When a buffer solution of sodium acetate and acetic acid is diluted with water, then pH

Options

A. Increases
B. Decreases
C. Unaffected
D. None of these

Answer

To solve the question of how the pH of a buffer solution of sodium acetate and acetic acid changes upon dilution with water, we can follow these steps:

### Step 1: Understand the Components of the Buffer
The buffer solution consists of:
- **Sodium Acetate (CH₃COONa)**: This is the salt derived from the weak acid acetic acid.
- **Acetic Acid (CH₃COOH)**: This is the weak acid component of the buffer.

### Step 2: Apply the Henderson-Hasselbalch Equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Where:
- \(\text{pK}_a\) is the negative logarithm of the acid dissociation constant of acetic acid.
- \([\text{Salt}]\) is the concentration of sodium acetate.
- \([\text{Acid}]\) is the concentration of acetic acid.

### Step 3: Analyze the Effect of Dilution
When the buffer solution is diluted with water:
- The number of moles of both the salt (sodium acetate) and the acid (acetic acid) remains the same.
- However, the total volume of the solution increases.

### Step 4: Concentration Changes
Let’s assume:
- Initially, we have a certain volume (V₁) and concentration (C₁) for both components.
- After dilution, the volume increases to V₂ (where V₂ > V₁), but the number of moles remains constant.

The concentration after dilution can be calculated as:
\[ [\text{Salt}] = \frac{n_{\text{Salt}}}{V_2} \]
\[ [\text{Acid}] = \frac{n_{\text{Acid}}}{V_2} \]
Where \(n_{\text{Salt}}\) and \(n_{\text{Acid}}\) are the number of moles of salt and acid, respectively.

### Step 5: Ratio of Salt to Acid
Since the number of moles of both components does not change, the ratio of the concentrations remains the same:
\[ \frac{[\text{Salt}]}{[\text{Acid}]} = \frac{n_{\text{Salt}}/V_2}{n_{\text{Acid}}/V_2} = \frac{n_{\text{Salt}}}{n_{\text{Acid}}} \]
Thus, the ratio of salt to acid remains constant even after dilution.

### Step 6: Conclusion on pH Change
Since the ratio \(\frac{[\text{Salt}]}{[\text{Acid}]}\) remains unchanged, the value of the logarithmic term in the Henderson-Hasselbalch equation does not change. Therefore, the pH of the buffer solution remains unaffected by the dilution with water.

### Final Answer
The pH of the buffer solution of sodium acetate and acetic acid remains **unaffected** upon dilution with water.

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