[untererde]

So, this is a bit of a long question, and I've worked out a solution, but have absolutely no idea whether I'm on the right track. Here's the question and my answer; could someone check over it? Thank you!

Q: 750 mL of an acetic acid-sodium acetate buffer with pH 4.50 is needed. Solid sodium acetate (NaC₂H₃O₂) and glacial acetic acid (HC₂H₃O₂) are available. Glacial acetic acid is 99% HC₂H₃O₂ by mass and has a density of 1.05 g/mL. If the buffer is to be 0.02M in  HC₂H₃O₂, how many grams of NaC₂H₃O₂ and how many milliliters of glacial acetic acid must be used?

A:
( for acetic acid)
.20 mol acetic/ L  * .75 L * 44.06 g acetic/ mol *1 mL acetic/ 1.05 g = 6.29 mL acetic acid 
and perhaps (??) I should take 6.29mL* 1.01 (to account for 99%) = 6.35 mL acetic acid

(for sodium acetate)
pH= pKa + log([Na-acetate]/[acetic])
so, 4.5= 4.74 + log ([acetate]/.20 M)
after the arithmetic
[acetate]= .117 M

then, .117 mol Na-acetate/ L  * .75 L * 82.04 g Na-acetate/ mol  = 7.19 g sodium acetate

There are about 12 places I fear I may have gone wrong, so any and all help will be appreciated.

[sdekivit]

you can calculate the ratio between HAc and Ac^- in a pH = 4.50 solution and calculate from there.

[Yggdrasil]

and perhaps (??) I should take 6.29mL* 1.01 (to account for 99%) = 6.35 mL acetic acid

Should be 6.29mL/0.99.  Think it like this:

(Volume of pure acetic acid) = (Volume of 99% acetic acid) * 0.99
Therefore
(Volume of 99% acetic acid) = (Volume of pure acetic acid)/0.99

As long as your arithmetic is fine for the rest of the steps, the answer should be fine. 

The question is worded in a weird way, however.  Usually when they ask this type of question, they specify the total molarity of the acetic acid and sodium acetate combined rather than specify the molarity of only one component.