0.1 M acetic acid solution is titrated against 0.1M NaOH solution. What would be the difference in pH between `1/4` and `3/4` stages of neutralisation of acid ?

Options

A. `2 "log"3/4`
B. `2 "log"1/4`
C. `"log"1/4`
D. `2 log 3`

Explanation

To solve the problem of finding the difference in pH between the 1/4 and 3/4 stages of neutralization of a 0.1 M acetic acid solution with a 0.1 M NaOH solution, we can follow these steps:

### Step 1: Understanding the Neutralization Reaction
The neutralization reaction between acetic acid (CH₃COOH) and sodium hydroxide (NaOH) can be represented as:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]

### Step 2: Calculate Moles of Acetic Acid and NaOH
Since both solutions are 0.1 M and we are considering equal volumes (let's assume 1 L for simplicity):
- Moles of acetic acid (CH₃COOH) = 0.1 M × 1 L = 0.1 moles
- Moles of NaOH = 0.1 M × 1 L = 0.1 moles

### Step 3: Determine the pH at 1/4 Neutralization
At 1/4 neutralization, 25% of acetic acid is neutralized:
- Moles of acetic acid neutralized = 0.1 moles × 0.25 = 0.025 moles
- Moles of acetic acid remaining = 0.1 - 0.025 = 0.075 moles
- Moles of acetate ion (CH₃COO⁻) formed = 0.025 moles

Using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
Where:
- \( \text{pK}_a \) of acetic acid ≈ 4.76
- \( [\text{A}^-] = 0.025 \, \text{moles} \) (acetate ion)
- \( [\text{HA}] = 0.075 \, \text{moles} \)

Calculating the concentrations:
- Total volume = 1 L (initial) + 0.025 L (added NaOH) = 1.025 L
- \( [\text{A}^-] = \frac{0.025}{1.025} \approx 0.02439 \, \text{M} \)
- \( [\text{HA}] = \frac{0.075}{1.025} \approx 0.07317 \, \text{M} \)

Now substituting into the Henderson-Hasselbalch equation:
\[ \text{pH}_{1/4} = 4.76 + \log \left( \frac{0.02439}{0.07317} \right) \]
Calculating the log term:
\[ \log \left( \frac{0.02439}{0.07317} \right) \approx -0.176 \]
Thus,
\[ \text{pH}_{1/4} \approx 4.76 - 0.176 \approx 4.584 \]

### Step 4: Determine the pH at 3/4 Neutralization
At 3/4 neutralization, 75% of acetic acid is neutralized:
- Moles of acetic acid neutralized = 0.1 moles × 0.75 = 0.075 moles
- Moles of acetic acid remaining = 0.1 - 0.075 = 0.025 moles
- Moles of acetate ion (CH₃COO⁻) formed = 0.075 moles

Using the Henderson-Hasselbalch equation again:
- \( [\text{A}^-] = 0.075 \, \text{moles} \)
- \( [\text{HA}] = 0.025 \, \text{moles} \)

Calculating the concentrations:
- Total volume = 1 L + 0.075 L = 1.075 L
- \( [\text{A}^-] = \frac{0.075}{1.075} \approx 0.06977 \, \text{M} \)
- \( [\text{HA}] = \frac{0.025}{1.075} \approx 0.02326 \, \text{M} \)

Now substituting into the Henderson-Hasselbalch equation:
\[ \text{pH}_{3/4} = 4.76 + \log \left( \frac{0.06977}{0.02326} \right) \]
Calculating the log term:
\[ \log \left( \frac{0.06977}{0.02326} \right) \approx 0.176 \]
Thus,
\[ \text{pH}_{3/4} \approx 4.76 + 0.176 \approx 4.936 \]

### Step 5: Calculate the Difference in pH
Now, we find the difference in pH between the two stages:
\[ \Delta \text{pH} = \text{pH}_{3/4} - \text{pH}_{1/4} \]
\[ \Delta \text{pH} = 4.936 - 4.584 = 0.352 \]

### Final Answer
The difference in pH between the 1/4 and 3/4 stages of neutralization is approximately **0.352**.

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