91Ó°ÊÓ

(NH₄)₂SO₄ is an inorganic salt with a variety of commercial applications. The most typical application is for soil fertilization. It has a nitrogen content of 21% and a sulphur content of 24%.

Let's write out what we know:

$$\begin{gathered} m\left({\left({\text{NH}}_4\right)}_2{\text{SO}}_4\right)=210\text{kg} \\ \%{\text{oxidized to NO}}_3^{-}=37\% \\ V\left(\text{water}\right)=1000{\text{m}}^3 \end{gathered}$$

The following equation can be used to describe the oxidation of${\text{NH}}_{\text{4}}^{\text{+}}$ to ${\text{NO}}_{\text{3}}^{\text{-}}$:

${\text{NH}}_4^{+}+2{\text{O}}_2→{\text{NO}}_3^{-}+{\text{H}}_2\text{O}+2{\text{H}}^{+}$

To compute the amount of ${\text{NO}}_3^{-}$ions in groundwater, we must first calculate the chemical amount of ${\text{NH}}_4^{+}$ ions, after which we can calculate the amount of${\text{NO}}_3^{-}$ ions.

$$\begin{gathered} m\left({\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}\right)=210\text{kg} \\ M\left({\left({\text{NH}}_4\right)}_2{\text{SO}}_4\right)=132.14{\text{gmol}}^{-1} \\ \left({\left({\text{NH}}_4\right)}_2{\text{SO}}_{\text{4}}\right)=\frac{m\left({\left({\text{NH}}_4\right)}_2{\text{SO}}_4\right)}{M\left({\left(N{\text{H}}_4\right)}_2{\text{SO}}_{\text{4}}\right)} \\ n\left({\left({\text{NH}}_4\right)}_2{\text{SO}}_4\right)={\frac{210·{10}^3\text{g}}{132.14\text{gmol}}}^{-1} \\ n\left({\left({\text{NH}}_4\right)}_2\text{SO}4\right)=1589\text{mol} \end{gathered}$$

Because each molecule of $\left({\left({\text{NH}}_4\right)}_2\text{SO}4\right)$ includes two${\text{NH}}_4^{+}$ ions, the chemical amount of ${\text{NH}}_4^{+}$ is as follows:

$$\begin{gathered} n\left({\text{NH}}_4^{+}\right)=2·n\left({\left({\text{NH}}_4\right)}_2{\text{SO}}_4\right) \\ n\left({\text{NH}}_4^{+}\right)=3178\text{mol} \end{gathered}$$

Hence, the chemical amount of$\left({\text{NH}}_4^{+}\right)$ is $3178\text{mol}$.

We can compute the amount of ${\text{NO}}_3^{-}$using equation (1)

$$\begin{gathered} n\left(N{O}_3^{-}\right)=n\left(N{H}_4^{+}\right) \\ n\left(N{O}_3^{-}\right)=3178\text{mol} \end{gathered}$$

However, we must consider the fact that only $37\%$of ${\text{NH}}_4^{+}$gets oxidized to ${\text{NO}}_3^{-}$.

$$\begin{gathered} n{\left({\text{NO}}_3^{-}\right)}_{37\%}=3178\text{mol}×\frac{37}{100} \\ n{\left({\text{NO}}_3^{-}\right)}_{37\%}=1176\text{mol} \end{gathered}$$

Finally, we can determine the mass of nitrate in the water as well as the level of contamination:

$$\begin{gathered} n{\left({\text{NO}}_3^{-}\right)}_{37\%}=1176\text{mol} \\ m\left({\text{NO}}_3^{-}\right)=n\left({\text{NO}}_3^{-}\right)·\text{M}\left({\text{NO}}_3^{-}\right) \\ m\left({\text{NO}}_3^{-}\right)=1176\text{mol}×62.0049{\text{gmol}}^{-1} \\ m\left({\text{NO}}_3^{-}\right)=7.29×{10}^4\text{g}=7.29·{10}^7\text{mg} \\ \text{V}\left(\text{water}\right)=1000{\text{m}}^3=1×{10}^6\text{L} \\ \left[{\text{NO}}_3^{-}\right]=\frac{m\left({\text{NO}}_3^{-}\right)}{V\left(\text{water}\right)}=\frac{7.29×{10}^7\text{mg}}{1×{10}^6\text{L}} \\ \left[{\text{NO}}_3^{-}\right]=73\text{mg}/\text{L} \end{gathered}$$

Therefore, the determined mass of nitrate in the water well as the level of contamination is $73\text{mg}/\text{L}$