Question

Two buffers are prepared by adding an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) to enough water to make \(1.00 \mathrm{~L}\) of solution. Buffer \(\mathrm{A}\) is prepared using \(1.00 \mathrm{~mol}\) each of formic acid and sodium formate. Buffer B is prepared by using \(0.010 \mathrm{~mol}\) of each. (a) Calculate the \(\mathrm{pH}\) of each buffer, and explain why they are equal. (b) Which buffer will have the greater buffer capacity? Explain. (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{M} \mathrm{HCl}\). (e) Discuss your answers for parts (c) and (d) in light of your response to part (b).

Short answer

In summary, both buffers A and B have the same pH of 3.75 initially, as they have equal concentrations of acid and conjugate base. Buffer A has a greater buffer capacity due to its higher concentration of acid and conjugate base. When adding the same amount of HCl to both buffers, Buffer A experiences a smaller change in pH, demonstrating its higher buffer capacity to resist changes in pH.

Step-by-step solution

Calculate the pH of each buffer

To calculate the pH of the two buffers, we will use the Henderson-Hasselbalch equation: pH = pKa + log(\(\frac{[A^-]}{[HA]}\)) For formic acid, the pKa value is 3.75. In both buffers, an equal number of moles of formic acid (HCOOH) and sodium formate (HCOONa) are used. As sodium formate dissociates completely in water, the concentrations of formic acid and formate ions in both solutions are equal. Therefore, the fractions in the logarithm become: \(\frac{[A^-]}{[HA]}\) = \(\frac{[HCOO^-]}{[HCOOH]}\) = 1 So, we can simplify the Henderson-Hasselbalch equation as follows: pH = pKa + log(1) = pKa Now, we can calculate the pH of both buffers: pH (Buffer A) = pH (Buffer B) = pKa = 3.75 The pH values of the buffers are equal because both solutions have equal concentrations of acid and conjugate base, resulting in the same pH value according to Henderson-Hasselbalch equation.

Determining the buffer with greater buffer capacity

Buffer capacity is the ability of a buffer solution to resist changes in pH when an acid or a base is added. A buffer with a higher concentration of acid and conjugate base has a greater buffer capacity. In this case, Buffer A has a higher number of moles (1.00 mol) of formic acid and sodium formate compared to Buffer B (0.010 mol). Therefore, Buffer A has a greater buffer capacity.

Calculate the change in pH upon the addition of HCl

We need to calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00 M HCl and 10 mL of 1.00 M HCl. We will first calculate the change in pH upon the addition of 1.0 mL of 1.00 M HCl. For Buffer A: Moles of HCl added: 1.00 L x 0.001 L = 0.001 mol New moles of formic acid: 1.00 mol + 0.001 mol = 1.001 mol New moles of formate ion: 1.00 mol - 0.001 mol = 0.999 mol For Buffer B: Moles of HCl added: 1.00 L x 0.001 L = 0.001 mol New moles of formic acid: 0.010 mol + 0.001 mol = 0.011 mol New moles of formate ion: 0.010 mol - 0.001 mol = 0.009 mol Now, we can calculate the new pH values for both buffers using the modified Henderson-Hasselbalch equation: pH = pKa + log(\(\frac{[A^-]}{[HA]}\)) Buffer A: pH = 3.75 + log(\(\frac{0.999}{1.001}\)) ≈ 3.75 Buffer B: pH = 3.75 + log(\(\frac{0.009}{0.011}\)) ≈ 3.70 Now, we will calculate the change in pH upon the addition of 10 mL of 1.00 M HCl. For Buffer A: Moles of HCl added: 1.00 L x 0.01 L = 0.010 mol New moles of formic acid: 1.00 mol + 0.010 mol = 1.010 mol New moles of formate ion: 1.00 mol - 0.010 mol = 0.990 mol For Buffer B: Moles of HCl added: 1.00 L x 0.01 L = 0.010 mol New moles of formic acid: 0.010 mol + 0.010 mol = 0.020 mol New moles of formate ion: 0.010 mol - 0.010 mol = 0.000 mol Buffer A: pH = 3.75 + log(\(\frac{0.990}{1.010}\)) ≈ 3.74 Buffer B: The solution is now just formic acid, so we must calculate the pH using the ionization constant equation for formic acid: \(H^+ = \sqrt{(1.8x10^{-4})(0.020)\} \) pH = -log(\(H^+\)) ≈ 2.18

Discuss the results

Comparing the pH change upon the addition of the same volume of HCl, it is clear that Buffer A is better at resisting pH change due to its higher buffer capacity. This is because Buffer A contains more moles of both the acid and its conjugate base, thus it can neutralize more added acid than Buffer B. Consequently, the pH change in Buffer A is significantly smaller than the change in Buffer B, which is consistent with our conclusion in step 2 that Buffer A has a greater buffer capacity.

Key concepts

Henderson-Hasselbalch Equation

Understanding the Henderson-Hasselbalch equation is essential for students dealing with buffer solutions in chemistry. This equation relates the pH of a solution to the pKa of the acid and the ratio of the concentrations of its conjugate base ([A^-]) to that of the acid ([HA]). It is represented mathematically as
\[ pH = pKa + \log\left(\frac{[A^-]}{[HA]}\right) \]
The equation assumes that the acid and its conjugate base are present in significant amounts and that the system adheres to the principles of equilibrium chemistry. In the case of a 1:1 molar ratio of sodium formateto formic acid, the ratio [A^-]/[HA] is equal to 1, which renders the logarithmic term zero, and thus the pH of such a buffer only depends on the pKa value of formic acid. This application allows students to predict and understand the pH of buffer systems with ease, a vital skill in both laboratory and industry settings where controlling pH is critical.

Buffer Capacity

Buffer capacity, another significant concept for students studying solutions, is defined as the measure of a buffer's ability to resist changes in pH upon the addition of an acid or base. High buffer capacity signifies that the buffer can neutralize significant amounts of added substances without drastically altering the pH of the solution. It directly depends on the absolute concentrations of the acid and its conjugate base in the buffer solution.
In practical terms, Buffer A, with 1.00 mol each of formic acid and sodium formate, showcases a higher buffer capacity compared to Buffer B which contains just 0.010 mol of each component. With regards to addition of hydrochloric acid, this means Buffer A can sustain its pH level more efficiently than Buffer B. Such knowledge plays a crucial role not just in academic exercises, but also in various applications including pharmaceuticals, where maintaining the correct pH is essential for the stability of drugs.

pH Calculation

The pH calculation is an integral part of understanding acid-base chemistry. Calculating pH goes beyond the Henderson-Hasselbalch equation when one component of the buffer is exhausted. This occurs when enough strong acid or base is added to shift the balance between a weak acid and its conjugate base.
For example, when 10 mL of 1 M HCl is added to Buffer B, all the formate ions react, leaving only formic acid in solution. The system no longer functions as a buffer, and the pH must be computed differently. Using the ionization constant (K_a) for the acid and the new concentration, students can calculate the pH of the resulting solution. Such complexities demonstrate the dynamic nature of chemistry, where students must adapt their calculation strategies to the context of the reaction. This illustrates the importance of a deep understanding of the principles that govern pH, which has vast implications in scientific research, environmental monitoring, and many industries.