A buffer solution contains 0.1 mole of sodium acetate in 1000 `cm^(3)` of 0.1 M acetic acid. To the above buffer solution, 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 equal to .......... .

Options

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

Explanation

To solve the problem, we will use the Henderson-Hasselbalch equation, which is used to calculate the pH of buffer solutions. The equation is given by:

\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]

### Step 1: Identify the components of the buffer solution
Initially, we have:
- Sodium acetate (Salt) = 0.1 moles
- Acetic acid (Acid) = 0.1 M in 1000 cm³ = 0.1 moles

### Step 2: Calculate the concentrations after adding more sodium acetate
After adding another 0.1 moles of sodium acetate, the total amount of sodium acetate becomes:
- Total sodium acetate = 0.1 moles (initial) + 0.1 moles (added) = 0.2 moles

The concentration of sodium acetate in the buffer solution (which still has a total volume of 1000 cm³) is:
\[ [\text{Sodium Acetate}] = \frac{0.2 \text{ moles}}{1 \text{ L}} = 0.2 \text{ M} \]

The concentration of acetic acid remains:
\[ [\text{Acetic Acid}] = 0.1 \text{ M} \]

### Step 3: Find the pK_a of acetic acid
The dissociation constant \( K_a \) for acetic acid is approximately \( 1.8 \times 10^{-5} \). To find \( pK_a \):
\[ pK_a = -\log(K_a) \]
\[ pK_a = -\log(1.8 \times 10^{-5}) \approx 4.74 \]

### Step 4: Substitute values into the Henderson-Hasselbalch equation
Now we can substitute the values into the Henderson-Hasselbalch equation:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{Sodium Acetate}]}{[\text{Acetic Acid}]} \right) \]
\[ \text{pH} = 4.74 + \log \left( \frac{0.2}{0.1} \right) \]
\[ \text{pH} = 4.74 + \log(2) \]
\[ \text{pH} = 4.74 + 0.301 \]
\[ \text{pH} \approx 5.04 \]

### Final Answer
The pH of the resulting buffer solution is approximately **5.04**.