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In the derivation of the Poisson-Boltzmann equation, my textbook arrives at the expression

$$ \rho_i = c_+ z_+ F + c_-z_-F = c_+^oz_+^oFe^{-z_+e\phi_i/kT} + c_-^oz_-Fe^{-z_-e\phi_i/kT} $$

where $c_\pm$ is the concentration (of positive or negative ions), $z_\pm$ is the valence of the ions, $F$ is the Faraday constant ($=eN_A$) and $c_\pm^o$ is the concentration in the bulk. The rest is standard notation.

The book now writes that

Because the average electrostatic interaction energy is small compared with $kT$, we may write

$$ \rho_i = (c_+^o z_+ + c_-^oz_-)F - (c_+^oz_+^2 + c_-^oz_-^2)\frac{F^2\phi_i}{kN_AT} + \cdots $$

How do they know that $ze\phi \ll kT$?

To make an estimate, I find the electric potential of a proton (or an $\text{Na}^+$ ion) in vacuum at 1 nm away.

$$ V^{vac} = \frac{1}{4\pi\epsilon_0}\frac{+1e}{1nm} \approx 1.43 \text{V}. $$

In water ($\epsilon = \epsilon_0\epsilon_r = \epsilon_078$),

$$ V^{sol} = \frac{1}{4\pi78}\frac{+1e}{1nm} \approx 0.02 \text{V}. $$

So in solution a $\text{Cl}^-$ ion would have the energy

$$ w=(-1e)V^{sol} \approx -3.2 \cdot 10^{-21}J $$

while at room temperature $kT \approx 3.9 \cdot 10^{-21}J$.

Is this roughly the idea behind the above assumption?

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1 Answer 1

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I see four reasons to make this assumption:

  • The first one is simply that we do not know really how to deal with the non linear problem in general cases and therefore we linearize but I agree that it is not a good justification although this is most of the time the hidden reason why people use it

  • The Poisson-Boltzmann equation is mostly used in aqueaous solutions and therefore you have a factor 80 (owing to the dielectric constant of bulk water) that appears in all your potentials...this more or less ensures that a monovalent ion does not generate a so high potential energy with other monovalent ions. If you have free ions in solution, it means that the thermal energy was enough to begin with to unbind them from the groups there were bound to

  • It can be rigorously shown that at large distances from an ion at any valency, the generated potential decays exponentially with the distance (or rather like a Yukawa potential) because of the screening owing to the mobile charges in solution (this result holds in the non linear Poisson-Boltzmann case but also beyond the Poisson-Boltzmann theory). Incindentally, a Yukawa potential is what you can expect from the field generated by an isolated ion in solution within the linearized Poisson-Boltzmann theory. Also, if you are a bit familiar with liquid theory, a particular closure of the Orstein-Zernike equation leads to an infinite summation over tree-like diagrams that leads to a total correlation function between two point ions in solutions that decays as a Yukawa as well which tells you more or less that although you linearize, you still encapsulate important many body effects

  • Because of the last point, there is a huge litterature on charge renormalization of charged particles in solution as a function of their valency, size and salt concentration. In the context of the Poisson-Boltzmann theory, charge renormalization refers to a mapping from the non linear theory to the linearized one via the use of effective charge paramaters

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The last two points are a bit beyond my background, but I find your second point appealing. But do you agree more or less to my numerical example? –  TMOTTM Apr 30 '13 at 9:01
    
The comparison between a bare Coulomb interaction in a solvent and $k_B T$ is usually done for monovalent ions via the Bjerrum length $l_B = e^2/4\pi \epsilon k_B T$. For a ionic molecule (e.g. NaCl) to dissolve in solution you roughly need that the distance between the ions in the molecule to be bigger than $l_B$ in that solution, then $k_B T$ dominates the Coulomb attraction and your molecule dissolves. [to be continued] –  gatsu Apr 30 '13 at 9:10
    
[continuation] The electrostatic potential cannot really be estimated the way you did because there is screening due to other charges. You need also to keep in mind that within the Poisson-Boltzmann theory for a symmetric 1:1 electrolyte, the electrostatic potential is chosen to be strictly zero in bulk because we have an homogeneous liquid. Overall, it depends on the question you ask to your system. –  gatsu Apr 30 '13 at 9:29
    
Ok, so at $l_B$, $kT \gg z_ie\phi$. So could I use a screened potential $\phi = 1/(4\pi\epsilon) \frac{q}{r} e^{-r/l_B}$ instead to calculate the interaction? I really would like to see what the number for $z_i e \phi$ would be. –  TMOTTM Apr 30 '13 at 9:39
    
that's a good try but the typical screened potential generated by an ion of charge $q$ and radius $a$ is $\phi(r) \approx \frac{q e^{\kappa a}}{4\pi \epsilon(1+\kappa a)}\frac{e^{-\kappa r}}{r}$ where $\kappa = \sqrt{8 \pi l_B n_b}$ is the inverse screening length and $n_b$ is the salt concentration in molecule per unit volume. This is within a linearized approximation, but as I said in my reply this functional form will always be true far enough from the ion (the prefactor only could change). –  gatsu Apr 30 '13 at 10:30

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