## 3 Results and Discussion
### Mathematical Model
\begin{array}{c}14\times f\times[\mathrm{C_{7}H_{8}}]\ =\ \left[\mathrm{H^{+}}\right]-\frac{ \mathrm{K_{w}}}{\left[\mathrm{H^{+}}\right]}-\left[\mathrm{NO_{3}^{-}}\right]- \left[\mathrm{NO_{2}^{-}}\right]+E+\left(2\times A+B+C\right)\times\frac{1}{ \left[\mathrm{Ca^{2+}}\right]}\\ \hskip 56.905512 ptA\ =\ \frac{\mathrm{K_{cal}}\times\left[\mathrm{H^{+}} \right]^{2}}{\mathrm{K_{H2 CO3}\times K_{H4 CO3}}}\\ \hskip 56.905512 ptB\ =\ \frac{\mathrm{K_{cal}}\times\left[\mathrm{H^{+}} \right]}{\mathrm{K_{HCOO-}}}\\ \hskip 56.905512 ptC\ =\ \frac{\mathrm{K_{cal}}\times\left[\mathrm{V-Va}\right] \left[\mathrm{H^{+}}\right]^{2}}{\mathrm{K_{HCOO3}\times K_{HCOO-}\times K_{H4 }\times K\times T\times Va}}\\ \hskip 56.905512 ptD\ =\ \frac{\left(\left[\mathrm{H^{+}}\right]-\frac{\mathrm{K_{w}}}{ \left[\mathrm{H^{+}}\right]}-\left[\mathrm{NO_{3}^{-}}\right]-\left[\mathrm{ NO_{2}^{-}}\right]+C\right)^{2}}{16}-\frac{\mathrm{K_{cal}}\times\left[\mathrm{H^{+}} \right]}{2\times\mathrm{K_{HCOO-}}}+\mathrm{K_{cal}}\\ \hskip 56.905512 ptE\ =\ \left[\mathrm{Na^{+}}\right]+\left[\mathrm{K^{+}} \right]+2\left[\mathrm{Mg^{2+}}\right]-\left[\mathrm{C^{-1}}\right]-2\left[ \mathrm{SO_{4}^{2-}}\right]\\ \hskip 56.905512 pt\left[\mathrm{Ca^{2+}}\right]\ =\ \frac{-\left[\mathrm{H^{+}} \right]+\frac{\mathrm{K_{w}}}{\left[\mathrm{H^{+}}\right]}+\left[\mathrm{NO_{ 3}^{-}}\right]-E}{4}+\sqrt{D}\end{array} \right. \tag{21}\]  
where [] is the ionic activity of the ions, mol/L. V is the total volume of air and solution, and Va is the volume of solution, L. \(f\) is the proportion of toluene consumed by microbial growth to the total toluene content. T is the temperature of the system, \({}^{\circ}\mathrm{C}\). The concentration of H\({}^{+}\) in the system can be calculated from the equation. d is defined as the proportion of nitrate consumed by the complete reduction of nitrogen to the total amount of nitrate consumed at a certain time. The simplified model can be expressed as:  
\[\left\{ \begin{array}{c}14\times f\times[\mathrm{C_{7}H_{8}}]\ =\ \left[\mathrm{H^{+}} \right]-\frac{\mathrm{K_{w}}}{\left[\mathrm{H^{+}}\right]}-\left(1+\mathrm{d} \right)\left[\mathrm{NO_{3}^{-}}\right]+E+\left(2\times A+B+C\right)\times \frac{1}{\left[\mathrm{Ca^{2+}}\right]}\\ \hskip 56.905512 ptA\ =\ \frac{\mathrm{K_{cal}}\times\left[\mathrm{H^{+}} \right]^{2}}{\mathrm{K_{HCOO3}\times K_{HCOO-}}}\\ \hskip 56.905512 ptB\ =\ \frac{\mathrm{K_{cal}}\times\left[\mathrm{H^{+}} \right]}{\mathrm{K_{HCOO-}}}\\ \hskip 56.905512 ptC\ =\ \frac{\mathrm{K_{cal}}\times\left(\mathrm{V-Va}\right) \times\left[\mathrm{H^{+}}\right]^{2}}{\mathrm{K_{HCOO3}\times K_{HCOO-}\times K _{H4}\times K\times T\times Va}}\\ \hskip 56.905512 ptD\ =\ \frac{\left(\left[\mathrm{H^{+}}\right]-\frac{\mathrm{K_{w}}}{ \left[\mathrm{H^{+}}\right]}-\left(1+\mathrm{d}\right)\left[\mathrm{NO_{3}^{-} }\right]+C\right)^{2}}{16}-\frac{\mathrm{K_{cal}}\times\left[\mathrm{H^{+}} \right]}{2\times\mathrm{K_{HCOO-}}}+\mathrm{K_{cal}}\\ \hskip 56.905512 ptE\ =\ \left[\mathrm{Na^{+}}\right]+\left[\mathrm{K^{+}} \right]+2\left[\mathrm{Mg^{2+}}\right]-\left[\mathrm{Cl^{-1}}\right]-2\left[ \mathrm{SO_{4}^{2-}}\right]\\ \hskip 56.905512 pt\left[\mathrm{Ca^{2+}}\right]\ =\ \frac{-\left[\mathrm{H^{+}} \right]+\frac{\mathrm{K_{w}}}{\left[\mathrm{H^{+}}\right]}+\left(1+\mathrm{d} \right)\left[\mathrm{NO_{3}^{-}}\right]-E}{4}+\sqrt{D}\end{array} \right. \tag{22}\]  
Due to the large number of parameters involved in the reaction, the following scenarios are proposed. The system is saturated with calcium carbonate. Based on the data from a contaminated site that we studied, the toluene concentration was 0.1 mmol/L. According to the theoretical stoichiometric relationship, the ratio of the nitrate concentration to the toluene concentration was 7.2:1. The specific discussion is as follows.