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 acetic acid with NaOH, we can follow these steps:

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

### Step 2: Define Initial Concentrations
Given that we have a 0.1 M solution of acetic acid and we are titrating with a 0.1 M NaOH solution, we can denote the initial concentration of acetic acid as \( A = 0.1 \, \text{M} \).

### Step 3: Calculate Concentrations at 1/4 Neutralization
At the 1/4 stage of neutralization, we have reacted 1/4 of the acetic acid with NaOH. Thus:
- Moles of acetic acid reacted = \( \frac{A}{4} \)
- Moles of acetic acid remaining = \( A - \frac{A}{4} = \frac{3A}{4} \)
- Moles of salt (sodium acetate) formed = \( \frac{A}{4} \)

The concentrations at this stage are:
- Concentration of acetic acid = \( \frac{3A}{4} = \frac{3 \times 0.1}{4} = 0.075 \, \text{M} \)
- Concentration of sodium acetate = \( \frac{A}{4} = \frac{0.1}{4} = 0.025 \, \text{M} \)

### Step 4: Calculate pH at 1/4 Neutralization
Using the Henderson-Hasselbalch equation for the pH of an acidic buffer:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Substituting the values:
\[ \text{pH}_1 = \text{pKa} + \log\left(\frac{0.025}{0.075}\right) \]
\[ \text{pH}_1 = \text{pKa} + \log\left(\frac{1}{3}\right) \]

### Step 5: Calculate Concentrations at 3/4 Neutralization
At the 3/4 stage of neutralization, we have reacted 3/4 of the acetic acid:
- Moles of acetic acid reacted = \( \frac{3A}{4} \)
- Moles of acetic acid remaining = \( A - \frac{3A}{4} = \frac{A}{4} \)
- Moles of salt (sodium acetate) formed = \( \frac{3A}{4} \)

The concentrations at this stage are:
- Concentration of acetic acid = \( \frac{A}{4} = 0.025 \, \text{M} \)
- Concentration of sodium acetate = \( \frac{3A}{4} = 0.075 \, \text{M} \)

### Step 6: Calculate pH at 3/4 Neutralization
Using the Henderson-Hasselbalch equation again:
\[ \text{pH}_2 = \text{pKa} + \log\left(\frac{0.075}{0.025}\right) \]
\[ \text{pH}_2 = \text{pKa} + \log(3) \]

### Step 7: Calculate the Difference in pH
Now, we find the difference in pH:
\[ \Delta \text{pH} = \text{pH}_2 - \text{pH}_1 \]
Substituting the values:
\[ \Delta \text{pH} = \left(\text{pKa} + \log(3)\right) - \left(\text{pKa} + \log\left(\frac{1}{3}\right)\right) \]
\[ \Delta \text{pH} = \log(3) - \log\left(\frac{1}{3}\right) \]
Using the property of logarithms:
\[ \Delta \text{pH} = \log(3) - (-\log(3)) = \log(3) + \log(3) = 2\log(3) \]

### Final Answer
The difference in pH between the 1/4 and 3/4 stages of neutralization is:
\[ \Delta \text{pH} = 2\log(3) \]

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