Farmers use ammonium sulfate as a fertilizer. In the soil. nitrifying bacteria oxidize \(\mathrm{NH}_{4}^{+}\) to \(\mathrm{NO}_{3}^{-},\) a groundwater contaminant that causes methemoglobinemia ("blue baby" syndrome). The World Health Organization standard for maximum \(\left[\mathrm{NO}_{3}^{-}\right]\) in groundwater is \(45 \mathrm{mg} / \mathrm{L}\). A farmer adds \(210 . \mathrm{kg}\) of \(\left(\mathrm{NH}_{4}\right)_{2} \mathrm{SO}_{4}\) to a field and \(37 \%\) is oxidized to \(\mathrm{NO}_{3}^{-}\). What is the groundwater \(\left[\mathrm{NO}_{3}^{-}\right]\) (in \(\mathrm{mg} / \mathrm{L}\) ) if \(1000 . \mathrm{m}^{3}\) of the water is contaminated?

Short Answer

Expert verified
The concentration of \(\text{NO}_{3}^{-}\) in the groundwater is 36.50 mg/L.

Step by step solution

01

Find Molar Mass of \(\text{(NH}_{4}\text{)}_{2}\text{SO}_{4}\)

Calculate the molar mass of \(\text{(NH}_{4}\text{)}_{2}\text{SO}_{4}\) by adding the atomic masses of its elements. The formula is: \[2 \times \text{N} + 8 \times \text{H} + 1 \times \text{S} + 4 \times \text{O} = 2 \times 14 + 8 \times 1 + 32 + 4 \times 16 = 132 \text{g/mol}\text{.}\]
02

Convert Mass to Moles

Convert the mass of \(\text{(NH}_{4}\text{)}_{2}\text{SO}_{4}\) to moles. Given mass = 210 kg = 210,000 grams. \[\text{Moles of (NH}_{4}\text{)}_{2}\text{SO}_{4} = \frac{210,000 \text{ g}}{132 \text{ g/mol}} = 1,590.91\text{ moles}\]
03

Calculate Moles of \(\text{NO}_{3}^{-}\)

Since 37% of \(\text{(NH}_{4}\text{)}_{2}\text{SO}_{4} \) is oxidized, calculate the moles of \(\text{NO}_{3}^{-}\) produced. \[\text{Moles of } \text{NO}_{3}^{-} = 1,590.91 \text{ moles} \times 0.37 = 588.64 \text{ moles}\]
04

Convert Moles of \(\text{NO}_{3}^{-}\) to Mass

Convert moles of \(\text{NO}_{3}^{-}\) to grams using its molar mass. Molar mass of \(\text{NO}_{3}^{-}\) is \(62 \text{ g/mol}\). \[\text{Mass of } \text{NO}_{3}^{-} = 588.64 \text{ moles} \times 62 \text{ g/mol} = 36,495.68 \text{ g}\]
05

Calculate Concentration in \(\text{mg}/\text{L}\)

Convert the mass of \(\text{NO}_{3}^{-}\) to milligrams and divide by the volume of water in liters. 1,000 m^3 equals 1,000,000 liters. \[\frac{36,495.68 \text{ g}}{1,000,000 \text{ L}} \times 1,000 = 36.50 \text{ mg/L}\]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Ammonium Sulfate Fertilizer
Ammonium sulfate \(\text{((NH}_4\text{)}_2\text{SO}_4)\) is a common nitrogen fertilizer used in agriculture. It provides essential nutrients, like nitrogen and sulfur, to crops. The nitrogen in ammonium sulfate is in the form of ammonium ions \(NH_{4}^{+}\), which plants can directly absorb and use for growth.
However, ammonium sulfate does more than just nourish plants. When applied to the soil, nitrifying bacteria convert ammonium \(NH_{4}^{+}\) into nitrate ions \(NO_{3}^{-}\). This conversion process is essential because nitrate is more easily absorbed by plant roots compared to ammonium.
But there is a downside. If not managed correctly, excess nitrate can leach into groundwater. Farmers must be mindful of how much and how often they apply fertilizers like ammonium sulfate to avoid pollution. Understanding how fertilizers impact both plant health and environmental health is crucial in modern agriculture.
Nitrifying Bacteria
Nitrifying bacteria play a key role in the nitrogen cycle. These bacteria convert ammonium \(NH_{4}^{+}\) into nitrites \(NO_{2}^{-}\) and then into nitrates \(NO_{3}^{-}\), through a process known as nitrification. This transformation has significant implications for both agriculture and environmental science.
The nitrifying process involves two main types of bacteria. First, \(\text{Nitrosomonas}\) convert ammonium into nitrites. Then, \(\text{Nitrobacter}\) convert nitrites into nitrates. Nitrates are the form of nitrogen that plants primarily use for growth.
However, this beneficial process comes with risks. The overuse of nitrogen fertilizers can lead to an excess of nitrates in the soil. These nitrates can travel through the soil and enter groundwater supplies. This contamination poses a risk to human health and the environment.
Understanding nitrifying bacteria helps us appreciate the balance needed in fertilizing crops. Proper management practices, such as controlled fertilizer application and crop rotation, can mitigate the negative impacts while promoting healthy plant growth.
Methemoglobinemia
Methemoglobinemia, also known as 'blue baby syndrome,' is a serious health condition caused by high levels of nitrate in drinking water. Nitrate \(NO_{3}^{-}\), once ingested, can be converted to nitrite \(NO_{2}^{-}\) in the human body. In infants, nitrite can interfere with the oxygen-carrying capacity of hemoglobin in the bloodstream.
Hemoglobin is responsible for transporting oxygen from the lungs to other parts of the body. Nitrites convert hemoglobin into methemoglobin, which cannot carry oxygen effectively. This reduction in oxygen transport can lead to symptoms such as bluish skin, hence the name 'blue baby syndrome.'
The World Health Organization (WHO) has set a maximum permissible level of 45 mg/L for nitrates in drinking water to protect against methemoglobinemia. Farmers must consider these guidelines when using fertilizers to prevent groundwater contamination.
Understanding the connection between agricultural practices and public health is critical. By managing the use of fertilizers like ammonium sulfate and monitoring nitrate levels, we can protect both food production and water quality.

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Most popular questions from this chapter

A key part of the carbon cycle is the fixation of \(\mathrm{CO}_{2}\) by photosynthesis to produce carbohydrates and oxygen gas. (a) Using the formula \(\left(\mathrm{CH}_{2} \mathrm{O}\right)_{n}\) to represent a carbohydrate, write a balanced equation for the photosynthetic reaction. (b) If a tree fixes \(48 \mathrm{~g}\) of \(\mathrm{CO}_{2}\) per day, what volume of \(\mathrm{O}_{2}\) gas measured at 1.0 atm and \(78^{\circ} \mathrm{F}\) does the tree produce per day? (c) What volume of air ( \(0.040 \mathrm{~mol} \% \mathrm{CO}_{2}\) ) at the same conditions contains this amount of \(\mathrm{CO}_{2} ?\)

Acid mine drainage (AMD) occurs when geological deposits containing pyrite (FeS \(_{2}\) ) are exposed to oxygen and moisture. AMD is generated in a multistep process catalyzed by acidophilic (acid-loving) bacteria. Balance each step and identify those that increase acidity: (1) \(\mathrm{FeS}_{2}(s)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\) (2) \(\mathrm{Fe}^{2+}(a q)+\mathrm{O}_{2}(g) \longrightarrow \mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l)\) (3) \(\mathrm{Fe}^{3+}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{Fe}(\mathrm{OH})_{3}(s)+\mathrm{H}^{+}(a q)\) (4) \(\operatorname{FeS}_{2}(s)+\mathrm{Fe}^{3+}(a q) \longrightarrow \mathrm{Fe}^{2+}(a q)+\mathrm{SO}_{4}^{2-}(a q)\)

The lead(IV) oxide used in car batteries is prepared by coating an electrode plate with \(\mathrm{PbO}\) and then oxidizing it to lead dioxide \(\left(P b O_{2}\right)\). Despite its name, \(P b O_{2}\) has a nonstoichiometric mole ratio of lead to oxygen of about \(1 / 1.98 .\) In fact, the holes in the \(\mathrm{PbO}_{2}\) crystal structure due to missing \(\mathrm{O}\) atoms are responsible for the oxide's conductivity. (a) What is the mole \(\%\) of \(\mathrm{O}\) missing from the \(\mathrm{PbO}_{2}\) structure? (b) What is the molar mass of the nonstoichiometric compound?

A blast furnace uses \(\mathrm{Fe}_{2} \mathrm{O}_{3}\) to produce \(8400 . \mathrm{t}\) of Fe per day. (a) What mass of \(\mathrm{CO}_{2}\), is produced each day? (b) Compare this amount of \(\mathrm{CO}_{2}\) with that produced by 1.0 million automobiles, each burning 5.0 gal of gasoline a day. Assume that gasoline has the formula \(\mathrm{C}_{\mathrm{s}} \mathrm{H}_{1 \mathrm{~s}}\) and a density of \(0.74 \mathrm{~g} / \mathrm{mL},\) and that it burns completely. (Note that U.S. gasoline consumption is over \(4 \times 10^{8}\) gal/day.)

The key reaction (unbalanced) in the manufacture of synthetic cryolite for aluminum electrolysis is $$ \mathrm{HF}(g)+\mathrm{Al}(\mathrm{OH})_{3}(s)+\mathrm{NaOH}(a q) \longrightarrow \mathrm{Na}_{3} \mathrm{AlF}_{6}(a q)+\mathrm{H}_{2} \mathrm{O}(l) $$ Assuming a \(95.6 \%\) yield of dried, crystallized product, what mass (in kg) of cryolite can be obtained from the reaction of \(365 \mathrm{~kg}\) of \(\mathrm{Al}(\mathrm{OH})_{3}, 1.20 \mathrm{~m}^{3}\) of \(50.0 \%\) by mass aqueous \(\mathrm{NaOH}(d=1.53 \mathrm{~g} / \mathrm{mL})\) and \(265 \mathrm{~m}^{3}\) of gaseous \(\mathrm{HF}\) at \(305 \mathrm{kPa}\) and \(91.5^{\circ} \mathrm{C}\) ? (Assume that the ideal gas law holds.

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