Calculate the pK_a of benzoic acid.

Step 1: Calculate the $\text{pK}_a$ of benzoic acid.
The $\text{K}_a$ for benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$) is $6.3 \times 10^{-5}$.

$$\text{pK}_a = -\log(\text{K}_a)$$ $$\text{pK}_a = -\log(6.3 \times 10^{-5})$$ $$\text{pK}_a \approx 4.201$$

Step 2: Use the Henderson-Hasselbalch equation to find the ratio of the conjugate base to the weak acid.
The target $\text{pH}$ is 4.25.

$$\text{pH} = \text{pK}_a + \log\left(\frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}\right)$$ $$4.25 = 4.201 + \log\left(\frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}\right)$$ $$4.25 - 4.201 = \log\left(\frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}\right)$$ $$0.049 = \log\left(\frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]}\right)$$ $$\frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]} = 10^{0.049}$$ $$\frac{[\text{C}_6\text{H}_5\text{COO}^-]}{[\text{C}_6\text{H}_5\text{COOH}]} \approx 1.119$$

Step 3: Calculate the initial moles of sodium benzoate ($\text{C}_6\text{H}_5\text{COO}^-$).
The volume of sodium benzoate solution is $5.0 \text{ L}$ and its concentration is $0.05 \text{ M}$.

$$\text{Moles of C}_6\text{H}_5\text{COO}^- = \text{Concentration} \times \text{Volume}$$ $$\text{Moles of C}_6\text{H}_5\text{COO}^- = 0.05 \text{ M} \times 5.0 \text{ L}$$ $$\text{Moles of C}_6\text{H}_5\text{COO}^- = 0.25 \text{ mol}$$

Step 4: Calculate the required moles of benzoic acid ($\text{C}_6\text{H}_5\text{COOH}$).
Since the volume is the same for both components in the buffer, the ratio of concentrations is equal to the ratio of moles.

$$\frac{\text{Moles of C}_6\text{H}_5\text{COO}^-}{\text{Moles of C}_6\text{H}_5\text{COOH}} = 1.119$$ $$$$