Problem 108 - Chemistry: The Central Science in SI Units, Global Edition

Question

A fertilizer railroad car carrying \(129,840 \mathrm{~L}\) of commercial aqueous ammonia (30\% ammonia by mass) tips over and spills. The density of the aqueous ammonia solution is \(0.88 \mathrm{~g} / \mathrm{cm}^{3}\) What mass of citric acid, \(\mathrm{C}(\mathrm{OH})(\mathrm{COOH})\left(\mathrm{CH}_{2} \mathrm{COOH}\right)_{2}\) (which contains three acidic protons) is required to neutralize the spill?

Step-by-Step Solution

Answer

To neutralize the ammonia spill, \(12,891,885.3\mathrm{~g}\) of citric acid is required.

Step 1: Find the mass of ammonia

In the given exercise, the ammonia solution has a volume of \(129,840 \mathrm{~L}\) and a density of \(0.88 \mathrm{~g} / \mathrm{cm}^{3}\). Considering the percentage of ammonia in the aqueous solution, we can find the mass of ammonia: First, convert the volume of the solution to cubic centimeters: \[1\mathrm{~L} = 1,000\mathrm{~cm}^3\] \[129,840\mathrm{~L} \times 1,000\mathrm{~cm}^3/\mathrm{L} = 12,984,000\mathrm{~cm}^3\] Next, find the mass of the solution: \[12,984,000\mathrm{~cm}^3 \times 0.88\mathrm{~g}/\mathrm{cm}^3 = 11,425,920\mathrm{~g}\] Now, find the mass of ammonia in the solution (30% by mass): \[11,425,920\mathrm{~g} \times 0.30 = 3,427,776\mathrm{~g}\]

Step 2: Find the moles of ammonia

Using the molar mass of ammonia (NH\(_{3}\)), \(17.031\mathrm{~g/mol}\), we will find the moles of ammonia: \[ \text{moles of ammonia} = \frac{3,427,776\mathrm{~g}}{17.031\mathrm{~g/mol}} = 201,260.2\mathrm{~mol}\]

Step 3: Determine the moles of citric acid needed

Citric acid has three acidic protons, so we need three moles of citric acid to neutralize one mole of ammonia. Thus: \[ \text{moles of citric acid} = \frac{201,260.2\mathrm{~mol}}{3} = 67,086.7\mathrm{~mol}\]

Step 4: Find the mass of citric acid required

Finally, using the molar mass of citric acid, \(192.124\mathrm{~g/mol}\), we will find the mass of citric acid required: \[ \text{mass of citric acid} = 67,086.7\mathrm{~mol} \times 192.124\mathrm{~g/mol} = 12,891,885.3\mathrm{~g}\] To neutralize the spill, \(12,891,885.3\mathrm{~g}\) of citric acid is required.

Key Concepts

Aqueous Ammonia

Aqueous ammonia is a solution of ammonia in water, often used in fertilizers and cleaning products. It's typically described by its concentration, such as 30% ammonia by mass in this scenario.

This concentration means that out of every 100 grams of the solution, 30 grams are ammonia. Aqueous ammonia is known for its alkaline nature, which allows it to react with acids in neutralization reactions. This property makes it useful for agricultural purposes, as it can improve the efficiency of fertilizers.

• Solution Composition: Consists mainly of ammonia (NH₃) and water.
• Physical Characteristics: Colorless with a sharp, penetrating odor.
• Purpose: Used in industrial and agricultural applications.

Understanding the role of aqueous ammonia is crucial when calculating how to neutralize spills, as its properties will dictate how much acid is needed to counteract its effects.

Citric Acid

Citric acid is an organic compound primarily found in citrus fruits. Known for its role in metabolism, it's represented by the formula C₆H₈O₇.

In chemistry, citric acid functions as a weak, triprotic acid, meaning it can donate three protons (hydrogen ions) in reactions. This property makes it effective in neutralizing basic substances like aqueous ammonia. Its role as a neutralizing agent is significant in spill scenarios to ensure that the impact on the environment is minimized.

• Chemical Formula: C(OH)(COOH)(CH₂COOH)₂
• Common Uses: Food preservation, cleaning agents, and as a pH buffer.
• Neutralization: Reacts with bases to form neutral compounds.

The ability of citric acid to neutralize multiple moles of base makes it incredibly valuable in various industrial and environmental applications.

Molar Mass Calculation

Molar mass is a fundamental concept in chemistry, allowing us to convert between the mass of a substance and the amount in moles. This conversion is essential for reacting substances in stoichiometrically correct proportions.

In the exercise, two molar masses are critical: ammonia (NH₃) with molar mass of 17.031 g/mol, and citric acid (C₆H₈O₇) with a molar mass of 192.124 g/mol. These figures let us compute the mass of each substance needed for a neutralization reaction.

• Ammonia Molar Mass Calculation: Add atomic masses – N (14.007) + 3 H (3 x 1.008).
• Citric Acid Molar Mass Calculation: Add atomic masses – C (6 x 12.011) + H (8 x 1.008) + O (7 x 15.999).
• Application: Use these molar masses to find the mass from the number of moles calculated.

Mastering molar mass calculations is essential for balancing chemical equations and understanding the relationships between reactants and products in any chemical process.