Transcript Chemistry: McMurry and Fay, 5th Edition

John E. McMurry • Robert C. Fay
C H E M I S T R Y
Fifth Edition
Chapter 11
Solutions and Their Properties
Lecture Notes
Alan D. Earhart
Southeast Community College • Lincoln, NE
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Solutions
Solution: A homogeneous mixture.
Solvent: The major component.
Solute: A minor component.
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Chapter 11/2
Energy Changes and the
Solution Process
Energy Changes and the
Solution Process
The sodium and chloride ions are hydrated.
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Chapter 11/4
Energy Changes and the
Solution Process
There is an entropy change for the solution process.
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Chapter 11/5
Energy Changes and the
Solution Process
There is an entropy change for the solution process.
Solution #1
Solution #2
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Chapter 11/6
Energy Changes and the
Solution Process
DG = DH - TDS
endothermic: +DH
exothermic: -DH
spontaneous: -DG
nonspontaneous: +DG
Energy Changes and the
Solution Process
The solute-solvent interactions
are greater than the sum of
the solute-solute and solventsolvent interactions.
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Chapter 11/8
Energy Changes and the
Solution Process
The solute-solvent interactions
are less than the sum of the
solute-solute and solventsolvent interactions.
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Chapter 11/9
Units of Concentration
Moles of solute
Molarity (M) =
Liters of solution
Moles of component
Mole Fraction (X) =
Total moles making up solution
Mass of component
Mass Percent =
x 100%
Total mass of solution
Moles of solute
Molality (m) =
Mass of solvent (kg)
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Chapter 11/10
Units of Concentration
Assuming that seawater is an aqueous solution of NaCl,
what is its molarity? The density of seawater is 1.025 g/mL at
20 °C, and the NaCl concentration is 3.50 mass %.
Assuming 100.00 g of solution, calculate the volume:
100.00 g solution
1 mL
1L
x
x
= 0.09756 L
1.025 g
1000 mL
Convert the mass of NaCl to moles:
3.50 g NaCl
1 mol
x
58.4 g
= 0.0599 mol
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Chapter 11/11
Units of Concentration
Then, calculate the molarity:
0.0599 mol
0.09756 L
= 0.614 M
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Chapter 11/12
Units of Concentration
What is the molality of a solution prepared by dissolving
0.385 g of cholesterol, C27H46O, in 40.0 g chloroform, CHCl3?
Convert the mass of cholesterol to moles:
0.385 g
1 mol
x
= 0.000 997 mol
386.0 g
Calculate the mass of chloroform in kg:
40.0 g
1 kg
x
= 0.0400 kg
1000 g
Calculate the molality of the solution:
0.000 997 mol
= 0.0249 m
0.0400 kg
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Chapter 11/13
Units of Concentration
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Chapter 11/14
Some Factors Affecting
Solubility
Saturated Solution: A solution containing the
maximum possible amount of dissolved solute at
equilibrium.
dissolve
Solute + Solvent
Solution
crystallize
Supersaturated Solution: A solution containing a
greater-than-equilibrium amount of solute.
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Chapter 11/15
Some Factors Affecting
Solubility
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Chapter 11/16
Some Factors Affecting
Solubility
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Chapter 11/17
Some Factors Affecting
Solubility
Henry’s Law
Solubility = k P
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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Chapter 11/19
Vapor-Pressure Lowering of
Solutions: Raoult’s Law
Raoult’s Law
Psoln = Psolv Xsolv
Vapor-Pressure Lowering of
Solutions: Raoult’s Law
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Chapter 11/21
Vapor-Pressure Lowering of
Solutions: Raoult’s Law
The vapor pressure of pure water at 25 °C is 23.76 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 x
15.0 mol
= 22.3 mm Hg
1.00 mol + 15.0 mol
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Chapter 11/22
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.
van’t Hoff Factor: i =
NaCl(aq)
moles of particles in solution
moles of solute dissolved
Na1+(aq) + Cl1-(aq)
For sodium chloride, the predicted value of i is 2. For a 0.5 m
solution of sodium chloride, the experimental value for i is 1.9.
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Chapter 11/23
Vapor-Pressure Lowering of
Solutions: Raoult’s Law
Ptotal = PA + PB = (P°A XA) + (P°B XB)
Boiling-Point Elevation and
Freezing-Point Depression
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Chapter 11/25
Boiling-Point Elevation and
Freezing-Point Depression
Nonelectrolytes
Electrolytes
DTb = Kb m
DTb = Kb m i
DTf = Kf m
DTf = Kf m i
Boiling-Point Elevation and
Freezing-Point Depression
DG = DH - TDS
At equilibrium, DG = 0.
T=
DH
DS
Boiling-Point Elevation and
Freezing-Point Depression
DG = DH - TDS
At equilibrium, DG = 0.
T=
DH
DS
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
x
1 mol
= 0.0777 mol
95.2 g
Calculate the molality of the solution:
0.0777 mol
110 g
x
1000 g
1 kg
= 0.71
mol
kg
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Chapter 11/29
Boiling-Point Elevation and
Freezing-Point Depression
Calculate the freezing point of the solution:
DTf = Kf m i = 1.86
°C kg
mol
x 0.71
mol
kg
x 2.7 = 3.6 °C
Tf = 0.0 °C - 3.6 °C = -3.6 °C
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Chapter 11/30
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 (P): The amount of pressure necessary
to cause osmosis to stop.
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Chapter 11/31
Osmosis and Osmotic
Pressure
P = MRT
Osmosis and Osmotic
Pressure
Calculate the osmotic pressure of a 1.00 M glucose
solution in water at 300 K.
P = MRT = 1.00
mol
L
x 0.08206
L atm
x 300 K = 24.6 atm
K mol
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Chapter 11/33
Some Uses of Colligative
Properties
Reverse Osmosis
Fractional Distillation of Liquid
Mixtures
Fractional Distillation of Liquid
Mixtures
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Chapter 11/36