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chapter11.ppt
- 3. Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 3
Solution Composition
A
A
moles of solute
Molarity ( ) =
liters of solution
mass of solute
Mass (weight) percent = 100%
mass of solution
moles
Mole fraction ( ) =
total moles of solution
moles of solute
Molality ( ) =
kilogram of s
M
m
olvent
- 7. Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 7
Consider separate solutions of NaOH and KCl
made by dissolving 100.0 g of each solute in 250.0
mL of solution. Calculate the concentration of
each solution in units of molarity.
10.0 M NaOH
5.37 M KCl
EXERCISE!
- 9. Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 9
What is the percent-by-mass concentration of
glucose in a solution made my dissolving 5.5 g of
glucose in 78.2 g of water?
6.6%
EXERCISE!
- 11. Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 11
A solution of phosphoric acid was made by
dissolving 8.00 g of H3PO4 in 100.0 mL of water.
Calculate the mole fraction of H3PO4. (Assume
water has a density of 1.00 g/mL.)
0.0145
EXERCISE!
- 13. Section 11.1
Solution Composition
Copyright © Cengage Learning. All rights reserved 13
A solution of phosphoric acid was made by
dissolving 8.00 g of H3PO4 in 100.0 mL of water.
Calculate the molality of the solution. (Assume
water has a density of 1.00 g/mL.)
0.816 m
EXERCISE!
- 14. Section 11.2
The Energies of Solution Formation
Formation of a Liquid Solution
1. Separating the solute into its individual components
(expanding the solute).
2. Overcoming intermolecular forces in the solvent to
make room for the solute (expanding the solvent).
3. Allowing the solute and solvent to interact to form the
solution.
Copyright © Cengage Learning. All rights reserved 14
- 15. Section 11.2
The Energies of Solution Formation
Steps in the Dissolving Process
Copyright © Cengage Learning. All rights reserved 15
- 16. Section 11.2
The Energies of Solution Formation
Steps in the Dissolving Process
Steps 1 and 2 require energy, since forces must be
overcome to expand the solute and solvent.
Step 3 usually releases energy.
Steps 1 and 2 are endothermic, and step 3 is often
exothermic.
Copyright © Cengage Learning. All rights reserved 16
- 17. Section 11.2
The Energies of Solution Formation
Enthalpy (Heat) of Solution
Enthalpy change associated with the formation of the
solution is the sum of the ΔH values for the steps:
ΔHsoln = ΔH1 + ΔH2 + ΔH3
ΔHsoln may have a positive sign (energy absorbed) or a
negative sign (energy released).
Copyright © Cengage Learning. All rights reserved 17
- 18. Section 11.2
The Energies of Solution Formation
Enthalpy (Heat) of Solution
Copyright © Cengage Learning. All rights reserved 18
- 19. Section 11.2
The Energies of Solution Formation
Explain why water and oil (a long chain hydrocarbon)
do not mix. In your explanation, be sure to address
how ΔH plays a role.
Copyright © Cengage Learning. All rights reserved 19
CONCEPT CHECK!
- 20. Section 11.2
The Energies of Solution Formation
The Energy Terms for Various Types of Solutes and Solvents
Copyright © Cengage Learning. All rights reserved 20
ΔH1 ΔH2 ΔH3 ΔHsoln Outcome
Polar solute, polar solvent Large Large Large, negative Small Solution forms
Nonpolar solute, polar solvent Small Large Small Large, positive No solution
forms
Nonpolar solute, nonpolar
solvent
Small Small Small Small Solution forms
Polar solute, nonpolar solvent Large Small Small Large, positive No solution
forms
- 21. Section 11.2
The Energies of Solution Formation
In General
One factor that favors a process is an increase in
probability of the state when the solute and solvent are
mixed.
Processes that require large amounts of energy tend not
to occur.
Overall, remember that “like dissolves like”.
Copyright © Cengage Learning. All rights reserved 21
- 22. Section 11.3
Factors Affecting Solubility
Structure Effects:
Polarity
Pressure Effects:
Henry’s law
Temperature Effects:
Affecting aqueous solutions
Copyright © Cengage Learning. All rights reserved 22
- 23. Section 11.3
Factors Affecting Solubility
Structure Effects
Hydrophobic (water fearing)
Non-polar substances
Hydrophilic (water loving)
Polar substances
Copyright © Cengage Learning. All rights reserved 23
- 24. Section 11.3
Factors Affecting Solubility
Pressure Effects
Little effect on solubility of solids or liquids
Henry’s law: C = kP
C = concentration of dissolved gas
k = constant
P = partial pressure of gas solute
above the solution
Amount of gas dissolved in a solution is directly proportional to
the pressure of the gas above the solution.
Copyright © Cengage Learning. All rights reserved 24
- 26. Section 11.3
Factors Affecting Solubility
Temperature Effects (for Aqueous Solutions)
Although the solubility of most solids in water increases
with temperature, the solubilities of some substances
decrease with increasing temperature.
Predicting temperature dependence of solubility is very
difficult.
Solubility of a gas in solvent typically decreases with
increasing temperature.
Copyright © Cengage Learning. All rights reserved 26
- 27. Section 11.3
Factors Affecting Solubility
The Solubilities
of Several Solids
as a Function of
Temperature
Copyright © Cengage Learning. All rights reserved 27
- 29. Section 11.4
The Vapor Pressures of Solutions
An Aqueous Solution and Pure Water in a Closed Environment
Copyright © Cengage Learning. All rights reserved 29
- 30. Section 11.4
The Vapor Pressures of Solutions
Liquid/Vapor Equilibrium
Copyright © Cengage Learning. All rights reserved 30
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- 31. Section 11.4
The Vapor Pressures of Solutions
Vapor Pressure Lowering: Addition of a Solute
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- 32. Section 11.4
The Vapor Pressures of Solutions
Vapor Pressures of Solutions
Nonvolatile solute lowers the vapor pressure of a
solvent.
Raoult’s Law:
Psoln = observed vapor pressure of solution
solv = mole fraction of solvent
= vapor pressure of pure solvent
Copyright © Cengage Learning. All rights reserved 32
soln solv solv
=
P P
solv
P
- 33. Section 11.4
The Vapor Pressures of Solutions
A Solution Obeying Raoult’s Law
Copyright © Cengage Learning. All rights reserved 33
- 34. Section 11.4
The Vapor Pressures of Solutions
Nonideal Solutions
Liquid-liquid solutions where both components are
volatile.
Modified Raoult’s Law:
Nonideal solutions behave ideally as the mole fractions
approach 0 and 1.
Copyright © Cengage Learning. All rights reserved 34
Total A A B B
= +
P P P
- 35. Section 11.4
The Vapor Pressures of Solutions
Vapor Pressure for a Solution of Two Volatile Liquids
Copyright © Cengage Learning. All rights reserved 35
- 36. Section 11.4
The Vapor Pressures of Solutions
Summary of the Behavior of Various Types of Solutions
Copyright © Cengage Learning. All rights reserved 36
Interactive Forces Between
Solute (A) and Solvent (B)
Particles
ΔHsoln
ΔT for Solution
Formation
Deviation
from
Raoult’s
Law
Example
A A, B B A B Zero Zero
None (ideal
solution)
Benzene-
toluene
A A, B B < A B
Negative
(exothermic)
Positive Negative
Acetone-
water
A A, B B > A B
Positive
(endothermic)
Negative Positive
Ethanol-
hexane
- 37. Section 11.4
The Vapor Pressures of Solutions
For each of the following solutions, would you expect it
to be relatively ideal (with respect to Raoult’s Law),
show a positive deviation, or show a negative deviation?
a) Hexane (C6H14) and chloroform (CHCl3)
b) Ethyl alcohol (C2H5OH) and water
c) Hexane (C6H14) and octane (C8H18)
Copyright © Cengage Learning. All rights reserved 37
CONCEPT CHECK!
- 38. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Colligative Properties
Depend only on the number, not on the identity, of the
solute particles in an ideal solution:
Boiling-point elevation
Freezing-point depression
Osmotic pressure
Copyright © Cengage Learning. All rights reserved 38
- 39. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling-Point Elevation
Nonvolatile solute elevates the boiling point of the
solvent.
ΔT = Kbmsolute
ΔT = boiling-point elevation
Kb = molal boiling-point elevation constant
msolute = molality of solute
Copyright © Cengage Learning. All rights reserved 39
- 40. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling Point Elevation: Liquid/Vapor Equilibrium
Copyright © Cengage Learning. All rights reserved 40
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- 41. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling Point Elevation: Addition of a Solute
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- 42. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Boiling Point Elevation: Solution/Vapor Equilibrium
Copyright © Cengage Learning. All rights reserved 42
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- 43. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing-Point Depression
When a solute is dissolved in a solvent, the freezing
point of the solution is lower than that of the pure
solvent.
ΔT = Kfmsolute
ΔT = freezing-point depression
Kf = molal freezing-point depression constant
msolute = molality of solute
Copyright © Cengage Learning. All rights reserved 43
- 44. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing Point Depression: Solid/Liquid Equilibrium
Copyright © Cengage Learning. All rights reserved 44
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- 45. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing Point Depression: Addition of a Solute
Copyright © Cengage Learning. All rights reserved 45
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- 46. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
Freezing Point Depression: Solid/Solution Equilibrium
Copyright © Cengage Learning. All rights reserved 46
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- 48. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
A solution was prepared by dissolving 25.00 g of
glucose in 200.0 g water. The molar mass of
glucose is 180.16 g/mol. What is the boiling point
of the resulting solution (in °C)? Glucose is a
molecular solid that is present as individual
molecules in solution.
100.35 °C
Copyright © Cengage Learning. All rights reserved 48
EXERCISE!
- 49. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
You take 20.0 g of a sucrose (C12H22O11) and NaCl
mixture and dissolve it in 1.0 L of water. The freezing
point of this solution is found to be -0.426°C.
Assuming ideal behavior, calculate the mass percent
composition of the original mixture, and the mole
fraction of sucrose in the original mixture.
72.8% sucrose and 27.2% sodium chloride;
mole fraction of the sucrose is 0.313
Copyright © Cengage Learning. All rights reserved 49
EXERCISE!
- 50. Section 11.5
Boiling-Point Elevation and Freezing-Point
Depression
A plant cell has a natural concentration of
0.25 m. You immerse it in an aqueous solution with a
freezing point of –0.246°C. Will the
cell explode, shrivel, or do nothing?
Copyright © Cengage Learning. All rights reserved 50
EXERCISE!
- 51. Section 11.6
Osmotic Pressure
Osmosis – flow of solvent into the solution through a
semipermeable membrane.
= MRT
= osmotic pressure (atm)
M = molarity of the solution
R = gas law constant
T = temperature (Kelvin)
Copyright © Cengage Learning. All rights reserved 51
- 55. Section 11.6
Osmotic Pressure
When 33.4 mg of a compound is dissolved in
10.0 mL of water at 25°C, the solution has an
osmotic pressure of 558 torr. Calculate the
molar mass of this compound.
111 g/mol
Copyright © Cengage Learning. All rights reserved 55
EXERCISE!
- 56. Section 11.7
Colligative Properties of Electrolyte Solutions
van’t Hoff Factor, i
The relationship between the moles of solute dissolved
and the moles of particles in solution is usually
expressed as:
Copyright © Cengage Learning. All rights reserved 56
moles of particles in solution
=
moles of solute dissolved
i
- 57. Section 11.7
Colligative Properties of Electrolyte Solutions
Ion Pairing
At a given instant a small percentage of the sodium and
chloride ions are paired and thus count as a single
particle.
Copyright © Cengage Learning. All rights reserved 57
- 58. Section 11.7
Colligative Properties of Electrolyte Solutions
Examples
The expected value for i can be determined for a salt by
noting the number of ions per formula unit (assuming
complete dissociation and that ion pairing does not
occur).
NaCl i = 2
KNO3 i = 2
Na3PO4 i = 4
Copyright © Cengage Learning. All rights reserved 58
- 59. Section 11.7
Colligative Properties of Electrolyte Solutions
Ion Pairing
Ion pairing is most important in concentrated solutions.
As the solution becomes more dilute, the ions are
farther apart and less ion pairing occurs.
Ion pairing occurs to some extent in all electrolyte
solutions.
Ion pairing is most important for highly charged ions.
Copyright © Cengage Learning. All rights reserved 59
- 61. Section 11.8
Colloids
A suspension of tiny particles in some medium.
Tyndall effect – scattering of light by particles.
Suspended particles are single large molecules or
aggregates of molecules or ions ranging in size from 1 to
1000 nm.
Copyright © Cengage Learning. All rights reserved 61