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{