pH of a buffer solution decreases by 0.02 units when 0.12g of acetic acid is added to 250mL of a buffer solution of acetic acid and potassium acetate at `27^(@)`C. The buffer capacity of the solution is

Options

A. `0.1`
B. 10
C. 1
D. `0.4`

Explanation

To find the buffer capacity of the solution, we can follow these steps:

### Step 1: Calculate the number of moles of acetic acid added.
Given:
- Mass of acetic acid (CH₃COOH) added = 0.12 g
- Molar mass of acetic acid = 60 g/mol

Using the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Substituting the values:
\[ \text{Number of moles} = \frac{0.12 \, \text{g}}{60 \, \text{g/mol}} = 0.002 \, \text{moles} = 2 \times 10^{-3} \, \text{moles} \]

### Step 2: Determine the volume of the buffer solution in liters.
Given:
- Volume of buffer solution = 250 mL

Convert milliliters to liters:
\[ \text{Volume in liters} = \frac{250 \, \text{mL}}{1000} = 0.250 \, \text{L} \]

### Step 3: Calculate the change in pH.
Given:
- Change in pH = 0.02 units

### Step 4: Calculate the buffer capacity.
The buffer capacity (β) can be calculated using the formula:
\[ \beta = \frac{\text{Number of moles of acid added}}{\text{Volume of solution (L)} \times \text{Change in pH}} \]

Substituting the values:
\[ \beta = \frac{2 \times 10^{-3} \, \text{moles}}{0.250 \, \text{L} \times 0.02} \]

Calculating the denominator:
\[ 0.250 \, \text{L} \times 0.02 = 0.005 \]

Now substituting back into the buffer capacity formula:
\[ \beta = \frac{2 \times 10^{-3}}{0.005} = 0.4 \, \text{mol/L/pH} \]

### Final Answer:
The buffer capacity of the solution is **0.4 mol/L/pH**.

---