0.1 M formic acid solution is titrated against 0.1 M NaOH solution. What would be the difference in pH between 1/5 and 4/5 stages of neutralization of acid?

Options

A. 2 log 3/4
B. 2 log 1/5
C. log 1/3
D. 2 log 4

Answer

To solve the problem of finding the difference in pH between the 1/5 and 4/5 stages of neutralization of a 0.1 M formic acid solution with 0.1 M NaOH, we can follow these steps:

### Step-by-Step Solution:

1. **Write the Neutralization Reaction**:
The neutralization reaction between formic acid (HCOOH) and sodium hydroxide (NaOH) can be written as:
\[ \text{HCOOH} + \text{NaOH} \rightarrow \text{HCOONa} + \text{H}_2\text{O} \]

2. **Identify the Stages of Neutralization**:
- At the **1/5 stage** of neutralization, 1 part of the acid reacts with 5 parts of the base. Thus, if we start with A moles of formic acid, at this stage we have:
- Acid remaining: \( A - \frac{A}{5} = \frac{4A}{5} \)
- Salt formed: \( \frac{A}{5} \)
- At the **4/5 stage** of neutralization, 4 parts of the acid react with 5 parts of the base. Thus, we have:
- Acid remaining: \( A - \frac{4A}{5} = \frac{A}{5} \)
- Salt formed: \( \frac{4A}{5} \)

3. **Calculate pH at 1/5 Stage**:
At the 1/5 stage, we have an acid buffer solution consisting of:
- Acid: \( \frac{4A}{5} \)
- Salt: \( \frac{A}{5} \)

The pH can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH}_1 = \text{pKa} + \log\left(\frac{\text{Salt}}{\text{Acid}}\right) = \text{pKa} + \log\left(\frac{\frac{A}{5}}{\frac{4A}{5}}\right) = \text{pKa} + \log\left(\frac{1}{4}\right) \]
Simplifying gives:
\[ \text{pH}_1 = \text{pKa} - \log(4) \]

4. **Calculate pH at 4/5 Stage**:
At the 4/5 stage, we have:
- Acid: \( \frac{A}{5} \)
- Salt: \( \frac{4A}{5} \)

Again using the Henderson-Hasselbalch equation:
\[ \text{pH}_2 = \text{pKa} + \log\left(\frac{\text{Salt}}{\text{Acid}}\right) = \text{pKa} + \log\left(\frac{\frac{4A}{5}}{\frac{A}{5}}\right) = \text{pKa} + \log(4) \]

5. **Calculate the Difference in pH**:
Now we can find the difference in pH between the two stages:
\[ \Delta \text{pH} = \text{pH}_2 - \text{pH}_1 \]
Substituting the expressions we found:
\[ \Delta \text{pH} = \left(\text{pKa} + \log(4)\right) - \left(\text{pKa} - \log(4)\right) \]
Simplifying gives:
\[ \Delta \text{pH} = \log(4) + \log(4) = 2\log(4) \]

### Final Answer:
The difference in pH between the 1/5 and 4/5 stages of neutralization is:
\[ \Delta \text{pH} = 2 \log(4) \]