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To solve this problem, we will use the Henderson-Hasselbalch equation, which relates the pH of a buffer solution to the concentration of the acid and its conjugate base:

\[
\text{pH} = \text{p}K_a + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
\]

### Given information:
- Volume of acetic acid (\( V \)) = \( 80.0 \text{ mL} \) (or \( 0.080 \text{ L} \))
- Concentration of acetic acid (\( [\text{Acid}] \)) = \( 0.462 \text{ M} \)
- Desired pH = \( 4.91 \)
- \( K_a \) for acetic acid = \( 1.8 \times 10^{-5} \)
- Sodium hydroxide (\( \text{NaOH} \)) is a strong base that will react with the acetic acid to form acetate (\( \text{CH}_3\text{COO}^- \))

### Step 1: Calculate \( \text{p}K_a \) of Acetic Acid
\[
\text{p}K_a = -\log(1.8 \times 10^{-5}) = 4.74
\]

### Step 2: Apply the Henderson-Hasselbalch Equation
\[
4.91 = 4.74 + \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
\]
\[
4.91 - 4.74 = \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
\]
\[
0.17 = \log \left(\frac{[\text{Base}]}{[\text{Acid}]}\right)
\]
\[
\frac{[\text{Base}]}{[\text{Acid}]} = 10^{0.17} \approx 1.48
\]

### Step 3: Determine the Required Moles of Base
Since \( [\text{Acid}] = 0.462 \text{ M} \), we can calculate the amount of acetic acid in moles:

\[
\text{moles of acetic acid} = 0.462 \times 0.080 = 0.03696 \text{ moles}
\]

Now, use the ratio of base to acid to find the moles of acetate (the conjugate base):

\[
[\text{

What mass of sodium hydroxide must be added to 80.0 mL of 0.462 M acetic acid in order to create a buffer with a pH of 4.91? Ka for acetic acid is 1.8 x 10^-5.

Learn how to calculate the mass of sodium hydroxide (NaOH) required to prepare a buffer solution with a specific pH of 4.91, using 80.0 mL of 0.462 M acetic acid. This solution involves applying the Henderson-Hasselbalch equation and understanding buffer chemistry, suitable for undergraduate chemistry students.

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