Do strong acids actually dissociate completely?

Question

Do strong acids actually dissociate completely (every single molecule dissociates), or are they just assumed to do so for the sake of simplicity? That seems odd, considering weak acids, many of which are very similar to strong acids, don't fully dissociate.

Answer 1

Yes, for sufficiently strong acids, in sufficiently dilute conditions, in sufficiently basic solvents. However, things are hazier than you might expect, and depending on your definition, "clear-cut" examples of complete dissociation only become possible when you start swatting flies with nuclear bombs.

The problem is that chemistry works on equilibria, and at some point it becomes difficult to measure tiny concentrations - in a haystack of one billion straws, how can you be sure there isn't at least one needle? And in one trillion? And what about in a septillion?

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First, let us perform a qualitative investigation by considering the generic acid dissociation equilibrium in water, as follows:

$$\ce{HA(aq)->H+(aq) + A-(aq)} \quad \quad \quad \mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}}=10^{-pK_a}}$$

To simplify the following discussion, we will make the very crude approximation where we equate the thermodynamic activity of a species with its molar concentration. In other words:

$$\mathrm{K_a=\frac{a_{H^+(aq)}a_{A^-(aq)}}{a_{HA(aq)}} \approx \frac{[H^+(aq)][A^-(aq)]}{[HA(aq)]}}$$

Our objective now is to virtually eliminate the presence of undissociated $\ce{HA}$ molecules. For clarity, we rearrange the above equation:

$$\mathrm{[HA(aq)] \approx \frac{[H^+(aq)][A^-(aq)]}{K_a}}$$

Clearly, to make $\ce{[HA]}$ as small as possible, the numerator must be very small, and the denominator must be very large. Let us set a goal of making $\mathrm{[HA]<10^{-27}\ mol\ L^{-1}}$, which implies less than $\mathrm{0.001}$ molecules of undissociated $\ce{HA}$ in a litre of solution.

At this point, I should mention that aqueous solutions of acids in ambient conditions must always have $\mathrm{[H^+(aq)]>10^{-7}\ mol\ L^{-1}}$, thanks to the self-ionization of water. We take the best-case scenario (which physically corresponds to very dilute solutions):

$$\mathrm{10^{-27} \gtrsim \frac{10^{-7}\ [A^-(aq)]}{K_a} \quad \longrightarrow \quad 10^{-20} \gtrsim \frac{[A^-(aq)]}{K_a}}$$

We now take a typical strong acid, hydrogen chloride ($\ce{HCl}$) which has an estimated pKa of -5.9 in water. In other words:

$$\mathrm{10^{-20} \gtrsim \frac{[Cl^-(aq)]}{10^{5.9}}=\frac{[Cl^-(aq)]}{790\ 000} \quad \longrightarrow \quad [Cl^-(aq)] \lesssim 8\times 10^{-15}\ mol\ L^{-1}}$$

We see therefore that meeting the criterion of full dissociation requires preparing an aqueous $\ce{HCl}$ solution with femtomolar concentration. Essentially every solution used in a chemistry laboratory will have a higher concentration, so every $\ce{HCl}$ solution anyone has ever touched will have undissociated $\ce{HCl}$ molecules. Strictly speaking, you could say the dissociation is never complete. However, if you have a mind to make a fully dissociated solution of $\ce{HCl}$, it is physically possible.

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Just diluting acid solutions into oblivion until every molecule dissociates is kind of boring. The more exciting prospect is, of course, to increase the strength of the acid. The acidity scale goes far, far beyond aqueous solutions of $\ce{HCl}$.

Interestingly, for sufficiently strong acids (typically $\mathrm{pK_a<-10}$, meaning acids at least 10 000 times stronger than aqueous $\ce{HCl}$), it is possible to isolate hydronium salts - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, where the cation is $\ce{H3O+}$ and the anion is supplied by the conjugate base of the acid. The stoichiometry can be made exact. Surely this represents full dissociation of the acid?

Well, again, it depends on your strictness. If you actually look at the crystal structures of many of these salts, it's clear that there is a strong, close interaction between the $\ce{H3O+}$ cation and the anion, and you could argue this still does not count as complete dissociation.

That said, you can go crazy and consider some of the strongest acids we can put in a bottle, such as carborane superacids. The corresponding hydronium salt of these superacids can have extremely weak interactions between the hydronium cation and the anion. For example, when the hydronium salt $\ce{(H3O+)(CHB11Cl11^-)}$ is prepared in a benzene solution, the solid that forms actually contains three molecules of benzene surrounding the hydronium cation, which is completely detached from the anion. If this doesn't imply "complete dissociation" of the carborane superacid in water, I don't know what would.

The acidity scale goes further still than even carborane superacids. There are a few known acids which will practically always fully dissociate, no matter what. The representatives of this class are rather exotic things such as $\ce{H_3^+}$, $\ce{HN2^+}$, $\ce{HNe^+}$, $\ce{HHe^+}$, and so forth. These acids can only be detected in the gas phase, as they fully dissociate by protonating virtually anything they come into contact with (including any anion, which is why they cannot be isolated as a neutral compound).

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The last way to ensure full dissociation of an acid in a solution is to change the solvent in which it is dissolved. All things considered, water is a pretty weak base. If, for example, ammonia (am) were used as the solvent, the relevant equilibrium for $\ce{HCl}$ would be:

$$\ce{HCl(am)->H+(am) + Cl-(am)} \quad \quad \quad \mathrm{K_a(am)=\frac{a_{H^+(am)}a_{Cl^-(am)}}{a_{HCl(am)}}=10^{-pK_a(am)}}$$

If I were to give a crude guess, I'd imagine that the $\mathrm{pK_a}$ for $\ce{HCl}$ in liquid ammonia would very roughly be somewhere around -15; that is, dissociation of $\ce{HCl}$ in liquid ammonia is about a billion times easier compared to water. This is because ammonia is a stronger base than water.

You know how we were talking about hydronium salts as if they were some unusual thing? Well, the equivalent concept in ammonia would be the formation of ammonium salts. And indeed, $\ce{HCl}$ does easily form a salt with ammonia in a exact 1:1 molecular ratio, namely ammonium chloride, $\ce{NH4Cl}$. Not exactly an unusual compound.

There's nothing stopping you from using an extremely strong base as a solvent. For example, in pure liquid phosphazene superbases, even very weak acids may be fully dissociated in ambient conditions, even in the most rigorous sense.

Answer 2

Fluoroantimonic acid is so strongly disassociating that molecules of $\ce{HSbF6}$ apparently have never been isolated. Only compounds in which a proton has been transferred to form ionic species are seen. (Replacing the fluorine with chlorine or bromine also gives an isolable compound, which may or may not be molecular depending on if water of hydration accepts a proton.) In its usual application as a solution in hydrofluoric acid solvent (the latter, by the way, is itself close to being a superacid, even without the fluoroantimonic acid solute), fluoroantimonic acid consists overwhelmingly of solvated $\ce{H_{n+1}F_n^+}$ cations and $\ce{Sb_nF_{5n+1}^-}$ anions, which equilibrate with $\ce{HF + SbF5}$ rather than with a molecular $\ce{HSbF6}$ species. When a "pure" fluoroantimonic acid compound is obtained from the solution, we still do not get association into neutral molecules:

Two related products have been crystallized from HF-SbF5 mixtures, and both have been analyzed by single crystal X-ray crystallography. These salts have the formulas $\ce{[H2F+][Sb2F_{11}−]}$ and $\ce{[H3F2+][Sb2F_{11}−]}$. In both salts, the anion is $\ce{Sb2F_{11}−}$.[1] As mentioned above, $\ce{SbF_6−}$ is weakly basic; the larger anion $\ce{Sb2F_{11}−}$ is expected to be still weaker.

Cited reference:

1. Mootz, Dietrich; Bartmann, Klemens (March 1988). "The Fluoronium Ions $\ce{H2F+}$ and $\ce{H3F2+}$: Characterization by Crystal Structure Analysis". Angewandte Chemie International Edition. 27 (3): 391–392. Link