has a density of 1.00 g/mL.
A 3.50 g sample of morphine hydrochloride is dissolved
in 750 mL of water.
13-15

16. 13-16

17. Calculate the concentration in %(w/w) of the following solutions.
Assume that water has a density of 1.00 g/mL.
30.0 mL of alcohol (density = 0.789 g/mL) is mixed into 370 mL of
water
13-17

18. 13-18

19. 13-19

20. 13-20

21. Examples
• What volume of 12.9% KOH solution
contains 40.0 grams of KOH? The
density of the solution is 1.11g/mL.
13-21

22. 13-22

23. 13-23

24. Solutions and Intermolecular Forces
When a solution forms, solute-solute attractions and
solvent-solvent attractions are replaced by solutesolvent attractions.
This can only occur if the forces within the solute and
solvent are similar to the forces that replace them.
13-24

25. Figure 13.2
Hydration shells around an Na+ ion.
Ion-dipole forces orient water molecules around an ion. In the innermost
shell here, six water molecules surround the cation octahedrally.
13-25

26. Dual Polarity and Effects on Solubility
• Alcohols are organic compounds that have dual
polarity.
– The general formula for an alcohol is CH3(CH2)nOH.
• The –OH group of an alcohol is polar.
– It interacts with water through H bonds and
– with hexane by through weak dipole-induced dipole forces.
• The hydrocarbon portion is nonpolar.
– It interacts through weak dipole-induced dipole forces with
water
– and by dispersion forces with hexane.
13-26

27. Figure 13.3
H 2O
Like dissolves like: solubility of methanol in water.
Methanol
CH3OH
Both H2O and methanol have H bonds between their molecules.
A solution of water and methanol.
13-27

28. Figure 13.18 Sodium acetate crystallizing from a supersaturated
solution.
A supersaturated solution contains more
than the equilibrium concentration of solute. It
is unstable and any disturbance will cause
excess solute to crystallize immediately.
13-28

29. Factors that affect Solubility
Temperature affects solubility.
Most solids are more soluble at higher temperatures.
Gases become less soluble as temperature increases.
Pressure affects the solubility of gases – they become
more soluble at higher pressure.
13-29

30. Figure 13.19
13-30
Relation between solubility and temperature for
several ionic compounds.

31. Table 13.5
Concentration Definitions
Concentration Term
Molarity (M)
Molality (m)
Ratio
amount (mol) of solute
volume (L) of solution
amount (mol) of solute
mass (kg) of solvent
Parts by mass
Parts by volume
mass of solute
mass of solution
volume of solute
volume of solution
Mole fraction (X)
amount (mol) of solute
amount (mol) of solute + amount (mol) of solvent
13-31

32. Molarity
• Moles of solute per liter of solution
• M = moles of solute
liters of solution
• Abbreviate: mol/L
• A 2.0M (molar) solution of NaCl
contains 2.0 moles of NaCl in 1.0 liter of
solution.
• Or 4.0 moles of NaCl in 2.0 liters of
solution etc.
13-32

33. 13-33

34. Examples
• Calculate the molarity of a solution
prepared by dissolving 12.5 g of
sulfuric acid (H2SO4) in sufficient
water to give 1.75L of solution.
• Determine the mass of Ca(NO3)2
required to prepare 3.50 L of a
solution that is 0.800M in Ca(NO3)2.
13-34

35. 13-35

36. 13-36

37. Examples
• The density of commercial
hydrochloric acid is 1.185g/mL. It is
36.31% HCl by mass. Determine its
molarity.
13-37

38. 13-38

39. 13-39

40. Using Solutions in Chemical
Reactions
• Na2SO4 + BaCl2
BaSO4 + 2NaCl
• Determine the volume of 0.500M
BaCl2 solution required for complete
reaction with 4.32g of Na2SO4
according to the reaction
13-40

41. 13-41

42. Using Solutions in Chemical
Reactions
• H2SO4 + 2NaOH
Na2SO4 + 2H2O
• Determine the molarity of the H2SO4
solution required for complete
reaction with 15.14 ml of 0.1022M
NaOH solution, if you started with
10.00ml of H2SO4 solution in the
erlenmeyer flask.
13-42

43. Using Solutions in Chemical
Reactions: Titration
13-43
The process, operation, or method of
determining the concentration of a
substance in solution by adding to it a
standard reagent of known
concentration in carefully measured
amounts until a reaction of definite
and known proportion is completed,
as shown by a color change or by
electrical measurement, and then
calculating the unknown
concentration.

44. Dilution of Solutions
• A more concentrated solution can be
diluted with solvent to produce a
solution that is more dilute. The
moles of solute will be the same in
both solutions. However, since the
volumes will be different the molarity
will change.
• M1V1 = M2V2
13-44

45. 13-45

46. 13-46

47. Examples
• Determine the volume of 18.0M
H2SO4 required to prepare 2.50 liters
of 2.40M H2SO4 solution.
• If 50.0 ml of 12.0M HCl is diluted to a
final total volume of 200.0 ml,
determine the final concentration of
the solution.
13-47

48. 13-48

49. 13-49

50. Molality
• Moles of solute per kilogram of
solvent
• m = moles of solute
•
kg of solvent
• A 1.0m (molal) solution of NaCl
contains 1.0 moles of NaCl in 1.0 kg
of water.
13-50

51. Sample Problem 13.3
Calculating Molality
PROBLEM: What is the molality of a solution prepared by dissolving
32.0 g of CaCl2 in 271 g of water?
PLAN: Molality is defined as moles of solute (CaCl2) divided by kg of
solvent (H2O). We convert the mass of CaCl2 to moles using
the molar mass, and then divide by the mass of H2O, being
sure to convert from grams to kilograms.
mass (g) of CaCl2
divide by M of (g/mol)
amount (mol) of CaCl2
divide by kg of water
molality (m) of CaCl2 solution
13-51

52. Sample Problem 13.3
SOLUTION:
32.0 g CaCl2
molality =
13-52
1 mol CaCl2
x
110.98 g CaCl2
0.288 mol CaCl2
1 kg
271 g H2O x
103 g
= 0.288 mol CaCl2
= 1.06 m CaCl2

53. Examples
13-53
• Determine the molality of a solution
prepared by dissolving 400 g of
NaOH in 500mL of water. The
density of water is 1.00g/mL. M.W.
of NaOH is 40.0 amu.
• Calculate the molarity and molality of
an aqueous solution that is 10.0%
glucose. The density of the solution
is 1.04g/mL. M.W. of glucose =
180.0amu.

54. 13-54

55. 13-55

56. Examples
• Determine the molality of a solution that
contains 7.25 g of benzoic acid,
C6H5COOH, in 200 mL of benzene,
C6H6. The density of benzene is
0.879g/mL and the molar mass of
benzoic acid = 122 amu.
13-56

57. 13-57

58. 13-58

59. Mole Fraction
• Moles of component A per total
moles of all components in the
solution.
• XA =
molesA
•
molesA + molesB + molesC
• XB =
• XA + XB + XC + ……… = 1
13-59

60. 13-60

61. Sample Problem 13.4
Expressing Concentrations in Parts by
Mass, Parts by Volume, and Mole Fraction
PROBLEM: (a) Find the concentration of calcium (in ppm) in a 3.50-g
pill that contains 40.5 mg of Ca.
(b) The label on a 0.750-L bottle of Italian chianti indicates
“11.5% alcohol by volume.” How many liters of
alcohol does the wine contain?
(c) A sample of rubbing alcohol contains 142 g of
isopropyl alcohol (C3H7OH) and 58.0 g of water. What
are the mole fractions of alcohol and water?
PLAN: (a) We convert mg to g of Ca2+, find the mass ratio of Ca2+ to
pill and multiply by 106.
(b) We know the volume % of the alcohol and the total
volume, so we can find the volume of alcohol.
(c) We convert g of solute and solvent to moles in order to
calculate the mole fractions.
13-61

62. Sample Problem 13.4
SOLUTION:
(a)
40.5 mg Ca2+ x
1g
103 mg
3.5 g
(b) 0.750 L chianti x
x 106
= 1.16x104 ppm Ca2+
11.5 L alcohol
100. L chianti
= 0.0862 L alcohol
(c)
moles isopropyl alcohol =
moles water =
13-62
58.0 g x
142 g x 1 mole = 2.36 mol C H OH
3 7
60.09 g
1 mole
= 3.22 mol H2O
18.02 g

63. Sample Problem 13.4
X C H OH =
3 7
XH O =
2
13-63
2.36 mol C2H8O2
2.36 mol C3H7OH + 3.22 mol H2O
= 0.423
3.22 mol H2O
= 0.577
2.36 mol C3H7OH + 3.22 mol H2O

64. Interconverting Concentration Terms
• To convert a term based on amount (mol) to one based
on mass, you need the molar mass.
• To convert a term based on mass to one based on
volume, you need the solution density.
• Molality involves quantity of solvent, whereas the other
concentration terms involve quantity of solution.
13-64

65. Sample Problem 13.5
Interconverting Concentration Terms
PROBLEM: Hydrogen peroxide is a powerful oxidizing agent used in
concentrated solution in rocket fuels and in dilute solution
as a hair bleach. An aqueous solution H2O2 is 30.0% by
mass and has a density of 1.11 g/mL. Calculate its
(a) Molality
(b) Mole fraction of H2O2
(c) Molarity
PLAN: (a) To find the mass of solvent we assume the % is per 100 g
of solution. Take the difference in the mass of the solute
and solution for the mass of peroxide.
(b) Convert g of solute and solvent to moles before finding X.
(c) Use the density to find the volume of the solution.
13-65

66. Sample Problem 13.5
SOLUTION:
(a) From mass % to molality:
g of H2O = 100. g solution - 30.0 g H2O2 = 70.0 g H2O
1 mol H2O2
30.0 g H2O2 x
= 0.882 mol H2O2
34.02 g H2O2
0.882 mol H2O2
molality =
70.0 g x 1 kg
103 g
13-66
= 12.6 m H2O2

67. Sample Problem 13.5
(b) From mass % to mole fraction:
1 mol H2O
70.0 g H2O x
= 3.88 mol H2O
18.02 g H2O
XH O =
2 2
0.882 mol H2O2
3.88 mol H2O + 0.882 mol H2O2
= 0.185
(c) From mass % and density to molarity:
1 mL
volume (mL) of solution = 100.0 g x
1.11 g
mol H2O2
0.882 mol H2O2
molarity =
=
L soln
90.1 mL x 1 L soln
103 mL
13-67
= 90.1 mL
= 9.79 M H2O2

68. Examples
• What are the mole fractions of
glucose and water in 10.0% glucose
solution.
13-68

69. 13-69

70. 13-70