236 Chapter 7. Entropic forces at work[[Student version, January 17, 2003]]
This expression has the feature thatV(0) = 0, so we need not add any extra constant toV.We
now check if we can choose values for the constantsBandx 0 in such a way that the proposed
solution solves the Poisson–Boltzmann equation, then choose a value for the parameterc 0 in that
equation so that our solution also satisfies the boundary condition (Equation 7.24).
SubstitutingBln(1 + (x/x 0 )) into Equation 7.23, we find that it works provided we takeB=2
andx 0 =1/
√
2 π Bc 0. With these choices, the boundary condition says 2/x 0 =4π B(σq/e). To
satisfy this condition, then, we must choosec 0 =2π B(σq/e)^2 .Our solution is then
V(x)=2
kBT
e ln(1 + (x/x^0 )) where x^0 =(2π Bσq/e)− (^1). (7.25)
Your Turn 7f
Find the equilibrium concentration profilec+(x)awayfrom the surface. Check your answer by
calculating the total surface density of counterions,
∫∞
0 dxc+(x)and verifying that the whole
system is electrically neutral.The solution you just found is sometimes called the “Gouy–Chapman layer.” It’s appropriate in
the neighborhood of a flat, charged surface in pure water.^7 Let us extract some physical conclusions
from the math.
First, we see from Your Turn 7f that indeed a diffuse layer forms, with thickness roughlyx 0 .As
argued physically in Section 7.4.1, the counterions are willing to pay some electrostatic potential
energy in order to gain entropy. More precisely, the counterions pull some thermal energy from
their environment to make this payment. They can do this because doing so lowers the entropic
part of their free energy more than it raises the electrostatic part. If we could turn off thermal
motion, that is sendT→0, the energy term would dominate and the layer would collapse. We see
this mathematically from the observation that then the Bjerrum length would go to infinity and
x 0 →0.
How much electrostatic energy must the counterions pay to dissociate from the planar surface?
Wecan roughly think of the layer as a planar sheet of charge hovering at a distancex 0 from
the surface. When two sheets of charge are separated we have a parallel-plate capacitor. Such a
capacitor, with areaA,stores electrostatic energyE=qtot^2 /(2C). Hereqtotis the total charge
separated; for our case it’sσqA.The capacitance of a parallel-plate capacitor is given by
C=εA/x 0. (7.26)Combining the preceding formulas gives an estimate for the density of stored electrostatic energy
perunit area for an isolated surface in pure water:
E/(area)≈kBT(σq/e). electrostatic self-energy, no added salt (7.27)That makes sense: The environment is willing to give up aboutkBTof energy per counterion. This
energy gets stored in forming the diffuse layer.
Is it a lot of energy? A fully dissociating bilayer membrane can have one unit of charge per lipid
headgroup, or roughly|σq/e|=0. 7 nm−^2 .Aspherical vesicle of radius 10μmthen carries stored
(^7) T 2 Or more realistically, a highly charged surface in a salt solution whose concentration is low enough; see
Section 7.4.3′on page 250 below.