Problem 88 - Chemistry: The Central Science in SI Units, Global Edition

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 A is prepared using 1.00 mol each of formic acid and sodium formate. Buffer B is prepared by using 0.010 mol of each. (a) Calculate the pH of each buffer. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\). (d) Calculate the change in \(\mathrm{pH}\) for each buffer upon the addition of \(10 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\).

Step-by-Step Solution

Answer

In summary, both Buffer A and Buffer B initially have a pH of 3.75. Buffer A has a greater buffer capacity, allowing it to resist changes in pH more effectively than Buffer B. When adding 1 mL of 1 M HCl, the pH of Buffer A remains approximately the same (change in pH ≈ 0), while the pH of Buffer B decreases to about 3.68 (change in pH ≈ -0.07). When adding 10 mL of 1 M HCl, the pH of Buffer A minimally decreases to approximately 3.74 (change in pH ≈ -0.01), while the pH of Buffer B drops significantly (change in pH ≈ -0.75, estimate).

Step 1: Calculate the pH of each buffer using the Henderson-Hasselbalch equation.

The Henderson-Hasselbalch equation is given by: \[pH = pK_a + \log{\frac{[\mathrm{A^-}]}{[\mathrm{HA}]}}\] where \(pH\) is the pH of the buffer, \(pK_a\) is the acid dissociation constant of the weak acid, and \([\mathrm{A^-}]\) and \([\mathrm{HA}]\) are the molar concentrations of the weak base and its conjugate acid, respectively. For formic acid, the \(pK_a\) value is 3.75. In both buffers, the molar concentrations of formic acid (\(HA\)) and sodium formate (\(A^-\)) are equal. Therefore, the Henderson-Hasselbalch equation simplifies to: \[pH = pK_a + \log{1} = pK_a\] Now, we can calculate the pH of each buffer: Buffer A: \(pH = 3.75\) Buffer B: \(pH = 3.75\)

Step 2: Determine which buffer has the greater buffer capacity

Buffer capacity refers to the ability of a buffer to resist changes in pH. A buffer with a higher buffer capacity will be able to resist a more significant change in pH upon the addition of an acid or base. In this case, Buffer A has more moles of the buffering species, formic acid and sodium formate, than Buffer B, meaning that it has a greater buffer capacity.

Step 3: Calculate the change in pH for each buffer upon the addition of 1.0 mL of 1.00 M HCl

We need to calculate the change in the moles of formic acid and sodium formate upon the addition of HCl. HCl will react with sodium formate to convert it to formic acid. The amount of HCl added is 1.0 mL x 1.00 M = 0.001 mol. Buffer A: Initial moles of formic acid: 1 mol Initial moles of sodium formate: 1 mol Moles of formic acid after HCl addition: 1 + 0.001 = 1.001 mol Moles of sodium formate after HCl addition: 1 - 0.001 = 0.999 mol Buffer B: Initial moles of formic acid: 0.010 mol Initial moles of sodium formate: 0.010 mol Moles of formic acid after HCl addition: 0.010 + 0.001 = 0.011 mol Moles of sodium formate after HCl addition: 0.010 - 0.001 = 0.009 mol Now, we can use the Henderson-Hasselbalch equation to calculate the new pH for each buffer: Buffer A: \(pH = 3.75 + \log{\frac{0.999}{1.001}} \approx 3.75\) Buffer B: \(pH = 3.75 + \log{\frac{0.009}{0.011}} \approx 3.68\) The change in pH for each buffer is: Buffer A: \(∆pH \approx 0\) Buffer B: \(∆pH \approx -0.07\)

Step 4: Calculate the change in pH for each buffer upon the addition of 10 mL of 1.00 M HCl

Similar to Step 3, we'll first calculate the change in the moles of formic acid and sodium formate. The amount of HCl added is 10 mL x 1.00 M = 0.010 mol. Buffer A: Initial moles of formic acid: 1.00 mol Initial moles of sodium formate: 1.00 mol Moles of formic acid after HCl addition: 1.00 + 0.010 = 1.010 mol Moles of sodium formate after HCl addition: 1.00 - 0.010 = 0.990 mol Buffer B: Initial moles of formic acid: 0.010 mol Initial moles of sodium formate: 0.010 mol Moles of formic acid after HCl addition: 0.010 + 0.010 = 0.020 mol Moles of sodium formate after HCl addition: 0.010 - 0.010 = 0 mol (all sodium formate has reacted) Now, we can use the Henderson-Hasselbalch equation to calculate the new pH for each buffer: Buffer A: \(pH = 3.75 + \log{\frac{0.990}{1.010}} \approx 3.74\) Buffer B: Cannot use the Henderson-Hasselbalch equation since there are no moles of sodium formate left. The solution is now predominantly formic acid, and the pH drops significantly. The change in pH for each buffer is: Buffer A: \(∆pH \approx -0.01\) Buffer B: \(∆pH \approx -0.75\) (estimate, since the exact value cannot be determined with the Henderson-Hasselbalch equation) In conclusion, (a) the pH of both buffers is 3.75 initially, (b) Buffer A has the greater buffer capacity, and the change in pH upon the addition of HCl is smaller for Buffer A compared to Buffer B in both cases of adding (c) 1 mL and (d) 10 mL of 1 M HCl.

Key Concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a mathematical formula used to estimate the pH of a buffer solution. It's an essential tool in chemistry, particularly useful for creating buffer solutions with a specific pH.
The equation is given by: \[ pH = pK_a + \log{\frac{[A^-]}{[HA]}} \] where:

• \( pK_a \) is the acid dissociation constant of the weak acid.
• \([A^-]\) represents the concentration of the conjugate base (like sodium formate).
• \([HA]\) represents the concentration of the weak acid (like formic acid).

This equation shows the relationship between the pH and the ratio of concentrations of the conjugate base to the weak acid.
When these concentrations are equal, the term \( \log{1} \) becomes 0, simplifying the equation to \( pH = pK_a \). This is why for solutions where equal moles of formic acid and sodium formate are used, both buffers have a pH equal to the \( pK_a \) value of formic acid.

Buffer capacity

Buffer capacity is an important concept, especially when dealing with solutions meant to maintain a stable pH. It measures a buffer's ability to resist changes in pH upon the addition of acids or bases.
In simple terms, the greater the buffer capacity, the more resistant the solution is to changes in pH. Several factors influence buffer capacity:

• The concentrations of the acid and its conjugate base: Higher concentrations result in greater buffer capacity.
• The ratio of the conjugate base to the acid: Ideally, this ratio should be close to 1 for maximum buffer capacity.

This is observed in the exercise where Buffer A, having 1.00 mol of both formic acid and sodium formate, exhibits greater buffer capacity compared to Buffer B, which contains only 0.010 mol of each component. With more buffering agents, Buffer A resists changes in pH better when HCl is added.

Formic acid

Formic acid (HCOOH) is a simple carboxylic acid commonly used in a variety of chemical reactions. It plays a key role in buffer solutions, particularly in those involving its conjugate base, sodium formate.
Formic acid is known for its moderate strength with a \( pK_a \) value of about 3.75. This value is critical when calculating the pH of a buffer solution using the Henderson-Hasselbalch equation.
When used appropriately in buffers, formic acid helps to stabilize pH levels by participating in equilibria where it can donate a hydrogen ion (proton) to neutralize added bases, while its conjugate formate ion can react with added acids.

Sodium formate

Sodium formate (HCOONa) is the sodium salt derived from formic acid. It acts as the conjugate base in formic acid-buffer systems.
In a buffer solution with formic acid, sodium formate works to absorb excess hydrogen ions by converting back to formic acid upon acid addition. This ability makes it an important part of maintaining a stable pH in the buffer system.
Within the context of buffer solutions, sodium formate, in combination with formic acid, uses the ideal action of buffers: balancing the pH by adjusting to the addition of acidic or basic elements. This highlights its crucial role in the resistance to pH change and its contribution to buffer capacity as discussed in the exercise.