Fixed volume of `0.1M` benzoic acid `(pK_(a) = 4.2)` solution is added into `0.2M` sodium benzote solution and formed a `300mL`, resulting acidic buffer solution. If `pH` of the resulting solution is `3.9`, then added volume of banzoic acid is

Options

A. `240mL`
B. `150mL`
C. `100mL`
D. None

Explanation

To solve the problem step by step, we will use the information provided about the concentrations of benzoic acid and sodium benzoate, the total volume of the solution, and the pH of the resulting buffer solution.

### Step 1: Define the Variables
Let:
- \( V \) = volume of benzoic acid solution added (in mL)
- The concentration of benzoic acid = \( 0.1 \, M \)
- The concentration of sodium benzoate = \( 0.2 \, M \)
- Total volume of the buffer solution = \( 300 \, mL \)
- pH of the buffer solution = \( 3.9 \)
- \( pK_a \) of benzoic acid = \( 4.2 \)

### Step 2: Calculate Millimoles of Benzoic Acid
The number of millimoles of benzoic acid can be calculated using the formula:
\[ \text{Millimoles of benzoic acid} = \text{Concentration} \times \text{Volume} = 0.1 \times V \]

### Step 3: Calculate Millimoles of Sodium Benzoate
The volume of sodium benzoate solution will be \( 300 - V \) mL. Thus, the millimoles of sodium benzoate can be calculated as:
\[ \text{Millimoles of sodium benzoate} = 0.2 \times (300 - V) \]

### Step 4: Use the Henderson-Hasselbalch Equation
For an acidic buffer, the Henderson-Hasselbalch equation is given by:
\[ \text{pH} = pK_a + \log \left( \frac{[\text{Salt}]}{[\text{Acid}]} \right) \]
Substituting the values we have:
\[ 3.9 = 4.2 + \log \left( \frac{0.2 \times (300 - V)}{0.1 \times V} \right) \]

### Step 5: Rearranging the Equation
Rearranging gives:
\[ 3.9 - 4.2 = \log \left( \frac{0.2 \times (300 - V)}{0.1 \times V} \right) \]
\[ -0.3 = \log \left( \frac{0.2 \times (300 - V)}{0.1 \times V} \right) \]

### Step 6: Convert Logarithmic Form to Exponential Form
Taking the antilog:
\[ \frac{0.2 \times (300 - V)}{0.1 \times V} = 10^{-0.3} \approx 0.5 \]
This simplifies to:
\[ 0.2 \times (300 - V) = 0.5 \times 0.1 \times V \]
\[ 0.2 \times (300 - V) = 0.05V \]

### Step 7: Solve for V
Expanding and rearranging gives:
\[ 60 - 0.2V = 0.05V \]
\[ 60 = 0.2V + 0.05V \]
\[ 60 = 0.25V \]
\[ V = \frac{60}{0.25} = 240 \, mL \]

### Final Answer
The added volume of benzoic acid is \( 240 \, mL \).

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