Given that the Ka of benzoic acid is 6.50 x 10-5, how would one prepare 0.500 L of a benzoic acid/sodium benzoate buffer of a desired pH of 4.00? The starting concentration of the benzoic acid is 3.00 M and the molecular weight of sodium benzoate is 144.10 g/mol

I think the easiest way to do this is in one step after the preliminary information is set. For simplicity, let's call benzoic acid HB (instead of C6H5COOH) and the salt will be NaB.

HB + NaOH ==> NaB + H2O
Now we have 3.00M HB and we want 0.5L of the buffer so we use all of the HB (which is 3.0M x 0.5L = 1.5 moles HB.
..............HB + NaOH ==> NaB + H2O
initial......1.5...0.........0.....0
add................x................
change.....1.5-x..-x.........+x.....+x
equil......1.5-x...0..........x......x

4.00 = pKa + log (B)/(HB)
4.00 = pKa + log (x)/(1.50-x)
Solve for x which equals moles NaOH (of course that is moles NaB but that's the same as moles NaOH. Then moles NaOH*molar mass NaOH = grams NaOH. If I didn't goof that should be close to 24 grams but you need to check that out closely. That's not an exact number.
I STRONGLY suggest you do one more thing, as I always do, and that is to check the final pH. Start with the 24 g NaOH, or whatever it turns out to be, convert to moles, plug into the ICE chart and plug into the HH equation and see if the pH turns out to be 4.00. If so you're in good shape, if not, there is an error somewhere.

To prepare a benzoic acid/sodium benzoate buffer of a desired pH, we need to use the Henderson-Hasselbalch equation:

pH = pKa + log ([A-]/[HA])

In this case, benzoic acid (HA) is the weak acid and sodium benzoate (A-) is the conjugate base. We want to achieve a pH of 4.00, and we are given the pKa of benzoic acid as 6.50 x 10^-5.

Step 1: Calculate the ratio of [A-]/[HA].
To obtain the desired pH, we need to calculate the ratio [A-]/[HA] using the Henderson-Hasselbalch equation.

4.00 = -log(Ka) + log ([A-]/[HA])

Rearranging the equation to solve for the ratio [A-]/[HA]:

[A-]/[HA] = 10^(pH - pKa)

[A-]/[HA] = 10^(4.00 - (-log(6.50 x 10^-5))) or simply [A-]/[HA] = 10^(4.00 + 4.19)

[A-]/[HA] = 10^8.19

[A-]/[HA] ≈ 8.38 x 10^8

Step 2: Determine the concentration of [HA].
We are given that the starting concentration of benzoic acid ([HA]) is 3.00 M.

Step 3: Calculate the concentration of [A-].
To find the concentration of sodium benzoate ([A-]), we first need to determine the number of moles of benzoic acid used, then convert it to moles of sodium benzoate based on the molecular weight ratio.

Number of moles of benzoic acid (HA) = Concentration (M) × Volume (L)
= 3.00 M × 0.500 L
= 1.50 moles

Number of moles of sodium benzoate (A-) = Number of moles of benzoic acid (HA)

Now, convert moles of sodium benzoate to its concentration using the volume of the solution (0.500 L):

Concentration of sodium benzoate ([A-]) = Number of moles of sodium benzoate / Volume (L)
= 1.50 moles / 0.500 L
= 3.00 M

Step 4: Determine the amount of sodium benzoate needed.
To find the mass of sodium benzoate needed, we need to use the molecular weight of sodium benzoate (144.10 g/mol) and the concentration of sodium benzoate ([A-]).

Mass of sodium benzoate (g) = Concentration (M) × Volume (L) × Molecular weight (g/mol)
= 3.00 M × 0.500 L × 144.10 g/mol

Mass of sodium benzoate = 108.08 g

Therefore, to prepare 0.500 L of a benzoic acid/sodium benzoate buffer with a desired pH of 4.00, you would need to mix 1.50 moles (or 108.08 grams) of sodium benzoate with 0.500 L of 3.00 M benzoic acid.