Question

Federal regulations set an upper limit of 50 parts per million \((\mathrm{ppm})\) of \(\mathrm{NH}_{3}\) in the air in a work environment \([\mathrm{that}\) 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 \(\mathrm{M}\) 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 g/L and an average molar mass of 29.0 \(\mathrm{g} / \mathrm{mol}\) under the conditions of the experiment.) (c) Is this manufacturer in compliance with regulations?

Short answer

The grams of \(\mathrm{NH}_{3}\) drawn into the acid solution is approximately 0.1656 g. The concentration of \(\mathrm{NH}_{3}\) in the air is 2350 ppm, which is higher than the federal regulations limit of 50 ppm. Therefore, the manufacturer is not in compliance with the regulations.

Step-by-step solution

Calculate moles of HCl initially in the solution

To find the moles of HCl initially in the solution, we use the formula: moles = concentration × volume. The concentration of HCl is 0.0105 M and the volume is \(1.00 \times 10^2\) mL. Make sure to convert mL into L (1 mL = 0.001 L) to keep the units consistent. Calculating the moles of HCl initially in the solution: moles of HCl = (0.0105 M) × (1.00 × 10² mL × 0.001 L/mL) = 0.0105 mol

Calculate moles of NaOH at the equivalence point

The equivalence point occurs when the moles of acid and base are equal. In this case, it's moles of HCl and NaOH. Use the same formula as earlier, moles = concentration × volume, where the concentration of NaOH is 0.0588 M and the volume needed is 13.1 mL. Calculating the moles of NaOH at the equivalence point: moles of NaOH = (0.0588 M) × (13.1 mL × 0.001 L/mL) = 0.00076828 mol

Calculate moles of NH₃ that reacted with HCl

Knowing the moles of HCl initially in the solution and the moles of NaOH at the equivalence point, we can now find the moles of NH₃ that reacted with HCl. From the balanced equation, the mole ratio of NH₃:HCl is 1:1. calculating moles of NH₃ that reacted with HCl: moles of NH₃ = moles of HCl_initial - moles of NaOH_equiv_point moles of NH₃ = 0.0105 mol - 0.00076828 mol = 0.00973172 mol

Calculate grams of NH₃ that reacted with HCl

Now convert the moles of NH₃ into grams using the molar mass of NH₃ (17.03 g/mol). calculating grams of NH₃ that reacted with HCl: grams of NH₃ = moles of NH₃ × molar mass of NH₃ grams of NH₃ = 0.00973172 mol × 17.03 g/mol ≈ 0.1656 g This is the answer to part (a), there were 0.1656 grams of NH₃ drawn into the acid solution.

Calculate the number of moles of air drawn through the solution

Use the air density (1.20 g/L) and molar mass of air (29.0 g/mol) to find the moles of air drawn through the solution. The volume of air drawn through the solution is given as 10.0 L/min for 10.0 min, which is a total of 100 L of air. Calculating the mass of air drawn: mass of air = (density of air) × (volume of air) = (1.20 g/L) × (100 L) = 120 g Calculating the number of moles of air drawn: moles of air = mass of air / molar mass_air = 120 g / 29.0 g/mol ≈ 4.14 mol

Calculate the concentration of NH₃ in ppm

Now, we can find the concentration of NH₃ in parts per million (ppm) using the moles of NH₃ and moles of air. Calculating the concentration of NH₃ in ppm: ppm of NH₃ = (moles of NH₃ / moles of air) × 10^6 = (0.00973172 mol / 4.14 mol) × 10^6 ≈ 2350 ppm This is the answer to part (b), the concentration of NH₃ in the air was 2350 ppm.

Check compliance with federal regulations

The federal regulations state that the maximum allowable concentration of NH₃ in the air is 50 ppm. We calculated the concentration to be 2350 ppm, which is much higher than the limit: Is the manufacturer in compliance? No, the concentration of NH₃ in the air (2350 ppm) is higher than the federal regulations limit (50 ppm). This is the answer to part (c), the manufacturer is not in compliance with federal regulations.

Key concepts

Stoichiometry in Acid-Base Titration

Stoichiometry is a fundamental concept in chemistry that helps us understand the quantitative aspects of chemical reactions.
When dealing with acid-base titration, stoichiometry becomes essential as it helps us calculate the exact amounts of reactants and products in a chemical reaction.
In acid-base titration, the goal is to determine the concentration of an unknown solution by reacting it with a solution of known concentration.

• The stoichiometry of the reaction provides us with the mole ratio, which is crucial for calculations. In our exercise, the reaction is between ammonia (\(\text{NH}_3\)) and hydrochloric acid (\(\text{HCl}\)), with a 1:1 mole ratio.
• This means that for every mole of \(\text{NH}_3\), one mole of \(\text{HCl}\) is required for the reaction to occur completely.

Understanding this 1:1 relationship allows us to calculate the moles of ammonia that reacted by subtracting the moles of leftover \(\text{HCl}\) (after titration) from the initial moles added. This stoichiometric relationship is crucial for accurate calculations both in industrial applications and laboratory settings.

Understanding Molar Mass

Molar mass is the mass of one mole of a substance and is a key concept in converting between mass and moles.
The molar mass allows us to transition from volumes and concentrations in a solution to the actual mass of a substance.
For ammonia (\(\text{NH}_3\)), the molar mass is calculated by adding the atomic masses of nitrogen and hydrogen:

• Nitrogen (N) has an atomic mass of approximately 14.01 g/mol.
• Hydrogen (H) has an atomic mass of approximately 1.01 g/mol, and since there are three hydrogen atoms in ammonia, their combined mass is 3.03 g/mol.
• Therefore, the molar mass of \(\text{NH}_3\) is 14.01 + 3.03 = 17.04 g/mol.

Once the molar amount of \(\text{NH}_3\) is determined through stoichiometry, converting it to grams using its molar mass is straightforward.
This step is necessary when expressing amounts in a form that can be practically assessed or measured, such as in industry for safety compliance as illustrated in the exercise.

Chemical Reactions and their Importance

Chemical reactions describe the process by which reactants are transformed into products.
They are fundamental in numerous applications, including industrial processes, environmental interactions, and biological systems.
In the given problem, we deal with a specific chemical reaction between ammonia (\(\text{NH}_3\)) and hydrochloric acid (\(\text{HCl}\)), which yields ammonium chloride (\(\text{NH}_4\text{Cl}\)).

• This reaction is an example of a neutralization reaction, where an acid reacts with a base to produce a salt and usually water.
• The balanced chemical equation is essential because it not only shows us what products are formed but also the necessary proportions of each reactant.
• In this example, understanding the reaction allows us to determine that any amount of ammonia introduced into the acid solution is completely captured due to this complete neutralization process.

Chemical reactions like this one are essential to ensure precise measurements and compliance with regulations, ensuring that environments meet safety standards as highlighted in the exercise scenario.