Lecture 43 (Slides) November 22

Report
Osmotic Pressure
For dilute solutions of nonelectrolytes:
πV = nRT
n
π=
RT = MRT
V
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 1 of 46
Osmotic Pressure
• 2. A 0.426 g sample of an organic compound is
dissolved in 225 mL of solvent at 24.5 0C to
produce a solution with an osmotic pressure of
3.36 mm Hg. What is the molar mass of the
organic compound?
• 3. What additional information would be
required to determine the molecular formula as
well as the molar mass?
Henry’s Law
• 4. At 0 0C a 1.00L aqueous solution contains
48.98 mL of dissolved oxygen when the O2(g)
pressure above the solution is 1.00 atm. What
would be the molarity of oxygen in the
solution if the oxygen gas pressure above the
solution were instead 4.15 atm?
Freezing-Point Depression and Boiling
Point Elevation of Nonelectrolyte
Solutions
• A Colligative property.
– Depends on the number of particles present.
• Vapor pressure is lowered when a solute is present.
– This results in boiling point elevation.
– Freezing point is also effected and is lowered.
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 4 of 46
ΔTf = -Kf  m
ΔTb = +Kb  m
FIGURE 13-19
Vapor-pressure lowering by a nonvolatile solute
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 5 of 46
ΔTf = -Kf  m
Copyright © 2011 Pearson
Canada Inc.
ΔTb = +Kb  m
General Chemistry: Chapter 13
Slide 6 of 46
Practical Applications
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 7 of 46
Solutions of Electrolytes
• Svante Arrhenius
–Nobel Prize 1903.
–Ions form when electrolytes dissolve
in solution.
–Explained anomalous colligative
properties.
Compare 0.0100 m aqueous urea to 0.0100 m NaCl (aq)
ΔTf = -Kf  m = -1.86°C m-1  0.0100 m = -0.0186°C
Freezing point depression for NaCl is -0.0361°C.
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 8 of 46
Van’t Hoff
i=
measured ΔTf
expected ΔTf
=
0.0361°C
= 1.98
0.0186°C
π = i  M  RT
ΔTf = -i  Kf  m
ΔTb = i  Kb  m
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 9 of 46
Interionic Interactions
•Arrhenius theory does not correctly predict the
conductivity of concentrated electrolytes.
Copyright © 2011 Pearson
Canada Inc.
General Chemistry: Chapter 13
Slide 10 of 46
Interionic Interactions
• The data on the previous slide show that
solutions of electrolytes behave less ideally as
(1) ionic charges increase and (2) electrolyte
concentraion increases. Are there
correspondences to nonideal gas behaviour?
The latter manifests itself at high P and low T
(high gas “concentration”) and as molecules
become more polar or polarizable.
Ideal van’t Hoff Factors (Aqueous
Solutions):
• We need to (a) identify covalent and ionic
substances by inspecting their chemical
formulas and (b) for ionic compounds (and
strong acids) determine how many ions are
formed in solution per formula unit of solute
dissolved. Examples are given on the next
slide.
Van’t Hoff “i Factors”:
Chemical Name
Chemical Formula
Ionic/Covalent ?
Ideal “i” value
Sucrose
C12H22O11
Covalent
1
Potassium nitrate
KNO3
Ionic
2
Urea
(NH2)2C=O
Covalent
1
Magnesium
chloride
MgCl2
Ionic
3
Aluminum nitrate
Al(NO3)3
Ionic
4
Lead (II) iodide
PbI2
Ionic
3
Copper (II) nitrate
hexahydrate
Cu(NO3)2∙6H2O
Ionic
Hydrochloric acid
HCl(aq)
Potassium
hydroxide
Boiling Pt. and Freezing Pt. Changes
• Ex. 2. If 3.30 g of ammonium nitrate were
dissolved in 475 g of water what would you
expect the boiling point and freezing point of
the solution to be? (Kb and Kf values for water
are 0.512 0C m-1 and 1.86 0C m-1 .) (Mention
different form of “freezing pt. equation” in
some texts?)
Freezing and Boiling Points
• Ex. 3. Which aqueous solution would have a
higher boiling point? (a) 1.00 m sucrose
(C12H22O11) or (b) 0.400 m calcium nitrate?
• Ex. 4. Which aqueous solution would have the
lowest freezing point? (a) 1.20 m sucrose, (b)
0.600 m sodium chloride (c) 0.400 m
magnesium nitrate or (d) 0.700 m copper (II)
sulfate? Why?

similar documents