I learned about acid strength, that the strength of an acid increases with it's degree of ionization when solvated. So, in water, a strong acid is one where $\ce{[H_3O^+]}$ is large, which is equal to a low pH: $\mathrm{pH=-log[H_3O^+]}$.

Considering extreme cases, such as superacids, I have found out that other methods are used to measure their acidity (methods I don't really understand). My question is why is it impossible to simply get super high concentrations of $\ce{[H_3O^+]}$ in aqueous solutions of superacids, and use this to determine the acid strength. Also, is pH used as a measure of acidity outside of aqueous solutions?

I have come over the leveling effect, but I don't think I fully understand it. The way I understand it (for the case with water as solvent) is that basically any acid in water will protolyze $\ce{H2O}$ to $\ce{H3O+}$, making this the effective acid. I don't understand why this would affect the measured pH, as it is $\ce{[H_3O^+]}$ you are measuring.


Any acid-base reaction is always an equilibrium:

$$\ce{HA^1 + (A^2)- <=> (A^1)- + HA2}\tag{1}$$

and for each pair of acids $\ce{HA^1}$ and $\ce{HA^2}$ you could calculate a $K_\mathrm{a}$ value to determine one acid’s strength with respect to the other. This $K_\mathrm{a}$ value is typically calculated according to equation $(2)$ if $\ce{(A^2)-}$ (which does not have to feature a negative charge; I just wanted to avoid different descriptions for the two acids) is the solvent.

$$K_\mathrm{a} = \frac{[\ce{HA^2}][\ce{(A^1)-}]}{[\ce{HA^1}]}\tag{2}$$

If $\ce{(A^2)-}$ is $\ce{H2O}$ (i.e. $\ce{HA^2}$ is $\ce{H3O+}$) and the acid $\ce{HA^1}$ is a strong acid or a superacid, then this equation is of little use; the equilibrium $(1)$ will be strongly shifted towards the product side, the concentration $[\ce{HA^1}]$ will be very small and thus $K_\mathrm{a}$ very large. With ‘very small’, I mean that $[\ce{HA^1}]\approx10^{-10}$ — acid $\ce{HA^1}$ is said to be fully dissociated.

Exact measurements of $K_\mathrm{a}$ become very hard if not impossible because you would need to determine extremely minute concentrations. The concentrations $[\ce{HA^2}]$ and $[\ce{(A^1)-}]$ are of little help since:

$$[\ce{HA^2}] = [\ce{(A^1)-}] \approx c_0(\ce{HA^1})\tag{3}$$

This is the levelling effect of water: no matter how strong the acid, there is no acidic species stronger than $\ce{H3O+}$ that can survive in aqueous solution for extended time. Most importantly, consider two very strong acids with different $K_\mathrm{a}$ — e.g. one with $\mathrm{p}K_\mathrm{a} = -15$, the other with $-10$. They have a very different acidity, yet the concentrations of all relevant species in aqueous solutions will be nigh identical.

Therefore, a new, more expandable way to measure and compare the strengths of very strong acids was required.


The reason we can differentiate weak acids in aqueous media is because the acids themselves are able to exist in the solution. Some dissociate more than others, so the same concentrations of different acids have different $\ce{[H3O+]}$.

Any acid that has a $\mathrm{p}K_\mathrm{a}$ lower than that of $\ce{H3O+}$ ($-1.74$), however, is said to be a strong acid, and cannot exist in aqueous solution. They dissociate completely, so for a solution of any strong acid ($\ce{HA}$), $\mathrm{pH}=-\log\ce{[HA]}$. This means that a $1\ \mathrm{M}$ solution of $\ce{HCl}$ ($\mathrm{p}K_\mathrm{a}=-6.3$, a strong acid) will have the same $\mathrm{pH}$ as a $1\ \mathrm{M}$ solution of trifluoromethanesulfonic acid ($\mathrm{p}K_\mathrm{a}=-14.7$, a superacid).

And to answer your other question, yes, $\mathrm{pH}$ is a quantity that is only used to measure aqueous solutions. A solution that doesn't contain water will always have $\ce{[H3O+]}=0$.

  • $\begingroup$ This gets a little fuzzy. Most people identify nitric acid as a strong acid and yet Wikipedia gives pKa=-1.4>-1.74. The "cutoff" for strong acids should reflect typical dilute concentrations instead of the total concentration of water, so should be $0$ (1M) or maybe even $+1$ (0.1M). $\endgroup$ – Oscar Lanzi Dec 21 '18 at 11:22

Acidity is solvent dependent, and water is but one of numerous solvents within which one may design an acidity scale.

More than 400 acidity scales have been developed over time, so you need to realise acidity in water is but one example! Aqueous acidity takes centre stage because of the importance of water in human life, and on planet earth, but it is not the only one!

Auto-protolysis of H2O being on the order of $10^{-14}$ gives rise to the aqueous pH scale of 14.

By definition, a superacid has what is called a "Hammett" acidity, $H_\mathrm o$, of $-12$ or lower. There is no equivalent of $H_\mathrm o = -12$ on the aqueous pH scale.

In water, the strongest acid you can generate, is $\ce{H3O+}$ (because $\ce{H2O}$ levels either to the acid, $\ce{H3O+}$, or the base, $\ce{HO-}$) . Any acid which is in fact stronger than the strongest acid in water, will simply level water to produce $\ce{H3O+}$, with an acid strength occupying the lower limit of pH in water (relative to the natural extent of $\ce{H3O+}$ or $\ce{HO-}$ as dictated by its auto-protolysis reaction) .

A superacid will simply be out of (off of) the scale of aqueous acidity, sort of in a class of its own, relatively speaking, with acidity strength which lays at lower pH levels than is accessible via aqueous pH,as a result of being a much stronger acid than $\ce{H3O+}$.

There is Brønsted-Lowry acidity in whatever solvent you want, aqueous and otherwise. Even liquid HCl (as its own solvent) has an auto-protolysis similar to that of $\ce{H2O}$.


You pretty much answered your own question, but perhaps I can help clarify: (1) Two very (but not equally) strong acids will both be effectively 100% dissociated in water, so the concentration of hydronium ion will be equal to the starting concentration of acid for either acid. Hence, the measured pH will be the same for the two acids. (2) Remember that the concept of pH as a scale of acidity from 0-14 becomes less relevant with increasing concentrations of acid (or other solutes) relative to water.

  • $\begingroup$ So, to differentiate between two strong acids (compounds which easily dissociate), you need a less alkaline solvent (which less easily take up protons) => the two acids protolyze to dissociate to different degrees? $\endgroup$ – Adroit Nov 28 '16 at 17:12
  • $\begingroup$ There you have it. $\endgroup$ – iad22agp Nov 28 '16 at 23:30

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