Q22.75CP - Chemistry: Molecular Nature Of Matter And Change

Question

Farmers use ammonium sulfate as a fertilizer. In the soil, nitrifying bacteria oxidize  $\mathbf{\text{N}}{{\mathbf{\text{H}}}_{\mathbf{\text{4}}}}^{\mathbf{\text{ + }}}$$\mathbf{\text{N}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{ - }}}$a groundwater contaminant that causes methemoglobinemia ("blue baby" syndrome). The World Health Organization standard for maximum $\left[\text{N}{{\text{O}}_{\text{3}}}^{\text{ - }}\right]$ in groundwater is $\mathbf{\text{45mg/L}}$ . A farmer adds  $\mathbf{\text{210.kg}}$ of  ${\left({\text{NH}}_{\text{4}}\right)}_{\mathbf{\text{2}}}$${\mathbf{\text{SO}}}_{\mathbf{\text{4}}}$ to a field and $\mathbf{\text{37\% }}$ is oxidized to $\mathbf{\text{N}}{{\mathbf{\text{O}}}_{\mathbf{\text{3}}}}^{\mathbf{\text{ - }}}$ What is the groundwater $\mathbf{\text{(inmg/L)}}$  if ${\mathbf{\text{1000.m}}}^{\mathbf{\text{3}}}$f the water is contaminated?

Step-by-Step Solution

Answer

The chemical amount of  ${\text{NH}}_4^{+}$ is $3178 \text{mol}$ .

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

Step 1: Concept introduction

(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%.

Step 2: Determining the chemical amount

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 ${\text{NO}}_{\text{3}}^{\text{ - }}$ :

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

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  $3178 \text{mol}$.

Step 3: Computing the equation

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 ${\text{NO}}_3^{-}$ .

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

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

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

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