You want to make an acetic acid/acetate buffer, pH 3.8, with a volume of 250 mL, and a final concentration of ([AcOH] + [AcO-]) = 0.1 M. You may only use acetic acid and sodium acetate (no strong acid or base). How many mL of glacial acetic acid would be needed (d = 1.05 g/mL)?

You need to look up pKa for acetic acid in your text or notes. I used pKa = 4.74.

And is that 3.8 or 3.80? It will make a difference in determining the number of significant figures.
pH = pKa + log[(base)/(acid)]
3.80 = 4.74 + log B/A
Solve for B/A which is equation 1.
Then A + B = 0.1 is equation 2.
Solve these two equation simultaneously for A and B which will give you the mols A and mols B. I assume you can take it from here.

using this method i calculated 1.03 mL, but this was not one of the options

the options were:

a. ) 1.28 mL
b. ) 1.41 mL
c. ) 0.15 mL
d. ) 2.18 mL

Sorry. I didn't correct for the volume.

You want 0.250 L (not a liter) of solution.
So A + B = 0.025
and B/A is the same as before. By the way, when I solved for B/A, I get 0.1136. That's too many s.f. but I carried it through. You can round at the end. I ended up with answer a.

I got 1.028mL instead of 1.28mL.

After finding the concentration of the acid, I multiplied by the molar mass. then, i multiplied by L/1050g to cancel out the g. Finally, I multiplied by 0.250L. Clearly, I'm making this too complicated.

B/A = 0.1136 or

B = 0.1136*A

A + B = 0.025
A + 0.1136A = 0.025
A = 0.025/1.1136 = 0.02245 moles
0.02245*60g/mol = 1.347 grams acetic acid.
1.347/1.05 = 1.282 which rounds to 1.28 mL.

To determine the volume of glacial acetic acid needed to prepare the acetic acid/acetate buffer, we need to use the Henderson-Hasselbalch equation:

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

Given that the desired pH is 3.8, we can rearrange the equation to solve for the ratio of [A-] to [HA]:

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

The pKa of acetic acid is 4.76, so we can substitute the values:

[A-]/[HA] = 10^(3.8 - 4.76) = 10^(-0.96) = 0.1102

Since the sum of [A-] and [HA] should be 0.1 M, we can assume that [A-] and [HA] are equal to 0.05 M each.

Now, we can calculate the moles of acetic acid and sodium acetate needed:

moles of acetic acid = 0.05 M x 0.25 L = 0.0125 moles

To find the volume of glacial acetic acid needed, we need to calculate the mass of glacial acetic acid required using its density:

mass of glacial acetic acid = volume x density = 0.0125 moles x 60.05 g/mol x 1 mL/1.05 g = 0.712 mL

Therefore, you would need approximately 0.712 mL of glacial acetic acid to prepare the acetic acid/acetate buffer.