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CM 4710, Biochemical Processes Fall 2007 Homework #8 Fri. 09 Nov., 2007
1. Adsorption Separation of Immunoglobulin G using Modified Dextran (adapted from Belter, Cussler, & Hu, pg 153).
The equilibrium between immunoglobulin G and a modified dextran (the adsorbent) can be
described by the Langmuir isotherm. Dextran will adsorb up to 7.8x10-6 moles of
immunoglobulin G per cm3 of adsorbent and the Langmuir constant, KL, is 1.9x10-5
moles/liter. The adsorbent density is 1 gram of adsorbent per cm3 of adsorbent and the void
fraction of a column packed with this adsorbent is 0.40.
a) What is the total adsorption capacity of a column packed with modified dextran if the feed
concentration of immunoglobulin G is 2x10-5 moles/liter? (use column data from b))
b) What is the mean retention time for immunoglobulin G in a column of diameter equal to 5
cm with a feed flow rate of 1 liter per minute, feed concentration of 2x10-5 moles/liter, and
a length of 1 meter?
c) How many columns would have to be employed to recover immunoglobulin G from a feed
tank of 10,000 Liter volume assuming the same feed concentration?
2. Travel Distance of Solutes A and B in a Chromatographic Column.
Problem 11.6 of the text, parts a and b. Assume that CA = 0.10 mg/ml and CB = 0.05 mg/ml in
the liquid phase of the column at equilibrium with the adsorbed solute. 3. Determining Time to Elute a Solutes A and B from a Chromatographic Column. Problem 11.7 of the text
Due Fri. 16 Nov., 2007.
CM 4710, Biochemical Processes Fall 2003Homework #7 Solution Fri. 7 Nov., 2003
3. Determining Time to Elute a Solutes A and B from a Chromatographic Column.
Problem 11.7 of the text
The fundamental eqn. For gel chromatography is
Ve = Vo + KD Vi
For A: Ve = 20 cm3 + 0.5(30 cm3) = 35 cm3
TA = 35 cm3/100 cm3/hr = 0.35 hr
For A: Ve = 20 cm3 + 0.15(30 cm3) = 24.5 cm3
TA = 24.5 cm3/100 cm3/hr = 0.245 hr
CM 4710, Biochemical Processes Fall 2003 Homework #8 Fri. 4 Nov., 2005
1. Adsorption Separation of Immunoglobulin G using Modified Dextran (adapted from Belter, Cussler, & Hu, pg 153).
The equilibrium between immunoglobulin G and a modified dextran (the adsorbent) can be
described by the Langmuir isotherm. Dextran will adsorb up to 7.8x10-6 moles of
immunoglobulin G per cm3 of adsorbent and the Langmuir constant, KL, is 1.9x10-5
moles/liter. The adsorbent density is 1 gram of adsorbent per cm3 of adsorbent and the void
fraction of a column packed with this adsorbent is 0.40.
a) What is the total adsorption capacity of a column packed with modified dextran if the feed
concentration of immunoglobulin G is 2x10-5 moles/liter? (use column data from b))
b) What is the mean retention time for immunoglobulin G in a column of diameter equal to 5
cm with a feed flow rate of 1 liter per minute, feed concentration of 2x10-5 moles/liter, and
a length of 1 meter?
c) How many columns would have to be employed to recover immunoglobulin G from a
feed tank of 10,000 Liter volume assuming the same feed concentration?
2. Travel Distance of Solutes A and B in a Chromatographic Column.
Problem 11.6 of the text, parts a and b. Assume that CA = 0.10 mg/ml and CB = 0.05 mg/ml in
the liquid phase of the column at equilibrium with the adsorbed solute. 3. Determining Time to Elute a Solutes A and B from a Chromatographic Column. Problem 11.7 of the text
Due Fri. 11 Nov., 2005.
1. Adsorption Separation of Immunoglobulin G using Modified Dextran (adapted
from Belter, Cussler, and Hu, pg 153) Information
( ) 40.0fraction void
1
/109.1
3mol/IG/cm^ 6-7.8E toup adsorb llDextran wi
isothermLangmuir
3
5
=
=
×=
=+
=
−
ε
ρcm
g
LmolK
C
CK
CCC
adsorbent
L
SMAX
LL
LSMAXS
Solution a) Adsorption capacity of the column
( )( )( )( )( )
( )
( )( ) molcm
molcm
cm
mol
L
mol
L
molL
mol
cm
mol
C
cmcmcm
C
L
molC
cmmL
cmD
S
S
L
33
63
36
55
53
6
32
5
10712.41045.196340.01:column theofCapacity
104102109.1
102108.7
5.19631002
5column theof Volume
column theof volume1column theofCapacity
adsorbenton ion concentrat soluteadsorbent of volumecolumn theofCapacity
102
1001
5
−−
−
−−
−−
−
×=
×−
×=×+×
×
×
=
=
=
−==
×=
===
π
ε
b) Mean retention time
( )
( )( )
min574.1154359.9740.0
40.0111
min324.127
100
4359.97102
097436.0
102109.1
109.1108.7102
min324.127
2
540.0
min1000
1
min1000
min1
100
11
3
3
3
3
35'
3255
53
6
5'
2'
2
3
3
3
'
=
−
+=
=
×
=
×+×
×
×
=
×
+=
+
=
=
==
=
==
=
−+=
−
−−
−−
−
ADSORBENT
SOLUTION
ADSORBENT
SOLUTION
ADSORBENT
SOLUTION
LL
LSMAX
LL
LSMAX
LL
i
adsorbent
Li
cm
cm
cm
gcm
cmt
cm
cm
L
molf
cm
L
L
mol
L
mol
L
mol
cm
mol
L
molf
CK
KC
CK
CC
dC
dCf
cm
cm
cm
A
FU
cm
g
cmLF
cmL
CfU
Lt
πε
ρ
εερ
c) Number of columns to recover product in 10000 L
( )columns 4344.42
10712.4
10210000columns ofNumber
3
5
≈=×
×
= −
−
mol
L
molL
2. Travel distance of solutes A and B in a chromatographic column. Information
Solution a) Position of each band in the column
( )
mlV
k
k
k
k
BAi
ck
ckcf
B
B
A
A
ii
iii
50
6cmA
0.35
150ml volumeBed
3000mg3gadsorbent of mass
liquid solute/ml mg 0.02
adsorbent gadsorbed/m B solute mg 0.05
liquid solute/ml mg 0.1
adsorbent gadsorbed/mA solute mg 0.2
,
m
2
2
1
2
1
2
1i
=∆=
==
======
=+
==
ε
( )( )
( )( )
( )
( )cm
mg
ml
ml
mg
mX
cm
mg
ml
ml
mg
mlX
ml
mgmgM
mg
mlmgf
mg
mlmgf
ck
kk
ck
ck
dc
dCf
M
CMfA
VX
adsorbent
B
adsorbent
A
adsorbent
adsorbent
ii
ii
ii
ii
iL
L
77494.2
204082.01335.06cm
50
21655.1
5.01335.06cm
50
13
.35-1
1503000
204082.0ml
B 05.0
BFor
5.0ml
A 1.0
AFor
(mg/ml)column of eunit volumper adsorbent ofamount theis
2
2
'
'
22
21
2
1'
'
=
+
=∆
=
+
=∆
=
=
=
=
+=
+
=
+∆=∆
ε
b)
117.0
ml
B 05.0
051.0
ml
B 1.0
438.0
'
'
=
+
=
=
+
=
=∆∆
=
mgfM
R
mgfM
R
X
X
L
L
fB
fA
B
A
B
A
ε
ε
ε
ε
3. Determining Time to Elute Solutes A and B from a Chromatographic Column Information
h
cmF
cmV
cmV
cmV
K
K
column
i
DB
DA
3
3
3
30
100
60
30
20
15.0
5.0
=
=
=
=
==
Solution Estimate exit time of A and B
( )( )
( )( )h
h
cm
cmcmt
h
h
cm
cmcmt
F
VKVt
B
A
iD
245.0
100
3015.020
35.0
100
305.020
3
33
3
33
0
=+=
=+=
+=
