Solutions: AP Notes
Use Pre-AP Notes for background solution info
Colligative Properties
Colligative Properties
Depend on NUMBER of solute particles, not the
kind of particles in a solution.
FOUR types
1. Boiling point elevation
2. Freezing point depression
3. Osmotic pressure
4. Vapor pressure lowering
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•
•
Boiling Point Elevation
Δ T = Kb mi
Kb is a proportionality constant that is given or
retrieved from a table. It is unique for a
particular solvent. (Usually °C/m)
Δ T is the change in temp, final – initial
(Usually °C)
•
i is the van’t Hoff factor, a measure of
ionization or dissociation. The simplest
assumption for i is to assume very dilute
solutions and assume that the ionic compound
completely dissociates.
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Freezing Point Depression
Δ T = Kf mi
•
Kf is a proportionality constant that is given or
retrieved from a table. It is unique for a particular
solvent. (Usually °C/m)
•
Δ T is the change in temp, final – initial
(Usually °C)
•
i is the van’t Hoff factor, a measure of ionization or
dissociation. The simplest assumption for i is to
assume very dilute solutions and assume that the
ionic compound completely dissociates.
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When i is not “ideal”…
+
+
K
+
+
+
+
-
For solutions which are not dilute, i is the ratio of the actual
colligative property to the value that would be observed if
no dissociation occurred.
•
Cl
-
-
+
+
K
+
+
-
Cl
+
+
-
Cl
K
+
+
-
+
+
+
+
-
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-
Calculating i for weak electrolytes
i=
ΔTf (actual)
= Kfmeffective = meffective
ΔTf (if nonelectrolyte) Kfmstated mstated
A problem that asks for the calculation of i will provide an actual
freezing or boiling point at the given concentration. First find
Δ T assuming that the solvent is pure water.
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Example: Calculating i
ΔTf (actual)
= Kfmeffective = meffective
ΔTf (if nonelectrolyte) Kfmstated mstated
The freezing point of a 0.100m Na2SO4 (aq)
solution is -0.0349°C. Calculate i.
Use the ratio above.
meffective = .0349°C = 0.188 m
1.86°C/m
meffective = 0.0188 m = 1.88
mstated
0.100 m
i=
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But I thought i = 3 for Na2SO4!
In very dilute solutions, no significant amount of ion
association occurs to lessen the effect of all three
available ions being separated from one another. This is
when i is assumed to be 3. Compare the following
example to the first one (0.100m Na2SO4 (aq)).
The freezing point of a 1.00m Na2SO4 (aq) solution is
-3.29°C. Calculate i.
meffective = 3.29°C = 1.77 m
1.86°C/m
meffective = 0.0188 m = 1.77
Value of i is smaller since
mstated 1.00 m
concentration increased.
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Osmotic Pressure
•
http://www.wasanlab.ubc.ca/assets/flash/osmosis.swf
• Click the above link for osmosis
simulation.
• Sugar solution is placed in the bulb of the
thistle tube and a semipermeable
membrane is placed over the end of this
tube. The tube is then submerged in
water.
• The sugar solution becomes diluted as
water moves into the tube, pushing the
solution level up until the pressure in the
tube is just sufficient to prevent solvent
flow from the pure solvent side of the
membrane to the solution side.
• This pressure is called osmotic pressure.
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Usefulness of Osmotic Pressure
•
This technique is a relatively simple way to
determine the molecular weight using very
little solute.
• Even a dilute solution may be used, so the
method is advantageous when:
1. Only a small amount of solute or solution is
available.
2. The solute is expensive
3. The solute does not dissolve well in water
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Calculating Osmotic pressure
Π= nRT
V
n = moles = M
V
L
Π =MRT
n= ΠV
RT
(M is molarity)
In a problem solving for molecular weight, solve for
n (moles), then divide a given mass by n to get
g/mol. This type of problem gives the mass,
volume, temperature and osmotic pressure.
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Example- Molecular weight from osmotic pressure
A sample of 2.05 g pf polystyrene was dissolved in enough
toluene to form 0.100 L of solution. The osmotic
pressure of the solution was found to be 1.21 kPa at 25 °
C. Find molar mass of polystyrene.
1.21 kPa |1
atm = 0.0119 atm
101.325 kPa
n =ΠV=0.0119 atm(0.100 L) = 4.84 x 10-5 mol
RT (0.0821 L-atm/mol-K)(298K)
Molar mass = 2.05 g
= 4.24 104 g__
4.84 x 10-5 mol
mol
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Vapor Pressure: Raoult’s Law
• Vapor = gas formed by the boiling or
evaporation of a liquid or a solid
• Vapor pressure of a liquid is the pressure
exerted by the vapor in equilibrium with its
liquid.
• A solution containing a nonvolatile solute
has a lower vapor pressure than the pure
solvent.
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Raoult’s Law
Psolvent = XsolventP°solvent
• Psolvent: Vapor pressure of solvent in the solution
• Xsolvent: Mole fraction of solvent in solution
• P°solvent: Vapor pressure of the pure solvent
• As the mole fraction (%solvent) goes up, its vapor
pressure also goes up proportionally.
• If the solute is nonvolatile, then Psolvent = Psolution
Lowering of vapor pressure is defined:
ΔPsolvent=P°solvent– Psolvent OR Δ Psolvent = XsoluteP°solvent
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Example 1: Raoult’s Law
At 25° C, determine the vapor pressure
lowering of 1.25 m sucrose solution that has
been made with 50.0 g C12H22O11 and 117 g
H2O. The vapor pressure of pure water is
23.8 torr at 25 ° C.
117 g H2O = 6.50 mol
50.0 g C12H22O11 = 1.46 mol
Δ Psolvent = XsoluteP°solvent
χsucrose = 1.46 mole/(1.46 + 6.50)moles =
0.0220
Δ Psolvent =0.0220(23.8 torr) = 0.524 torr
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Example 2: Raoult’s Law
At 40° C, the v.p. of pure hexane is 92.0 torr and the v.p. of pure octane is
31.0 torr. In a solution containing 1.00 mole heptane and 4.00 moles
octane, calculate the vapor pressure of each component and the v.p.
above the solution.
Psolvent = XsolventP°solvent
χheptane = 1.00 mole/(1.00 + 4.00)moles = 0.200
χoctane = 1- χheptane = 0.800
Poctane=XsolventP°solvent =0.800(31.0 torr)=24.8 torr
Pheptane=XsoluteP°solute=0.200(92.0 torr)=18.4 torr
Ptotal = 18.4 + 24.8 torr = 43.2 torr
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Henry’s Law
Solubility of gas in liquid
• The solubility of a gas is directly proportional to
the partial pressure of that gas on the surface of
the liquid.
• Soda bottle:
– High pressure at the surface while the bottle is closed,
so lots of CO2 in the liquid
– Open bottle, pressure on surface lowers to room
atmosphere and CO2 leaves the liquid
• High pressure = High gas concentration
• Low pressure = low gas concentration
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