Simulated Vinegar One way to make vinegar (not the preferred way) is to prepare a solution of acetic acid, the sole acid component of vinegar, at the proper pH (see Fig. 2-15) and add appropriate flavoring agents. Acetic acid (Mr 60) is a liquid at 25 °C, with a density of 1.049 g/mL. Calculate the volume that must be added to distilled water to make 1 L of simulated vinegar (see Fig. 2-16).

Explanation:

Well, to make it simple, you want to make a solution of acetic acid with pH of 3 using pure acetic acid. We have the dissociation K which makes the life easier:

CH3COOH <--> CH3COO- + H+

k= [CH3COO-]*[H+]/[CH3COOH]

In the target solution [H+]=[CH3COO-]=10^-3. (Based upon pH=-log[H+])

So, all we have to know is how many starting mols of acetic acid (x) can end up with this concentration of H+.

At equilibrium and using the K you've mentioned we can write:

(10^-3)*(10^-3)/(X-10^-3)=1.74*10^-5

(Remeber that if we have a 10^-3 mol concentration of [H+], it means that 10^-3 mol of our starting acid has already dissociated to make it, hence that X-10^-3 instead of a sole X)

If we solve the above equation, we will see that X (the strarting concentration of acetic acid) is about 0.058 mol/lit. Now life gets even easier:

If a mol of acid acetic weights 60 gr, then 0.058 mols should weight... let me see... yes, 3.48 gr... And now, we have the weight, and all we need to calculate the volume, is the density (density=weight/volume), and fortunately, we have that piece of information as well... With a density of 1.049 g/ml, to have 3.49 gr of acid at hand we need to take 3.317 ml of acetic acid which can be rounded to 3.32 ml which you mentioned as the answer!

All that is left to find is that "appropriate flavouring agent" which in my opinion will eventually pose the main problem! (An ester will probably do it...)