Diffusion of solvent through a membrane
For this experiment you can design your own experiment where a membrane is separated
between a solution and a pure solvent (Read the section on The Thermodynamics of Osmosis
given in page 2). You can measure the permeability of the solvent through the membrane and/or
other thermodynamics properties of the system.
The membrane supplied in the CME lab is dialysis membrane tubing which is a selectively
permeable regenerated cellulose commonly used to demonstrate the fundamentals of osmosis
and diffusion. Pores in the membrane permit the passage of water, most ions, and small
molecules in solution. Particles of high molecular weight, such as polysaccharides, starch, and
protein are restricted. When the membrane is utilized in laboratory dialysis or ultra filtration,
normal molecular weight cut-off is reported to be about 13,000. However, molecular weight is
not the only determining factor. Other factors affecting molecular transfer or membrane
permeability include molecular size and shape, pH of the solution used, and variations in the
porosity produced by tension on the membrane. Once thoroughly wetted, the membrane must not
be allowed to dry. Wetted membrane may be stored in water containing 0.1% benzoic acid.
(Dialysis membrane tubing can be obtained from Amico Scientific Corp, 7231-A Garden Grove
Blvd., Garden Grove CA 92941, (714) 894-6633)
Diffusion Through a Semi-permeable Membrane
Soak the needed length of dialysis tubing in de-ionized water for about 30 seconds. Open the
tubing by rolling it between thumb and index finger; immediately proceed as follows.
To demonstrate the effect of water diffusion (osmotic) pressure, use about 20 cm (8 in.) of
tubing. Knot one end tightly and fill to about 5 cm from the top with an 80% sugar solution.
Insert the end of a graduated 1-ml pipet into the opening of the tube and secure tightly in place
with a rubber band. Position the filled tube in a vessel of water and hold the pipet upright with a
clamp on a support stand. Recond and plot the rise of solution in the pipe at 2 to 5-minute
intervals. Rate of rise is proportional to the osmotic pressure generated by the solution in the
dialysis tubing.
Report Format
1) Title Page (See page 9) + Table of Contents (10%)
2) Research and summarize in your report the permeability of solvent to various
solutions. Summarize this research with proper referencing.(10%)
3) Permeability measurement: Research & summarize in your report methods for
measuring the permeability of solvent. Use proper referencing.(15%)
4) Experimental Plan: Your report should include an appropriate drawing and
description of the procedure that you used to determine the permeability of solvent in
various solutions. Your experimental plan should include studying the following:
different solutions and temperatures. What about the effect of temperature? What were
the ambient pressure and the pressure at both sides of the membrane? Is it important to
know these pressures & the temperature? Explain how you assured that temperature was
1
held constant or you accounted for the variation in temperature during an experiment in
your data analysis. (15%)
5) Tabulate your experimental data (an appropriate measurement & temperature
& ambient pressure vs time) in a spreadsheet & apply appropriate regression analysis.
Based on this statistical analysis what is the +/-% uncertainty in your reported
permeability of the solvent in various solutions. (35%)
6) Recommendation: Describe how you could improve your experimental
procedure. (15%)
2
1. The Thermodynamics of Osmosis
We will consider the equilibrium state of liquid mixtures in two regions separated by a
membrane that is permeable to some of the species present and impermeable to others. This
situation is illustrated in Figure 1.-1 where a semi-permeable membrane separates regions A that
contains a nondiffusing solute and region B that contains only water.
Water and
non-diffusable
solute
A
h
water
B
PA
PB
Figure 1.-1 Osmotic pressure  = PA  PB  Agh
Water will diffuse from region B into region A until the chemical potential or fugacity of water
on each side of the membrane is the same. This phenomenon is called osmosis and the pressure
difference between regions A and B at equilibrium is the osmotic pressure of region A. The
chemical potential of species i , i, is defined by the following relations
 G 
 H 
 U 
 A 




i = 
= 
= 
= 
 ni T ,P ,n
 ni  P ,S ,n
 ni  S ,V ,n  ni V ,T ,n
j
j
j
j
In these expressions, the subscript j denotes the moles of every species except i is a constant. The
definition can be obtained from the following diagram
G
P
Great Physicists Have Study
Under Very Able Teachers
S
T
A
H
V
U
3
We now want an expression that gives us the solution osmotic pressure as a function of the
solute concentration. At equilibrium
f WA (T, PA, xs) = f WB (T, PB)
(1.-1)
where f WB (T, PB) is the fugacity of water as a pure component and f WA (T, PA, xs) is the fugacity
of water as it exists in solution with solute at mole fraction xs. A similar equation is not written
for the solute since it cannot diffuse through the membrane. Equation (1.-1) can be expressed in
terms of the pure water fugacity using the activity coefficient  WA
 WA xWA f WA (T, PA) = f WB (T, PB)
(1.-2)
The fugacity is a thermodynamic function defined by
 G(T , P )  G IG (T , P ) 
 1
f(T, P) = Pexp 
 = Pexp 
RT
 RT



P
0
RT

V 
P

 
dP 
 
where G(T, P) is the molar Gibbs free energy and GIG(T, P) is the molar Gibbs free energy as the
fluid approached ideal gas state. The water fugacities at states (T, PA) and (T, PB) are then
ln
1
f (T , P A )
=
A
RT
P
ln
1
f (T , P B )
=
B
RT
P

PA

PB

PB
0
0
RT

V 
P


dP

RT

V 
P


dP

Since PA > PB
1
f (T , P A )
ln
=
A
RT
P
ln
0
1
RT 

V 
dP +
RT
P 

1
f (T , P A )
f (T , P B )
=
ln
+
A
B
RT
P
P

PA
P

PA
PB
V dP 
B
RT 

V 
dP
P 


PA
PB
dP
P
V is the molar volume of water, an incompressible liquid
ln
f (T , P A )
PA
f (T , P B ) V ( P A  P B )
=
ln
+

ln
PA
PB
RT
PB
ln
f (T , P A )
f (T , P B ) V ( P A  P B )
=
ln
+
PA
PA
RT
4
V ( P A  P B ) 
f (T , P A )
f (T , P B )
=
exp


PA
PA
RT


From the equality of fugacity, equation (1.-2)  WA xWA f WA (T, PA) = f WB (T, PB), we have

A
W
x
V WL ( P A  P B ) 
B
B
f (T, P ) exp 
 = f W (T, P )
RT


B
A
W
A
W
Since f WA (T, PB) = f WB (T, PB) = pure water fugacity

A
W
  V WL ( P A  P B ) 
x = exp 

RT


A
W
The osmotic pressure is then
 = P A  PB = 
RT
ln  WA xWA 
L
VW
(1.-3)
For an ideal aqueous solution at 298oK with xW = 0.98, W = 1, the osmotic pressure is
 = P A  PB = 
RT
ln(xW)
L
VW

bar  m 3 
 8.314  10 5
( 298o K )
o
mol  K 
=  
ln(0.98) = 27.8 bar
18  10 6 m 3 / mol
For ideal solution and small solute concentration, xWA  1, and ln( xWA )   (1  xWA )
Hence
=
x SA =
RT
RT
RT
ln( xWA )  L (1  xWA ) = L x SA
L
VW
VW
VW
Moles
x SA
=
L
VW
(1.-4)
Moles solute
Moles solute

solute  Moles solvent
Moles solvent
Moles
( Moles
solute
Moles solute
=
= CS
 Volume solvent  Volume solvent

solvent )
 Moles solvent 
5
The ideal dilute solution osmotic pressure, described by equation (1.-4), is known as van’t Hoff’s
law. This equation can also be written in terms of the mass concentration S

RT
 = L x SA = RT CS = RT S
(1.-5)
Mw S
VW
Mass solute
and MwS = molecular weight of solute. Equation (1.-5) can be used
Volume solvent
to determine solvent activity coefficient in a solvent-solute system provided a semi-permeable
membrane can be found.
where S =
 = P A  PB = 
RT
ln  WA xWA 
L
VW
Osmotic pressure measurements are more commonly used to determine the molecular weights of
proteins and other macromolecules using an osmometer shown in Figure 1.-2. At equilibrium the
osmotic pressure  is equal to gh, where  is the solution density and h is the difference in
liquid heights. Equation (1.-5) is then solved for the molecular weight of the solute.
MwS = RT
S

Solvent
solute mixture
h
Solvent
Semipermeable
membrane
Figure 1.-2 A graphical depiction of a simple osmometer.
Example 1.-1. ---------------------------------------------------------------------------------(Sandler, Chemical and Engineering Thermodynamics, Wiley, 1999, p.605)
The polymer polyvinyl chloride (PVC) is soluble in solvent cyclohexanone. At 25oC it is found
that if a 2 g of a specific batch of PVC per liter of solvent is placed in an osmometer, the height h
to which the pure cyclohexanone rises is 0.85 cm. Use this information to estimate the molecular
weight of the PVC polymer. Density of cyclohexanone is 0.98 g/cm3.
Solution ----------------------------------------------------------------------------------------- = gh = 980
kg
m
9.81 2 8.510-3 m = 81.72 Pa
3
m
s
6
MwS = RT
S

g
Pa  m 3
298.15 K2,000 3 /81.72 Pa = 60,670 g/mol
m
mol  K
---------------------------------------------------------------------------------------------------
MwS = 8.314
If the dilute solution contains N ideal solutes then
N
 = RT  C S ,i
i 1
The term osmole is defined as one mole of a nondiffusing and nondissociating substance. One
mole of a dissociating substance such as NaCl is equivalent to two osmoles. The number of
osmoles per liter of solution is called osmolarity. For physiological solutions, it is convenient to
work in terms of milliosmoles (mOsm) or milliosmolar (mOsM). The number of particles formed
by a given solute determines osmotic pressure. Each nondiffusing particle in the solution
contributes the same amount to the osmotic pressure regardless of the size of the particle.
The osmotic pressure difference between the interstitial and plasma fluids is due to the plasma
proteins since the proteins do not readily pass through the capillary wall. The osmotic pressure
created by the proteins is given the special name of colloid osmotic pressure or oncotic pressure.
For human plasma, the colloid osmotic pressure is about 28 mmHg; 19 mmHg caused by the
plasma proteins and 9 mmHg caused by the cations within the plasma that are retained through
electrostatic interaction with the negative surface charges of the proteins.
Figure 1.-3 Osmosis of water through red blood cell5.
If a cell such as red blood cell is placed in a hypotonic solution that has a lower concentration of
solutes or osmolarity, then the establishment of osmotic equilibrium requires the osmosis of
water into the cell resulting in swelling of the cell. If the cell is placed in a hypertonic solution
with a higher concentration of solutes or osmolarity, then osmotic equilibrium requires osmosis
of water out of the cell resulting in shrinkage of the cell. An isotonic solution has the same
osmolarity of the cell and will not cause any osmosis of water as shown in Figure 1.-3. A 0.9
5
Seeley R.R, Stephens T.D., Tate P., Anatomy & Physiology, McGraw Hill, 2003
7
weight percent solution of sodium chloride or a 5 weight percent solution of glucose is just about
isotonic with respect to a cell.
Example 1.-26. ---------------------------------------------------------------------------------Experiments show that at 0oC a 0.2 molarity sucrose solution has an osmotic pressure (relative to
pure water) of 4.76 atm whereas a 0.2 molarity NaCl solution has an osmotic pressure of 8.75
atm. Estimate the fraction of the NaCl molecules that are dissociated at this temperature and
concentration.
Solution ------------------------------------------------------------------------------------------
h
Sucrose
solution
h
NaCl
solution
Water
Semipermeable
membrane
Assume that all the sucrose dissolves and let  be the dissociated fraction of NaCl. When a salt
dissociates each ion contributes to the osmotic pressure
NaCl  Na+ + Cl
If  is the fraction of NaCl that is dissociated then 1   is the fraction that is not dissociated.
Since each dissociated NaCl molecule contributes 2 ions and each undissociated molecules
contributes 1 molecules, the osmolarity of the NaCl solution is
N
C
NaCl ,i
= [2 + (1   )]CNaCl
i 1
The osmolarity of the sucrose solution is
N
C
Sucrose,i
= CSucrose
i 1
Therefore,
6
Weiss T.F., Cellular Biophysics Transport, MIT Press, 1996, p. 258
8
 NaCl
RT (  1)C NaCl
8.75
=
=+1=
= 1.838
4.76
 Sucrose
RTC Sucrose
Hence  = 0.838
Table 1.-1 lists the osmotic pressure of aqueous sucrose solution at 30oC calculated using
equation (1.-3) for non-ideal solution and equation (1.-4) for ideal solution. The values for nonideal model agree well with the experimental data.
RT
ln  WA xWA 
L
VW
(1.-3)
RT A
x S = RT CS
L
VW
(1.-4)
= 
=
Table 1.-1 Osmotic pressure (atm) of aqueous sucrose solution at 30oC
CS (mol/liter)
Eq. (1.-4)
Eq. (1.-3)
Exp. Data
0.991
20.3
26.8
27.2
1.646
30.3
47.3
47.5
2.366
39.0
72.6
72.5
3.263
47.8
107.6
105.9
4.108
54.2
143.3
144.0
5.332
61.5
199.0
204.3
At a physiological temperature of 37oC, Equation (1.-4) may be written as follows to give the
osmotic pressure in mmHg when the solute concentration of each non-diffusing species is
expressed in mOsM:
 = 19.33 CS
(1.-5)
9
Title Page
California State Polytechnic University, Pomona
Chemical and Materials Engineering
CHE 313- MASS TRANSPORT SPRING QUARTER, 2012
EXPERIMENT #3
Measurement of Permeability through balloon wall
INSTRUCTOR:
T. K. Nguyen
SECTION 01
GROUP #1
MEMBERS: Lau, Lee
McCaffrey, Charles
DUE DATE: May 10 2013
Title Page
(10) ______
Research binary gaseous diffusivity
(10) ______
Diffusivity measurement
(15) ______
Experimental Plan
(15) ______
Tabulated Data & Analysis
(35) ______
Recommendation
(15) ______
TOTAL
(100) ______
10