Problem 115 - Chemistry: The Central Science

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 \(\mathrm{HCl}\) 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 \(10.0 \mathrm{~min}\) 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.) (c) Is this manufacturer in compliance with regulations?

Step-by-Step Solution

Answer

(a) The grams of NH3 drawn into the acid solution is \(5.62\times10^{-3}\,\text{g}\). (b) The concentration of NH3 in the air is 79.7 ppm. (c) The manufacturer is not in compliance with the regulations, as the calculated concentration of NH3 in the air is greater than the limit of 50 ppm.

Step 1: Calculate the initial moles of HCl

First, we need to find the initial moles of HCl present in the solution. This can be calculated using the formula: Moles = Volume × Concentration where Volume is the volume of the solution and Concentration is the concentration of HCl. Initial moles of HCl = \(1\times10^2\,\text{mL}\) × \(0.0105\,\text{M}\) = \(1.05\times10^{-3}\,\text{mol}\)

Step 2: Calculate the moles of NaOH used for titration

Next, we need to find the moles of NaOH required to titrate the remaining HCl. This can be calculated using the formula: Moles = Volume × Concentration where Volume is the volume of NaOH and Concentration is the concentration of NaOH. Moles of NaOH = \(13.1\,\text{mL}\) × \(0.0588\,\text{M}\) = \(7.70\times10^{-4}\,\text{mol}\)

Step 3: Calculate the moles of NH3

The moles of NH3 that reacted with HCl can be found by calculating the difference between the initial moles of HCl and the moles of NaOH used for titration. Moles of NH3 = Initial moles of HCl - Moles of NaOH = \(1.05\times10^{-3}\,\text{mol}\) - \(7.70\times10^{-4}\,\text{mol}\) = \(3.30\times10^{-4}\,\text{mol}\)

Step 4: (a) Calculate the grams of NH3

Now that we have the moles of NH3, we can find the grams of NH3 by using its molar mass (17.03 g/mol): Grams of NH3 = Moles of NH3 × Molar Mass of NH3 = \(3.30\times10^{-4}\,\text{mol}\) × \(17.03\,\text{g/mol}\) = \(5.62\times10^{-3}\,\text{g}\)

Step 5: Calculate the total moles of air

Next, we need to find the total moles of air in the \(10.0\,\text{min}\) time frame. First, find the total liters of air drawn through the solution by multiplying the flow rate and the time. Total liters of air = flow rate × time = \(10.0\,\text{L/min}\) × \(10.0\,\text{min}\) = \(100\,\text{L}\) Now, find the total mass of air by multiplying the density of air by the total liters of air: Mass of air = Density × Total liters of air = \(1.20\,\text{g/L}\) × \(100\,\text{L}\) = \(120\,\text{g}\) Finally, find the total moles of air by dividing the mass of the air by the average molar mass of air: Total moles of air = Mass of air / Average molar mass of air = \(120\,\text{g}\) / \(29.0\,\text{g/mol}\) = \(4.14\,\text{mol}\)

Step 6: (b) Calculate ppm of NH3 in the air

To find the concentration of NH3 in ppm, first, find the proportion of NH3 in the air by dividing the moles of NH3 by the total moles of air: Proportion of NH3 = Moles of NH3 / Total moles of air = \(3.30\times10^{-4}\,\text{mol}\) / \(4.14\,\text{mol}\) = \(7.97\times10^{-5}\) Now, convert this proportion to ppm: NH3 concentration (ppm) = Proportion of NH3 × 1,000,000 = \(7.97\times10^{-5}\) × 1,000,000 = 79.7 ppm

Step 7: (c) Check if the manufacturer meets the regulations

The manufacturer is in compliance with the regulations if the concentration of NH3 in the air is less than or equal to the limit of 50 ppm. As the calculated concentration of NH3 in the air is 79.7 ppm, which is greater than the limit of 50 ppm, the manufacturer is not in compliance with the regulations.

Key Concepts

Molar Concentration

Molar concentration, often represented by the symbol 'M', is a measure of the amount of a solute that is dissolved in a given volume of solution. It is commonly expressed in moles per liter (\text{mol/L}). Understanding molar concentration is essential in chemistry as it allows for the quantification of substances involved in chemical reactions.

For instance, in the exercise, the concentration of hydrochloric acid (\text{HCl}) in the solution is 0.0105 M. This indicates that for every liter of solution, there are 0.0105 moles of \text{HCl}. To determine the initial moles of \text{HCl}, which is a crucial step towards calculating the amount of ammonia (\text{NH}\(_3\)) in the air, you would use the formula:
\text{Initial moles of HCl = Volume} \times \text{Concentration}
This calculation initiates the process of quantifying \text{NH}\(_3\), a suspect pollutant, to assess compliance with air quality regulations.

Gas Phase Titration

Gas phase titration is a technique used to determine the concentration of a gaseous analyte by reacting it with a known volume of a titrant until an endpoint is reached. In the context of air quality monitoring, gas phase titration can be utilized to find the amount of a gaseous pollutant, such as ammonia (\text{NH}\(_3\)), in the atmosphere.

The provided exercise involves drawing air containing \text{NH}\(_3\) through a known concentration of \text{HCl}. The ammonia reacts with the acid and forms ammonium chloride (\text{NH}\(_4\text{Cl}\)). The amount of \text{NH}\(_3\) drawn into the acid solution can be deduced by titrating the remaining \text{HCl} with sodium hydroxide (\text{NaOH}) and using stoichiometry to find the moles of \text{NH}\(_3\) that reacted.

Stoichiometry

Stoichiometry is the quantitative relationship between reactants and products in a chemical reaction. It is based on the conservation of mass and the concept of molar ratios as dictated by the balanced chemical equation.

In the exercise's context, stoichiometry is applied after the \text{NH}\(_3\) present in the air reacts with \text{HCl}, and the remaining \text{HCl} is titrated with \text{NaOH}. The formula for stoichiometry used in the solution steps is:
\text{Moles of NH}\(_3\) \text{ = Initial moles of HCl - Moles of NaOH}
Using stoichiometry, the moles of \text{NH}\(_3\) that reacted with \text{HCl} are found, which later helps in calculating the concentration of \text{NH}\(_3\) in the air, thus assessing the compliance with air quality regulations.

Molar Mass

Molar mass is the weight of one mole of a substance and is usually expressed in grams per mole (g/mol). It plays a crucial role in converting between moles and grams, which is a common requirement in various stoichiometric calculations.

In our exercise, the molar mass of ammonia (\text{NH}\(_3\)) is used to convert moles of \text{NH}\(_3\) into grams. The calculation is necessary to determine the mass of \text{NH}\(_3\) that the air from the manufacturing operation contains:\text{ Grams of NH}\(_3\) \text{ = Moles of NH}\(_3\) \times \text{Molar Mass of NH}\(_3\).

Application in Air Quality

Understanding molar mass is crucial since air quality regulations are often expressed in mass units (like ppm), requiring conversion from moles for compliance evaluation.