Question

Calculate the \(\mathrm{pH}\) after 0.010 mole of gaseous \(\mathrm{HCl}\) is added to \(250.0 \mathrm{mL}\) of each of the following buffered solutions. a. \(0.050 M \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 M \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their pH or their capacity? What advantage is there in having a buffer with a greater capacity?

Short answer

The pH of both buffered solutions after adding 0.010 mol of HCl is approximately 8.98. The two original buffered solutions have the same pH but differ in capacity. The 0.50 M NH₃ / 1.50 M NH₄Cl buffered solution has a greater capacity compared to the 0.050 M NH₃ / 0.15 M NH₄Cl buffered solution, which makes it more efficient and resistant to pH changes when more acid or base is added to the system.

Step-by-step solution

Calculate the initial moles of NH₃ and NH₄Cl

In a 250 mL solution: Moles of NH₃ = Molarity × Volume = 0.050 M × 0.250 L = 0.0125 mol Moles of NH₄Cl = Molarity × Volume = 0.15 M × 0.250 L = 0.0375 mol

Reaction between HCl and NH₃

When 0.010 mol of gaseous HCl is added, it will react with NH₃ to form NH₄⁺ ions. As the amount of HCl is less than NH₃, 0.010 mol of NH₃ will react with 0.010 mol of HCl to form 0.010 mol of NH₄⁺. Moles of NH₃ after reaction = 0.0125 - 0.010 = 0.0025 mol Moles of NH₄⁺ after reaction = 0.0375 + 0.010 = 0.0475 mol

Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is: pH = pK_a + log ([NH₃]/[NH₄⁺]) The pK_a of NH₄⁺ is 9.25. Hence, the pH = 9.25 + log (0.0025/0.0475) = 9.25 + log(0.05263) ≈ 8.98 #b. 0.50 M NH₃ / 1.50 M NH₄Cl buffered solution#

Calculate the initial moles of NH₃ and NH₄Cl

In a 250 mL solution: Moles of NH₃ = Molarity × Volume = 0.50 M × 0.250 L = 0.125 mol Moles of NH₄Cl = Molarity × Volume = 1.50 M × 0.250 L = 0.375 mol

Reaction between HCl and NH₃

When 0.010 mol of gaseous HCl is added, it will react with NH₃ to form NH₄⁺ ions. As the amount of HCl is less than NH₃, 0.010 mol of NH₃ will react with 0.010 mol of HCl to form 0.010 mol of NH₄⁺. Moles of NH₃ after reaction = 0.125 - 0.010 = 0.115 mol Moles of NH₄⁺ after reaction = 0.375 + 0.010 = 0.385 mol

Use the Henderson-Hasselbalch equation to find pH

The Henderson-Hasselbalch equation is: pH = pK_a + log ([NH₃]/[NH₄⁺]) Hence, the pH = 9.25 + log (0.115/0.385) = 9.25 + log(0.2987) ≈ 8.98

Comparison of pH and capacity between the two buffered solutions

The two original buffered solutions have the same pH values (8.98), but their capacities are different. A buffer with greater capacity contains more moles of NH₃ and NH₄⁺ ions, allowing it to maintain the pH even when more acid or base is added. The advantage of having a buffer with greater capacity is to maintain the pH under a broader range of conditions, making it more efficient and resistant to pH changes when more acid or base is added to the system. In this case, the 0.50 M NH₃ / 1.50 M NH₄Cl buffered solution has a greater capacity compared to the 0.050 M NH₃ / 0.15 M NH₄Cl buffered solution.

Key concepts

Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is a fundamental tool in understanding buffer solutions. It's used to calculate the pH of a buffer solution. The equation is: \[\text{pH} = \text{p}K_a + \log \left( \frac{[\text{Base}]}{[\text{Acid}]} \right)\] In our problem, ammonia \((\text{NH}_3)\) acts as the base, while ammonium chloride \((\text{NH}_4^+\)) represents the acid. Here,

• \(\text{p}K_a\) is a constant that depends on the weak acid involved. For ammonium, \(\text{p}K_a = 9.25\).
• The equation helps you find the pH by using the concentrations (or moles) of the base and acid in the buffer.

The Henderson-Hasselbalch equation simplifies the calculations, making it easier to predict how the buffer will respond to the addition of acids or bases.

pH Calculation

Calculating pH is an essential part of understanding how buffers work. In the given example, we used the initial moles of ammonia \((\text{NH}_3)\) and ammonium \((\text{NH}_4^+)\) after adding \(\text{HCl}\).

• The purpose of calculating pH is to see how close to neutral (pH = 7) a solution is.
• By using the Henderson-Hasselbalch equation, we determined that the final pH in both buffer solutions is 8.98.

The pH calculation helps in assessing how well a buffer can maintain a stable pH even when external acids or bases are added.

Buffer Capacity

Buffer capacity tells us how well a buffer can resist changes in pH. A buffer with a high capacity contains more moles of the buffering components (like \(\text{NH}_3\) and \(\text{NH}_4^+\)).

• A higher buffer capacity implies that the solution can handle larger amounts of added acid or base before the pH changes significantly.
• In the example, the 0.50 M \(\text{NH}_3\) / 1.50 M \(\text{NH}_4\text{Cl}\) solution has a greater capacity than the 0.050 M \(\text{NH}_3\) / 0.15 M \(\text{NH}_4\text{Cl}\) solution.

Having a buffer with a greater capacity is advantageous when you need to maintain stable pH conditions in a solution, especially during chemical reactions or biological processes.

Acid-Base Reactions

Acid-base reactions play a crucial role in buffer solutions. When an acid like \(\text{HCl}\) is added to the buffer containing ammonia \((\text{NH}_3)\), it reacts to form ammonium ions \((\text{NH}_4^+)\). This reaction showcases:

• The buffer’s ability to neutralize acids, thanks to the presence of \(\text{NH}_3\).
• How the balance between \(\text{NH}_3\) and \(\text{NH}_4^+\) is restored, maintaining an almost constant pH.

Understanding these reactions helps in grasping how buffers protect against drastic pH changes, ensuring the stability of the environment in which they are used.