0.01 mole of NaOH is added to 1 litre of a buffer solution which contains 0.1 M acetic acid and 0.1 M sodium acetate If the pKa value of acetic acid is 4.76, the pH of the solution is

Options

A. `4.76`
B. ` lt 4.76`
C. `7.6`
D. ` gt 4.76`

Explanation

To find the pH of the buffer solution after adding 0.01 moles of NaOH, we can follow these steps:

### Step 1: Understand the buffer system
The buffer solution consists of acetic acid (CH₃COOH) and sodium acetate (CH₃COONa). The pKa of acetic acid is given as 4.76.

### Step 2: Use the Henderson-Hasselbalch equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
In this case, the salt is sodium acetate and the acid is acetic acid.

### Step 3: Identify initial concentrations
Initially, the concentrations of acetic acid and sodium acetate are both 0.1 M:
- \([\text{Acid}] = 0.1 \, \text{M}\)
- \([\text{Salt}] = 0.1 \, \text{M}\)

### Step 4: Calculate initial pH
Substituting the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.76 + \log\left(\frac{0.1}{0.1}\right) \]
Since \(\frac{0.1}{0.1} = 1\), we have:
\[ \log(1) = 0 \]
Thus, the initial pH is:
\[ \text{pH} = 4.76 + 0 = 4.76 \]

### Step 5: Determine the effect of adding NaOH
When we add 0.01 moles of NaOH to the buffer solution, it will react with the acetic acid to form more sodium acetate:
\[ \text{CH}_3\text{COOH} + \text{NaOH} \rightarrow \text{CH}_3\text{COONa} + \text{H}_2\text{O} \]
This means that:
- The amount of acetic acid decreases by 0.01 moles.
- The amount of sodium acetate increases by 0.01 moles.

### Step 6: Calculate new concentrations
After adding NaOH, the new concentrations in 1 L of solution will be:
- New concentration of acetic acid:
\[ [\text{Acid}] = 0.1 - 0.01 = 0.09 \, \text{M} \]
- New concentration of sodium acetate:
\[ [\text{Salt}] = 0.1 + 0.01 = 0.11 \, \text{M} \]

### Step 7: Calculate new pH
Now we can substitute the new concentrations into the Henderson-Hasselbalch equation:
\[ \text{pH} = 4.76 + \log\left(\frac{0.11}{0.09}\right) \]
Calculating the ratio:
\[ \frac{0.11}{0.09} \approx 1.222 \]
Now, calculate the logarithm:
\[ \log(1.222) \approx 0.087 \]
Thus, the new pH is:
\[ \text{pH} = 4.76 + 0.087 \approx 4.847 \]

### Final Answer
The pH of the solution after adding 0.01 moles of NaOH is approximately **4.85**.