Problem 154 - Chemistry The Molecular Science

Question

The mass percent of ammonia in a commercial window cleaner was determined by titration with HCl. A \(9.360-\mathrm{g}\) sample of the window cleaner was diluted with \(35.778 \mathrm{~g}\) water and \(4.188 \mathrm{~g}\) of this diluted solution was titrated using \(13.58 \mathrm{~mL}\) of \(0.1093-\mathrm{M} \mathrm{HCl}\) to reach the equivalence point. Calculate the mass percent of ammonia in the window cleaner.

Step-by-Step Solution

Answer

The mass percent of ammonia in the window cleaner is approximately 2.92%.

Step 1: Understand the Reaction

The reaction between ammonia (\( \text{NH}_3 \)) and hydrochloric acid (\( \text{HCl} \)) is a neutralization reaction: \( \text{NH}_3 + \text{HCl} \rightarrow \text{NH}_4\text{Cl} \). This means one mole of \( \text{NH}_3 \) reacts with one mole of \( \text{HCl} \).

Step 2: Calculate Moles of HCl Used

The moles of \( \text{HCl} \) used in the titration is calculated from its concentration and volume. Use the formula: \( \text{moles} = \text{molarity} \times \text{volume} \). Here, the volume must be converted to liters: \( 13.58 \text{ mL} = 0.01358 \text{ L} \). Thus, the moles of \( \text{HCl} \) are \( 0.1093 \text{ M} \times 0.01358 \text{ L} = 0.001485 \text{ moles} \).

Step 3: Calculate Moles of Ammonia

Since the reaction is one-to-one, the moles of \( \text{NH}_3 \) are equal to the moles of \( \text{HCl} \). Therefore, the moles of \( \text{NH}_3 \) in the 4.188 g of diluted solution is also 0.001485 moles.

Step 4: Find Mass of Ammonia

Use the molar mass of \( \text{NH}_3 \) (17.03 g/mol) to find the mass of \( \text{NH}_3 \) in the sample: \( 0.001485 \text{ moles} \times 17.03 \text{ g/mol} = 0.0253 \text{ g} \).

Step 5: Calculate Mass Percent in Diluted Solution

Calculate the mass percent of ammonia in the 4.188 g of the diluted solution: \( \frac{0.0253 \text{ g}}{4.188 \text{ g}} \times 100 \approx 0.604 \% \).

Step 6: Calculate Total Diluted Solution Volume

The total mass of the diluted solution is the sum of the window cleaner and water: \( 9.360 \text{ g} + 35.778 \text{ g} = 45.138 \text{ g} \).

Step 7: Determine Mass of Ammonia in Original Sample

Use the proportion to determine the mass of ammonia in the original window cleaner sample. From the 4.188 g of diluted solution, covering the entire 45.138 g, we know: \( \frac{0.0253 \text{ g}}{4.188 \text{ g}} = \frac{x \text{ g}}{45.138 \text{ g}} \) which gives \( x \approx 0.273 \text{ g} \).

Step 8: Calculate Mass Percent in Original Sample

Calculate the mass percent of ammonia in the original window cleaner sample: \( \frac{0.273 \text{ g}}{9.360 \text{ g}} \times 100 \approx 2.92 \% \).

Key Concepts

Neutralization Reaction

A neutralization reaction occurs when an acid reacts with a base to form water and a salt. In this context, hydrochloric acid (HCl) reacts with ammonia (NH₃). Both substances are in opposite realms of chemical nature: HCl is a strong acid, and NH₃ is a weak base. Together, they produce ammonium chloride (NH₄Cl), a salt.
The general equation is:

• NH₃ + HCl → NH₄Cl

This equation illustrates a 1-to-1 mole ratio relationship between NH₃ and HCl, which simplifies calculations since equal moles of each neutralize each other. Keep in mind that understanding this mole relationship is key in finding out how much of one reactant is required to react completely with a given amount of the other.

Molarity

Molarity is a measure of the concentration of a solution. It tells us how many moles of a solute are present in one liter of solution. Molarity is expressed in moles per liter (M). This is crucial when you are calculating how much of a solution you should use in a reaction.
In this exercise, the molarity of HCl is 0.1093 M, which means there are 0.1093 moles of HCl in every liter of solution. To calculate the number of moles in any volume, use the formula:

• number of moles = molarity × volume (in liters)

This becomes really practical when you convert the provided volume of HCl from milliliters to liters and execute the calculation. Understanding molarity and its calculation is fundamental to comprehensively solving problems involving solutions and titrations.

Equivalence Point

The equivalence point in a titration is the point at which the amount of acid equals the amount of base, signifying the complete neutralization. In our problem, when the equivalence point is reached, the moles of HCl added matches exactly with the moles of NH₃ in the solution.
This is often indicated by a sudden change in property of the solution, frequently determined using a pH indicator or a pH meter. Reaching the equivalence point signifies that the amounts of the reactants are in the exact stoichiometric ratio they need to form the products. By knowing the volume of titrant required to reach this point, you can back-calculate the unknown concentration of one reactant if you know the other, or determine unknown characteristics like mass percent as we are doing here.

Titration Analysis

Titration analysis is a method used to determine the concentration of an identified analyte by reacting it with a corresponding reagent of known concentration. In this exercise, titration allows us to find out how much ammonia is present in the cleaner solution by using hydrochloric acid as the titrant.
Here's how it works in steps:

• Prepare the sample by diluting it appropriately.
• Add a measured amount of the titrant (known concentration) to the analyte until the equivalence point is reached.
• Use the known concentration and volume of titrant to calculate the concentration (or amount) of the analyte.

Given these steps, titration helps to determine the precise concentration of a given solution and is a very effective analytic technique in many fields of chemistry.