[DutchGirl13]

I am doing research, and one of my professors gave some numbers for an acetic acid/acetate buffer and asked me to check them. The buffer we'll be using will have a concentration of 0.01 M and be at pH 6. Here is my work:

Henderson-Hasselbach
6=4.74 + log (acetate/acetic acid)

And assuming
0.01= [acetate] + [acetic acid]

Solving for [acetic acid] and substituting into H-H yields
6=4.74 + log (acetate/(0.01-acetate)).
I simplified this equation to 1.24= log[acetate]-log(0.01)+log[acetate]), and I solved it. My acetate concentration, however turned out to be 0.009455 rather than the 6.4 e-3 M given. Where did I go wrong?

[Borek]

log(0.01)+log[acetate]

Think it over.

Who uses acetate buffer at pH 6.0? Rule of thumb is pKa±1, plenty of better buffers for pH 6.

And check Buffer Maker (see my signature).

[ajay.p.mamtora]

Your calculation is correct.
However, (acetate) '6.4 e-3' M equals to '0.319' M, which seems incorrect.
If you are asked to prepare 0.01 M (acetic acid/acetate) buffer of pH 6.0, you need to mix equal volumes of acetate (0.0094555 M) and acetic acid (0.0005445 M)

[Borek]

Your calculation is correct.
However, (acetate) '6.4 e-3' M equals to '0.319' M, which seems incorrect.

So it is correct, but it is incorrect?

To clarify - it is incorrect. And it was already pointed why.