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 \(\mathrm{HCl}\) as follows: $$\mathrm{NH}_{3}(a q)+\mathrm{HCl}(a q) \longrightarrow \mathrm{NH}_{4} \mathrm{Cl}(a q)$$ 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) 4.755 x 10^-3 g of NH3 were drawn into the acid solution. (b) 67.6 ppm of NH3 were in the air. (c) The manufacturer is not in compliance with regulations, as the calculated ppm of NH3 (67.6 ppm) is greater than the federal regulation limit (50 ppm).

Step 1: Calculate moles of HCl initially present

Determine the moles of HCl in the initial solution. moles = M * V where M is the molarity and V is the volume. moles_HCl_initial = (0.0105 mol/L) * (1.00 x 10^2 mL) moles_HCl_initial = 1.05 x 10^-3 mol

Step 2: Calculate moles of NaOH used in titration

Calculate the moles of NaOH used to titrate the remaining HCl. moles = M * V moles_NaOH = (0.0588 mol/L) * (13.1 mL) moles_NaOH = 7.707 x 10^-4 mol

Step 3: Determine moles of NH3

Find the moles of NH3 by comparing the moles of HCl before and after the reaction. moles_NH3 = moles_HCl_initial - moles_NaOH moles_NH3 = (1.05 x 10^-3) - (7.707 x 10^-4) moles_NH3 = 2.793 x 10^-4 mol

Step 4: Convert moles of NH3 to grams

Convert moles of NH3 to grams using the molar mass of NH3. molar_mass_NH3 = 17.03 g/mol grams_NH3 = moles_NH3 * molar_mass_NH3 grams_NH3 = (2.793 x 10^-4 mol) * (17.03 g/mol) grams_NH3 = 4.755 x 10^-3 g

Step 5: Calculate ppm of NH3 in the air

Calculate the ppm of NH3 in the air based on the grams of NH3 and the total air passed through the solution. Total air passed through = rate * time Total air = (10.0 L/min) * (10.0 min) Total air = 100.0 L Total mass of air = density * volume Total mass of air = (1.20 g/L) * (100.0 L) Total mass of air = 120.0 g moles_air = mass_air / molar_mass_air moles_air = 120.0 g / 29.0 g/mol moles_air = 4.138 mol From the definition of ppm, we have: ppm_NH3 = (moles_NH3 / moles_total) * 10^6 ppm_NH3 = (2.793 x 10^-4 / 4.138) * 10^6 ppm_NH3 = 67.6 ppm

Step 6: Determine compliance with regulations

Compare the calculated ppm of NH3 to the federal regulation limit (50 ppm). Since 67.6 ppm > 50 ppm, the manufacturer is not in compliance with the regulations.

Key Concepts

Chemical Reaction Calculations

Understanding chemical reaction calculations is essential for determining the amount of reactants and products involved in a chemical reaction. This involves using the mole concept to relate quantities of different substances. For instance, in the provided exercise, the reaction between ammonia () and hydrochloric acid () to form ammonium chloride () is the basis of the calculation.

First, knowing the initial volume and concentration of HCl, the initial moles of HCl are calculated. It's critical to understand that the molarity (), a measure of concentration, is defined as moles of solute per liter of solution. This can be combined with volume, measured in milliliters (converted to liters), to find the mole quantity: ). Step by step, we can use molarity () and the volume () of HCl to find the initial number of moles. Next, the exercise provides the volume and molarity of NaOH used to titrate the remaining HCl. Again, molarity and volume reveal the moles of NaOH, showing us how much HCl was unused and thus the amount of ammonia that reacted.

Finally, to understand how much ammonia was in the air, we convert moles of ammonia to grams using its molar mass (1 mol of weighs 17.03 g). Such systematic, mole-based calculations are indispensable for quantitative descriptions of chemical reactions.

PPM Concentration

PPM, or parts per million, is a unit of measurement that describes the concentration of a substance in another substance. It's particularly useful in scenarios like monitoring pollution levels, where concentrations are typically very low. In the context of the exercise, it helps in expressing the amount of ammonia in the air of a work environment.

To find ppm concentration, we first need to convert the quantities of interest into comparable units, here, into moles. Since there's already a known volume of air that the ammonia is mixed with, and we've calculated the moles of ammonia present, the ppm concentration is found by dividing the moles of ammonia by the total moles of air and then multiplying by (a million). The exercise requires understanding the density and molar mass of air to calculate its total number of moles.

This is a prime example of how ppm can provide a clear and understandable quantitative measure even when dealing with extremely small quantities, a topic that is often seen as complex but is essential in fields such as environmental science and health and safety regulations.

Titration Analysis

Titration analysis is a laboratory method of quantitative chemical analysis used to determine the concentration of a known reactant. Through this technique, a solution of known concentration (titrant) is used to react with a solution of unknown concentration (analyte) until the reaction reaches its endpoint, which can be indicated by a color change due to a pH indicator or by other means.

In our exercise, titration analysis comes into play after the air (containing ammonia) has reacted with HCl. The remaining HCl, which did not react with ammonia, is then titrated using NaOH, a strong base, until the equivalence point is reached. Here, stoichiometry allows us to subtract the moles of NaOH used from the initial moles of HCl to find the moles of ammonia that were in the air sample.

The titration process hinges on the concept of mole-to-mole relationships dictated by the balanced chemical equations, highlighting the importance of stoichiometry to derive accurate, quantitative information about the analyte. Understanding titration analysis is fundamental for various applications, from pharmaceuticals to environmental testing.