8.4. Self-assembly of amphiphiles[[Student version, January 17, 2003]] 281
Wecan interpret McBain’s results with a simplified model. Suppose that the soap he used,
potassium oleate, dissociates fully into potassium ions and oleate amphiphiles. The potassium ions
contribute to the osmotic pressure by the ideal (van ’t Hoff) relation. But the remaining oleate
amphiphiles will instead be assumed to be in thermodynamic equilibrium between individual ions
and aggregates ofNions.Nis an unknown parameter, which we will choose to fit the data. It will
turn out to be just a few dozen, justifying our picture of micelles as mesoscopic objects.
To work out the details, apply the Mass Action rule (Equation 8.16) to the reaction (N
monomers)(one aggregate). Thus we find that the concentrationc 1 of free monomers in solution
is related to that of micelles,cN,by
cN/(c 1 )N=Kˆeq, (8.29)
whereKˆeqis a second unknown parameter of the model. (Kˆeqequals the dimensionless equilibrium
constant for aggregation,Keq,divided by (c 0 )N−^1 .) The total concentration of all monomers is
thenctot=c 1 +NcN.
Example Find the relation between the total number of amphiphile molecules in solution,ctot,
and the number that remain unaggregated,c 1.
Solution:
ctot=c 1 +NcN=c 1(
1+NKˆeq(c 1 )N−^1)
. (8.30)
Wecould stop at this point, but it’s more meaningful to express the answer not in
terms ofKˆeq,but rather in terms of the CMC,c∗.Bydefinitionc∗is the value ofctot
at which half the monomers are free and half are assembled into micelles. In other
words, whenctot=c∗thenc 1 ,∗=NcN,∗=^12 c∗. Substituting into Equation 8.29
gives
( 1
2 N
c∗)( 1
2 c∗)−N
=Kˆeq. (8.31)Wenow solve to findNKˆeq=(
2 /c∗)N− 1
and substitute into Equation 8.30, findingctot=c 1 (1+(2c 1 /c∗)N−^1 ). (8.32)Once we have chosen values for the parametersNandc∗,wecan solve Equation 8.32 to getc 1
in terms of the total amount of surfactantctotstirred into the solution. Though this equation
has no simple analytical solution, we can understand its limiting behavior. At low concentrations,
ctot c∗,the first term dominates and we getctot≈c 1 :Essentially all the surfactants are loners.
But well above the CMC, the second term dominates and we instead getctot≈NcN;nowessentially
all the surfactants are accounted for by the micelles.
Wecan now find the osmotic pressure. The contribution from the Na+ions is simplyctotkBT
as usual. The contribution from the amphiphiles resembles Equation 8.32 with one key difference:
Each micelle counts as just one object, not asNobjects.
Your Turn 8h
Show that the total osmotic pressure relative to the value 2ctotkBTin this model is
1
2(
1+
1+N−^1 (2c 1 /c∗)N−^1
1+(2c 1 /c∗)N−^1)
. (8.33)
Touse this formula, solve Equation 8.32 numerically forc 1 and substitute into Equation 8.33 to