#### Answer

The necessary volume of acetate is equal to 18.2 mL; therefore, the correct answer is (d).

#### Work Step by Step

1. Calculate the pKa value for acetic acid:
$pKa = -log(Ka)$
$pKa = -log( 1.8 \times 10^{- 5})$
$pKa = 4.74$
2. Use the Henderson-Hasselbalch equation to find the concentration of the acetate ion:
** Since we are analyzing a single solution, the concentration ratio is equal to the amount (moles) ratio, because they are in the same volume, so:
** 100 mL = 0.100 L
Amount of acetic acid: $0.10M \times 0.100L = 0.010 $ moles.
So:
$pH = pK_a + log\frac{[Base]}{[Acid]}$
$4.00 = 4.74 + log\frac{(Acetate)}{0.010}$
$-0.74 = log(Acetate) - log(0.010)$
$-0.74 = log(Acetate) + 2.00$
$log(Acetate) = -2.74$
$(Acetate) = 10^{-2.74} = 1.82 \times 10^{-3}$ moles.
3. Conver that number to volume:
$1.82 \times 10^{-3} moles \times \frac{1 L}{0.10 mol} = 1.82 \times 10^{-2} L = 18.2 mL $