What volume of 0.10 M sodium formate solution should be added to 50 mL of 0.05 M formic acid to produce a buffer solution of pH = 4.0 ? `pK_a` for formic acid = 3.80?

Explanation

To solve the problem of determining the volume of 0.10 M sodium formate solution that should be added to 50 mL of 0.05 M formic acid to create a buffer solution with a pH of 4.0, we can follow these steps:

### Step-by-Step Solution:

1. **Identify Given Values**:
- Concentration of formic acid (HCOOH), \( [HA] = 0.05 \, M \)
- Volume of formic acid, \( V_{HA} = 50 \, mL = 0.050 \, L \)
- Concentration of sodium formate (HCOONa), \( [A^-] = 0.10 \, M \)
- Desired pH of the buffer solution, \( pH = 4.0 \)
- \( pK_a \) of formic acid, \( pK_a = 3.80 \)

2. **Use the Henderson-Hasselbalch Equation**:
The Henderson-Hasselbalch equation is given by:
\[ pH = pK_a + \log\left(\frac{[A^-]}{[HA]}\right) \]
Plugging in the values we have:
\[ 4.0 = 3.80 + \log\left(\frac{[A^-]}{[HA]}\right) \]

3. **Rearranging the Equation**:
Subtract \( pK_a \) from both sides:
\[ 4.0 - 3.80 = \log\left(\frac{[A^-]}{[HA]}\right) \]
\[ 0.20 = \log\left(\frac{[A^-]}{[HA]}\right) \]

4. **Convert Logarithmic Form to Exponential Form**:
Converting from logarithmic to exponential form gives:
\[ \frac{[A^-]}{[HA]} = 10^{0.20} \]
Calculate \( 10^{0.20} \):
\[ \frac{[A^-]}{[HA]} \approx 1.58 \]

5. **Calculate the Moles of Formic Acid**:
The moles of formic acid can be calculated as:
\[ \text{Moles of } HCOOH = [HA] \times V_{HA} = 0.05 \, M \times 0.050 \, L = 0.0025 \, moles \]

6. **Set up the Ratio**:
Let \( V \) be the volume of sodium formate solution added in liters. The moles of sodium formate will be:
\[ \text{Moles of } HCOONa = [A^-] \times V = 0.10 \, M \times V \]
Using the ratio from the Henderson-Hasselbalch equation:
\[ \frac{0.10 \times V}{0.0025} = 1.58 \]

7. **Solve for V**:
Rearranging gives:
\[ 0.10 \times V = 1.58 \times 0.0025 \]
\[ V = \frac{1.58 \times 0.0025}{0.10} = 0.0395 \, L = 39.5 \, mL \]

### Final Answer:
The volume of 0.10 M sodium formate solution that should be added is approximately **39.5 mL**.