Chemistry 434/534
Final Exam, 1996.
G. Marangoni
December 12, 1996.
Answer the following questions in the time allotted. CAREFULLY FOLLOW THE
INSTRUCTIONS IN EACH QUESTION.
Cheat sheets are permissible. There are no tricks on this exam, so don’t look for any.
Marks for each question are indicated in the [ ].
Chemistry 434 Students are to write the 3 hr. written section ONLY. Chemistry 534
students are to complete this section as well as a take home section.
Question
1
2
3
4
5
6
7
8
9
10
Total
Marks
/14
/10
/8
/14
/5
/8
/14
/12
/10
/10
/105
Chemistry 434/534 Final Exam, 1996.
1.
2
Given the following data on interfacial tensions (mN/m) (note: VA = valeric acid)
[14 marks]
Interface
air/water
 /(mN m-1)
72
Interface
Hg/water
 /(mN m-1)
415
air/VA
air/hexane
air/Hg
25
18
485
Hg/VA
Hg/hexane
Water/hexane
329
378
50
a)
b)
c)
d)
e)
f)
Will hexane spread at the air/water surface?
Will hexane spread at the air/Hg surface?
Will hexane spread at the Hg/water interface?
Will VA spread at the air/water interface?
Will VA spread at the Hg/H2O interface?
If the answer to e) is yes, what will be the orientation of VA at the
Hg/water interface? (note: assume that d (VA) = d (H2O) = 22 mN/m).
a)


S   H 2O   Hex   Hex / H 2O  72  18  50
4
mN
m
mN
 spontaneou s spreading
m
b)
S   Hg   Hex / Hg   Hex   485  378  18
 89
mN
m
mN
 spontaneou s spreading
m
c)


S   Hg / H 2O   Hex / Hg   Hex / H 2O  415  378  50
 13
mN
 non  spontaneous spreading
m
mN
m
Chemistry 434/534 Final Exam, 1996.
d)

3

 2
S   H 2O   VA / H 2O   VA  need to calculate  VA / H 2O
 VA / H 2O   VA   H 2O

 72  25  222 x 22
1/ 2
 S  72  53  25
 6
d
VA
 dH 2O

1/ 2
mN
 mN
 53
m
m
mN
m
mN
 non  spontaneous spreading
m
e)


S   Hg / H 2O   VA / Hg   VA / H 2O  415  329  53
 33
f)
mN
m
mN
 spontaneou s spreading
m
Note since the VA/H2O is significantly smaller than VA/Hg, it is quite likely that the
orientation of the VA at the water/Hg interface is one where the carboxylic acid
groups are in contact with water, but the chain is essentially flat at the interface
(i.e., since the interfacial energy between VA/Hg is large, having the chain in
contact with the Hg significantly increases the Gibbs energy of the system).
Chemistry 434/534 Final Exam, 1996.
2.
4
The Gibbs energy change for micelle formation is generally written in the form for
ionic 1:1 surfactants [10 marks]
 mic G o  2RT ln( X cmc )
a)
b)
a)
Use the following equation derived from the data by Burrows et al.i to
estimate the thermodynamic parameters of micelle formation at 298.15 K
and 308.15 K. Comment on the results.
ln( X cmc )  0.6401  0.05565 T  9.49x10 5 T 2
One may naively expect that the process of micelle formation would be
accompanied by a decrease in entropy (i.e., micS is negative). Is this
what we observe? Explain.
i)
Burrows, J.C.; Flynn, D.J.; Kutay, S.M.; Leriche, T.G.; Marangoni,
D.G. Langmuir 1995, 11, 3388-3394.
 mic G o  2RT ln( X cmc )  2R  0.6401T  0.05565 T 2  9.49x10 5 T 3 
  mic G o 
o
o
5
2

   mic S   mic S  2R 0.6401  0.1130 T  28.47 x10 T 
 T  p
 mic H o   mic G o  T mic So
Temp. /K
298
308
298
308
298
308
Property
micG
micG
micH
micH
micS
micS
-43.62 kJ/mole
-44.98 kJ/mole
-1.34 kJ/mole
-4.15 kJ/mole
142 J/(K mole)
132 J/(K mole)
Comment:
even though the Gibbs energy change is essentially invariant with
temperature, there are large changes in the value of the enthalpy and entropy of
micelle formation and these changes are compensatory in nature.
b)
For the surfactant, we would expect a large decrease in entropy as we transfer the
monomer from the bulk of the solution to the aggregated state (the micellar
interior). However, since we see large, positive entropy changes upon micelle
formation, the partial molar entropy change for the water must be very large and
positive. This is consistent with the dissolution of the surfactant molecules in
water having an ordering effect on the water molecules (i.e., the ‘iceberg effect’).
Chemistry 434/534 Final Exam, 1996.
3.
a)
5
Calculate the thickness of the diffuse double layer for a negatively charged solid
in contact with aqueous solutions of the following concentrations at 25oC. [8
marks]
a)
0.010 M KCl;
b)
0.0010 M KCl;
c)
0.0010 M K2SO4;
d)
0.0010 M MgCl2.
 
1/ 2
  3.29 x10 9 cZ 2 m 1
1
1
 Debye Length 
9

3.29x10 0.010Mx12


1/ 2
m  3.0x10 9 m
b)
1
1

9
 3.29x10 0.0010Mx12

m  9.6x10 9 m
1
1

9
 3.29x10 0.0020Mx12

m  6.8x10 9 m
1
1

 3.29x10 9 0.0010Mx 2 2


1/ 2
c)

1/ 2
d)

1/ 2
m  4.8x10 9 m
Chemistry 434/534 Final Exam, 1996.
6
4.
In an electrophoresis experiment, spherical colloidal particles of diameter 0.5 m
are dispersed in a solution of 0.10 mol/L NaCl. If the particles are observed to
cover a distance of 120 m in 5.0 s, under a potential gradient of 10.0 V/cm,
calculate: [14 marks]
a)
the electrophoretic mobility of the particles;
b)
an estimate of the zeta potential () of the particles;
c)
an estimate of the charge density at the surface of shear;
d)
an estimate of the error in the electrophoresis measurement arising from
the Brownian displacement of the particle that occurs during the
experiment.
a)
check a

a  3.29x10 9 0.10Mx12
b)

1/ 2
m 1 x 0.25x10 6 m  260
 Smoluchows ki equation is valid

120 x10 6 mx 5.0s
m2
uE  E 
 2.4x10 8
V
cm
E
Vs
10.0
x100
cm
m
Using the Smoluchowski equation
m
0.891x10 3 Pa s x 2.4x10 8
u E
s  0.0308V  30.8mV


F
78.5 o
78.5x8.854x10 12
m
c)
C2 
o
(note : assume  d   )
d
o
 C 2 x   o

C
1/ 2 F
 o  2.280.10
x 0.0308V  0.0222 2
2
m
m
C2 
d)
x  2Dt 
1/ 2
J
1.381x10  23 298K
2
k BT
13 m
K
D


9
.
81
x
10
6a 60.891x10 3 Pa s x 0.25x10 -6 m
s


m2
 x  2Dt    2 x 9.81x10 13
5.0s 
s


6
3.13x10 m
% error 
x100%  2.61%
120x10 6 m
1/ 2
1/ 2
 3.13x10 6 m
Chemistry 434/534 Final Exam, 1996.
5.
True or False. If false, explain why. [6 marks]
a)
All colloidal systems are thermodynamically unstable.
b)
All foams consist of a gas dispersed in a liquid.
c)
Electro-osmotic flow is related directly to the streaming potential
developed in the capillary.
a)
False. Micellar systems. Microemulsions, and a wide variety of
polymer/surfactant complexes all form spontaneously (G < 0 kJ/mole).
False. An excellent example is Styrofoam (a gas dispersed in solid).
True
b)
c)
7
Chemistry 434/534 Final Exam, 1996.
6.
8
DO EITHER A OR B
What do we mean by the terms mass average and number average
molecular weights of a colloidal system? How does this define the
polydispersity of the colloidal system? [8 marks]
A certain polymer has 15% of its molecules having a mass of 10 kg/mol,
20% of it molecules having a mass of 20 kg/mol, 50% of its molecules
with a mass of 25 kg/mol, and 15% of its molecules having a mass of 35
kg/mol. Is the system monodisperse? Explain. [8 marks]
a)
b)
a)
M r mass  


n M 
 n j M 2r , j
M r number  
j
j
j

 n jM r, j
j
r, j
n j 

j
Note: the ratio of the mass average to the number averaged molecular weight is a
measure of the polydispersity of the colloidal system.
b)
M r mass  


n M 
 n jM
j
j
 25.4 kg
j
2
r, j
r, j
2
2
2
mole
M r number  

 n jM r, j
j
n j 
j
 23 kg
2
  20x 20 kg
  50 x 25 kg
  15x 35 kg

15x10 kg
mole
mole
mole
mole 








  20x 20 kg
  50 x 25 kg
  15x 35 kg

15x10 kg
mole 
mole 
mole 
mole 





  20x 20 kg
  50x 25 kg
  15x 35 kg

15x10 kg
mole 
mole 
mole 
mole 





100
mole
M r mass 
25.4
 polydisper sity index 
 1.09
M r number 
23
 monodisper se system
Chemistry 434/534 Final Exam, 1996.
7.
9
A protein having a molecular weight of 60 kg/mol and a sedimentation coefficient
of 4.1 Svedbergs (note: 1 Svedberg = 10-13 / s) has a density of 1.31 g/cm3. Given
that the density and viscosity of water are 0.997 g/cm3 and 0.890 x 10-3 kg/(m s)
at 298.2 K, respectively, calculate [14 marks]
a)
the frictional coefficient of the protein from the sedimentation coefficient;
b)
the ideal frictional coefficient, fo;
c)
can we deduce something about the shape of the protein in solution?
a)
S
MD1  
SRT
D

RT
M 1  
D  7.09 x10 11 m
Df  k B T  f 
J
298K
K mole
kg
60
x 1  0.997
1.31
mole
4.1x10 13 / s 8.314


2
s
1.381x10  23 J
x 298K
K
 5.81x10 11 kg
s
11 m 2
7.09 x10
s
b)
f o  6a  assume the protein particles are spherical
1.31 g
cm 3  2.183x10 5 mole
cm 3
60 x10 3 g
mole
3
4.580x10 5 cm
mole  7.61x10  20 cm 3
V

23
molecule
molecule
6.022x10 mole 1
4
3
V  a 3  7.61x10  20 cm 3 x
 a3
3
4
a  2.63x10 7 cm  2.63x10 9 m
kg
9
fo  6 a  6 0.891x10  3
ms x 2.63x10 m
kg
fo  4.42 x10 11
s
note :
c)
f
5.81x10 11

 1.31  oblate spheroid
f o 4.42x10 11
Chemistry 434/534 Final Exam, 1996.
8.
10
A Cahn DCA 312 was used to scan the surface of a new polymer that may be used
to make extended-wear contact lenses. The experiment was carried out using
artificial tears as the wetting liquid. The DCA force-distance profile is given
below. [12 marks]
a)
Does the artificial tears solution wet the surface of the polymer?
b)
Estimate the contact angle hysteresis for the wetting liquid.
c)
Given that the surface tension of the wetting liquid is 45 mN/m, calculate
the Wimm and the Gibbs energy of wetting of the polymer.
d)
Will contact lenses made from this new polymer be comfortable enough
for extended wear lenses? Explain briefly.
Wetting Experiment for Artifical Tears
Solution on a Contact Lens Polymer
Force / mg
450. 0
rec = 12.52
300. 0
150. 0
avd = 46.56
0.0
0
5
10
15
Immersion Depth / mm
a)
b)
c)
d)
Note, since the advancing contact angle for the tears solution on the polymer is
>> 0, this solution does not wet the polymer well.
 =adv - rec = 46.56 - 12.52 =34.04
Wimm = LG Cos (adv) = 45 mN/m x cos (46.56) = 30.94 mN/m
immG = -Wimm x A = - 0.0309 J/m2
No! For a comfortable contact lens, the contact angle should be as small as
possible (ideally 0). Since the artificial tears solution does not wet the polymer
well, this polymer would not be a good candidate for contact lens manufacturing.
Chemistry 434/534 Final Exam, 1996.
9.
11
It is well known that the electrical double layer contributes to the stability of
inorganic colloids like alumina and zirconia. The following curves represent the
internal energy of attraction, repulsive, and the overall stabilisation energy of the
charged colloid as a function of interparticle separation. [10 marks]
- + - + - - +
- + - + - - +
+ - + + + + + + - + + + + -+ -- +
+ - +
+ +
+
+
+
+ + - +
- +
+ - + + + +
+ - + + + +
- - +
- - +
+
+
+
+
- r
- -
b
E
c
r
a
a)
b)
c)
Identify the curves represented by a, b, and c.
How will the addition of electrolyte to the solution affect the shape of the
curves. Explain.
Identify the region of maximum stability for the charged colloid.
a)
Curve a 
Curve b 
Curve c 
b)
The addition of electrolyte has little effect on the shape of the attractive curve,
however, the repulsive curve is changed in shape and moves closer to the origin.
The net effect is that the attractive forces beginning to dominate. With sufficient
electrolyte addition, the colloid will precipitate.
attractive curve
repulsive curve
overall potential energy curve
Chemistry 434/534 Final Exam, 1996.
10.
12
DO EITHER A OR B
a)
At 298 K, a solution of polystyrene in methyl ethyl ketone solution has a
(dn/dc)o = 0.231 mL/g, I/Io = 0.9865, and no =1.377 at a wavelength of
436 nm. Calculate [10 marks]
i)
the scattering constant for this solution (b = 1.00 cm).
ii)
If 1.00 L of the solution contains 1.07 g of polymer, what is the
molar mass of the polystyrene?
b)
Shibita et al.(1) determined the  -A curves for a series of perfluorinated
carboxylic acids (FC-10  FC18) and tetradecanoic acid in 0.10 mol/L
HCl at 298.2 K. [10 marks]
i)
Identify the transitions indicated on the curves for the HC14
(tetradecanoic acid) and perfluorododecanoic acid (FC-12).
ii)
Calculate an estimate of the molar mass of the FC-12 using the
surface pressure data at 1.0 nm2, given that 9.91 x 10-4 g of acid
was spread per m2.
iii)
Give a plausible explanation as to why the transition to the
condensed film occur at higher surface areas for the perfluorinated
acids versus tetradecanoic acid.
(1)
Shibata, O.; Yamamoto, S.K.; Lee, S.; Sugihara, G.
J Colloid Interface Sci. 1996, 184, 201-208.
Chemistry 434/534 Final Exam, 1996.
13
a)
Kc 1
I

note   0.9865  e  l
M
R 90o
Io
  ln 0.9865
R 90o 
 1.359 m 1
3
 0.0811 m -1
16
2
2
3
2 2 1.377 
 dn 
 5 mole m


m

231

9
.
178
x
10
 
4 
kg 
kg 2
 dc 
6.022 x10 23 mole 1 436 x10 9 m 
0.0811/m

 826 kg
2
Kc
mole
kg 
mole m 
9.177x10 -5
1
.
07


kg 2 
m3 
K
2 2 n o2
N A 4
M
R 90o
b)
0.01 m
i)
2
2


Transitions
(1)
(2)
(3)
gaseous  liquid expanded
liquid expanded  condensed
condensed  close-packed
ii)
g
1
m2
 9.91x10  4 2  A m  1.010 x10 4
Am
g
m
8.314 J
x 298.2K
RT
K mole
N
  0.0040
M

 0.614 kg
 614 g
2
m
mole
mole
A m
m
0.040 N x1.010 x10 4
m
g
A m M  RT from the curve
iii)

Larger electronegativity of the F atoms compared to the H atoms leads to
significant repulsions in the surface layer. Hence, we observe much larger than
expected film areas for the FC film vs. the HC film.
Have fun. Merry Christmas and Happy New Year.