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. \(0.050 \mathrm{M} \mathrm{NH}_{3} / 0.15 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) b. \(0.50 \mathrm{M} \mathrm{NH}_{3} / 1.50 \mathrm{M} \mathrm{NH}_{4} \mathrm{Cl}\) Do the two original buffered solutions differ in their \(\mathrm{pH}\) or their capacity? What advantage is there in having a buffer with a greater capacity?

Short answer

The pH of the first buffered solution after adding HCl is 7.35, and the pH of the second buffered solution after adding HCl is 8.25. Both original buffered solutions have the same pH, as their ratio of \(NH_{3}\) to \(NH_{4}^{+}\) was the same. However, the second buffered solution has a higher capacity. A buffer with a greater capacity can neutralize a larger amount of added acid or base before the solution's pH changes significantly, providing more stable pH control.

Step-by-step solution

a. Calculate the moles and \(NH_{3}\): \(NH_{4}^{+}\) ratio in the original buffered solution#a.

To begin, we will determine the amount of moles for ammonia (\(NH_{3}\)) and ammonium (\(NH_{4}^{+}\)) ions in the original buffered solution. Moles of \(NH_{3} = 0.050\,\text{M} \times 0.250\,\text{L} = 0.0125\,\text{moles}\) Moles of \(NH_{4}^{+} = 0.15\text{M} \times 0.250\,\text{L} = 0.0375\,\text{moles}\) Now let's calculate the ratio of \(NH_{3}\) to \(NH_{4}^{+}\) in the solution: Ratio \(= \frac{0.0125\,\text{moles}}{0.0375\,\text{moles}} = 0.333\) As the ratio remains constant, this value will be used in our calculations later.

b. Calculate the moles after adding HCl to the first buffered solution

After adding HCl, it reacts with the present \(NH_{3}\) immediately, converting it to \(NH_{4}^{+}\). Therefore, the moles of \(NH_{3}\) and \(NH_{4}^{+}\) in the solution will change. Moles of \(NH_{3}(new) = 0.0125\,\text{moles} - 0.010\text{moles} = 0.0025\,\text{moles}\) Moles of \(NH_{4}^{+}(new) = 0.0375\,\text{moles} + 0.010\text{moles} = 0.0475\,\text{moles}\)

c. Determine the pH of the first buffered solution after adding HCl

Using the Henderson-Hasselbalch equation, we can determine the new pH of the solution: \(pH = pK_{a} + \log_{10} \left(\frac{[NH_{3}]}{[NH_{4}^{+}]}\right) \) The \(pK_{a}\) of \(NH_{4}^{+}\) is approximately 9.25. Substituting the new concentrations, we get: \(pH = 9.25 + \log_{10}\left(\frac{0.0025\,\text{moles}}{0.0475\,\text{moles}}\right) = 9.25 - 1.90 = 7.35\)

d. Repeat steps a, b, and c for the second buffered solution

Calculate moles: Moles of \(NH_{3} = 0.50\,\text{M} \times 0.250\,\text{L} = 0.125\,\text{moles}\) Moles of \(NH_{4}^{+} = 1.50\,\text{M} \times 0.250\,\text{L} = 0.375\,\text{moles}\) Calculate the moles after adding HCl: Moles of \(NH_{3}(new) = 0.125\,\text{moles} - 0.010\text{moles} = 0.115\,\text{moles}\) Moles of \(NH_{4}^{+}(new) = 0.375\,\text{moles} + 0.010\text{moles} = 0.385\,\text{moles}\) Determine the pH: \(pH = 9.25 + \log_{10}\left(\frac{0.115\,\text{moles}}{0.385\,\text{moles}}\right) = 9.25 - 1.00 = 8.25\) Now we can compare the results: Both the original buffered solutions have the same pH, as their ratio of \(NH_{3}\) to \(NH_{4}^{+}\) was the same. However, the second buffered solution has a higher capacity, as it uses larger amounts of \(NH_{3}\) and \(NH_{4}^{+}\). Having a buffer with a greater capacity means that it can neutralize a larger amount of added acid or base before the pH of the solution changes significantly. This can be especially important in situations where the system requires to maintain a stable pH over an extended period or when large amounts of acid/base might be introduced.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is crucial when studying chemistry, particularly when you need to calculate the pH of buffer solutions. It simplifies the task of assessing the impact of adding an acid or base to a buffered system.

This equation is represented as:
\[\text{pH} = \text{p}K_a + \log_{10} \left(\frac{[\text{base}]}{[\text{acid}]}\right)\]
What this formula does is directly relate the pH of a solution to the ratio of the concentration of the base form over the acid form of a buffer pair. The base in this context refers to the weak base, and the acid is its conjugate acid form. The pKa is the negative log of the acid dissociation constant, which is a unique value for every acid determining its strength and tendency to release a proton.

In the problem at hand, when HCl gas was added to the ammonia/ammonium chloride buffer, the NH3 (base) got converted to NH4+ (conjugate acid). The Henderson-Hasselbalch equation allowed us to calculate the new pH after this reaction took place.

Buffered Solutions

Buffered solutions are a vital concept in Chemistry and are known for their resistance to changes in pH upon the addition of small amounts of an acid or base. This resilience is due to the presence of a weak conjugate acid-base pair. In our exercise, the buffer consists of ammonia (NH3), the weak base, and ammonium chloride (NH4Cl), the conjugate acid.

The capacity of a buffer solution is directly related to the concentrations of these conjugate pairs. A higher concentration means that the buffer can neutralize more added acid or base, making it more robust, as shown in example B of our exercise.

When choosing a buffer for any particular use, not only the pH range is important but also the buffer capacity. A buffered solution with a high capacity is particularly advantageous when the solution needs to withstand large amounts of added acids or bases without a significant shift in pH, which can be critical in many biological and chemical applications.

Acid-Base Neutralization

Acid-base neutralization is a fundamental reaction in Chemistry where an acid and a base react to form water and a salt. This reaction is the core of buffering action. When a buffer solution encounters an acid or base, it undergoes a neutralization reaction that minimizes pH changes.

In the exercise provided, the addition of HCl (a strong acid) to the buffered solutions results in the neutralization reaction with NH3, forming NH4+ (ammonium ions). This reaction demonstrates the buffer's ability to maintain a stable pH despite the addition of an acid.

The buffering capacity becomes critically important in systems subject to fluctuating acidic or basic conditions. Ensuring that pH remains stable could be essential to the proper functioning of biological systems, such as the human body, or chemical processes, such as fermentations in biotechnology. An effective buffer will allow these systems to resist extreme pH changes, maintaining the optimum conditions for the processes taking place.