Fixed volume of 0.1 M benzoic acid `(pK_(a)=4.2)` solution is added into 0.2 M sodium benzoate solution and formed a 300 mL, resultant acidic buffer solution. If pH of this buffer solution is 4.5 then find added volume of benzoic acid :

Options

A. 100 mL
B. 150 mL
C. 200 mL
D. None of these

Answer

To solve the problem step by step, we will use the Henderson-Hasselbalch equation for the buffer solution, which relates the pH of the solution to the pKa and the concentrations of the acid and its conjugate base.

### Step 1: Understand the given information
- **Concentration of benzoic acid (C6H5COOH)** = 0.1 M
- **pKa of benzoic acid** = 4.2
- **Concentration of sodium benzoate (C6H5COONa)** = 0.2 M
- **Final volume of buffer solution** = 300 mL
- **pH of buffer solution** = 4.5

### Step 2: Set up the equation
Let the volume of benzoic acid solution added be \( V \) mL. Therefore, the volume of sodium benzoate solution will be \( 300 - V \) mL.

### Step 3: Calculate the moles of each component
- Moles of benzoic acid (C6H5COOH) = \( 0.1 \, \text{M} \times \frac{V}{1000} \)
- Moles of sodium benzoate (C6H5COO^-) = \( 0.2 \, \text{M} \times \frac{300 - V}{1000} \)

### Step 4: Write the Henderson-Hasselbalch equation
The Henderson-Hasselbalch equation is given by:
\[ \text{pH} = \text{pKa} + \log\left(\frac{[\text{Base}]}{[\text{Acid}]}\right) \]
Substituting the known values:
\[ 4.5 = 4.2 + \log\left(\frac{0.2 \times (300 - V)/1000}{0.1 \times V/1000}\right) \]

### Step 5: Simplify the equation
This simplifies to:
\[ 4.5 = 4.2 + \log\left(\frac{0.2(300 - V)}{0.1V}\right) \]
\[ 0.3 = \log\left(\frac{0.2(300 - V)}{0.1V}\right) \]

### Step 6: Convert logarithmic form to exponential form
From the logarithmic equation:
\[ 10^{0.3} = \frac{0.2(300 - V)}{0.1V} \]
Calculating \( 10^{0.3} \) gives approximately \( 2 \):
\[ 2 = \frac{0.2(300 - V)}{0.1V} \]

### Step 7: Cross-multiply and solve for V
Cross-multiplying gives:
\[ 2 \times 0.1V = 0.2(300 - V) \]
\[ 0.2V = 60 - 0.2V \]
Adding \( 0.2V \) to both sides:
\[ 0.4V = 60 \]
Dividing both sides by \( 0.4 \):
\[ V = \frac{60}{0.4} = 150 \, \text{mL} \]

### Conclusion
The volume of benzoic acid added is **150 mL**.