CM 4710, Biochemical Processes
Homework #8
Fall 2007
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.
Problem 1
Problem 2
CM 4710, Biochemical Processes
Homework #7 Solution
Fall 2003
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
Homework #8
Fall 2003
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
Langmuir isotherm
CS =
C SMAX C L
K L + CL
Dextran will adsorb up to 7.8E - 6 mol/IG/cm^3 = C SMAX
K L = 1.9 × 10 −5 mol / L
g
cm 3
ε (void fraction ) = 0.40
Solution
a) Adsorption capacity of the column
D = 5cm
ρ adsorbent = 1
L = 1m = 100cm
mol
C L = 2 × 10 −5
L
Capacity of the column = (volume of adsorbent )(solute concentration on adsorbent )
Capacity of the column = (1 − ε )(volume of the column )(C S )
2
 5cm 
3
Volume of the column = π 
 (100cm ) = 1963.5cm
 2 
mol 

− 6 mol 
2 × 10 −5
 7.8 × 10

3 
mol
L 
cm 

CS =
= 4 × 10 −6
mol
mol
cm 3
1.9 × 10 −5
+ 2 × 10 −5
L
L
mol 

Capacity of the column : (1 − 0.40) 1963.5cm 3  4 × 10 −6
= 4.712 × 10 −3 mol
3 
cm 

(
)
b) Mean retention time

L 
1− ε  '
t=
1 + ρ
 f (C L )

Ui 
 ε 

L = 100cm
L
cm 3
F =1
= 1000
min
min
g
ρ adsorbent = 1 3
cm
cm 3
1000
F
cm
min
Ui =
=
= 127.324
2
εA
min
  5cm  

0.40 π 

  2  


f ' (C L ) =

C
K
 = SMAX L 2
 (K L + C L )
mol 

− 6 mol 
1.9 × 10 −5
 7.8 × 10

3 
L
mol  
L 
cm 
= 0.097436 3SOLUTION
=
2
L  
cm ADSORBENT
mol 
− 5 mol
+ 2 × 10 −5
1.9 × 10

L
L 

d
dC L

f '  2 × 10 −5

 C SMAX C L

 K L + CL
mol 
cm 3 SOLUTION

f '  2 × 10 −5
 = 97.4359 3
L 
cm ADSORBENT

cm 3 SOLUTION
100cm   g  1 − 0.40 

t=
1 + 1
 97.4359 3

cm   cm 3  0.40 
cm ADSORBENT
127.324
min

 = 115.574 min

c) Number of columns to recover product in 10000 L
Number of columns =
(10000 L ) 2 × 10 −5 mol 
L 

= 42.44 ≈ 43 columns
−3
4.712 × 10 mol
2. Travel distance of solutes A and B in a chromatographic column.
Information
k c
m i = f i (c ) = 1i i
k 2i + ci
i = A, B
k1 A = 0.2 mg solute A adsorbed/mg adsorbent
k 2 A = 0.1 mg solute/ml liquid
k1B = 0.05 mg solute B adsorbed/mg adsorbent
k 2 B = 0.02 mg solute/ml liquid
mass of adsorbent = 3g = 3000mg
Bed volume = 150ml
ε = 0.35
A = 6cm 2
∆V = 50ml
Solution
a) Position of each band in the column
∆X =
∆V
A ε + Mf ' (C L )
(
)
M is the amount of adsorbent per unit volume of column (mg/ml)
f ' (C L ) =
d
dci
 k1i ci

 k 2i + ci

k1i k 2i
 =
2
 (k 2i + ci )
For A
ml
 mg A 
f '  0.1
 = 0.5
ml 
mg adsorbent

For B
mg B 
ml

f '  0.05
 = 0.204082
ml 
mg adsorbent

3000mg
mg
= 13
M =
ml
 150 


 1 - .35 
50ml
∆X A =
= 1.21655cm




mg
ml



6cm 2  0.35 + 13
 0.5
mg adsorbent  
 ml 

50m
∆X B =
= 2.77494cm




mg
ml


 
6cm 2  0.35 + 13
 0.204082

ml
mg


adsorbent



b)
L A ∆X A
=
= 0.438
LB ∆X B
R fA =
R fB =
(
)
(
)
ε
  mg B  
ε + M  f '  0.1

ml  
 
= 0.051
ε
 
mg B  
ε + M  f '  0.05

ml  
 
= 0.117
3. Determining Time to Elute Solutes A and B from a Chromatographic Column
Information
K DA = 0.5
K DB = 0.15
V0 = 20cm 3
Vi = 30cm 3
Vcolumn = 60cm 3
cm 3
h
Solution
Estimate exit time of A and B
F = 100
V0 + K DVi
F
20cm 3 + (0.5) 30cm 3
= 0.35h
tA =
cm 3
100
h
3
20cm + (0.15) 30cm 3
= 0.245h
tB =
cm 3
100
h
t=
(
)
(
)