Problem 115 - Chemistry The Central Science In Si Units

Question

Federal regulations set an upper limit of 50 parts per million (ppm) of \(\mathrm{NH}_{3}\) in the air in a work environment [that is, 50 molecules of \(\mathrm{NH}_{3}(g)\) for every million molecules in the air]. Air from a manufacturing operation was drawn through a solution containing \(1.00 \times 10^{2} \mathrm{~mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\). The \(\mathrm{NH}_{3}\) reacts with HCl according to: $$ \mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q) $$ After drawing air through the acid solution for 10.0 min at a rate of \(10.0 \mathrm{~L} / \mathrm{min},\) the acid was titrated. The remaining acid needed \(13.1 \mathrm{~mL}\) of \(0.0588 \mathrm{M} \mathrm{NaOH}\) to reach the equivalence point. (a) How many grams of \(\mathrm{NH}_{3}\) were drawn into the acid solution? (b) How many ppm of \(\mathrm{NH}_{3}\) were in the air? (Air has a density of \(1.20 \mathrm{~g} / \mathrm{L}\) and an average molar mass of \(29.0 \mathrm{~g} / \mathrm{mol}\) under the conditions of the experiment.) \((\mathbf{c})\) Is this manufacturer in compliance with regulations?

Step-by-Step Solution

Answer

The air contains 67.89 ppm \( \mathrm{NH}_3 \), so the manufacturer is not compliant with regulations.

Step 1: Calculate the initial moles of HCl

Calculate the initial moles of HCl present in the solution using the formula: \( \text{moles of HCl} = M \times V \). Given: \( M = 0.0105 \text{ M} \) and \( V = 0.100 \text{ L} \), we have: \[ \text{moles of HCl} = 0.0105 \times 0.100 = 0.00105 \text{ moles} \]

Step 2: Calculate the moles of NaOH used

Calculate the moles of NaOH used in the titration to neutralize the remaining HCl after reaction with \( \mathrm{NH_3} \). Using the formula: \( \text{moles of NaOH} = M \times V \). Given: \( M = 0.0588 \text{ M} \) and \( V = 0.0131 \text{ L} \), we have: \[ \text{moles of NaOH} = 0.0588 \times 0.0131 = 0.00076908 \text{ moles} \]

Step 3: Calculate the moles of HCl that reacted with NH3

The moles of HCl that reacted with \( \mathrm{NH_3} \) are the initial moles of HCl minus the moles of HCl neutralized by NaOH. \[ \text{moles of HCl that reacted} = 0.00105 - 0.00076908 = 0.00028092 \text{ moles} \]Since \( \text{NH}_3 \) and HCl react in a 1:1 molar ratio, the moles of \( \mathrm{NH_3} \) are also \( 0.00028092 \text{ moles} \).

Step 4: Calculate the grams of NH3 drawn into the solution

Find the mass of \( \mathrm{NH_3} \) using the molar mass of \( \mathrm{NH_3} \) which is approximately \( 17.03 \text{ g/mol} \).\[ \text{mass of } \mathrm{NH}_3 = 0.00028092 \text{ moles} \times 17.03 \text{ g/mol} = 0.004784266 \, \text{grams} \]

Step 5: Calculate the total volume of air drawn through the solution

Calculate the total air volume passed through the solution: \( \text{Volume} = \text{rate} \times \text{time} \). Given: \( \text{rate} = 10.0 \text{ L/min} \) and \( \text{time} = 10.0 \text{ min} \), we have:\[ \text{total volume of air} = 10.0 \times 10.0 = 100.0 \text{ L} \]

Step 6: Calculate moles of air

Calculate the moles of air using the formula: \( \text{moles} = \frac{\text{mass}}{\text{molar mass}} \). Calculate mass of air:\[ \text{mass} = \text{density} \times \text{volume} = 1.20 \text{ g/L} \times 100.0 \text{ L} = 120 \text{ g} \] \[ \text{moles of air} = \frac{120}{29.0} = 4.137931 \text{ moles} \]

Step 7: Convert to parts per million (ppm)

Calculate ppm of \( \mathrm{NH_3} \) in the air: \[ \text{ppm} = \left(\frac{0.00028092 \text{ moles of } \mathrm{NH}_3 \times \frac{1}{4.137931 \text{ moles of air}}}{1,000,000} \right) \times 1,000,000 = 67.89 \text{ ppm} \]

Step 8: Determine compliance with regulations

Since the calculated ppm of \( \mathrm{NH}_3 \) in the air is 67.89 ppm, which is greater than the federal limit of 50 ppm, the manufacturer is not in compliance with federal regulations.

Key Concepts

Ammonia Detection

Ammonia detection in air is a crucial process, especially in industrial environments where workers' safety is at stake. Detecting ammonia involves monitoring air quality to ensure safe levels are maintained.
Ammonia \(NH_3\) is a common byproduct in manufacturing processes and can be harmful when inhaled in high concentrations. Its detection is usually performed through chemical reactions, where captured air is bubbled through a liquid that reacts with ammonia and changes properties. This change can be monitored and quantified to determine ammonia levels in the air.
In our exercise, air was sampled through a hydrochloric acid \(HCl\) solution, reacting with ammonia to form ammonium chloride \(NH_4Cl\). The extent of this reaction helps us quantify the ammonia content in the air. Understanding the chemical basis of detection is key in ensuring workplace safety.

Federal Regulations

Federal regulations set limits to maintain safety and environmental standards. These regulations, particularly in industrial settings, dictate the permissible levels of chemicals like ammonia in the air.
Governments and regulatory bodies establish these limits to protect workers' health. For ammonia, the limit is often set at 50 parts per million (ppm).
Compliance with these regulations involves regular monitoring and ensuring that ammonia concentrations do not exceed safe levels.

• Regular monitoring helps preemptively tackle any increase in hazardous concentrations.
• Regular audits and checks ensure that safety measures are not only established but are effectively maintained.

In our example, the ppm was found to be over the allowed limit, indicating a need for improved safety measures.

Stoichiometry

Stoichiometry is key in calculating the amounts of reactants and products in a chemical reaction. It involves using balanced chemical equations to find quantitative relationships between substances.
In the context of ammonia detection, stoichiometry is used to calculate how much ammonia reacts with hydrochloric acid:\[\text{NH}_3(aq) + \text{HCl}(aq) \rightarrow \text{NH}_4Cl(aq)\]
Given the molar relationships, you can calculate how much ammonia was present in the air based on the amount of hydrochloric acid that reacted. Our step-by-step breakdown showed that the molarity of remaining acid after reaction, and the amount neutralized by NaOH, helped determine the exact ammonia amount.

Solution Titration

Solution titration is a precise analytical method used to determine concentration. In titration, a solution of known concentration, called the titrant, is added to a solution of unknown concentration until the reaction reaches its equivalence point.
In our scenario, after allowing the ammonia to react with hydrochloric acid, the remaining HCl was titrated with a sodium hydroxide \(NaOH\) solution to determine the leftover \(HCl\). Counted using:

• The titrant's concentration and volume aid in calculating the amount of analyte initially present.
• By knowing how much \(NaOH\) was needed to neutralize the remaining \(HCl\), one can back-calculate to find out the ammonia amount initially reacted.

The balance and precision of solution titration make it invaluable in determining concentrations, crucial in assessing regulatory compliance. Thus, it's a central technique in environmental monitoring and quality control.