# link between acidity and [H3O+]

Suppose I have an aqueous solution (1) of $\ce{CH3COOH}$. The chemical species in the solution will be $\ce{H2O, H3O+, CH3COO-}$ and, since it's a weak acid, some $\ce{CH3COOH}$ will remain.

Suppose I have another solution (2) containing initially water and $\ce{HCl}$. The chemical species in the solution will be $\ce{H2O, H3O+}$ and $\ce{Cl-}$ since it's a strong acid. Suppose the initial quantity of $\ce{HCl}$ was calculated so that the obtained concentration of $\ce{H3O+}$ in solution (2) is the same than in solution (1). But in solution (1), in addition to the $\ce{H3O+}$ ions, we have some molecules of another acid, the remaining $\ce{CH3COOH}$. Then the $\mathrm{pH} = - \log [\ce{H3O+}]$ is the same for (1) and (2), while (1) contains more "acidic" molecules.

How is pH a measure of acidity then ? It seems to me that if for practical purposes I had to use one of these solutions and wanted to avoid protons being exchanged, I would not consider these two solutions equally.

• Please use the \ce{...} environment for chemical formulae. Type $\ce{SO4^2-}$ for $\ce{SO4^2-}$ or $\ce{CH3COO-}$ for $\ce{CH3COO-}$. It not only makes typing much easier but also uses an upright font. Check out the help center and this meta-post for more information.
– Jan
Nov 18, 2015 at 14:18
• $\mathrm{pH}$ is a measure of acidity of a solution, not of a substance. One of the measures for the latter would be the substance's $\mathrm{pK_a}$. If you look at the formula for the $\mathrm{pK_a}$, you can see that a type of normalization is going on; you divide the amount of protons and deprotonated acid in a solution of the acid by the amount of still protonated acid. If you really want to compare substance acidities using $\mathrm{pH}$, you also need to know the concentration of the acid, which ultimately allows one to calculate the $\mathrm{pK_a}$, even if only implicitly. Nov 18, 2015 at 14:42
• I was considering pH as a measure of acidity of a solution in my question. What puzzles me is that two solutions having the same concentration of $\ce{H3O+}$, but one having on top of that some $\ce{CH3COOH}$ in it are considered equally acid, as far as pH is concerned. Nov 18, 2015 at 14:48

$\mathrm{pH}$ is a measure of acidity but defining acidity as the concentration of dissoluted protons in solution. This is quite a bit different from defining acidity as the amount of acid in solution.

Taking your example, the appropriate concentration of $\ce{HCl}$ would be something like $1~\mathrm{mM}$ while the corresponding acetic acid concentration is maybe $1~\mathrm{M}$. (These numbers are most certainly not correct, and maybe even the dimension is off by one or two orders of magnitute, but they can still serve well as an example.) So you could say ‘but there’s one thousand times as many acid molecules in the acetic acid solution!’

When considering the actual effects of an acid, the concentration of the original acidic species is irrelevant. Vinegar is, for example, usually $5~\%$ acetic acid and can be added to food (tough guys might even drink it) — I wouldn’t try that with $5~\%~\ce{HCl}$ since that contains a much higher active proton concentration.

What you can do, though, is consider the higher concentration of weak acids a type of buffer: Dilute acetic acid and the $\mathrm{pH}$ will hardly change while diluting $\ce{HCl}$ will have an immediate effect on $\mathrm{pH}$. So a weak acid can supply the same weak source of protons for a longer time.

And even though real-world analogies should be avoided in any case since they often give a wrong impression: If you jump down the staircase of a high building two steps at a time there will be likely no effect (small effect but buffered — acetic acid) just like if you only jump the two steps once (small effect unbuffered — strongly diluted $\ce{HCl}$), but if you try to jump the whole staircase (large effect unbuffered — $\ce{HCl}$ in the same concentration as acetic acid) you won’t remain unaffected.

• Is it correct to say that "acidity is defined as the concentration of dissoluted protons in solution" because we can neglect the presence of the other, non dissoluted protons still attached to the acid ? The argument in that case would be that if the extra $\ce{CH3COOH}$ were to matter in any reaction based on proton exchange, they would have done it already with the water they are in ? Nov 18, 2015 at 14:52
• @aidaGoG Well, it depends on what you want to do with your protons. If acidic protons are merely catalytic, it doesn’t matter at all. If you need stoichiometric protons, you need to add the appropriate concentration of the acid (e.g. 2 equivalents of $\ce{CH3COOH}$), not the appropriate concentration of free protons. And in that case you would keep the overall free proton count at a specific time more or less constant while maintaining a constant flow of new ones.
– Jan
Nov 18, 2015 at 15:56