Problem 25 - Chemistry and Chemical Reactivity

Question

You require \(36.78 \mathrm{mL}\) of \(0.0105 \mathrm{M} \mathrm{HCl}\) to reach the equivalence point in the titration of 25.0 mL of aqueous ammonia. (a) What was the concentration of \(\mathrm{NH}_{3}\) in the original ammonia solution? (b) What are the concentrations of \(\mathrm{H}_{3} \mathrm{O}^{+}, \mathrm{OH}^{-},\) and \(\mathrm{NH}_{4}^{+}\) at the equivalence point? (c) What is the pH of the solution at the equivalence point?

Step-by-Step Solution

Answer

(a) 0.0154 M NH₃, (b) [NH₄⁺] = 0.00625 M, (c) pH = 5.2

Step 1: Calculate Moles of HCl

To find the moles of HCl used, we use the formula: \( ext{moles} = M imes V \). Given that the concentration \( M \) of \( ext{HCl} \) is \( 0.0105 \mathrm{M} \) and the volume \( V \) is \( 36.78 \mathrm{mL} \) (or \( 0.03678 \text{L} \)), we calculate: \[ \text{moles of HCl} = 0.0105 \mathrm{M} \times 0.03678 \text{L} = 3.8619 \times 10^{-4} \text{ moles}\]

Step 2: Determine Moles of NH3

Since the reaction between \( ext{HCl} \) and \( ext{NH}_3 \) at equivalence point is \( ext{HCl} + ext{NH}_3 \rightarrow ext{NH}_4 ext{Cl} \) and it proceeds on a 1:1 mole basis, the moles of \( ext{NH}_3 \) are also \( 3.8619 \times 10^{-4} \).

Step 3: Calculate Concentration of NH3

Using the moles of \( ext{NH}_3 \) found and the initial volume of the ammonia solution (\( 25.0 \mathrm{mL} = 0.025 \text{L} \)), calculate the concentration: \[ \text{Concentration of } \text{NH}_3 = \frac{3.8619 \times 10^{-4} \text{ moles}}{0.025 \text{L}} = 0.0154476 \mathrm{M}\]

Step 4: Determine Concentrations at Equivalence Point

At equivalence, the concentration of \( \text{NH}_4^+ \) equals the moles of \( \text{NH}_3 \) initially since each mole of \( \text{NH}_3 \) gives one mole of \( \text{NH}_4^+ \). Thus, \[ \text{Concentration of } \text{NH}_4^+ = \frac{3.8619 \times 10^{-4}}{0.06178 \text{L}} = 0.00625 \mathrm{M} \] For the equilibrium, considering water autoionization \( K_w = [\text{H}_3\text{O}^+][\text{OH}^-] \), we need \( K_b \) of \( \text{NH}_3 \) to find \( [\text{OH}^-] \). But, equivalence point largely features \( \text{NH}_4^+ \) in slightly acidic condition, hence calculating directly using Ka.

Step 5: Calculate pH at Equivalence Point

Using Ka for \( ext{NH}_4^+ ightarrow ext{NH}_3 + ext{H}^+ \, (Ka = 5.56 \times 10^{-10}) \), and assuming \([\text{H}_3\text{O}^+] \approx \sqrt{Ka \times C} \) since solution highly dilute:\[ [\text{H}_3\text{O}^+] = \sqrt{5.56 \times 10^{-10} \times 0.00625} \approx 5.9 \times 10^{-6} \] Hence\[\text{pH} = -\log([\text{H}_3\text{O}^+]) \approx 5.2 \]

Key Concepts

Understanding the Equivalence Point

The equivalence point in titration is a fundamental concept in chemistry. It is the stage where the number of moles of titrant added equals the number of moles of substance in the sample being analyzed. At this point, the reaction between the titrant and the analyte is complete. In this particular exercise, an aqueous ammonia solution
is titrated with hydrochloric acid (HCl). The equivalence point is reached when the moles of HCl added match the moles of ammonia (NH extsubscript{3}).

This means that all the ammonia has reacted with the HCl, forming ammonium chloride (NH extsubscript{4}Cl). To calculate the equivalence point, we determine the moles of HCl used, and since the reaction is 1:1, the moles of NH extsubscript{3} present initially are equal to these moles.

Properties of Ammonia Solution

Ammonia solution is a commonly used weak base in chemistry. It consists of ammonia gas dissolved in water, and is characterized by its pungent smell. In this exercise, the original concentration of the ammonia solution can be found by using a simple calculation:

Once we know how many moles of NH extsubscript{3} were present (the same number as the moles of HCl added), and knowing the initial volume of the ammonia solution, we can find its concentration by dividing moles by volume:

• Moles of NH extsubscript{3} = 3.8619 × 10\(^{-4}\) moles
• Volume of NH extsubscript{3} solution = 0.025 L

This allows us to calculate the concentration of NH extsubscript{3}, which is vital in understanding the strength of the solution before titration occurs.

Concentration of Hydrochloric Acid and its Role

Hydrochloric acid (HCl) is a strong acid often used in titration because of its ability to fully dissociate in water. In this case, we know the concentration and volume of HCl used, allowing us to calculate the moles of HCl. This is critical because titration calculations center around the precise stoichiometry of HCl and NH extsubscript{3}:

• HCl concentration = 0.0105 M
• Volume of HCl used at equivalence = 36.78 mL or 0.03678 L

With these values, we calculate the moles of HCl using the formula: moles = concentration × volume. Finding the moles of HCl used helps us determine the number of moles of NH extsubscript{3} in the original solution and sets the stage for further calculations, including finding the pH at the equivalence point, which is 5.2 in our example. Understanding HCl concentration is thus key in solving this titration problem.