A buffer solution contains 0.1 mole of sodium acetate dissolved in `1000cm^(3)` of 0.1 M acetic acid. To the above buffer solution, 0.1 mole of sodium acetate is further added and dissolved. The pH of the resulting buffer is :

Options

A. `pK_(a)`
B. ) `pK_(a)+2`
C. `pK_(a)- log2`
D. `pK_(a)+ log2`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Identify the components of the buffer solution
The buffer solution consists of:
- Sodium acetate (CH₃COONa)
- Acetic acid (CH₃COOH)

Initially, we have:
- 0.1 moles of sodium acetate
- 1000 cm³ (1 L) of 0.1 M acetic acid

### Step 2: Calculate the initial concentrations
1. **Concentration of Sodium Acetate (CH₃COONa)**:
- Since we have 0.1 moles in 1 L, the concentration is:
\[ [CH₃COONa] = \frac{0.1 \text{ moles}}{1 \text{ L}} = 0.1 \text{ M} \]

2. **Concentration of Acetic Acid (CH₃COOH)**:
- Given as 0.1 M.

### Step 3: Add additional sodium acetate
We are adding 0.1 moles of sodium acetate to the existing solution. Therefore, the new amount of sodium acetate becomes:
\[ \text{Total moles of } CH₃COONa = 0.1 + 0.1 = 0.2 \text{ moles} \]
The concentration of sodium acetate in the same volume (1 L) is:
\[ [CH₃COONa] = \frac{0.2 \text{ moles}}{1 \text{ L}} = 0.2 \text{ M} \]

### Step 4: Write the Henderson-Hasselbalch equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Salt}]}{[\text{Acid}]}\right) \]
Where:
- \([\text{Salt}] = [CH₃COONa] = 0.2 \text{ M}\)
- \([\text{Acid}] = [CH₃COOH] = 0.1 \text{ M}\)

### Step 5: Substitute the values into the equation
Substituting the concentrations into the equation gives:
\[ \text{pH} = \text{pKa} + \log\left(\frac{0.2}{0.1}\right) \]
This simplifies to:
\[ \text{pH} = \text{pKa} + \log(2) \]

### Step 6: Conclusion
The pH of the resulting buffer solution is:
\[ \text{pH} = \text{pKa} + \log(2) \]