1. Chapter 11: Solutions and Their Properties
4/17/2017
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Copyright © 2010 Pearson Prentice Hall, Inc.

2. Chapter 11: Solutions and Their Properties
4/17/2017
Solutions
Solution: A homogeneous mixture.
Solvent: The major component.
Solute: A minor component.
Students will sometimes equate solute with “solid” and solvent with “water.”
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

3. Chapter 11: Solutions and Their Properties
4/17/2017
Solutions
Copyright © 2010 Pearson Prentice Hall, Inc.

4. Energy Changes and the Solution Process
Chapter 11: Solutions and Their Properties
4/17/2017
Energy Changes and the Solution Process
Except for gas-gas solutions (air, for example), understanding intermolecular forces is crucial.
You break the attractions for both the solvent-solvent and solute-solute particles. This requires energy. You form attractions between the solute and solvent. This releases energy.
The energy difference is the net difference between the two processes.
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5. Energy Changes and the Solution Process
Chapter 11: Solutions and Their Properties
4/17/2017
Energy Changes and the Solution Process
The sodium and chloride ions are hydrated.
Solvated: any solvent.
Hydrated: water is the solvent.
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6. Energy Changes and the Solution Process
Chapter 11: Solutions and Their Properties
4/17/2017
Energy Changes and the Solution Process
There is an entropy change for the solution process.
The sodium chloride solid is very ordered. The ions are less ordered when hydrated. The water molecules become more ordered due to the ion-dipole forces. The entropy change is the net change.
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7. Energy Changes and the Solution Process
Chapter 11: Solutions and Their Properties
4/17/2017
Energy Changes and the Solution Process
There is an entropy change for the solution process.
Solution #1
Solution #2
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8. Energy Changes and the Solution Process
Chapter 11: Solutions and Their Properties
4/17/2017
Energy Changes and the Solution Process
∆G = ∆H - T∆S
Cold packs are endothermic while hot packs are exothermic.
KCl dissolved in water is endothermic and the temperature will drop.
LiCl dissolved in water is exothermic and the temperature will go up (drastically!).
endothermic:
exothermic:
+∆H
-∆H
spontaneous:
nonspontaneous:
-∆G
+∆G
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Copyright © 2010 Pearson Prentice Hall, Inc.

9. Energy Changes and the Solution Process
Chapter 11: Solutions and Their Properties
4/17/2017
Energy Changes and the Solution Process
Compare all the forces to each other.
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10. Units of Concentration
Chapter 11: Solutions and Their Properties
4/17/2017
Units of Concentration
Liters of solution
Moles of solute
Molarity (M) =
Total moles making up solution
Moles of component
Mole Fraction (X) =
Total mass of solution
Mass of component
x 100%
Mass Percent =
It’s easy to confuse molarity with molality due to the 1 letter difference in the words!
Molarity calculations were first introduced in Chapter 6.
Mass of solvent (kg)
Moles of solute
Molality (m) =
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Copyright © 2010 Pearson Prentice Hall, Inc.

11. Units of Concentration
Chapter 11: Solutions and Their Properties
4/17/2017
Units of Concentration
Assuming that seawater is an aqueous solution of NaCl, what is its molarity? The density of seawater is g/mL at 20 °C, and the NaCl concentration is 3.50 mass %.
Assuming g of solution, calculate the volume:
g solution
1.025 g
1 mL
1000 mL
1 L x x
= L
While any mass of solution may be chosen, mass % suggests 100 g.
Convert the mass of NaCl to moles:
3.50 g NaCl
58.4 g
1 mol x
= mol
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Copyright © 2010 Pearson Prentice Hall, Inc.

12. Units of Concentration
Chapter 11: Solutions and Their Properties
4/17/2017
Units of Concentration
Then, calculate the molarity: L
mol
= M
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

13. Units of Concentration
Chapter 11: Solutions and Their Properties
4/17/2017
Units of Concentration
What is the molality of a solution prepared by dissolving g of cholesterol, C27H46O, in 40.0 g chloroform, CHCl3?
Convert the mass of cholesterol to moles:
0.385 g
386.0 g
1 mol x
= mol
Calculate the mass of chloroform in kg:
40.0 g
1000 g
1 kg x
= kg
Calculate the molality of the solution: kg
mol
= m
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

14. Units of Concentration
Chapter 11: Solutions and Their Properties
4/17/2017
Units of Concentration
Mole fraction, mass %, and molality are all temperature-independent since they are all based on mass or moles.
Volume can change with temperature so molarity is temperature-dependent.
Molarity is solution volume (in L) while molality is solvent mass (in kg).
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15. Some Factors Affecting Solubility
Chapter 11: Solutions and Their Properties
4/17/2017
Some Factors Affecting Solubility
Saturated Solution: A solution containing the maximum possible amount of dissolved solute at equilibrium.
dissolve
crystallize
Solution
Solute + Solvent
Supersaturated Solution: A solution containing a greater-than-equilibrium amount of solute.
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

16. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

17. Some Factors Affecting Solubility
Chapter 11: Solutions and Their Properties
4/17/2017
Some Factors Affecting Solubility
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18. Some Factors Affecting Solubility
Chapter 11: Solutions and Their Properties
4/17/2017
Some Factors Affecting Solubility
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19. Some Factors Affecting Solubility
Chapter 11: Solutions and Their Properties
4/17/2017
Some Factors Affecting Solubility
Henry’s Law
Solubility = k P
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20. Physical Behavior of Solutions: Colligative Properties
Chapter 11: Solutions and Their Properties
4/17/2017
Physical Behavior of Solutions: Colligative Properties
Colligative Properties: Properties that depend on the amount of a dissolved solute but not on its chemical identity.
Vapor-Pressure Lowering
Boiling-Point Elevation
Freezing-Point Depression
Osmotic Pressure
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Copyright © 2010 Pearson Prentice Hall, Inc.

21. Vapor-Pressure Lowering of Solutions: Raoult’s Law
Chapter 11: Solutions and Their Properties
4/17/2017
Vapor-Pressure Lowering of Solutions: Raoult’s Law
Raoult’s Law
Psoln = Psolv Xsolv
This assumes a nonvolatile solute.
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22. Vapor-Pressure Lowering of Solutions: Raoult’s Law
Chapter 11: Solutions and Their Properties
4/17/2017
Vapor-Pressure Lowering of Solutions: Raoult’s Law
The vapor pressure of pure water at 25 °C is mm Hg. What is the vapor pressure of a solution made from 1.00 mol glucose in 15.0 mol of water at 25 °C? Glucose is a nonvolatile solute.
Psoln = Psolv Xsolv
23.76 mm Hg
15.0 mol = x
= 22.3 mm Hg
1.00 mol mol
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

23. Vapor-Pressure Lowering of Solutions: Raoult’s Law
Chapter 11: Solutions and Their Properties
4/17/2017
Vapor-Pressure Lowering of Solutions: Raoult’s Law
Solutions of ionic substances often have a vapor pressure significantly lower than predicted, because the ion-dipole forces between the dissolved ions and polar water molecules are so strong.
moles of solute dissolved
moles of particles in solution
van’t Hoff Factor:
i =
The smaller than expected value for i is due to incomplete dissociation of ionic compounds. As the concentration decreases, the experimental value for i increases until it reaches the theoretical maximum (decreased occurrence of ion-pairs in solution also plays a factor).
Na1+(aq) + Cl1-(aq)
NaCl(aq)
For sodium chloride, the predicted value of i is 2. For a 0.05 m solution of sodium chloride, the experimental value for i is 1.9.
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

24. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

25. Vapor-Pressure Lowering of Solutions: Raoult’s Law
Chapter 11: Solutions and Their Properties
4/17/2017
Vapor-Pressure Lowering of Solutions: Raoult’s Law
Ptotal = PA + PB = (P°A XA) + (P°B XB)
This assumes volatile components and Dalton’s law of partial pressures.
There can be significant deviations from Raoult’s law.
Copyright © 2010 Pearson Prentice Hall, Inc.

26. Boiling-Point Elevation and Freezing-Point Depression
Chapter 11: Solutions and Their Properties
4/17/2017
Boiling-Point Elevation and Freezing-Point Depression
Copyright © 2010 Pearson Prentice Hall, Inc.

27. Boiling-Point Elevation and Freezing-Point Depression
Chapter 11: Solutions and Their Properties
4/17/2017
Boiling-Point Elevation and Freezing-Point Depression
Nonelectrolytes
Electrolytes
∆Tb = Kb m
∆Tb = Kb m i
∆Tf = Kf m
∆Tf = Kf m i
Change in freezing or boiling temperature.
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28. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

29. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

30. Boiling-Point Elevation and Freezing-Point Depression
Chapter 11: Solutions and Their Properties
4/17/2017
Boiling-Point Elevation and Freezing-Point Depression
What is the freezing point (in °C) of a solution prepared by dissolving 7.40 g of MgCl2 in 110 g of water? The van’t Hoff factor for MgCl2 is i = 2.7.
Calculate the moles of MgCl2:
7.40 g
1 mol x
= mol
95.3 g
Calculate the molality of the solution:
mol
1000 g
mol kg x
= 0.71
110 g
1 kg
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

31. Boiling-Point Elevation and Freezing-Point Depression
Chapter 11: Solutions and Their Properties
4/17/2017
Boiling-Point Elevation and Freezing-Point Depression
Calculate the freezing point of the solution:
mol
°C kg kg
mol
∆Tf = Kf m i = 1.86 x
0.71
x 2.7 = 3.6 °C
Tf = 0.0 °C °C = -3.6 °C
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Copyright © 2010 Pearson Prentice Hall, Inc.

32. Osmosis and Osmotic Pressure
Chapter 11: Solutions and Their Properties
4/17/2017
Osmosis and Osmotic Pressure
Osmosis: The passage of solvent through a semipermeable membrane from the less concentrated side to the more concentrated side.
Osmotic Pressure (): The amount of pressure necessary to cause osmosis to stop.
Solvent flows from a high solvent concentration to a low solvent concentration. Or, solvent flows from a low solute concentration to a high solute concentration.
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

33. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

34. Chapter 11: Solutions and Their Properties
4/17/2017
 = MRT
Osmotic pressure can be calculated using molarity, since it is calculated at a specific temperature and thus there is no need for temperature-independent units.
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35. Osmosis and Osmotic Pressure
Chapter 11: Solutions and Their Properties
4/17/2017
Osmosis and Osmotic Pressure
Calculate the osmotic pressure of a 1.00 M glucose solution in water at 300 K.
mol L
L atm
K mol
 = MRT = 1.00 x
x 300 K = 24.6 atm
Copyright © 2010 Pearson Prentice Hall, Inc.
Copyright © 2010 Pearson Prentice Hall, Inc.

36. Some Uses of Colligative Properties
Chapter 11: Solutions and Their Properties
4/17/2017
Some Uses of Colligative Properties
Reverse Osmosis
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37. Fractional Distillation of Liquid Mixtures
Chapter 11: Solutions and Their Properties
4/17/2017
Fractional Distillation of Liquid Mixtures
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38. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.

39. Chapter 11: Solutions and Their Properties
4/17/2017
Copyright © 2010 Pearson Prentice Hall, Inc.