Colligative Properties
of Nonvolatile Solutes
01
•
Colligative Properties: Depend on the amount not
on the identity
•
There are four main colligative properties:
1.
2.
3.
4.
Colligative Properties
of Nonvolatile Solutes
•
02
When solute molecules displace solvent molecules
at the surface, the vapor pressure drops since
fewer gas molecules are needed to equalize the
escape rate and capture rates at the liquid surface.
Vapor pressure lowering
Freezing point depression
Boiling point elevation
Osmotic pressure
Colligative Properties
of Nonvolatile Solutes
Colligative Properties
of Nonvolatile Solutes
03
04
Xsolv = moles of solvent
Total moles
Raoult’s Law:
Van’t Hoff factor
Psoln = PsolvXsolv
i = moles of particles in solution
moles of solute dissolved
*Think of total moles in the above equation as total moles of particles in
solution
For example: If we dissolve 0.313 moles of NaCl in 27.75
moles of water, what is the mole fraction of water?
i = 2 for NaCl (if completely dissociated)
0.313moles of NaCl x 2 = 0.626 total moles of particles of NaCl
XH2O =
27.75moles H2O
= 0.978
27.75 moles of H2O + 0.626 moles of NaCl
1
Vapor Pressure of a Solution
and a Nonvolatile Solute
Colligative Properties of a
Volatile Solute
05
Why is the vapor pressure of a solution with a nonvolatile
solute lower than for the pure solvent? ⇒ Entropy
06
•
What happens if both components are volatile
(have measurable vapor pressures)?
•
The vapor pressure has a value intermediate between the
vapor pressures of the two liquids.
P T = P A + PB
=
XAP°A + XBP°B
= XAP°A + (1 – XA)P°B
Colligative Properties of a
Volatile Solute
•
The following diagram shows a close-up view of
part of the vapor–
pressure curves
for two pure liquids
and a mixture of
the two. Which
curves represent
pure liquids, and
which the mixture?
Boiling-Point Elevation
07
•
08
Boiling-Point Elevation (∆Tb): The boiling point of
the solution (Tb) minus the boiling point of the pure
solvent (T°b):
∆Tb = Tb – T°b
∆Tb is proportional to concentration:
∆Tb = Kb mi
Kb = molal boiling-point elevation constant.
2
Freezing-Point Depression
•
09
Freezing-Point Depression (∆Tf): The freezing point
of the pure solvent (T°f) minus the freezing point of
the solution (Tf).
∆Tf = T°f – Tf
∆Tf is proportional to concentration:
∆Tf = Kf mi
10
van’t Hoff Factor, i: This factor equals the number
of ions produced from each molecule of a
compound upon dissolving.
i = 1 for CH3OH
i = 3 for CaCl2
i = 2 for NaCl
i = 5 for Ca3(PO4)2
For compounds that dissociate on dissolving, use:
Kf = molal freezing-point depression constant.
Boiling-Point Elevation and
Freezing-Point Depression
Boiling-Point Elevation and
Freezing-Point Depression
∆Tb = Kb mi
11
∆Tf = Kf mi
Entropy of a Solution
12
3
Why is the Boiling Point of a
Solution Elevated?
Why is the Freezing Point of a
Solution Depressed?
13
ENTROPY!
Boiling-Point Elevation and
Freezing-Point Depression
14
ENTROPY!
Boiling-Point Elevation and
Freezing-Point Depression
15
•
16
The phase diagram shows a close-up of the liquid–vapor
phase transition boundaries for pure chloroform.
a)
Estimate the boiling
point of pure
chloroform.
a)
Estimate the molal
concentration of the
nonvolatile solute.
(See Table 11.4 for Kb).
4
Osmosis and Osmotic Pressure
17
•
Osmosis: The selective passage of solvent molecules
through a porous membrane from a dilute solution to a more
concentrated one.
•
Osmotic pressure (! or ∏): The pressure required to stop
osmosis (obtain equilibrium)
Osmosis and Osmotic Pressure
18
Example
20
! = i⋅MRT
M = molar concentration of solute particles
R = 0.08206 (L⋅atm)/(mol⋅K)
Osmosis and Osmotic Pressure
19
The average osmotic pressure of seawater is about 30.0 atm at 25°C.
Calculate the molar concentration of an aqueous solution of urea
[(NH2)2CO].
! = iMRT
M = !/iRT = 30.0atm/(1(0.08206Latm/molK)(298K))
=1.23 mol/L
5
Example
21
What is the osmotic pressure (in atm) of a 0.884 M sucrose
Uses of Colligative Properties
⇒Reverse Osmosis
22
Desalination:
solution at 16°C?
T = 16°C = 289K
! =(0.884mol/L)(0.08206Latm/molK)(289K)
= 20.96 atm
Uses of Colligative Properties
⇒ Molecular Mass Calculation
•
•
•
23
Any of the four colligative properties can be used to
calculate molecular mass
Example
24
A 202 ml benzene solution containing 2.47 g of an organic polymer has
an osmotic pressure of 8.63 mm Hg at 21°C. Calculate the molar mass
of the polymer.
In lab we used freezing point depression, but the
degree of depression is very small
M = !/iRT =(8.63mmHg x 1atm/760mm Hg) = 4.71 x 10-4 mol/L
Most accurate value of molecular mass is to use
osmotic-pressure measurements
4.71 x 10-4 mol/L = x / 0.202L benzene x = 9.514 x 10-5 moles of polymer
(1)(0.08206Latm/molK)(294K)
MM organic polymer = 2.47g/ 9.514 x 10-5 moles = 25961 g/mol
6
Example
Uses of Colligative Properties
25
What is the molar mass of sucrose if a solution of 0.822 g of sucrose in
300.0 mL of water has an osmotic pressure of 149 mm Hg at 298 K?
M = !/iRT = (149mm Hg x 1 atm/760mm Hg) = 0.00801mol/L
•
26
Fractional Distillation is the separation of volatile
liquid mixtures into fractions of different
composition.
(1)(0.08206Latm/molK)(298K)
0.00801mol/L = x / 0.300L
x = 0.0024moles of sucrose
MM sucrose = 0.822g/0.0024moles = 341.8g/mol
Uses of Colligative Properties
•
Fractional distillation can be represented on a
phase diagram by plotting temperature against
composition.
27
Example
28
Two miscible liquids, A and B, have vapor pressures of 250 mm Hg and 450 mm
Hg, respectively. They were mixed in equal molar amounts. What is the total
vapor pressure of the mixture and what are their mole fractions in the vapor
phase?
Pmix = P°AXA + P°BXB = (250mm Hg)(0.5) + (450mm Hg)(0.5) = 350 mm Hg
PA = P°AXA = (250mm Hg)(0.5) = 125mm Hg
XA = 125mm Hg/350mm Hg = 0.357
PB = P°BXB = (450mm Hg)(0.5) = 225mm Hg
XB = 225mm Hg/350mm Hg = 0.643
7