Question

Calculate the pH of each of the following buffered solutions. $$ \begin{array}{l}{\text { a. } 0.10 M \text { acetic acid0.25 } M \text { sodium acetate }} \\ {\text { b. } 0.25 M \text { acetic acid/0.10 } M \text { sodium acetate }}\end{array} $$ $$ \begin{array}{l}{\text { c. } 0.080 M \text { acetic acid/0.20 } M \text { sodium acetate }} \\ {\text { d. } 0.20 M \text { acetic acid/0.080 } M \text { sodium acetate }}\end{array} $$

Short answer

The pH values for each buffered solution are: 1. 5.15 2. 4.35 3. 5.15 4. 4.35

Step-by-step solution

Buffered solution 1: 0.10 M acetic acid/0.25 M sodium acetate

We have the following concentrations: - (HA) acetic acid: 0.10 M - (A^-) sodium acetate: 0.25 M Now we can plug these values into the Henderson-Hasselbalch equation: \[ pH = 4.75 + \log \frac{0.25}{0.10} \] Calculate the pH: \[ pH = 4.75 + \log 2.5 \] \[ pH = 4.75 + 0.40 \] \[ pH = 5.15 \]

Buffered solution 2: 0.25 M acetic acid/0.10 M sodium acetate

We have the following concentrations: - (HA) acetic acid: 0.25 M - (A^-) sodium acetate: 0.10 M Now we can plug these values into the Henderson-Hasselbalch equation: \[ pH = 4.75 + \log \frac{0.10}{0.25} \] Calculate the pH: \[ pH = 4.75 - \log 2.5 \] \[ pH = 4.75 - 0.40 \] \[ pH = 4.35 \]

Buffered solution 3: 0.080 M acetic acid/0.20 M sodium acetate

We have the following concentrations: - (HA) acetic acid: 0.080 M - (A^-) sodium acetate: 0.20 M Now we can plug these values into the Henderson-Hasselbalch equation: \[ pH = 4.75 + \log \frac{0.20}{0.080} \] Calculate the pH: \[ pH = 4.75 + \log 2.5 \] \[ pH = 4.75 + 0.40 \] \[ pH = 5.15 \]

Buffered solution 4: 0.20 M acetic acid/0.080 M sodium acetate

We have the following concentrations: - (HA) acetic acid: 0.20 M - (A^-) sodium acetate: 0.080 M Now we can plug these values into the Henderson-Hasselbalch equation: \[ pH = 4.75 + \log \frac{0.080}{0.20} \] Calculate the pH: \[ pH = 4.75 - \log 2.5 \] \[ pH = 4.75 - 0.40 \] \[ pH = 4.35 \] The pH values for each buffered solution are: 1. 5.15 2. 4.35 3. 5.15 4. 4.35

Key concepts

Henderson-Hasselbalch equation

The Henderson-Hasselbalch equation is a powerful tool used to calculate the pH of buffered solutions. It relates the pH of a solution to the pKa of the acidic component and the ratio of the concentrations of the conjugate base and the acid itself. The equation is expressed as:

\[ pH = pK_a + \log \left(\frac{[A^-]}{[HA]}\right) \]

In this equation:

• \( pH \) is the measure of acidity or basicity of the solution.
• \( pK_a \) is the negative logarithm of the acid dissociation constant of the weak acid.
• \([A^-]\) represents the concentration of the conjugate base.
• \([HA]\) represents the concentration of the acid.

The logarithmic term focuses on the ratio of the conjugate base to the acid, which gives us a straightforward way to calculate pH from known concentrations in buffered solutions. It is essential to note that this equation assumes the solution behaves ideally and is diluted sufficiently.

acetic acid

Acetic acid is a simple carboxylic acid with the chemical formula \( CH_3COOH \). In water, it partially dissociates to form acetate ions \( CH_3COO^- \) and hydrogen ions \( H^+ \). This partial dissociation qualifies it as a weak acid, a critical property for it to be used in buffered solutions. Essential characteristics of acetic acid:

• It has a relatively low ionization, meaning it doesn't fully dissociate in water.
• The \( pK_a \) value of acetic acid is approximately 4.75, which is the point where half of the acid is dissociated.

In the context of the Henderson-Hasselbalch equation, acetic acid often acts as the acid \([HA]\), pairing with its conjugate base acetate \([A^-]\) to maintain a stable pH across the solution.

sodium acetate

Sodium acetate is the sodium salt of acetic acid and is represented by the formula \( CH_3COONa \). It fully dissociates in water to release acetate ions \( CH_3COO^- \), which is the conjugate base needed for buffered solutions. Key features of sodium acetate include:

• It provides a ready supply of acetate ions \([A^-]\) to balance the addition of acids or bases in a solution.
• As a salt, it aids in maintaining the pH by interacting with the acetic acid.

In the buffered solutions calculation, the concentration of sodium acetate is used to find the ratio \( \frac{[A^-]}{[HA]} \) in the Henderson-Hasselbalch equation. This balanced interaction between acetic acid and sodium acetate creates the buffering effect, stabilizing the pH.

buffered solutions

Buffered solutions are specially designed mixtures that resist drastic changes in pH when small amounts of acids or bases are added. They typically consist of a weak acid and its conjugate base or a weak base and its conjugate acid. Understanding how they work:

• The presence of both an acid and its conjugate base allows for the neutralization of added acid or base.
• When an acid is added, the conjugate base in the buffer will neutralize it, reducing its effect.
• Conversely, if a base is added, the weak acid component of the buffer will neutralize the base.

These properties make buffered solutions useful in a variety of chemical and biological processes where maintaining a stable pH is crucial. For example, the human body uses buffered solutions in the blood to help maintain homeostasis.