Hello!

I have a chemistry question (unfortunately I forgot a lot about what I learned in high-school )

I have bought distilled white vinegar that is labelled as 14% acid, which I assume means 140g of acetic acid in 1L of solution.

Most A/B teks that I saw use standard 5% acid vinegar, so I wanted to calculate how much I should dilute my 14% vinegar in order to get a solution with a pH of 4. Note that I know that I could simply buy a pH tester and do it experimentally (which I will probably end up doing just to be safe), but I would be interested to know the chemistry.

I used two different approaches to calculate the dilution, but they gave me two wildly different answers, so I was hoping that someone more well versed in chemistry could shed some light on this.

First, I calculate the molarity of my initial 14% solution: the molar mass of acetic acid is 60.052g/mol, so 140g/L is 2.33mol/L.

Then, I used this website to calculate the pH of my 14% solution, and it said that the pH of 2.33M acetic acid is 2.20.

Then, using the same website, I reduced the molarity until it said that the pH is 4.0, and this happened with a molarity of 0.000666mol/L. This is 3500 times more diluted than the original 2.33mol/L! This seemed really weird to me and I don't trust this result, but I can't tell what did I do wrong? I think it might have to do with the fact that acetic acid is a weak acid but I'm not sure.

Then I tried another method using this other website. I set the original pH value as 2.2, the acid amount as 1, and I played with the water amount until I got to a pH of 4. I found that with 63 volumes of water for 1 volume of acid, the pH got to 4. This seems more correct intuitively.

Can anyone help me understand?

Me too, i forgot most of what i learned in school long time ago. So i went to wikipedia to look it up..


I hope one of the professional or amateur chemists will chime in with some semi simple trick

Anyway, i suspect you want pH 4 in an extraction. So you can't forget about the plant matter and its contents, that will influence pH one way or another. You'll probably have to measure, or just use the amounts mentioned in the tek.

This seems useful.

Also, if you want to to end up with 5% acetic acid from 14%, you could just add 1.8 liters of water per liter of vinegar.

Thanks for the answer! Unfortunately I can't just use the amounts of the tek since it uses 5% vinegar and I have 14%. But yes I will measure. I was asking because I would like to understand.



Very useful indeed, but it does not really explain anything and it's not the same result as the other two methods I tried ^^' this one says that the ratio is 1 volume of 14% vinegar for 4100 volumes of water, doesn't that seem like a really low concentration? For instance if you calculate for 5% vinegar then it would be 1 for 1500, which is a much lower ratio compared to what the teks use.

As I said I will do it with a pH tester anyway just to be safe, the reason why I am asking is because I would like to learn how it works and understand why all methods do not give the same result

Diluting 14% vinegar to 5% is a matter of fairly simple arithmetic.

14% w/v == 14g vinegar per 100mL solution.

Therefore we work out how many mL contains 5g of vinegar (it must be 5/14 of that volume):

(5/14)*100 = 71.4285/2 = 35.7142 mL

This volume (35.7-ish mL) is then diluted using distilled water to make up a volume of 100mL.
[NB - this might not be the same as adding 64.3mL of water because there may be a volume change due to the entropy of mixing.]

Putting it another way, 0.357mL of 14% w/v acid can be considered equivalent to 1mL of 5% acid. Multiply the amounts in the 'tek' by 0.357 and adjust the final volume accordingly; that said , it's likely the exact pH of the extraction is not absolutely crucial so you needn't worry about it too much.

The difference between the two websites mentioned in the OP may be down to how much each one takes account of the degree of dissociation in an acetic acid solution of any given strength. Acetic acid is only partially dissociated so at higher concentrations it exhibits a higher pH than a stronger acid would.

The pH scale is logarithmic, thus with 10^1.8 = 63.095734 we can see that the second website is merely applying a linear formula for the dilution whereas the first is taking account of the non-linearity of the dissociation of weak acids.