Debye–Hückel equation

D&H note that the Guldberg–Waage formula for electrolyte species in chemical reaction equilibrium in classical form is[2]^: 221

$${\displaystyle \prod _{i=1}^{s}x_{i}^{\nu _{i}}=K,}$$

where

• ${\textstyle \prod }$ is a notation for multiplication,
• ${\displaystyle i}$ is a dummy variable indicating the species,
• ${\displaystyle s}$ is the number of species participating in the reaction,
• ${\displaystyle x_{i}}$ is the mole fraction of species ${\displaystyle i}$,
• ${\displaystyle \nu _{i}}$ is the stoichiometric coefficient of species ${\displaystyle i}$,
• K is the equilibrium constant.

D&H say that, due to the "mutual electrostatic forces between the ions", it is necessary to modify the Guldberg–Waage equation by replacing ${\displaystyle K}$ with ${\displaystyle \gamma K}$, where ${\displaystyle \gamma }$ is an overall activity coefficient, not a "special" activity coefficient (a separate activity coefficient associated with each species)—which is what is used in modern chemistry as of 2007.

The relationship between ${\displaystyle \gamma }$ and the special activity coefficients ${\displaystyle \gamma _{i}}$ is[2]^: 248

$${\displaystyle \log(\gamma )=\sum _{i=1}^{s}\nu _{i}\log(\gamma _{i}).}$$

D&H use the Helmholtz and Gibbs free entropies ${\displaystyle \Phi }$ and ${\displaystyle \Xi }$ to express the effect of electrostatic forces in an electrolyte on its thermodynamic state. Specifically, they split most of the thermodynamic potentials into classical and electrostatic terms:

$${\displaystyle \Phi =S-{\frac {U}{T}}=-{\frac {A}{T}},}$$

where

• ${\displaystyle \Phi }$ is Helmholtz free entropy,
• ${\displaystyle S}$ is entropy,
• ${\displaystyle U}$ is internal energy,
• ${\displaystyle T}$ is temperature,
• ${\displaystyle A}$ is Helmholtz free energy.

D&H give the total differential of ${\displaystyle \Phi }$ as[2]^: 222

$${\displaystyle d\Phi ={\frac {P}{T}}\,dV+{\frac {U}{T^{2}}}\,dT,}$$

where

• ${\displaystyle P}$ is pressure,
• ${\displaystyle V}$ is volume.

By the definition of the total differential, this means that

$${\displaystyle {\frac {P}{T}}={\frac {\partial \Phi }{\partial V}},}$$

$${\displaystyle {\frac {U}{T^{2}}}={\frac {\partial \Phi }{\partial T}},}$$

which are useful further on.

As stated previously, the internal energy is divided into two parts:[2]^: 222

$${\displaystyle U=U_{k}+U_{e}}$$

where

• ${\displaystyle k}$ indicates the classical part,
• ${\displaystyle e}$ indicates the electric part.

Similarly, the Helmholtz free entropy is also divided into two parts:

$${\displaystyle \Phi =\Phi _{k}+\Phi _{e}.}$$

D&H state, without giving the logic, that[2]^: 222

$${\displaystyle \Phi _{e}=\int {\frac {U_{e}}{T^{2}}}\,dT.}$$

It would seem that, without some justification,

$${\displaystyle \Phi _{e}=\int {\frac {P_{e}}{T}}\,dV+\int {\frac {U_{e}}{T^{2}}}\,dT.}$$

Without mentioning it specifically, D&H later give what might be the required (above) justification while arguing that ${\displaystyle \Phi _{e}=\Xi _{e}}$, an assumption that the solvent is incompressible.

The definition of the Gibbs free entropy ${\displaystyle \Xi }$ is[2]^: 222–3

$${\displaystyle \Xi =S-{\frac {U+PV}{T}}=\Phi -{\frac {PV}{T}}=-{\frac {G}{T}},}$$

where ${\displaystyle G}$ is Gibbs free energy.

D&H give the total differential of ${\displaystyle \Xi }$ as[2]^: 222

$${\displaystyle d\Xi =-{\frac {V}{T}}\,dP+{\frac {U+PV}{T^{2}}}\,dT.}$$

At this point D&H note that, for water containing 1 mole per liter of potassium chloride (nominal pressure and temperature aren't given), the electric pressure ${\displaystyle P_{e}}$ amounts to 20 atmospheres. Furthermore, they note that this level of pressure gives a relative volume change of 0.001. Therefore, they neglect change in volume of water due to electric pressure, writing[2]^: 223

$${\displaystyle \Xi =\Xi _{k}+\Xi _{e},}$$

and put

$${\displaystyle \Xi _{e}=\Phi _{e}=\int {\frac {U_{e}}{T^{2}}}\,dT.}$$

D&H say that, according to Planck, the classical part of the Gibbs free entropy is[2]^: 223

$${\displaystyle \Xi _{k}=\sum _{i=0}^{s}N_{i}(\xi _{i}-k_{\text{B}}ln(x_{i})),}$$

where

• ${\displaystyle i}$ is a species,
• ${\displaystyle s}$ is the number of different particle types in solution,
• ${\displaystyle N_{i}}$ is the number of particles of species i,
• ${\displaystyle \xi _{i}}$ is the particle specific Gibbs free entropy of species i,
• ${\displaystyle k_{\text{B}}}$ is the Boltzmann constant,
• ${\displaystyle x_{i}}$ is the mole fraction of species i.

Species zero is the solvent. The definition of ${\displaystyle \xi _{i}}$ is as follows, where lower-case letters indicate the particle specific versions of the corresponding extensive properties:[2]^: 223

$${\displaystyle \xi _{i}=s_{i}-{\frac {u_{i}+Pv_{i}}{T}}.}$$

D&H don't say so, but the functional form for ${\displaystyle \Xi _{k}}$ may be derived from the functional dependence of the chemical potential of a component of an ideal mixture upon its mole fraction.[4]

D&H note that the internal energy ${\displaystyle U}$ of a solution is lowered by the electrical interaction of its ions, but that this effect can't be determined by using the crystallographic approximation for distances between dissimilar atoms (the cube root of the ratio of total volume to the number of particles in the volume). This is because there is more thermal motion in a liquid solution than in a crystal. The thermal motion tends to smear out the natural lattice that would otherwise be constructed by the ions. Instead, D&H introduce the concept of an ionic atmosphere or cloud. Like the crystal lattice, each ion still attempts to surround itself with oppositely charged ions, but in a more free-form manner; at small distances away from positive ions, one is more likely to find negative ions and vice versa.[2]^: 225

Electroneutrality of a solution requires that[2]^: 233

$${\displaystyle \sum _{i=1}^{s}N_{i}z_{i}=0,}$$

where

• ${\displaystyle N_{i}}$ is the total number of ions of species i in the solution,
• ${\displaystyle z_{i}}$ is the charge number of species i.

To bring an ion of species i, initially far away, to a point ${\displaystyle P}$ within the ion cloud requires interaction energy in the amount of ${\displaystyle z_{i}q\varphi }$, where ${\displaystyle q}$ is the elementary charge, and ${\displaystyle \varphi }$ is the value of the scalar electric potential field at ${\displaystyle P}$. If electric forces were the only factor in play, the minimal-energy configuration of all the ions would be achieved in a close-packed lattice configuration. However, the ions are in thermal equilibrium with each other and are relatively free to move. Thus they obey Boltzmann statistics and form a Boltzmann distribution. All species' number densities ${\displaystyle n_{i}}$ are altered from their bulk (overall average) values ${\displaystyle n_{i}^{0}}$ by the corresponding Boltzmann factor ${\displaystyle e^{-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}}}$, where ${\displaystyle k_{\text{B}}}$ is the Boltzmann constant, and ${\displaystyle T}$ is the temperature.[5] Thus at every point in the cloud[2]^: 233

$${\displaystyle n_{i}={\frac {N_{i}}{V}}e^{-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}}=n_{i}^{0}e^{-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}}.}$$

Note that in the infinite temperature limit, all ions are distributed uniformly, with no regard for their electrostatic interactions.[2]^: 227

The charge density is related to the number density:[2]^: 233

$${\displaystyle \rho =\sum _{i}z_{i}qn_{i}=\sum _{i}z_{i}qn_{i}^{0}e^{-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}}.}$$

When combining this result for the charge density with the Poisson equation from electrostatics, a form of the Poisson–Boltzmann equation results:[2]^: 233

$${\displaystyle \nabla ^{2}\varphi =-{\frac {\rho }{\varepsilon _{r}\varepsilon _{0}}}=-\sum _{i}{\frac {z_{i}qn_{i}^{0}}{\varepsilon _{r}\varepsilon _{0}}}e^{-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}}.}$$

This equation is difficult to solve and does not follow the principle of linear superposition for the relationship between the number of charges and the strength of the potential field. It has been solved by the Swedish mathematician Thomas Hakon Gronwall and his collaborators physicical chemists V. K. La Mer and Karl Sandved in a 1928 article from Physikalische Zeitschrift dealing with extensions to Debye–Huckel theory, which resorted to Taylor series expansion.

However, for sufficiently low concentrations of ions, a first-order Taylor series expansion approximation for the exponential function may be used (${\displaystyle e^{x}\approx 1+x}$ for ${\displaystyle 0<x\ll 1}$) to create a linear differential equation (Hamann, Hamnett, and Vielstich. Electrochemistry. Wiley-VCH. section 2.4.2). D&H say that this approximation holds at large distances between ions,[2]^: 227 which is the same as saying that the concentration is low. Lastly, they claim without proof that the addition of more terms in the expansion has little effect on the final solution.[2]^: 227 Thus

$${\displaystyle -\sum _{i}{\frac {z_{i}qn_{i}^{0}}{\varepsilon _{r}\varepsilon _{0}}}e^{-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}}\approx -\sum _{i}{\frac {z_{i}qn_{i}^{0}}{\varepsilon _{r}\varepsilon _{0}}}\left(1-{\frac {z_{i}q\varphi }{k_{\text{B}}T}}\right)=-\left(\sum _{i}{\frac {z_{i}qn_{i}^{0}}{\varepsilon _{r}\varepsilon _{0}}}-\sum _{i}{\frac {z_{i}^{2}q^{2}n_{i}^{0}\varphi }{\varepsilon _{r}\varepsilon _{0}k_{\text{B}}T}}\right).}$$

The Poisson–Boltzmann equation is transformed to[2]^: 233

$${\displaystyle \nabla ^{2}\varphi =\sum _{i}{\frac {z_{i}^{2}q^{2}n_{i}^{0}\varphi }{\varepsilon _{r}\varepsilon _{0}k_{\text{B}}T}},}$$

because the first summation is zero due to electroneutrality.[2]^: 234

Factor out the scalar potential and assign the leftovers, which are constant, to ${\displaystyle \kappa ^{2}}$. Also, let ${\displaystyle I}$ be the ionic strength of the solution:[2]^: 234

$${\displaystyle \kappa ^{2}=\sum _{i}{\frac {z_{i}^{2}q^{2}n_{i}^{0}}{\varepsilon _{r}\varepsilon _{0}k_{\text{B}}T}}={\frac {2Iq^{2}}{\varepsilon _{r}\varepsilon _{0}k_{\text{B}}T}},}$$

$${\displaystyle I={\frac {1}{2}}\sum _{i}z_{i}^{2}n_{i}^{0}.}$$

So, the fundamental equation is reduced to a form of the Helmholtz equation:[6]

$${\displaystyle \nabla ^{2}\varphi =\kappa ^{2}\varphi .}$$

Today, ${\displaystyle \kappa ^{-1}}$ is called the Debye screening length. D&H recognize the importance of the parameter in their article and characterize it as a measure of the thickness of the ion atmosphere, which is an electrical double layer of the Gouy–Chapman type.[2]^: 229

The equation may be expressed in spherical coordinates by taking ${\displaystyle r=0}$ at some arbitrary ion:[7][2]^: 229

$${\displaystyle \nabla ^{2}\varphi ={\frac {1}{r^{2}}}{\frac {\partial }{\partial r}}\left(r^{2}{\frac {\partial \varphi (r)}{\partial r}}\right)={\frac {\partial ^{2}\varphi (r)}{\partial r^{2}}}+{\frac {2}{r}}{\frac {\partial \varphi (r)}{\partial r}}=\kappa ^{2}\varphi (r).}$$

The equation has the following general solution (keep in mind that ${\displaystyle \kappa }$ is a positive constant):[2]^: 229

$${\displaystyle \varphi (r)=A{\frac {e^{-{\sqrt {\kappa ^{2}}}r}}{r}}+A'{\frac {e^{{\sqrt {\kappa ^{2}}}r}}{2r{\sqrt {\kappa ^{2}}}}}=A{\frac {e^{-\kappa r}}{r}}+A''{\frac {e^{\kappa r}}{r}}=A{\frac {e^{-\kappa r}}{r}},}$$

where ${\displaystyle A}$, ${\displaystyle A'}$, and ${\displaystyle A''}$ are undetermined constants

The electric potential is zero at infinity by definition, so ${\displaystyle A''}$ must be zero.[2]^: 229

In the next step, D&H assume that there is a certain radius ${\displaystyle a_{i}}$, beyond which no ions in the atmosphere may approach the (charge) center of the singled out ion. This radius may be due to the physical size of the ion itself, the sizes of the ions in the cloud, and any water molecules that surround the ions. Mathematically, they treat the singled out ion as a point charge to which one may not approach within the radius ${\displaystyle a_{i}}$.[2]^: 231

The potential of a point charge by itself is

$${\displaystyle \varphi _{\text{pc}}(r)={\frac {1}{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {z_{i}q}{r}}.}$$

D&H say that the total potential inside the sphere is[2]^: 232

$${\displaystyle \varphi _{\text{sp}}(r)=\varphi _{\text{pc}}(r)+B_{i}={\frac {1}{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {z_{i}q}{r}}+B_{i},}$$

where ${\displaystyle B_{i}}$ is a constant that represents the potential added by the ionic atmosphere. No justification for ${\displaystyle B_{i}}$ being a constant is given. However, one can see that this is the case by considering that any spherical static charge distribution is subject to the mathematics of the shell theorem. The shell theorem says that no force is exerted on charged particles inside a sphere (of arbitrary charge).[8] Since the ion atmosphere is assumed to be (time-averaged) spherically symmetric, with charge varying as a function of radius ${\displaystyle r}$, it may be represented as an infinite series of concentric charge shells. Therefore, inside the radius ${\displaystyle a_{i}}$, the ion atmosphere exerts no force. If the force is zero, then the potential is a constant (by definition).

In a combination of the continuously distributed model which gave the Poisson–Boltzmann equation and the model of the point charge, it is assumed that at the radius ${\displaystyle a_{i}}$, there is a continuity of ${\displaystyle \varphi (r)}$ and its first derivative. Thus[2]^: 232

$${\displaystyle \varphi (a_{i})=A_{i}{\frac {e^{-\kappa a_{i}}}{a_{i}}}={\frac {1}{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {z_{i}q}{a_{i}}}+B_{i}=\varphi _{\text{sp}}(a_{i}),}$$

$${\displaystyle \varphi '(a_{i})=-{\frac {A_{i}e^{-\kappa a_{i}}(1+\kappa a_{i})}{a_{i}^{2}}}=-{\frac {1}{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {z_{i}q}{a_{i}^{2}}}=\varphi _{\text{sp}}'(a_{i}),}$$

$${\displaystyle A_{i}={\frac {z_{i}q}{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {e^{\kappa a_{i}}}{1+\kappa a_{i}}},}$$

$${\displaystyle B_{i}=-{\frac {z_{i}q\kappa }{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {1}{1+\kappa a_{i}}}.}$$

By the definition of electric potential energy, the potential energy associated with the singled out ion in the ion atmosphere is[2]^{: 230 & 232}

$${\displaystyle u_{i}=z_{i}qB_{i}=-{\frac {z_{i}^{2}q^{2}\kappa }{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {1}{1+\kappa a_{i}}}.}$$

Notice that this only requires knowledge of the charge of the singled out ion and the potential of all the other ions.

To calculate the potential energy of the entire electrolyte solution, one must use the multiple-charge generalization for electric potential energy:[2]^{: 230 & 232}

$${\displaystyle U_{e}={\frac {1}{2}}\sum _{i=1}^{s}N_{i}u_{i}=-\sum _{i=1}^{s}{\frac {N_{i}z_{i}^{2}}{2}}{\frac {q^{2}\kappa }{4\pi \varepsilon _{r}\varepsilon _{0}}}{\frac {1}{1+\kappa a_{i}}}.}$$