Problem 148 - General Chemistry

Question

A sample of \(\mathrm{NH}_{4} \mathrm{Cl}\) is prepared for titration by dissolving the salt in water, making \(100.0 \mathrm{~mL}\) of solution. A second sample of \(\mathrm{NH}_{4} \mathrm{Cl}\) of identical mass is used to prepare \(200.0 \mathrm{~mL}\) of solution. A solution of \(0.10 \mathrm{M}\) \(\mathrm{NaOH}\) or \(0.10 \mathrm{M} \mathrm{HCl}\) may be used to perform the titration. a. Which solution, the \(0.10 \mathrm{M} \mathrm{NaOH}\) or \(0.10 \mathrm{M} \mathrm{HCl}\) should be used to perform the titration? Justify your answer? b. Compare the volume of titrant selected in part a that you would need to reach the equivalence point of the titrations for each of the samples. c. Compare the \(\mathrm{pH}\) of the two solutions at the equivalence point of the titrations (assume additive volumes).

Step-by-Step Solution

Answer

Use 0.10 M NaOH as the titrant. The 200 mL solution needs twice as much NaOH as the 100 mL solution. The 100 mL solution is more basic at equivalence.

Step 1: Analyze the Nature of NH4Cl

Ammonium chloride (\(\text{NH}_4\text{Cl}\)) is a salt derived from a strong acid (hydrochloric acid, \(\text{HCl}\)) and a weak base (ammonia, \(\text{NH}_3\)). Therefore, \(\text{NH}_4\text{Cl}\) will react with a strong base like \(\text{NaOH}\) to form the weak base \(\text{NH}_3\) and water. Consequently, \(\text{NaOH}\) is the appropriate titrant as it will neutralize \(\text{NH}_4^+\) ions.

Step 2: Solution Preparation and Molarity Calculation

For both 100.0 mL and 200.0 mL solutions prepared, the mass of \(\text{NH}_4\text{Cl}\) is the same, implying the number of moles of \(\text{NH}_4\text{Cl}\) are identical in each case. However, the molarity differs: Molarity for 100.0 mL = \(\frac{\text{moles of } \text{NH}_4\text{Cl}}{0.100}\).Molarity for 200.0 mL = \(\frac{\text{moles of } \text{NH}_4\text{Cl}}{0.200}\). Thus, the molarity is halved in the 200.0 mL solution.

Step 3: Volume of NaOH Needed for Neutralization

Using \(c_1V_1 = c_2V_2\), where \(c_1\) and \(V_1\) are the molarity and volume of \(\text{NH}_4\text{Cl}\), and \(c_2\) and \(V_2\) are the molarity and volume of \(\text{NaOH}\):For 100 mL solution: \(c_{\text{NH}_4\text{Cl}} \times 0.100 = 0.10 \times V_2\)For 200 mL solution: \(c_{\text{NH}_4\text{Cl}} \times 0.200 = 0.10 \times V_2\)Since \(c_{\text{NH}_4\text{Cl}}\) is halved for the 200 mL solution, \(V_2\) will double, meaning twice the volume is needed.

Step 4: Comparing the pH at Equivalence Point

At the equivalence point of an \(\text{NH}_4\text{Cl}\) and \(\text{NaOH}\) titration, \(\text{NH}_3\) is formed, a weak base. In both solutions, completion leaves only \(\text{NH}_3\), so the \(\text{pH}\) at equivalence is determined by its concentration and affects the \(\text{pH}\) through hydrolysis: \(\text{NH}_3 + \text{H}_2\text{O} \rightleftharpoons \text{NH}_4^+ + \text{OH}^-\).The concentration of \(\text{NH}_3\) is lower in the 200.0 mL solution, making its \(\text{pH}\) closer to neutral compared to the 100.0 mL solution, which would be more basic.

Key Concepts

Ammonium Chloride

Ammonium chloride (\(\mathrm{NH}_4\mathrm{Cl}\)) is a fascinating chemical compound that plays a crucial role in various chemical reactions. It is a salt composed of the ammonium ion (\(\mathrm{NH}_4^+\)) and the chloride ion (\(\mathrm{Cl}^-\)). This compound is derived from the combination of a strong acid, hydrochloric acid (\(\mathrm{HCl}\)), and a weak base, ammonia (\(\mathrm{NH}_3\)).
The presence of these ions indicates that \(\mathrm{NH}_4\mathrm{Cl}\) will engage in chemical reactions, especially with strong bases. During a titration with a strong base such as sodium hydroxide (\(\mathrm{NaOH}\)), \(\mathrm{NH}_4\mathrm{Cl}\) will act as an acid.
In this reaction, \(\mathrm{NH}_4^+\) ions are neutralized by hydroxide ions (\(\mathrm{OH}^-\)) from the base, resulting in the formation of ammonia and water. This concept is integral in understanding how \(\mathrm{NH}_4\mathrm{Cl}\) behaves in titration setups.

NaOH Solution

Sodium hydroxide (\(\mathrm{NaOH}\)) is a fundamental reagent frequently used in chemical titrations. It is a strong alkali, which means it can completely disassociate in water, releasing hydroxide ions (\(\mathrm{OH}^-\)). These ions are critical in neutralizing acids.
In the case of \(\mathrm{NH}_4\mathrm{Cl}\) titration, \(\mathrm{NaOH}\) is used because it reacts with the \(\mathrm{NH}_4^+\) ions present in the solution, forming water and ammonia, effectively neutralizing the acidity. The use of \(\mathrm{NaOH}\) allows the determination of the amount of ammonium ion present, making it an appropriate titrant choice.
Choosing \(\mathrm{NaOH}\) over other titrants like \(\mathrm{HCl}\) is beneficial in this scenario because the reaction with \(\mathrm{NH}_4\mathrm{Cl}\) results in a more straightforward measurement of the equivalence point.

pH Equivalence Point

The pH at the equivalence point in a titration is where the amount of titrant added is stoichiometrically equivalent to the amount of substance being titrated. For ammonium chloride when titrated with sodium hydroxide, the equivalence point signifies the complete conversion of \(\mathrm{NH}_4^+\) ions to ammonia.
This transformation means that at the equivalence point, any further addition of \(\mathrm{NaOH}\) will result in the solution containing a weak base, ammonia, affecting the pH level.
Generally, the pH is above 7 due to the weak basic nature of \(\mathrm{NH}_3\). However, the concentration of ammonia formed will dictate the exact pH.
In the example provided, when the titration involves a 100 mL solution, the concentration of resulting ammonia will be higher compared to a more diluted 200 mL solution. Thus, the pH at equivalence would be higher in the 100 mL setup.

Molarity Calculation

Calculating molarity is a crucial step in understanding the strength of a solution that is about to undergo titration. Molarity is defined as the number of moles of a solute divided by the volume of the solution in liters.
Given a constant mass of \(\mathrm{NH}_4\mathrm{Cl}\), the molarity differs depending on the volume it is dissolved in. For instance, in a solution prepared with 100 mL, the molarity would be \(\frac{\text{moles of } \mathrm{NH}_4\mathrm{Cl}}{0.1}\), whereas it's \(\frac{\text{moles of } \mathrm{NH}_4\mathrm{Cl}}{0.2}\) for a 200 mL solution.
This implies that the molarity is halved in the more diluted 200 mL solution. Knowing the molarity is vital because it directly influences how much \(\mathrm{NaOH}\) is required to reach the equivalence point.

• 100 mL solution requires a smaller volume of \(\mathrm{NaOH}\) since it is more concentrated.
• The 200 mL, being more diluted, needs double the amount of \(\mathrm{NaOH}\) to achieve the same neutralization.

Understanding these differences in concentrations allows for precise control and prediction of titration outcomes.