Study of phosphate sorption on rendzina soil by heterogeneous isotopic exchange: a biexponential kinetic model

Abstract

Phosphate sorption on rendzina soil was studied by P-32 heterogeneous isotopic exchange under a steady-state. There are two types of sorbed phosphate, namely strongly and weakly bonded phosphate, the latter being able to exchange with phosphate (H₂³²PO₄⁻) ions in the soil solution. The experimental kinetic data was not fitted by the one exponential kinetic model. Starting from this observation, a new kinetic model is established by assuming two types of weakly bonded phosphate, which take part in the two parallel exchange processes. A biexponential kinetic equation is obtained, which fits the experimental data much better than the one exponential equation.

Graphical abstract

Introduction

Phosphorus is an essential nutrient for plants, which is taken up from the soil solution in the form of phosphate ions [1, 2]. Phosphate ions can be sorbed on a soil by different mechanisms, namely weakly and tightly sorbed phosphate [3, 4], the tightly sorbed phosphate being significant. Weakly sorbed phosphate can get back to the solution via desorption, so it is potentially available for plants, while tightly sorbed phosphate is hardly desorbed.

Weakly sorbed phosphate ions can be exchanged by phosphate ions in the soil solution, kinetic behavior can be examined by radioisotope (³²P) heterogeneous isotopic exchange [1]. In this process the phosphate ions sorbed on the solid phase exchange with the radiolabelled phosphate ions in the soil solution, the driving force is the increase of mixing entropy, while there is no change in enthalpy. The main advantage of this method is that kinetics can be examined even if there is steady-state for sorption and desorption. In this case the total amounts of phosphate ions in the soil solution and the solid phase do not change, but the amounts of radiolabelled phosphate in the soil solution is deminishing because of the exchange process, and the rate of heterogeneous isotopic exchange is proportional to the steady-state rate of sorption and desorption. Models and equations describing the kinetics of heterogeneous isotopic exchange (Table 1) can be classified into three main groups, including empiric equations (such as the series of exponential terms [5], the two constant equation by Edgington [5] and the empiric formula used by Fardeau et al. [1, 6]), compartmental models (including the modified Elovich equation by Atkinson et al. [7] and the one exponential model by Kónya et al. [3]) and mechanistic exchange models (for example the Paneth exchange and recrystallization model by Imre [8] and the surface reaction-solid state diffusion model by Barrow et al. [9]).

Table 1 Kinetic equations for heterogeneous isotopic exchange ($x$ is the dimensionless relative radioactivity of soil, $t$ is time in min)

The aim of this research is to establishe a compartmental model for heterogeneous isotopic exchange on rendzina soil under a steady-state condition. Kinetic data has been applied at several phosphate concentrations.

Experimental

The studied soil was rendzina [Miskolc (Hungary), collected in 2019, clay and silt content: 28.36%, CaCO₃ content: 11.51%, humus content: 14.52%, pH = 7.49, conductivity = 695 µS/cm], the detailed description of experiments can be found elsewhere [3]. P was added to the soil samples in amounts of 0, 40, 80, 160 and 320 µgP/g in the form of KH₂PO₄ solution, and they were incubated for 1 week, 3 weeks and 3 months at room temperature at 60% of maximum water holding capacity of soil (0 µgP/g were the control samples, in those cases destilled water was added instead of KH₂PO₄ solution). Heterogeneous (³²P) isotopic exchange experiments were carried out at room temperature (25 °C) by using 1.288 g of soil sample (1 g air dried material) stirred with 210 cm³ of distilled water for 60 min—during this time reaction mixture has reached steady-state. Then, 10 cm³ sample was taken from the soil solution and analyzed for phosphorous concentration by ammonium molybdate method by spectrophotometric technique [10]. 100 µl of carrier free H₃³²PO₄ radiotracer (carrier free means there it contains only ³²P, concentration of ³²P < 10^–12 gP/experiment) was added to the remaining 200 cm³ soil solution, and 1.8–1.8 cm³ samples were taken after 2, 4, 6, 8, 10, 15, 20, 25, 30, 45, 60, 90 and 120 min. To 1–1 cm³-s of the samples 4–4 cm³ of scintillation cocktail (composition see in [3]) was added, and radioactive intensities were measured by TriCarb 4810 TR Liquid Scintillation Analyser, radioactivity of ³²P is limited by the half-life. In order to determine the total radioactivity in the system a zero solution was also established, in which to 200 mL of destilled water 100 µL of carrier free H₃³²PO₄ radiotracer was added. 1 cm³ of the zero solution was also taken (zero sample) and the radioactive intensities were determined as seen before.

Parameter evaluation according to the one exponential equation

First, relative radioactivities of soil were plotted versus time for each reaction mixture, and data points were fitted by the one exponential equation [3]:

$$x=A\cdot \left[1-{\text{exp}}(-B\cdot t)\right]$$

(1)

where $x$ is the relative radioactivity of soil (which is dimensionless because it is the ratio of the radioactivity of the soil and the total radioactivity). $x$ can be calculated as following: $x=1-\frac{{I}_{{\text{l}}}}{{I}_{{\text{tot}}}}$ where ${I}_{{\text{l}}}$ and ${I}_{{\text{tot}}}$ are the intesities of the soil solution and the zero sample. ${\text{A}}=\frac{{m}_{2}}{{m}_{1}+{m}_{2}}$ and ${\text{B}}=\frac{C}{{m}_{1}}\cdot \frac{{m}_{1}+{m}_{2}}{{m}_{2}}$. $C$ is the steady-state rate (in µg/min), ${m}_{1}$ and ${m}_{2}$ are the amounts of phosphorus in the soil soluton and weakly sorbed on the solid phase (in µg). Fitted curves of soil samples incubated for 1 week, 3 weeks and 3 months can be seen in the Appendix in Figs. 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20.

As seen, all fitted curves show systematic deviation from experimental data points. In the middle period (in the range of 20–60 min) the experimental data are below, while at longer reaction times (over 60 min) they are above the fitted curves (Fig. 1). This reveals that more than one sites of weakly sorbed/isotopically exchangeable phosphate exist. There should be at least two different isotope exchange processes with different rates. At the initial period (up to 20 min) the faster exchange process is almost completed. According to this, a new kinetic model is established. Another compartmental model is the modified Elovich equation [7], but it’s application for soil and other natural systems is not relevant, because these systems do not meet the boundary condition that the activation energy changes linearly with surface coverage.

Fig. 1

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 0 µgP/g for 1 week

Establishment of a biexponential kinetic model

The new kinetic model assumes two types of weakly sorbed phosphate, which take part in the following sorption and desorption processes:

$$S_1 + P{\mathop \rightleftarrows \limits^{C_1 }} S_1 - P$$

(2)

$$S_2 + P{\mathop \rightleftarrows \limits^{C_2 }} S_2 - P$$

(3)

where S₁ and S₂ are the bonding sites for the two different exchangeable phosphate compartments, and P is phosphate species (at soil solution pH it is mostly dihydrogen phosphate ion). It should be emphasised that there is no statement about the type of bonding sites or the mechanism of sorption and desorption—it can be either adsorption or ion exchange or even precipitation. The only assumption is that it must be a reversible process in order that steady-state can occur. That is why it can be said that the model is nonmechanistic.

In steady state, sorption and desorption rates are equal at both bonding sites:

$$\frac{{\text{d}}{m}_{{\text{P}};\mathrm{ L}\to {S}_{1} }}{{\text{d}}t}=\frac{{\text{d}}{m}_{{\text{P}}; {S}_{1}\to \mathrm{L }}}{{\text{d}}t}={C}_{1}$$

(4)

$$\frac{{\text{d}}{m}_{{\text{P}};\mathrm{ L}\to {S}_{2} }}{{\text{d}}t}=\frac{{\text{d}}{m}_{{\text{P}}; {S}_{2}\to \mathrm{L }}}{{\text{d}}t}={C}_{2}$$

(5)

where t is time (in minutes), and C₁ and C₂ are steady state rates for sorption and desorption at bonding sites S₁ and S₂ (in µg/min) and L means the soil solution. Let us be ${m}_{2{\text{a}}}$ and ${m}_{2b}$ the mass of phosphorus (in µg) sorbed at bonding sites S₁ and S₂ in steady state, ${m}_{1}$ the mass of phosphorus (in µg) in the soil solution and ${m}_{2a}^{*}$, ${m}_{2b}^{*}$ and ${m}_{1}^{*}$ are the masses of the radioactive forms (in µg). When kinetic isotopic effect is negligible, radioactive phosphorus is exchanged in the same manner as inactive one. This means, at a given t time $\frac{{m}_{2a}^{*}}{{m}_{2a}}\cdot {C}_{1}$ is the desorption rate for radioactive phosphorus at bonding site S₁. The same is valid for S₂ and for sorption rates, so the following differential equation system can be constructed:

$$\frac{{\text{d}}{m}_{2a}^{*}}{{\text{d}}t}=\frac{{m}_{1}^{*}}{{m}_{1}}\cdot {C}_{1}-\frac{{m}_{2a}^{*}}{{m}_{2a}}\cdot {C}_{1}$$

(6)

$$\frac{{\text{d}}{m}_{2b}^{*}}{{\text{d}}t}=\frac{{m}_{1}^{*}}{{m}_{1}}\cdot {C}_{2}-\frac{{m}_{2b}^{*}}{{m}_{2b}}\cdot {C}_{2}$$

(7)

After radiotracer has been added to reaction mixture, the system is closed for radioactive phosphorus, so Eq. (8) can be written:

$${m}^{*}={m}_{1}^{*}+{m}_{2a}^{*}+{m}_{2b}^{*}$$

(8)

where ${m}^{*}$ is the total mass of radioactive phosphorus (in µg).

From Eqs. (6–8), the following differential equation system is obtained:

$$\frac{{\text{d}}{m}_{2a}^{*}}{{\text{d}}t}=-\left(\frac{{C}_{1}}{{m}_{1}}+\frac{{C}_{1}}{{m}_{2a}}\right)\cdot {m}_{2a}^{*}-\frac{{C}_{1}}{{m}_{1}}\cdot {m}_{2b}^{*}+\frac{{C}_{1}}{{m}_{1}}\cdot {m}^{*}$$

(9)

$$\frac{{\text{d}}{m}_{2b}^{*}}{{\text{d}}t}=-\frac{{C}_{2}}{{m}_{1}}\cdot {m}_{2a}^{*}-\left(\frac{{C}_{2}}{{m}_{1}}+\frac{{C}_{2}}{{m}_{2b}}\right)\cdot {m}_{2b}^{*}+\frac{{C}_{2}}{{m}_{1}}\cdot {m}^{*}$$

(10)

After arrangement of Eqs. (9) and (10), and parameters ${\text{A}}=\frac{{C}_{1}}{{m}_{1}}$; ${\text{B}}=\frac{{C}_{1}}{{m}_{2a}}$; ${\text{D}}=\frac{{C}_{2}}{{m}_{1}}$; ${\text{E}}=\frac{{C}_{2}}{{m}_{2b}}$ and functions ${x}_{1}=\frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}$, ${x}_{2}=\frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}$ are introduced, a linear differential equation system is obtained:

$$\left(\begin{array}{c}\frac{{\text{d}}{x}_{1}}{{\text{d}}t}\\ \frac{{\text{d}}{x}_{2}}{{\text{d}}t}\end{array}\right)=\left[\begin{array}{cc}-\left(A+B\right)& -A\\ -D& -\left(D+E\right)\end{array}\right]\cdot \left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)+\left(\begin{array}{c}A\\ D\end{array}\right)$$

(11)

At $t=0\to {x}_{1}, {x}_{2}=0$, the solution of the equation system is:

$${x}_{1}=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot A\cdot \mathit{exp}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot \frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot \mathit{exp}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(12)

$${x}_{2}=\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot \frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot D\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(13)

The relative radioactivity of soil is the sum of relative radioactivities at bonding sites S₁ and S₂ (they are dimensionless):

$$x={x}_{1}+{x}_{2}$$

(14)

So, the relative radioactivity of soil versus time according to the biexponential model is:

$$x=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}+\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot \left(A+\frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot \left(D+\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(15)

where,

${K}_{1}=\frac{\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}}{A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}}$ and

${K}_{2}=\frac{\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot B\cdot D}{A\cdot E+B\cdot D+B\cdot E}}{A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}}$

Thus, when there are two compartments of weakly sorbed phosphate with two different steady-state rates, a biexponential kinetic equation is obtained. There are 4 independent fitting parameters namely A, B, D and E, which are derived from the 5 original parameters m₁, m_2a, m_2b, C₁ and C₂. A, B, D and E can be obtained from fitting, while m₁ is calculated from photometry. In the next session, the application of the new model will be discussed.

Results and discussion

The phosphate concentrations in the soil solution determined by spectrophotometry, fitting parameters of the biexponential kinetic equation and their standard deviations and phosphate concentrations are summarized in Table 2. From the phosphate concentration in the soil solution (c_P in mol/dm³) m₁ (in µg) is calculated. The fitting parameters (which are all in min⁻¹) are the following: $A=\frac{{C}_{1}}{{m}_{1}}$; $B=\frac{{C}_{1}}{{m}_{2a}}$; $D=\frac{{C}_{2}}{{m}_{1}}$; $E=\frac{{C}_{2}}{{m}_{2b}}$, from here the steady state rates and masses of weakly sorbed phosphate at bonding sites S₁ and S₂ can be calculated as

$$C_1 = A \cdot m_1 ,\;m_{2a} = \frac{A}{B} \cdot m_1 ,\;C_2 = D \cdot m_1 ,\;m_{2b} = \frac{D}{E} \cdot m_1$$

and summarized in Table 3. The kinetic curves can be seen in Figs. 2, 3 and 4. The differences of relative radioactivities calculated from the fitted biexponential equation and measured values (d, which is also dimensionless) versus time are also plotted, which can be found in the Appendix (Figs. 21, 22, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33). In Fig. 2 it is clearly seen that the order of kinetic curves from the top to the bottom is 0, 40, 160 and 320 µgP/g soil, which is in accordance with the fact that the more phosphate is in the system, the less is the ratio of the amount of weakly bonded phosphate to the amount of phosphate in the soil solution (saturation effect).

Table 2 Phosphate concentrations in the soil solution and fitting parameters of the biexponential kinetic equation

Table 3 Steady-state rates and masses of sorbed P at the two different bonding sites

Fig. 2

Biexponential kinetic curves of heterogeneous isotopic exchange on rendzina soil samples (1 week incubation time)

Fig. 3

Biexponential kinetic curves of heterogeneous isotopic exchange on rendzina soil samples (3 weeks incubation time)

Fig. 4

Biexponential kinetic curves of heterogeneous isotopic exchange on rendzina soil samples (3 months incubation time)

It can be seen (Fig. 5) that the new, biexponential kinetic equation obtained from the biexponential model fits much better experimental data, than the one exponential equation by Kónya et al. Data points fit well even in the middle period and at long times, where the one exponential model showed systematic deviation. The differences of calculated and measured relative radioactivities (d) also have to be examined (Figs. 6, 22, 21, 23, 24, 25, 26, 27, 28, 29, 30, 31, 32, 33). As for the one exponential model, d versus time shows a maximum, which means, there is a systematic deviation. As for the biexponential model, d values scatter randomly around 0, and their values are smaller compared to one exponential fitting, which means biexponential equation obtained from the kinetic model assuming two types of weakly bonded phosphate describes appropriately the kinetics of heterogeneous isotopic exchange. The R² values further strengthen this assumption (Table 4). As for the biexponential model, R² values are all above 0.99, while in the case of one exponential model, they are between 0.95 and 0.99. This means that there are at least two different types of weakly sorbed phosphate on rendzina soil, which exchange at two different rates. However, it cannot be excluded that there are more than two types of weakly sorbed phosphate, but kinetically they can be separated only to two compartments.

Fig. 5

Comparing the one exponential and the biexponential model (0 µgP/g 1 week)

Fig. 6

Differences (d) of calculated and measured relative radioactivities (x) versus time (1 week incubation, 0 µgP/g)

Table 4 R² values for the 1 and 2 exponential equation fitting

As for 1 week incubation time, the amounts of weakly sorbed phosphorus at the two different bonding sites (m_2a and m_2b) obtained from the new biexponential model have been plotted versus c_P (the concentration of phosphorus in the solution in µg/dm³) and fitted by Langmuir isotherm [Eq. (16), Fig. 7]. Fitting parameters can be found in Table 5.

$${a}_{{\text{P}}}=\frac{K\cdot M\cdot {c}_{{\text{P}}}}{1+K\cdot {c}_{{\text{P}}}}$$

(16)

where ${a}_{{\text{P}}}$ is the amount of sorbed phosphorus on the soil surface sites S₁ and S₂ (in µgP/g), ${c}_{{\text{P}}}$ is the concentration of phosphorus in the soil solution (in µg/dm³), $M$ is the maximum value of sorbed phosphorus at bonding sites S₁ and S₂ (in µgP/g) and $K$ is the sorption constant for bonding sites S₁ and S₂ (in dm³/µg). In the case of 3 weeks and 3 months incubation time, m_2a and m_2b versus c_P cannot be fitted by the Langmuir or any other isotherm—data points scatter on a very large scale. The reason for this can be that some secondary processes might occur. This can also explain the contradictious fact that the phosphate concentration in the soil solution is smaller at 80 µgP/g than at 0 or 40 µgP/g incubation (Table 2, lines 6–8).

Fig. 7

Langmuir plots for soil samples incubated for 1 week (amounts of phosphorus at surface sites 2a and 2b calculated from the biexponential model)

Table 5 Langmuir parameters at the two different surface sites for 1 week incubation time

Conclusions

It is clearly seen that heterogeneous isotopic exchange on rendzina soil can be described by the biexponential kinetic equation obtained from a new model, which assumes two types of exchangeable phosphate. There were such biexponential kinetic equations before too, however, they were only empirical ones [5, 8]. The only exception we know about is the biexponential model by Imre [11] established for the heterogeneous isotopic exchange of lead and actinium ions sorbed on barium sulphate, but the meaning of the parameters was different, and it contained also some mechanistic considerations. There is also a biexponential model by Darbee et al. similar to ours, but it was established for homogeneous isotopic exchange [12]. Earlier Kosmulski et al. already proved, that if there are several parallel exchange processes, the kinetics of heterogeneous isotopic exchange can be described by a series of exponential terms [13]. The novelty of the biexponential equation derived in the present work is that it is obtained from a kinetic model, so the parameters have physical meaning, and can be related to the two exchange processes, however, there is no assumption about the details of the mechanism of sorption and desorption, so it can be used universally. From the fitting parameters the amounts of weakly sorbed phosphorus at the two different bonding sites can be determined, and sorption isotherms can be obtained.

Normally the concentration of phosphate ions in the soil solution increases with the amount of the added phosphate during the incubation period (0, 40, 80, 160 and 320 µgP/g soil), and the kinetic curves follow from the top to the bottom in that order—as it can be observed for 1 week incubation time (Fig. 2). When this is the case, the Langmuir isotherm can be applied (Fig. 7). However, there are cases, when the concentration of phosphate ions in the soil solution does not correlate with the amount of added phosphorus during incubation (3 weeks and 3 months incubation time, Figs. 3 and 4), in these cases the order of kinetic curves is also not unambiguous, and the Langmuir isotherm cannot be applied. The reasons for this phenomenon might be some secondary processes. It can happen, for example, that at higher amounts of added phosphorus precipitation takes places, which deminishes the amount of phosphate in the soil solution. To clear up these secondary processes, further research is demanded.

The new model makes possible to understand better the sorption and desorption of phosphate on soil, which is important to improve fertilizing strategies in agriculture. From the kinetic parameters of the heterogeneous isotopic exchange on soil it becomes possible to identify which components of the soil are responsible for weak phosphate bonding, and strategies can be established to prevent the transformation of phosphate to unavailable forms.

Appendices

Appendix 1

See Table 

Table 6 List of physical quantities and their units

6.

Appendix 2

Complete mathematical deduction of biexponential kinetic model

Sorption–desorption processes at the two types of bonding sites:

$$S_1 + P{\mathop \rightleftarrows \limits^{C_1 }} S_1 - P$$

$$S_2 + P{\mathop \rightleftarrows \limits^{C_2 }} S_2 - P$$

Let us be ${m}_{2{\text{a}}}$ and ${m}_{2{\text{b}}}$ the mass of phosphorus sorbed at bonding sites S₁ and S₂ in steady state, and ${m}_{1}$ the mass of phosphorus in the soil solution, which is in equillibrium with the solid state. In steady state, sorption and desorption rates are equal at both bonding sites, so the following equations can be written:

$$\frac{{\text{d}}{m}_{{\text{P}}; L\to {S}_{1} }}{{\text{d}}t}=\frac{{\text{d}}{m}_{{\text{P}}; {S}_{1}\to L }}{{\text{d}}t}={C}_{1}$$

(17)

$$\frac{{\text{d}}{m}_{{\text{P}}; L\to {S}_{2} }}{{\text{d}}t}=\frac{{\text{d}}{m}_{{\text{P}}; {S}_{2}\to L }}{{\text{d}}t}={C}_{2}$$

(18)

where t is time, and C₁ and C₂ are steady state rates for sorption and desorption at bonding sites S₁ and S₂, and L means the liquid phase. Adding radioactive H₃³²PO₄ to the liquid phase, it will get into three forms:

${m}_{2{\text{a}}}^{*}$, ${m}_{2{\text{b}}}^{*}$ and ${m}_{1}^{*}$ are the masses of radioactive phosphorus at a given t time at bonding sites S₁, S₂ and in the liquid phase (soil solution), and ${m}^{*}$ is the total amount of radioactive phosphorus added to the solution. The mass balance is valid for them:

$${m}^{*}={m}_{1}^{*}+{m}_{2{\text{a}}}^{*}+{m}_{2{\text{b}}}^{*}$$

(19)

At a given t time, $\frac{{m}_{1}^{*}}{{m}_{1}}$, $\frac{{m}_{2{\text{a}}}^{*}}{{m}_{2{\text{a}}}}$ and $\frac{{m}_{2{\text{b}}}^{*}}{{m}_{2{\text{b}}}}$ are the proportions of radioactive phosphorus at bonding sites S₁, S₂ and in the liquid phase. Radioactive phosphorus goes to bonding sites S₁ and S₂ via sorption, and leaves the bonding sites via desorption, so the following differential equation system can be written:

$$\frac{{\text{d}}{m}_{2{\text{a}}}^{*}}{{\text{d}}t}=\frac{{m}_{1}^{*}}{{m}_{1}}\cdot {C}_{1}-\frac{{m}_{2{\text{a}}}^{*}}{{m}_{2{\text{a}}}}\cdot {C}_{1}$$

(20)

$$\frac{{\text{d}}{m}_{2{\text{b}}}^{*}}{{\text{d}}t}=\frac{{m}_{1}^{*}}{{m}_{1}}\cdot {C}_{2}-\frac{{m}_{2{\text{b}}}^{*}}{{m}_{2{\text{b}}}}\cdot {C}_{2}$$

(21)

${m}_{1}^{*}$ can be obtained from (19) as following:

$${m}_{1}^{*}={m}^{*}-{m}_{2{\text{a}}}^{*}-{m}_{2{\text{b}}}^{*}$$

(22)

So Eqs. (20) and (21) get the forms:

$$\frac{{\text{d}}{m}_{2{\text{a}}}^{*}}{{\text{d}}t}=\frac{{m}^{*}-{m}_{2{\text{a}}}^{*}-{m}_{2{\text{b}}}^{*}}{{m}_{1}}\cdot {C}_{1}-\frac{{m}_{2{\text{a}}}^{*}}{{m}_{2{\text{a}}}}\cdot {C}_{1}$$

(23)

$$\frac{{\text{d}}{m}_{2{\text{b}}}^{*}}{{\text{d}}t}=\frac{{m}^{*}-{m}_{2{\text{a}}}^{*}-{m}_{2{\text{b}}}^{*}}{{m}_{1}}\cdot {C}_{2}-\frac{{m}_{2{\text{b}}}^{*}}{{m}_{2{\text{b}}}}\cdot {C}_{2}$$

(24)

From (23) and (24) the following expressions can be obtained:

$$\frac{{\text{d}}{m}_{2{\text{a}}}^{*}}{{\text{d}}t}=\frac{{m}^{*}}{{m}_{1}}\cdot {C}_{1}-\frac{{m}_{2{\text{a}}}^{*}}{{m}_{1}}\cdot {C}_{1}-\frac{{m}_{2{\text{b}}}^{*}}{{m}_{1}}\cdot {C}_{1}-\frac{{m}_{2{\text{a}}}^{*}}{{m}_{2{\text{a}}}}\cdot {C}_{1}$$

(25)

$$\frac{{\text{d}}{m}_{2{\text{b}}}^{*}}{{\text{d}}t}=\frac{{m}^{*}}{{m}_{1}}\cdot {C}_{2}-\frac{{m}_{2{\text{a}}}^{*}}{{m}_{1}}\cdot {C}_{2}-\frac{{m}_{2{\text{b}}}^{*}}{{m}_{1}}\cdot {C}_{2}-\frac{{m}_{2{\text{b}}}^{*}}{{m}_{2{\text{b}}}}\cdot {C}_{2}$$

(26)

After rearrangement:

$$\frac{{\text{d}}{m}_{2{\text{a}}}^{*}}{{\text{d}}t}=\frac{{C}_{1}}{{m}_{1}}\cdot {m}^{*}-\frac{{C}_{1}}{{m}_{1}}\cdot {m}_{2{\text{a}}}^{*}-\frac{{C}_{1}}{{m}_{1}}\cdot {m}_{2{\text{b}}}^{*}-\frac{{C}_{1}}{{m}_{2{\text{a}}}}\cdot {m}_{2{\text{a}}}^{*}$$

(27)

$$\frac{{\text{d}}{m}_{2{\text{b}}}^{*}}{{\text{d}}t}=\frac{{C}_{2}}{{m}_{1}}\cdot {m}^{*}-\frac{{C}_{2}}{{m}_{1}}\cdot {m}_{2{\text{a}}}^{*}-\frac{{C}_{2}}{{m}_{1}}\cdot {m}_{2{\text{b}}}^{*}-\frac{{C}_{2}}{{m}_{2{\text{b}}}}\cdot {m}_{2{\text{b}}}^{*}$$

(28)

Via reduction of (27) and (28) the following expressions are obtained:

$$\frac{{\text{d}}{m}_{2{\text{a}}}^{*}}{{\text{d}}t}=-\left(\frac{{C}_{1}}{{m}_{1}}+\frac{{C}_{1}}{{m}_{2{\text{a}}}}\right)\cdot {m}_{2{\text{a}}}^{*}-\frac{{C}_{1}}{{m}_{1}}\cdot {m}_{2{\text{b}}}^{*}+\frac{{C}_{1}}{{m}_{1}}\cdot {m}^{*}$$

(29)

$$\frac{{\text{d}}{m}_{2{\text{b}}}^{*}}{{\text{d}}t}=-\frac{{C}_{2}}{{m}_{1}}\cdot {m}_{2{\text{a}}}^{*}-\left(\frac{{C}_{2}}{{m}_{1}}+\frac{{C}_{2}}{{m}_{2{\text{b}}}}\right)\cdot {m}_{2{\text{b}}}^{*}+\frac{{C}_{2}}{{m}_{1}}\cdot {m}^{*}$$

(30)

After division by ${m}^{*}$:

$$\frac{1}{{m}^{*}}\cdot \frac{{\text{d}}{m}_{2{\text{a}}}^{*}}{{\text{d}}t}=-\left(\frac{{C}_{1}}{{m}_{1}}+\frac{{C}_{1}}{{m}_{2{\text{a}}}}\right)\cdot \frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}-\frac{{C}_{1}}{{m}_{1}}\cdot \frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}+\frac{{C}_{1}}{{m}_{1}}$$

(31)

$$\frac{1}{{m}^{*}}\cdot \frac{{\text{d}}{m}_{2{\text{b}}}^{*}}{{\text{d}}t}=-\frac{{C}_{2}}{{m}_{1}}\cdot \frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}-\left(\frac{{C}_{2}}{{m}_{1}}+\frac{{C}_{2}}{{m}_{2{\text{b}}}}\right)\cdot \frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}+\frac{{C}_{2}}{{m}_{1}}$$

(32)

$\frac{1}{{m}^{*}}$ is constant, so it can be taken in the derivatives:

$$\frac{{\text{d}}\frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}}{{\text{d}}t}=-\left(\frac{{C}_{1}}{{m}_{1}}+\frac{{C}_{1}}{{m}_{2{\text{a}}}}\right)\cdot \frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}-\frac{{C}_{1}}{{m}_{1}}\cdot \frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}+\frac{{C}_{1}}{{m}_{1}}$$

(33)

$$\frac{{\text{d}}\frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}}{{\text{d}}t}=-\frac{{C}_{2}}{{m}_{1}}\cdot \frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}-\left(\frac{{C}_{2}}{{m}_{1}}+\frac{{C}_{2}}{{m}_{2{\text{b}}}}\right)\cdot \frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}+\frac{{C}_{2}}{{m}_{1}}$$

(34)

${x}_{1}=\frac{{m}_{2{\text{a}}}^{*}}{{m}^{*}}$ and ${x}_{2}=\frac{{m}_{2{\text{b}}}^{*}}{{m}^{*}}$ are the relative radioactivities of phosphorus sorbed at sites S₁ and S₂, in such way the following equations can be obtained:

$$\frac{{\text{d}}{x}_{1}}{{\text{d}}t}=-\left(\frac{{C}_{1}}{{m}_{1}}+\frac{{C}_{1}}{{m}_{2{\text{a}}}}\right)\cdot {x}_{1}-\frac{{C}_{1}}{{m}_{1}}\cdot {x}_{2}+\frac{{C}_{1}}{{m}_{1}}$$

(35)

$$\frac{{\text{d}}{x}_{2}}{{\text{d}}t}=-\frac{{C}_{2}}{{m}_{1}}\cdot {x}_{1}-\left(\frac{{C}_{2}}{{m}_{1}}+\frac{{C}_{2}}{{m}_{2{\text{b}}}}\right)\cdot {x}_{2}+\frac{{C}_{2}}{{m}_{1}}$$

(36)

The following notation can be introduced:

$$A = \frac{C_1 }{{m_1 }};\;B = \frac{C_1 }{{m_{2a} }};\;D = \frac{C_2 }{{m_1 }};\;E = \frac{C_2 }{{m_{2b} }}$$

(37)

With these notations the equation system can be written in the following form:

$$\frac{{\text{d}}{x}_{1}}{{\text{d}}t}=-\left(A+B\right)\cdot {x}_{1}-A\cdot {x}_{2}+A$$

(38)

$$\frac{{\text{d}}{x}_{2}}{{\text{d}}t}=-D\cdot {x}_{1}-\left(D+E\right)\cdot {x}_{2}+D$$

(39)

It can be also rewritten in matrix representation:

$$\left(\begin{array}{c}\frac{{\text{d}}{x}_{1}}{{\text{d}}t}\\ \frac{{\text{d}}{x}_{2}}{{\text{d}}t}\end{array}\right)=\left[\begin{array}{cc}-\left(A+B\right)& -A\\ -D& -\left(D+E\right)\end{array}\right]\cdot \left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)+\left(\begin{array}{c}A\\ D\end{array}\right)$$

(40)

First, the eigenvalues of coefficient matrix should be determined:

$$\left|\begin{array}{cc}-\left(A+B\right)-\lambda & -A\\ -D& -\left(D+E\right)-\lambda \end{array}\right|=\left[-\left(A+B\right)-\lambda \right]\cdot \left[-\left(D+E\right)-\lambda \right]-\left(-A\right)\cdot \left(-D\right)=0$$

(41)

$$\left(A+B\right)\cdot \left(D+E\right)+\left(A+B\right)\cdot \lambda +\left(D+E\right)\cdot \lambda +{\lambda }^{2}-A\cdot D=0$$

(42)

$${\lambda }^{2}+A\cdot D+A\cdot E+B\cdot D+B\cdot E+\left(A+B\right)\cdot \lambda +\left(D+E\right)\cdot \lambda -A\cdot D=0$$

(43)

$${\lambda }^{2}+\left(A+B+D+E\right)\cdot \lambda +\left(A\cdot E+B\cdot D+B\cdot E\right)=0$$

(44)

Using the solving formula for quadratic equations:

$${\lambda }_{1}=\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(45)

$${\lambda }_{2}=\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(46)

The $\underline{u}$ eigenvector can be found as following:

$$\left[\begin{array}{cc}-\left(A+B\right)-{\lambda }_{1}& -A\\ -D& -\left(D+E\right)-{\lambda }_{1}\end{array}\right]\cdot \left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

(47)

That means:

$$\left[-\left(A+B\right)-{\lambda }_{1}\right]\cdot {u}_{1}-A\cdot {u}_{2}=0$$

(48)

$$-D\cdot {u}_{1}+\left[-\left(D+E\right)-{\lambda }_{1}\right]\cdot {u}_{2}=0$$

(49)

$$-\left(A+B\right)-{\lambda }_{1}=\frac{-\left(A+B\right)\cdot 2}{2}-\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{-2\cdot A-2\cdot B+A+B+D+E-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(50)

$$-\left(D+E\right)-{\lambda }_{1}=\frac{-\left(D+E\right)\cdot 2}{2}-\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{-2\cdot D-2\cdot E+A+B+D+E-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{\left(A+B\right)-\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(51)

From (48)–(51) Eqs. (52) and (53) can be derived:

$$\frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {u}_{1}-A\cdot {u}_{2}=0$$

(52)

$$-{\text{D}}\cdot {u}_{1}+\frac{\left(A+B\right)-\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {u}_{2}=0$$

(53)

From (52) ${u}_{2}$ can be expressed:

$${u}_{2}=\frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2\cdot A}\cdot {u}_{1}$$

(54)

So one possible $\underline{u}$ eigenvector is:

$$\underline{u}=\left(\begin{array}{c}{u}_{1}\\ {u}_{2}\end{array}\right)=\left(\begin{array}{c}A\\ \frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\end{array}\right)$$

(55)

Simillary, for $\underline{v}$ eigenvector:

$$\left[\begin{array}{cc}-\left(A+B\right)-{\lambda }_{2}& -A\\ -D& -\left(D+E\right)-{\lambda }_{2}\end{array}\right]\cdot \left(\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

(56)

From this Eqs. 57 and 58 can be obtained:

$$\left[-\left(A+B\right)-{\lambda }_{2}\right]\cdot {v}_{1}-A\cdot {v}_{2}=0$$

(57)

$$-D\cdot {v}_{1}+\left[-\left(D+E\right)-{\lambda }_{2}\right]\cdot {v}_{2}=0$$

(58)

$$-\left(A+B\right)-{\lambda }_{2}=\frac{-\left(A+B\right)\cdot 2}{2}-\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{-2\cdot A-2\cdot B+A+B+D+E+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{-\left(A+B\right)+D+E+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(59)

$$-\left(D+E\right)-{\lambda }_{2}=\frac{-\left(D+E\right)\cdot 2}{2}-\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{-2\cdot D-2\cdot E+A+B+D+E+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}=\frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(60)

From (57)–(60) Eqs. (61) and (62) can be derived:

$$\frac{-\left(A+B\right)+D+E+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {v}_{1}-A\cdot {v}_{2}=0$$

(61)

$$-D\cdot {v}_{1}+\frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {v}_{2}=0$$

(62)

From (62) ${v}_{1}$ can be expressed:

$${v}_{1}=\frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2\cdot D}\cdot {v}_{2}$$

(63)

So one possible $\underline{v}$ eigenvector is:

$$\underline{v}=\left(\begin{array}{c}{v}_{1}\\ {v}_{2}\end{array}\right)=\left(\begin{array}{c}\frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\\ D\end{array}\right)$$

(64)

The general solution of a homogenous linear differential equation system is:

$${\underline{x}}_{H}={K}_{1}\cdot \underline{u}\cdot {\text{exp}}\left({\lambda }_{1}\cdot t\right)+{K}_{2}\cdot \underline{v}\cdot {\text{exp}}\left({\lambda }_{2}\cdot t\right)$$

(65)

So the homogenous solution of our system is:

$$\left(\begin{array}{c}{x}_{1H}\\ {x}_{2H}\end{array}\right)={K}_{1}\cdot \left(\begin{array}{c}A\\ \frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\end{array}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot \left(\begin{array}{c}\frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\\ D\end{array}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(66)

In order to get the solution of the inhomogenous system, the proof function method is used. It can be supposed, that there is a particular solution of the inhomogenous system in the following form:

$$\left(\begin{array}{c}{x}_{1}\\ {x}_{2}\end{array}\right)=\left(\begin{array}{c}{P}_{1}\\ {P}_{2}\end{array}\right)$$

(67)

where ${P}_{1}$ and ${P}_{2}$ are constant. So, the differential equiation system can be written:

$$\left(\begin{array}{c}\frac{{\text{d}}{P}_{1}}{{\text{d}}t}\\ \frac{{\text{d}}{P}_{2}}{{\text{d}}t}\end{array}\right)=\left[\begin{array}{cc}-\left(A+B\right)& -A\\ -D& -\left(D+E\right)\end{array}\right]\cdot \left(\begin{array}{c}{P}_{1}\\ {P}_{2}\end{array}\right)+\left(\begin{array}{c}A\\ D\end{array}\right)=\left(\begin{array}{c}0\\ 0\end{array}\right)$$

(68)

Therefore:

$$-\left(A+B\right)\cdot {P}_{1}-A\cdot {P}_{2}+A=0$$

(69)

$$-D\cdot {P}_{1}-\left(D+E\right)\cdot {P}_{2}+D=0$$

(70)

${P}_{2}$ can be expressed from (69):

$$P_2 = \frac{{A - \left( {A + B} \right) \cdot P_1 }}{A}$$

(71)

Then it can be substitued for ${P}_{2}$ in (70):

$$-D\cdot {P}_{1}-\left(D+E\right)\cdot \frac{A-\left(A+B\right)\cdot {P}_{1}}{A}+D=0$$

(72)

$$\frac{-D\cdot A\cdot {P}_{1}}{A}-\frac{\left(D+E\right)\cdot \left[A-\left(A+B\right)\cdot {P}_{1}\right]}{A}+\frac{A\cdot D}{A}=0$$

(73)

$$\frac{-D\cdot A\cdot {P}_{1}-\left(D+E\right)\cdot \left[A-\left(A+B\right)\cdot {P}_{1}\right]+A\cdot D}{A}=0$$

(74)

$$\frac{-D\cdot A\cdot {P}_{1}-\left(D+E\right)\cdot A+\left(D+E\right)\cdot \left(A+B\right)\cdot {P}_{1}+A\cdot D}{A}=0$$

(75)

$$\frac{-D\cdot A\cdot {P}_{1}-A\cdot D-A\cdot E+D\cdot A\cdot {P}_{1}+D\cdot B\cdot {P}_{1}+E\cdot A\cdot {P}_{1}+E\cdot B\cdot {P}_{1}+A\cdot D}{A}=0$$

(76)

$$-A\cdot E+D\cdot B\cdot {P}_{1}+E\cdot A\cdot {P}_{1}+E\cdot B\cdot {P}_{1}=0$$

(77)

$$\left(A\cdot E+B\cdot D+B\cdot E\right)\cdot {P}_{1}=A\cdot E$$

(78)

$$P_1 = \frac{A \cdot E}{{A \cdot E + B \cdot D + B \cdot E}}$$

(79)

Then ${P}_{2}$ can be obtained from (71):

$${P}_{2}=\frac{A-\left(A+B\right)\cdot \frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}}{A}=1-\frac{\left(A+B\right)\cdot E}{A\cdot E+B\cdot D+B\cdot E}=\frac{A\cdot E+B\cdot D+B\cdot E-A\cdot E-B\cdot E}{A\cdot E+B\cdot D+B\cdot E}=\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}$$

(80)

So the particular solution of the system is:

$${\underline{x}}_{{\text{P}}}=\left(\begin{array}{c}{P}_{1}\\ {P}_{2}\end{array}\right)=\left(\begin{array}{c}\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\\ \frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\end{array}\right)$$

(81)

After that the general solution of a linear system is:

$$\underline{x}={\underline{x}}_{{\text{P}}}+{\underline{x}}_{{\text{H}}}$$

(82)

So the general solution of the system can be written as:

$$\underline{x}=\left(\begin{array}{c}\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\\ \frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\end{array}\right)+{K}_{1}\cdot \left(\begin{array}{c}A\\ \frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\end{array}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot \left(\begin{array}{c}\frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\\ D\end{array}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(83)

Boundary condition is: $t=0 \leftrightarrow {x}_{1}, {x}_{2}=0$, in consequence:

$$0=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot A+{K}_{2}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(84)

$$0=\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{2}\cdot D$$

(85)

$$0=\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot A\cdot D+{K}_{2}\cdot D\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(86)

$$0=\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{1}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{2}\cdot D\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(87)

Substracting (87) from (86) (88) is got:

$$\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}-\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{1}\cdot A\cdot D-{K}_{1}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot \left[-\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right]=0$$

(88)

$$\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}-\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{1}\cdot A\cdot D+{K}_{1}\cdot {\left[\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right]}^{2}=0$$

(89)

$$\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}-\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{1}\cdot \left[A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}\right]=0$$

(90)

From (90) ${K}_{1}$ can be expressed:

$${K}_{1}=\frac{\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}}{A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}}$$

(91)

Determination of ${K}_{2}$:

$$0=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{1}\cdot A\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{2}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$$

(92)

$$0=\frac{A\cdot B\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot A\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}+{K}_{2}\cdot A\cdot D$$

(93)

Substracting (93) from (92):

$$0=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot B\cdot D}{A\cdot E+B\cdot D+B\cdot E}-{K}_{2}\cdot {\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}-{K}_{2}\cdot A\cdot D$$

(94)

From (94) ${K}_{2}$ can be expressed:

$${K}_{2}=\frac{\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot B\cdot D}{A\cdot E+B\cdot D+B\cdot E}}{A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}}$$

(95)

The total relative radioactivity of the solid phase ($x$) can be obtained as tha sum of relative radioactivities at the two sorption sites:

$$x={x}_{1}+{x}_{2}$$

(96)

$${x}_{1}=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot A\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot \frac{\left(A+B\right)-\left(D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(97)

$${x}_{2}=\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot \frac{-\left(A+B\right)+\left(D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot D\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(98)

From (96), (97) and (98) the kinetic equation of heterogenous isotopic exchange is obtained:

$$x=\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}+\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}+{K}_{1}\cdot \left(A+\frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t+{K}_{2}\cdot \left(D+\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)\cdot {\text{exp}}\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\cdot t$$

(99)

where,

${K}_{1}=\frac{\frac{B\cdot D}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot E\cdot D}{A\cdot E+B\cdot D+B\cdot E}}{A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}}$ and

${K}_{2}=\frac{\frac{A\cdot E}{A\cdot E+B\cdot D+B\cdot E}\cdot \frac{-\left[\left(A+B\right)-\left(D+E\right)\right]-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}-\frac{A\cdot B\cdot D}{A\cdot E+B\cdot D+B\cdot E}}{A\cdot D+{\left(\frac{\left[\left(A+B\right)-\left(D+E\right)\right]+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}\right)}^{2}}$

It can be prooved, that expression under squer root ${\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)$ is always positive:

$${\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)={A}^{2}+{B}^{2}+{D}^{2}+{E}^{2}+2AB+2AD+2AE+2BD+2BE+2DE-4AE-4BD-4BE={A}^{2}+{B}^{2}+{D}^{2}+{E}^{2}+2AB+2AD-2AE-2BD-2BE+2DE$$

(100)

$${\left(-A-B+D+E\right)}^{2}={A}^{2}+{B}^{2}+{D}^{2}+{E}^{2}+2AB-2AD-2AE-2BD-2BE+2DE$$

(101)

Comparing (100) and (101) the following expression is obtained:

$${\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)={\left(-A-B+D+E\right)}^{2}+4AD>0$$

(102)

It is always positive because A and D are positive quantities—therefore, expression under squre root is always positive, so exponent is always a real number. As a consequence, (98) is a biexponential kinetic equation. It can also be shown that exponents are negative real numbers:

$\frac{-\left(A+B+D+E\right)-\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$ is negative because A, B, D and E and are positive and squre root is also positive. As for $\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$,

$\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}<\sqrt{{\left(A+B+D+E\right)}^{2}}=(A+B+D+E)$, consequently.

$\frac{-\left(A+B+D+E\right)+\sqrt{{\left(A+B+D+E\right)}^{2}-4\cdot \left(A\cdot E+B\cdot D+B\cdot E\right)}}{2}$ is always negative, too. So both exponential members have negative exponents, as a result, they are deminishing with time. As a consequence, relative radioactivity of soil shows a saturating tendency with time.

Appendix 3

See Figs.

Fig. 8

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 0 µgP/g for 1 week

8,

Fig. 9

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 40 µgP/g for 1 week

9,

Fig. 10

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 160 µgP/g for 1 week

10,

Fig. 11

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 320 µgP/g for 1 week

11,

Fig. 12

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 0 µgP/g for 3 weeks

12,

Fig. 13

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 40 µgP/g for 3 weeks

13,

Fig. 14

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 80 µgP/g for 3 weeks

14,

Fig. 15

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 160 µgP/g for 3 weeks

15,

Fig. 16

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 0 µgP/g for 3 months

16,

Fig. 17

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 40 µgP/g for 3 months

17,

Fig. 18

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 80 µgP/g for 3 months

18,

Fig. 19

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 160 µgP/g for 3 months

19,

Fig. 20

One exponential kinetic curve of heterogeneous isotopic exchange on rendzina soil sample treated with 320 µgP/g for 3 months

20,

Fig. 21

Differences (d) of calculated and measured relative radioactivities (x) versus time (1 week incubation, 0 µgP/g)

21,

Fig. 22

Differences (d) of calculated and measured relative radioactivities (x) versus time (1 week incubation, 40 µgP/g)

22,

Fig. 23

Differences (d) of calculated and measured relative radioactivities (x) versus time (1 week incubation, 160 µgP/g)

23,

Fig. 24

Differences (d) of calculated and measured relative radioactivities (x) versus time (1 week incubation, 320 µgP/g)

24,

Fig. 25

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 weeks incubation, 0 µgP/g)

25,

Fig. 26

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 weeks incubation, 40 µgP/g)

26,

Fig. 27

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 weeks incubation, 80 µgP/g)

27,

Fig. 28

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 weeks incubation, 160 µgP/g)

28,

Fig. 29

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 months incubation, 0 µgP/g)

29,

Fig. 30

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 months incubation, 40 µgP/g)

30,

Fig. 31

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 months incubation, 80 µgP/g)

31,

Fig. 32

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 months incubation, 160 µgP/g)

32,

Fig. 33

Differences (d) of calculated and measured relative radioactivities (x) versus time (3 months incubation, 320 µgP/g)

33.