7.4. A repulsive interlude[[Student version, January 17, 2003]] 237
free energy≈ 4 π(10μm)^2 ×(0. 7 /nm^2 )kBTr≈ 109 kBT. It’s a lot! We’ll see how to harness this
stored energy in Section 7.4.5 below.
Forsimplicity, the calculations above assumed that a dissociating surface was immersed in pure
water. In real cells, on the contrary, the cytosol is anelectrolyte,orsalt solution. In this case, the
density of counterions at infinity is not zero, the counterions originally on the surface have less to
gain entropically by escaping, and so the diffuse layer will hug the surface more tightly than it does
in Equation 7.25. That is,
Increasing salt in the solution shrinks the diffuse layer. (7.28)T 2 Section 7.4.3′on page 250 solves the Poisson–Boltzmann equation for a charged surface in
asalt solution, arriving at the concept of the Debye screening length and making Equation 7.28
quantitative.
7.4.4 The repulsion of like-charged surfaces arises from compressing their ion clouds
their ion clouds
Now that we know what it’s like near a charged surface, we’re ready to go farther and compute an
entropicforcebetween charged surfaces in solution. Figure 7.8b shows the geometry. One may be
tempted to say, “Obviously two negatively charged surfaces will repel.” But wait: Each surface,
together with its counterion cloud, is an electricallyneutralobject! Indeed, if we could turn off
thermal motion the mobile ions would collapse down to the surfaces, rendering them neutral. Thus
the repulsion between like-charged surfaces can only arise as an entropic effect. As the surfaces get
closer than twice their Gouy–Chapman length, their diffuse counterion clouds get squeezed; they
then resist with an osmotic pressure. Here are the details.
Forsimplicity let’s continue to suppose that the surrounding water has no added salt, and
hence no ions other than the counterions dissociated from the surface.^8 This time we’ll measure
distance from the midplane betweentwosurfaces, which are located atx=±D.We’ll suppose
each surface has surface charge density−σq.Wechoose the constant inV so thatV(0) = 0, and
so the parameterc 0 =c+(0) is the unknown concentration of counterions at the midplane. V(x)
will then be symmetric about the midplane, so Equation 7.25 won’t work. Keeping the logarithm
idea, though, this time we tryV(x)=Aln cos(βx), whereAandβare unknown. Certainly this
trial solution is symmetric and equals zero at the midplane, wherex=0.
The rest of the procedure is familiar. Substituting the trial solution into the Poisson–Boltzmann
equation (Equation 7.23), gives thatA=2andβ=
√
2 π Bc 0 .The boundary condition atx=−D
is again Equation 7.24. Imposing the boundary conditions on our trial solution gives a condition
fixingβ:
4 π B(σq/e)=2βtan(Dβ). (7.29)
Given the surface charge density−σq,wesolve Equation 7.29 forβas a function of the spacing
2 D;then the desired solution is
V(x)=2ln cos(βx), or c+(x)=c 0 (cosβx)−^2. (7.30)(^8) T 2 This is not as restrictive as it sounds. Even in the presence of salt, our result will be accurate if the surfaces
are highly charged, since Section 7.4.3′on page 250 shows that in this case the Gouy-Chapman thickness is less than
the Debye screening length.