What is the difference in `pH` for `1//3` and `2//3` stages of neutralization of `0.1 M CH_(3)COOH` with `0.1 M NaOH`?

Options

A. `2log""(1)/(4)`
B. `2 log 3`
C. `0.9542`
D. `0.3010`

Answer

To solve the problem of finding the difference in pH for the 1/3 and 2/3 stages of neutralization of 0.1 M acetic acid (CH₃COOH) with 0.1 M sodium hydroxide (NaOH), we can follow these steps:

### Step 1: Understand the Neutralization Process
Neutralization is the reaction between an acid and a base to form a salt and water. In this case, acetic acid (CH₃COOH) is the acid and sodium hydroxide (NaOH) is the base.

### Step 2: Define the Stages of Neutralization
- **1/3 Stage**: At this stage, 1/3 of the acetic acid is left unreacted, and 2/3 of the salt (sodium acetate, CH₃COONa) is formed.
- **2/3 Stage**: At this stage, 2/3 of the acetic acid is left unreacted, and 1/3 of the salt (sodium acetate) is formed.

### Step 3: Use the Henderson-Hasselbalch Equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

### Step 4: Calculate pH at the 1/3 Stage
At the 1/3 stage:
- Concentration of salt (sodium acetate) = 2/3
- Concentration of acetic acid = 1/3

Using the Henderson-Hasselbalch equation:
\[ \text{pH}_{1/3} = \text{pKa} + \log \left( \frac{2/3}{1/3} \right) \]
\[ = \text{pKa} + \log (2) \]

### Step 5: Calculate pH at the 2/3 Stage
At the 2/3 stage:
- Concentration of salt (sodium acetate) = 1/3
- Concentration of acetic acid = 2/3

Using the Henderson-Hasselbalch equation:
\[ \text{pH}_{2/3} = \text{pKa} + \log \left( \frac{1/3}{2/3} \right) \]
\[ = \text{pKa} + \log \left( \frac{1}{2} \right) \]
\[ = \text{pKa} - \log (2) \]

### Step 6: Find the Difference in pH
Now, we need to find the difference in pH between the 1/3 and 2/3 stages:
\[ \text{Difference} = \text{pH}_{1/3} - \text{pH}_{2/3} \]
Substituting the values we calculated:
\[ \text{Difference} = \left( \text{pKa} + \log (2) \right) - \left( \text{pKa} - \log (2) \right) \]
\[ = \log (2) + \log (2) \]
\[ = 2 \log (2) \]

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