How much of each solution will you need to mix to obtain 100 – 120 ml of 20 mM buffer pH 4.0?

A.
35 ml sodium acetate + 90 ml acetic acid

B.
100 ml sodium acetate + 16 ml acetic acid

C.
16 ml sodium acetate + 100 ml acetic acid

D.
90 ml sodium acetate + 35 ml acetic acid

Use the Henderson-Hasselbalch equation.

1. pH = pKa + lg (base)/(acid)
Substitute the numbers and solve for base/acid ratio. That's equation 1.

2. The second equation is acid + base = 0.020M

3. Solve equations 1 and 2 simultaneously to obtain (acid) and (base)

4a. You know M acid and L acid solve for mols and convert to grams.
4b. You know M base and L base solve for mols and convert to grams.

Post your work if you get stuck. These problem always look formidable but they are relatively easy once you get the hang of it.

Would I have to substitute the base/acid with the number given in the answers?

No. You can work out the problem and pick the answer that's the closest. Show your work if you have trouble and I can help you through it. Post your work up to where you don't know what to do then explain in detail why you're stuck.

This is the information given to me in the question.

-Acetic acid
pKa values(s)
4,8

MW
60.1

I replace the pKa with whichever of the two, 4 or 8, given pKa?

pH = pKa + lg (base)/(acid)
4= 8 + lg (base)/(acid)

So then
ratio of base/acid= lg(-4)?

Saraly, I think you have misread the problem. The pH is 4.0 and the pKa (there is only one listed) is 4.8

So 4.0 = 4.8 + log base/acid
and base/acid = ?

I simply do not understand how I am going to get one of those answers. This problem comes after a video explaining pH and buffers, only I didn't understand at all, to be sincere, the work shown with the few given examples.

That information is part of a table with more pka's, and i'm sure it's not a decimal in between the 4 and 8. I will try it with the decimal anyway, maybe it was a mistake.

OK. Forget about the video and follow my instructions. There are only 5 or 6 steps.

One equation you need is that last one I wrote.
4.0 is the pH you want the buffer to be.
4.8 is the pKa given to you in the problem. Substitute into the HH equation to obtain
4.0 = 4.8 + log(base)/(acid)
and solve for (base)/(acid).
What do you get for that?

For whatever it's worth, the pKa of acetic acid is 4.76 according to a text I have so I suspect the author of your problem just rounded it to 4.8. And acetic acid has only one pKa value, not two.

10^-0.8= .158/1 ?