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 \(1.00 \mathrm{~mol}\) 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. (b) Which buffer will have the greater buffer capacity? (c) Calculate the change in pH for each buffer upon the addition of \(1.0 \mathrm{~mL}\) of \(1.00 \mathrm{MHCl}\). (d) Calculate the change in pH for each buffer upon the addition of \(10 \mathrm{~mL}\). of \(1.00 \mathrm{M} \mathrm{HCl}\).

Short answer

(a) The pH of both buffer A and buffer B is 3.75. (b) Buffer A has the greater buffer capacity. (c) Upon the addition of 1.0 mL of 1.00 M HCl, the change in pH is negligible for buffer A, and it decreases by about 0.2 units for buffer B. (d) Upon the addition of 10 mL of 1.00 M HCl, the change in pH is small for buffer A (about 0.01 units decrease) and significant for buffer B (decrease by about 0.76 units).

Step-by-step solution

Find pKa value

For formic acid, the pKa value is approximately 3.75. We will use this value in the calculations for both buffer systems. Step 2: Calculate the pH of buffer A

Using Henderson-Hasselbalch for Buffer A

For buffer A, we have 1.00 mol of formic acid (HCOOH) and 1.00 mol of sodium formate (HCOONa) in 1.00 L of the solution. The equilibrium concentrations of HCOOH and HCOO- will be equal, so we can apply the Henderson-Hasselbalch equation: \(pH_A = pK_a + \log{\frac{[HCOO^-]}{[HCOOH]}} = 3.75 + \log{\frac{1.00}{1.00}} = 3.75 + \log{1} = 3.75\) The pH of buffer A is 3.75. Step 3: Calculate the pH of buffer B

Using Henderson-Hasselbalch for Buffer B

For buffer B, we have 0.010 mol of formic acid (HCOOH) and 0.010 mol of sodium formate (HCOONa). The equilibrium concentrations of HCOOH and HCOO- will be equal, so we can apply the Henderson-Hasselbalch equation: \(pH_B = pK_a + \log{\frac{[HCOO^-]}{[HCOOH]}} = 3.75 + \log{\frac{0.010}{0.010}} = 3.75+\log{1} = 3.75\) The pH of buffer B is also 3.75. Step 4: Compare the buffer capacities

Buffer capacities comparison

The buffer capacity depends on the concentrations of acidic and basic components. Higher concentrations yield higher buffer capacities. Since buffer A has a higher concentration of formic acid and formate ions, it has the greater buffer capacity compared to buffer B. Step 5: Calculate the change in pH upon the addition of 1.00 mL of 1.00 M HCl

Change in pH with 1.00 mL HCl

The acid added will react with the sodium formate, reducing the concentration of the base in the solution. For Buffer A: The amount of HCl added is \(1.0 \times 10^{-3} mol\), making the new concentrations of HCOOH and HCOO- equal to \([HCOOH] = 1.001 \mathrm{mol/L}\) and \([HCOO^-] = 0.999 \mathrm{mol/L}\). \(pH_{A'} = 3.75 + \log{\frac{[HCOO^-]}{[HCOOH]}} = 3.75 + \log{\frac{0.999}{1.001}} \approx 3.75\) For Buffer B: The new concentrations of HCOOH and HCOO- after the addition of HCl are \([HCOOH] = 0.011 \mathrm{mol/L}\) and \([HCOO^-] = 0.009 \mathrm{mol/L}\). \(pH_{B'} = 3.75 + \log{\frac{[HCOO^-]}{[HCOOH]}} = 3.75+\log{\frac{0.009}{0.011}} \approx 3.55\) The change in pH upon the addition of 1.0 mL HCl is negligible for buffer A and decreases by about 0.2 units for buffer B. Step 6: Calculate the change in pH upon the addition of 10 mL of 1.00 M HCl

Change in pH with 10.0 mL HCl

The acid added will react with the sodium formate, reducing the concentration of the base in the solution. In this case, 0.01 mol of HCl is added to both systems. For Buffer A: The new concentrations of HCOOH and HCOO- after the addition of HCl are \([HCOOH] = 1.010 \mathrm{mol/L}\) and \([HCOO^-] = 0.990 \mathrm{mol/L}\). \(pH_{A''} = 3.75 + \log{\frac{[HCOO^-]}{[HCOOH]}} = 3.75 + \log{\frac{0.990}{1.010}} \approx 3.74\) For Buffer B: The addition of 0.01 mol of HCl consumes all the 0.01 mol of sodium formate in buffer B, and the buffering capacity is exceeded. The pH will become that of a solution of \(1.0 \times 10^{-3} mol\) of HCl diluted to a final volume of \(1010 \mathrm{mL}\). \[pH_{B''} = -\log{([H^+])} = -\log{\frac{1.0 \times 10^{-3}}{1.01}} \approx 2.99\] The change in pH upon the addition of 10.0 mL HCl is small for buffer A (about 0.01 units decrease) and significant for buffer B (decrease by about 0.76 units).

Key concepts

Understanding the Henderson-Hasselbalch Equation

When it comes to understanding buffer solutions and their pH levels, the Henderson-Hasselbalch equation is a pivotal tool. It reveals the relationship between the pH of a buffer solution and the concentrations of its acid and base components. Let's break it down for better comprehension:

The equation is expressed as: \[ pH = pK_a + \text{log} \left( \frac{[A^-]}{[HA]} \right) \]Here, \( pK_a \) is the acid dissociation constant's negative logarithm, \( [A^-] \) represents the concentration of the conjugate base, and \( [HA] \) denotes the concentration of the acid. By substituting the known values, we can easily calculate the pH of a buffer solution. The beauty of this equation is that when the concentrations of the acid \( (HA) \) and its conjugate base \( (A^-) \) are equal, the log term becomes zero, and the pH is equal to the \( pK_a \). This explains why both buffers A and B share the same initial pH value in our exercise — their acid and base components are present in equal molar amounts.

Deciphering Buffer Capacity

Buffer capacity is a measure of a buffer solution's ability to maintain its pH when small amounts of acid or base are added. It's a critical concept for many biological and chemical processes, as it ensures stability within a system. Essentially, the greater the buffer capacity, the more resistant the solution is to changes in pH upon adding an acid or base.

Factors that influence buffer capacity include:

• The total concentration of the buffer components (acid and its conjugate base).
• The pH of the solution relative to the \( pK_a \) of the buffer components.

The solution is most effective as a buffer at pH levels close to the acid's \( pK_a \). In our example, buffer A exhibits a higher buffer capacity than buffer B because it contains more moles of the buffering agents. More relevantly, this is why, upon adding HCl, buffer A shows negligible changes in pH, while buffer B displays more significant shifts.

Navigating Acid-Base Reactions in Buffers

Acid-base reactions are fundamental to the function of buffers. Buffers are typically made up of a weak acid and its conjugate base (or conversely, a weak base and its conjugate acid). Their role is to resist changes in pH when acids or bases are added to the solution. This is possible because of the acid-base neutralization that takes place.

Consider the addition of HCl to our buffers. The HCl (hydrochloric acid) donates a proton (H+), which then reacts with the conjugate base of the buffer (in this case, the formate ion, \( HCOO^- \)). This neutralization process forms the weak acid, formic acid \( (HCOOH) \), and water.Through these reactions, the buffer solution resists drastic changes in pH. However, as seen in buffer B, if the quantity of added acid surpasses the buffer's capacity, the solution can no longer resist the change, leading to a noticeable shift in pH, moving it closer to that of the added strong acid. Hence, understanding acid-base reactions in the context of buffer solutions enables us to predict and control pH changes within various systems.