Problem 28 - Chemistry: An Atoms-Focused Approach

Question

What masses of acetic acid and sodium acetate are needed to prepare \(125 \mathrm{mL}\) of \(\mathrm{pH}=5.00\) buffer if the total concentration of the two components is \(0.500 M ?\)

Step-by-Step Solution

Answer

Answer: To prepare the buffer solution, 1.37 g of acetic acid and 3.25 g of sodium acetate are required.

Step 1: Write the Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation relates the pH of a buffer solution to the concentrations of the acidic and basic components: \(pH = pK_a + \log{\frac{[A^-]}{[HA]}}\) Where: \(pH\) is the pH of the buffer solution \(pK_a\) is the negative logarithm of the acid dissociation constant (\(K_a\)) of the weak acid \([A^-]\) is the concentration of the conjugate base \([HA]\) is the concentration of the weak acid

Step 2: Find the pKa value of acetic acid

Acetic acid (\(CH_3COOH\)) has an acid dissociation constant (\(K_a\)) of \(1.75 \times 10^{-5}\). To find the \(pK_a\), we take the negative logarithm of \(K_a\): \(pK_a = -\log{(1.75 \times 10^{-5})} \approx 4.76\)

Step 3: Use the Henderson-Hasselbalch equation to find the ratio of concentrations

We are given that the pH of the buffer solution is 5.00. Using the Henderson-Hasselbalch equation and the pKa value we calculated previously, we can find the ratio of the concentrations of the sodium acetate to the acetic acid: \(5.00 = 4.76 + \log{\frac{[A^-]}{[HA]}}\) \(\log{\frac{[A^-]}{[HA]}} = 5.00 - 4.76 = 0.24\) We can now solve the equation to find the ratio of the concentrations: \(\frac{[A^-]}{[HA]} = 10^{0.24} \approx 1.74\)

Step 4: Find the individual concentrations of acetic acid and sodium acetate

We are given that the total concentration of the two components is 0.500 M. We can set up equations based on the total concentration and the ratio of the concentrations: \(x = [HA], y = [A^-]\) \(x + y = 0.500\) (1) \(y = 1.74x\) (2) Substituting Eq. (2) into Eq. (1), we get: \(x + 1.74x = 0.500\) \(2.74x = 0.500\) \(x = 0.182 M\) Now, using Eq. (2), we can find the concentration of the conjugate base: \(y = 1.74 \times 0.182 M \approx 0.318 M\)

Step 5: Calculate the masses of acetic acid and sodium acetate

We have the concentrations and we are asked to find the masses required for 125 mL of the buffer solution. We use the following formulas: \(mass = volume \times concentration \times molar\ mass\) For acetic acid (\(CH_3COOH\)), the molar mass is approximately 60.05 g/mol: \(mass_{HA} = 0.125L \times 0.182 M \times 60.05 \frac{g}{mol} \approx 1.37\ g\) For sodium acetate (\(CH_3COONa\)), the molar mass is approximately 82.03 g/mol: \(mass_{A^-} = 0.125L \times 0.318 M \times 82.03 \frac{g}{mol} \approx 3.25\ g\) So, to prepare the buffer solution, we need 1.37 g of acetic acid and 3.25 g of sodium acetate.

Key Concepts

Understanding the Henderson-Hasselbalch Equation

The Henderson-Hasselbalch equation is essential for determining the pH of buffer solutions. Buffers are solutions that resist changes in pH when small amounts of acid or base are added. The equation is written as:

\[pH = pK_a + \log{\frac{[A^-]}{[HA]}}\]

Here's what each term means:

• pH: The measure of acidity or basicity in the solution.
• pK_a: The negative logarithm of the acid dissociation constant. It indicates the strength of the acid, with lower values showing stronger acids.
• [A^-]: Concentration of the conjugate base (in this case, sodium acetate).
• [HA]: Concentration of the weak acid (such as acetic acid).

This equation is particularly useful for calculating the pH of a buffer solution or finding the ratio of concentrations between the acid and its conjugate base.

In the provided exercise, the target pH is 5.00, making it crucial to apply the Henderson-Hasselbalch equation to find the necessary balance between acetic acid and sodium acetate.

The Role of Acetic Acid in Buffer Solutions

Acetic acid, with the chemical formula \(CH_3COOH\), is a common weak acid found in vinegar. It plays a central role in buffer solutions due to its ability to partially ionize in water, releasing hydrogen ions (H⁺).

Here's why acetic acid is suitable for buffers:

• Partial Ionization: It doesn't fully dissociate in water, allowing it to interact gently, which is ideal for maintaining the solution's pH.
• pK_a Value: The pK_a of acetic acid is approximately 4.76, making it effective for buffers with pH around this range.

Acetic acid's limited dissociation makes it an excellent candidate for forming buffer solutions, such as in our exercise, where its concentration needs to be balanced with sodium acetate to achieve a desired pH.

In the example, acetic acid's concentration was found to be 0.182 M, showing how to carefully proportion the quantities to create an effective buffer.

Understanding Sodium Acetate and Its Buffering Action

Sodium acetate, \(CH_3COONa\), is the conjugate base of acetic acid. When dissolved in water, it releases acetate ions (CH₃COO^-). This plays a crucial role in buffer solutions by balancing the effect of added acids or bases.

Here's how sodium acetate contributes:

• Conjugate Base Action: It can capture added hydrogen ions, reducing sudden pH changes.
• Solubility: Highly soluble in water, meaning it easily mixes and dissociates into ions.
• Complementary to Acetic Acid: Works in tandem with acetic acid to stabilize the buffer's pH.

In the exercise, the concentration of sodium acetate in the buffer was calculated to be 0.318 M. This highlights its vital role in maintaining the buffer's pH at the target level, demonstrating its effectiveness in real-world applications.