updates midterms are feb. 08 and mar. 15 at 7pm… anyone with a night class or other midterm will...
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Updates
• Midterms are Feb. 08 and Mar. 15 at 7pm… anyone with a night class or other midterm will write it at 6 pm (notify me at least 1 week prior if you have one of these conflicts)
• Assignment 01 is posted and is due (in class) Wed., January 24
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Physical Properties of Solutions
Chapter 13
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Effect of pressure on gas solubility
• The solubilities of solids and liquids are not affected appreciably by pressure
• When the pressure of a gas is increased, as in (b), the rate at which gas molecules enter the solution increases
• The concentration of solute molecules at equilibrium increases in proportion to the pressure
• So the solubility of a gas increases with pressure
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Pressure and Solubility of Gases
13.5
Solubility decreases as pressure decreases
• Soft drink bottled under CO2 pressure greater than 1 atm
• When the bottle is opened, partial pressure of CO2 above the solution decreases
• Solubility of CO2 decreases -> bubbles
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Pressure and Solubility of Gases
13.5
Solubility decreases as pressure decreases
In many consumer beverages such as soft drinks, carbonation is used to give "bite". Contrary to popular belief, the fizzy taste is caused by the dilute carbonic acid inducing a slight burning sensation, and is not caused by the presence of bubbles. This can be shown by drinking a fizzy drink in a hyperbaric chamber at the same pressure as the beverage. This gives much the same taste, but the bubbles are completely absent.
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Pressure and Solubility of Gases
13.5
The solubility of a gas in a liquid is proportional to the pressure of the gas over the solution (Henry’s law).
c = kPc is the concentration (M) of the dissolved gasP is the pressure of the gas over the solutionk is a constant (mol/L•atm) that depends only on temperature
low P
low c
high P
high c
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SAMPLE EXERCISE A Henry’s Law Calculation
Calculate the concentration of CO2 in a soft drink that is bottled with a partial pressure of CO2 of 4.0 atm over the liquid at 25°C. The Henry’s law constant for CO2 in water at this temperature is 3.1 10–2 mol/L-atm.
PRACTICE EXERCISECalculate the concentration of CO2 in a soft drink after the bottle is opened and equilibrates at 25°C under a CO2 partial pressure of 3.0 10–4 atm.
SolutionAnalyze: We are given the partial pressure of CO2, and the Henry’s law constant, k, and asked to calculate the concentration of CO2 in the solution.
Plan: With the information given, we can use Henry’s law to calculate the solubility.
Solve: Check: The units are correct for solubility, and the answer has two significant figures consistent with both the partial pressure of CO2 and the value of Henry’s constant.
2
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Blood gases and deep sea diving• Solubility increases as pressure increases• Divers who use compressed gases must be concerned about the solubility of the gases in
their blood • At depth, the blood contains higher concentrations of dissolved gases• Ascension, if too rapid, will cause the blood to fizz similar to 7-UP when opened!• This is called decompression sickness, or “the bends”, which is painful and can be fatal
because the bubbles affect things like nerve impulses
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Temperature and Solubility
13.4
Solubility of most solid solutes in water increases with increasing temperature
In contrast, solubility of gases in waterdecreases with increasing temperature
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Temperature and Solubility
13.4
In contrast, solubility of gases in waterdecreases with increasing temperature
• Carbonated beverages go “flat” as they warm due to a decreased solubility of dissolved CO2
• Bubbles form on the inside wall of a cooking pot when water is heated even though the temperature is well below boiling
• Thermal pollution of lakes and streams causes low oxygen levels in deeper layers
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Colligative Properties
• Changes in colligative properties depend only on the number of solute particles present, not on the identity of the solute particles.
• Among colligative properties areVapor pressure lowering Boiling point elevationMelting point depressionOsmotic pressure
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Vapor Pressure
Because of solute-solvent intermolecular attraction, higher concentrations of nonvolatile solutes make it harder for solvent to escape to the vapor phase.
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Colligative Properties of Nonelectrolyte Solutions
Vapor-Pressure Lowering
Raoult’s law
If the solution contains only one solute:
X1 = 1 – X2
P 10 - P1 = P = X2 P 1
0
P 10 = vapor pressure of pure solvent
X1 = mole fraction of the solvent
X2 = mole fraction of the solute13.6
P1 = X1 P 10
Nonvolatile solutes lower the vapor pressure of a solventby an amount proportional to the solute mole fraction.
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SAMPLE EXERCISE Calculation of Vapor-Pressure Lowering
Glycerin (C3H8O3) is a nonvolatile nonelectrolyte with a density of 1.26 g/mL at 25°C. Calculate the vapor pressure at 25°C of a solution made by adding 50.0 mL of glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr.
SolutionAnalyze: Our goal is to calculate the vapor pressure of a solution, given the volumes of solute and solvent and the density of the solute.Plan: We can use Raoult’s law to calculate the vapor pressure of a solution. The mole fraction of the solvent in the solution, XA, is the ratio of the number of moles of solvent (H2O) to total solution (moles C3H8O3 + moles H2O).
Solve: To calculate the mole fraction of water in the solution, we must determine the number of moles of C3H8O3 and H2O:
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PRACTICE EXERCISEThe vapor pressure of pure water at 110°C is 1070 torr. A solution of ethylene glycol and water has a vapor pressure of 1.00 atm at 110°C. Assuming that Raoult’s law is obeyed, what is the mole fraction of ethylene glycol in the solution?
Answer: 0.290
SAMPLE EXERCISE continued
The vapor pressure of the solution has been lowered by 0.6 torr relative to that of pure water.
We now use Raoult’s law to calculate the vapor pressure of water for the solution:
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PA = XA P A0
PB = XB P B0
PT = PA + PB
PT = XA P A0 + XB P B
0
13.6
If both components of the solution are volatile, the vaporpressure of the solution is the sum of the individual partial pressures.
Pure toluene Pure benzene1:1 mix
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Boiling-Point Elevation
Tb = Tb – T b0
Tb > T b0 Tb > 0
T b is the boiling point of the pure solvent
0
T b is the boiling point of the solution
Tb = Kb m
m is the molality of the solution
Kb is the molal boiling-point elevation constant (0C/m)
13.6
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Freezing-Point Depression
Tf = T f – Tf0
T f > Tf0 Tf > 0
T f is the freezing point of the pure solvent
0
T f is the freezing point of the solution
Tf = Kf m
m is the molality of the solution
Kf is the molal freezing-point depression constant (0C/m)
13.6
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13.6
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What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.
Tf = Kf m
m =moles of solute
mass of solvent (kg)= 2.41 m=
3.202 kg solvent
478 g x 1 mol62.01 g
Kf water = 1.86 0C/m
Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C
Tf = T f – Tf0
Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C
13.6
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Osmotic Pressure ()
13.6
Osmosis is the selective passage of solvent molecules through a porous membrane from a dilute solution to a more concentrated one.
A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.
Osmotic pressure () is the pressure required to stop osmosis.
dilutemore
concentrated
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HighP
LowP
Osmotic Pressure ()
= MRT
M is the molarity of the solution
R is the gas constant
T is the temperature (in K) 13.6
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A cell in an:
isotonicsolution
hypotonicsolution
hypertonicsolution
13.6
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Colligative Properties of Nonelectrolyte Solutions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
13.6
Vapor-Pressure Lowering P1 = X1 P 10
Boiling-Point Elevation Tb = Kb m
Freezing-Point Depression Tf = Kf m
Osmotic Pressure () = MRT
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Colligative Properties of Electrolyte Solutions
13.7
0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions
Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.
0.1 m NaCl solution 0.2 m ions in solution
van’t Hoff factor (i) = actual number of particles in soln after dissociation
number of formula units initially dissolved in soln
nonelectrolytesNaCl
CaCl2
i should be
12
3
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Boiling-Point Elevation Tb = i Kb m
Freezing-Point Depression Tf = i Kf m
Osmotic Pressure () = iMRT
Colligative Properties of Electrolyte Solutions
13.7