Kinetics simulation of transmembrane transport of ions and molecules through a semipermeable membrane

Abstract

We have developed a model to study the kinetics of the redistribution of ions and molecules through a semipermeable membrane in complex mixtures of substances penetrating and nonpenetrating through a membrane. It takes into account the degree of dissociation of these substances, their initial concentrations in solutions separated by a membrane, and volumes of these solutions. The model is based on the assumption that only uncharged particles (molecules or ion pairs) diffuse through a membrane (and not ions as in the Donnan model). The developed model makes it possible to calculate the temporal dependencies of concentrations for all processing ions and molecules at system transition from the initial state to equilibrium. Under equilibrium conditions, the ratio of ion concentrations in solutions separated by a membrane obeys the Donnan distribution. The Donnan effect is the result of three factors: equality of equilibrium concentrations of penetrating molecules on each side of a membrane, dissociation of molecules into ions, and Le Chatelier’s principle. It is shown that the Donnan distribution (irregularity of ion distribution) and accordingly absolute value of the Donnan membrane potential increases if: (i) the nonpenetrating salt concentration (in one of the solutions) and its dissociation constant increases, (ii) the total penetrating salt concentration and its dissociation constant decreases, and (iii) the volumes ratio increases (between solutions with and without a nonpenetrating substance). It is shown also that only a slight difference between the degrees of dissociation of two substances can be used for their membrane separation.

Appendix

We will show that our kinetic model based on the assumption that only molecules (uncharged particles) penetrate through a membrane, and the Donnan model, based on the assumption that only ions can penetrate through a membrane, are equivalent. For this purpose, we use the “classical” thermodynamic approach and consider two states of the membrane system: the initial and equilibrium states.

In the initial state, the substance KB at concentration C₁ and infinitely large dissociation constant K₁, is located in the i-cell (i.e., only K⁺ and B⁻ ions are present) while salt KA with concentration C₂ and relatively small dissociation constant K₂ is located in the e-cell. It should be taken into account that K⁺ and A⁻ ions are in the i-cell simultaneously with the undissociated KA molecules. The cell volumes are equal to V_i = V_e.

To calculate the concentrations of ions and molecules in the e-cell, we designate z = [K⁺] = [A⁻] and write down an equation for the dissociation constant K₂ (taking into account Eq. (8)):

$$ {K}_2=\frac{z^2}{C_2-z} $$

(A1)

Solving the quadratic equation in z, we get:

$$ z=\frac{-{K}_2+\sqrt{K_2^2+4{K}_2{C}_2}}{2} $$

(A2)

Thereby the initial state of the system under investigation may be presented as:

[K⁺]_i = C₁ | [K⁺]_e = $\frac{-{K}_2+\sqrt{K_2^2+4{K}_2{C}_2}}{2}$.

[B⁻]_i = C₁ | [A⁻]_e = $\frac{-{K}_2+\sqrt{K_2^2+4{K}_2{C}_2}}{2}$

| [KA]_e = $\frac{2{C}_2+{K}_2-\sqrt{K_2^2+4{K}_2{C}_2}}{2}$

Firstly, we will calculate the equilibrium state using the Donnan approach, i.e. we assume that only K⁺ and A⁻ ions penetrate through the membrane.

As a result of permeable K⁺ and A⁻ ions passing through the membrane, some amount of salt moves from the e-cell into the i-cell, taking into account that the electroneutrality condition requires transferring equal amounts of K⁺ and A⁻ ions. As a result, the total salt concentration in the e-cell decreases by x and is then C₂ – x. In accordance with Eq. (A1) in which C₂ – x has to be written in the denominator instead of C₂, it is possible to calculate the concentrations of ions and molecules in the e-cell at equilibrium.

Some of the ions that penetrate into the i-cell form molecules whose concentration is denoted by y. Then the concentrations of K⁺ cations and A⁻ anions may be written as C₁ + x – y and x – y, respectively. The concentrations of ions and undissociated molecules are related by the following equation:

$$ {K}_2=\frac{\left({C}_1+x-y\right)\left(x-y\right)}{y} $$

(A3)

with which y can be calculated:

$$ y=\frac{\left({C}_1+{K}_2\right)+2x-\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}}{2} $$

(A4)

Thereby the equilibrium state of the system under investigation may be presented as:

$$ {\displaystyle \begin{array}{ll}{\left[{K}^{+}\right]}_i=\frac{C_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4x{K}_2}}{2}& \mid {\left[{K}^{+}\right]}_e=\frac{-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}}{2}\\ {}{\left[{A}^{-}\right]}_i=\frac{-{C}_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4x{K}_2}}{2}& \mid {\left[{A}^{-}\right]}_e=\frac{-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}}{2}\\ {}{\left[ KA\right]}_i=\frac{\left({C}_1+{K}_2\right)+2x-\sqrt{{\left({C}_1+{K}_2\right)}^2+4x{K}_2}}{2}& \mid {\left[ KA\right]}_e=\frac{2\left({C}_2-x\right)+{K}_2-\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}}{2}\\ {}{\left[{B}^{-}\right]}_i={C}_1& \mid \end{array}} $$

Now, let us assume that only undissociated KA molecules penetrate through the membrane like in our kinetic approach. In the process of transition of the system from the initial state to the equilibrium state, the total salt concentration in the e-cell will decrease by x, and these x moles of salt KA will move to the solution located in the i-cell. In this case, the concentrations of ions and molecules in the e-cell at equilibrium are as follows:

$$ {\left[{K}^{+}\right]}_e={\left[{A}^{-}\right]}_e=\frac{-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}}{2} $$

(A5)

$$ {\left[\mathrm{KA}\right]}_e=\frac{2\left({C}_2-x\right)+{K}_2-\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}}{2} $$

(A6)

Some KA molecules that have moved to the i-cell dissociate into ions. Let us denote the concentration of resulting ions as y. Then the concentrations of K⁺ cations and A⁻ anions may be written as C₁ + y and y, respectively The concentrations of ions and undissociated molecules are related by the following equation:

$$ {K}_2=\frac{y\left({C}_1+y\right)}{x-y} $$

(A7)

with which y can be calculated:

$$ y=\frac{-\left({C}_1+{K}_2\right)+\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}}{2} $$

(A8)

The resulting expression is the concentration of the A⁻ anion in i-cell ([A⁻]_i). Now, it is possible to calculate [K⁺]_i and [KA]_i:

$$ {\left[{K}^{+}\right]}_i={C}_1+y=\frac{C_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}}{2} $$

(A9)

$$ {\left[\mathrm{KA}\right]}_i=x-y=\frac{\left({C}_1+{K}_2\right)+2x-\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}}{2} $$

(A10)

As you can see, both approaches lead to identical results, suggesting their equivalence. Thus, regardless of whether undissociated molecules or individual ions are transported through the membrane (in quantities guaranteeing the maintenance of electrical neutrality), the system reaches the same equilibrium state (from an identical initial state).

We also show that both approaches lead to the equality of equilibrium concentrations of undissociated salt molecules in solutions separated by a semipermeable membrane.

At equilibrium, the penetrating ions obey the Donnan distribution, so the following equations may be written:

$$ \lambda =\frac{{\left[{K}^{+}\right]}_i}{{\left[{K}^{+}\right]}_e}=\frac{C_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}}{-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}} $$

(A11)

$$ \lambda =\frac{{\left[{A}^{-}\right]}_e}{{\left[{A}^{-}\right]}_i}=\frac{-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}}{-{C}_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}} $$

(A12)

Let us compare the concentrations of undissociated KA molecules in the i- and e-cells:

$$ \frac{{\left[ KA\right]}_i}{{\left[ KA\right]}_e}=\frac{\left({C}_1+{K}_2\right)+2x-\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}}{2\left({C}_2-x\right)+{K}_2-\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}} $$

(A13)

It is difficult to determine from Eq. (A13) if the above ratio equals one. So let us focus on Eqs. (A11) and (A12): their ratio equals one.

$$ \frac{{\left[{K}^{+}\right]}_i{\left[{A}^{-}\right]}_i}{{\left[{K}^{+}\right]}_e{\left[{A}^{-}\right]}_e}=\frac{\left({C}_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}\right)\left(-{C}_1-{K}_2+\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}\right)}{\left(-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}\right)\left(-{K}_2+\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}\right)}=1 $$

(A14)

After transformations we get:

$$ \frac{{\left[{K}^{+}\right]}_i{\left[{A}^{-}\right]}_i}{{\left[{K}^{+}\right]}_e{\left[{A}^{-}\right]}_e}=\frac{2{K}_2\left[\left({C}_1+{K}_2\right)+2x-\sqrt{{\left({C}_1+{K}_2\right)}^2+4{xK}_2}\right]}{2{K}_2\left[2\left({C}_2-x\right)+{K}_2-\sqrt{K_2^2+4{K}_2\left({C}_2-x\right)}\right]}=1 $$

(A15)

The comparison between Eqs. (A13) and (A15) clearly shows that at equilibrium the concentrations of undissociated molecules in the cells separated by a membrane are equal:

$$ \frac{{\left[ KA\right]}_i}{{\left[ KA\right]}_e}=\frac{{\left[{K}^{+}\right]}_i{\left[{A}^{-}\right]}_i}{{\left[{K}^{+}\right]}_e{\left[{A}^{-}\right]}_e}=1 $$

(A16)