How much volume of 0.1 M Hac should be added to 50 mL of 0.2 M NaAc solution if we want to prepare a buffer solution of pH 4.91. Given `pK_(a)` for acetic acid is 4.76.

Explanation

To solve the problem of how much volume of 0.1 M acetic acid (HAc) should be added to 50 mL of 0.2 M sodium acetate (NaAc) to prepare a buffer solution of pH 4.91, we can follow these steps:

### Step 1: Understand the Buffer Equation
The pH of a buffer solution can be calculated using the Henderson-Hasselbalch equation:
\[ \text{pH} = \text{pK}_a + \log \left( \frac{[\text{A}^-]}{[\text{HA}]} \right) \]
where:
- \([\text{A}^-]\) is the concentration of the base (sodium acetate),
- \([\text{HA}]\) is the concentration of the acid (acetic acid).

### Step 2: Identify Given Values
From the problem, we have:
- Desired pH = 4.91
- \(pK_a\) for acetic acid = 4.76
- Volume of NaAc solution = 50 mL
- Concentration of NaAc = 0.2 M
- Concentration of HAc = 0.1 M

### Step 3: Set Up the Henderson-Hasselbalch Equation
Substituting the known values into the Henderson-Hasselbalch equation:
\[ 4.91 = 4.76 + \log \left( \frac{[\text{NaAc}]}{[\text{HAc}]} \right) \]

### Step 4: Rearrange to Find the Ratio of Concentrations
Subtract \(pK_a\) from both sides:
\[ 4.91 - 4.76 = \log \left( \frac{[\text{NaAc}]}{[\text{HAc}]} \right) \]
\[ 0.15 = \log \left( \frac{[\text{NaAc}]}{[\text{HAc}]} \right) \]

### Step 5: Convert Logarithmic Equation to Exponential Form
Converting the logarithmic equation to its exponential form:
\[ \frac{[\text{NaAc}]}{[\text{HAc}]} = 10^{0.15} \]

### Step 6: Calculate the Value of \(10^{0.15}\)
Using a calculator:
\[ 10^{0.15} \approx 1.41 \]
Thus,
\[ \frac{[\text{NaAc}]}{[\text{HAc}]} = 1.41 \]

### Step 7: Express Concentrations in Terms of Volume
Let \(V\) be the volume of 0.1 M HAc added. The concentration of NaAc in the final solution after adding \(V\) mL of HAc is:
\[ [\text{NaAc}] = \frac{0.2 \times 50}{50 + V} \]
And the concentration of HAc is:
\[ [\text{HAc}] = \frac{0.1 \times V}{50 + V} \]

### Step 8: Set Up the Ratio
Substituting these into the ratio:
\[ \frac{\frac{0.2 \times 50}{50 + V}}{\frac{0.1 \times V}{50 + V}} = 1.41 \]
This simplifies to:
\[ \frac{0.2 \times 50}{0.1 \times V} = 1.41 \]
\[ \frac{10}{V} = 1.41 \]

### Step 9: Solve for \(V\)
Rearranging gives:
\[ V = \frac{10}{1.41} \approx 7.09 \text{ mL} \]

### Step 10: Final Calculation
Thus, the volume of 0.1 M HAc that should be added is approximately:
\[ V \approx 7.09 \text{ mL} \]

### Summary
To prepare a buffer solution of pH 4.91, you need to add approximately **7.09 mL** of 0.1 M acetic acid to 50 mL of 0.2 M sodium acetate.