Do solid acids in reality dissociate totally (every solitary molecule dissociates), or space they just assumed to perform so for the services of simplicity? That appears odd, considering weak acids, countless of which space very comparable to strong acids, don"t totally dissociate.

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Yes, because that sufficiently strong acids, in saturated dilute conditions, in sufficiently straightforward solvents. However, things space hazier 보다 you might expect, and depending on her definition, "clear-cut" examples of complete dissociation only become feasible when you start swatting flies with nuclear bombs.

The difficulty is that jajalger2018.org works on equilibria, and at some allude it becomes an overwhelming to measure tiny concentrations - in a haystack of one exchange rate straws, how deserve to you be sure there isn"t at least one needle? and in one trillion? and what around in a septillion?

First, allow us perform a qualitative investigation by considering the generic mountain dissociation equilibrium in water, together follows:

$$\ceHA(aq)->H+(aq) + A-(aq) \quad \quad \quad \mathrmK_a=\fraca_H^+(aq)a_A^-(aq)a_HA(aq)=10^-pK_a$$

To leveling the adhering to discussion, we will make the very crude approximation whereby we equate the thermodynamic activity of a types with that molar concentration. In other words:

$$\mathrmK_a=\fraca_H^+(aq)a_A^-(aq)a_HA(aq) \approx \frac$$

Our objective currently is come virtually get rid of the existence of undissociated $\ceHA$ molecules. Because that clarity, we rearrange the above equation:

$$\mathrm \approx \fracK_a$$

Clearly, to do $\ce$ as tiny as possible, the numerator need to be very small, and the denominator need to be very large. Let us collection a score of do $\mathrm{, which implies much less than $\mathrm0.001$ molecules of undissociated $\ceHA$ in a litre that solution.

At this point, ns should cite that aqueous solutions of acids in ambient conditions must constantly have $\mathrm>10^-7\ mol\ L^-1$, many thanks to the self-ionization that water. Us take the best-case script (which physically corresponds to very dilute solutions):

$$\mathrm10^-27 \gtrsim \frac10^-7\ K_a \quad \longrightarrow \quad 10^-20 \gtrsim \fracK_a$$

We now take a typical solid acid, hydrogen chloride ($\ceHCl$) which has an estimated pKa that -5.9 in water. In other words:

$$\mathrm10^-20 \gtrsim \frac10^5.9=\frac790\ 000 \quad \longrightarrow \quad \lesssim 8\times 10^-15\ mol\ L^-1$$

We see thus that meeting the criterion of full dissociation needs preparing one aqueous $\ceHCl$ systems with femtomolar concentration. Essentially every solution offered in a jajalger2018.org activities will have actually a greater concentration, therefore every $\ceHCl$ solution anyone has ever touched will have actually undissociated $\ceHCl$ molecules. Strictly speaking, you might say the dissociation is never complete. However, if you have a mind to make a fully dissociated equipment of $\ceHCl$, it is physical possible.

Just diluting mountain solutions into oblivion till every molecule dissociates is type of boring. The an ext exciting possibility is, the course, to rise the stamin of the acid. The acidity scale goes far, far beyond aqueous services of $\ceHCl$.

Interestingly, because that sufficiently solid acids (typically $\mathrm{pK_a, definition acids at least 10 000 times more powerful than aqueous $\ceHCl$), it is feasible to isolation hydronium salt - when the strong acid is mixed with water in a 1:1 molecular ratio, a solid salt is formed, wherein the cation is $\ceH3O+$ and also the anion is gave by the conjugate basic of the acid. The stoichiometry have the right to be do exact. Certain this represents complete dissociation that the acid?

Well, again, it counts on your strictness. If you in reality look in ~ the crystal frameworks of countless of these salts, it"s clear the there is a strong, near interaction between the $\ceH3O+$ cation and also the anion, and you could argue this still does not count together complete dissociation.

That said, you deserve to go crazy and consider some of the strongest acids we have the right to put in a bottle, such as carborane superacids. The matching hydronium salt of this superacids deserve to have extremely weak interactions between the hydronium cation and the anion. Because that example, as soon as the hydronium salt $\ce(H3O+)(CHB11Cl11^-)$ is prepared in a benzene solution, the hard that creates actually includes three molecule of benzene neighboring the hydronium cation, i beg your pardon is totally detached from the anion. If this doesn"t indicate "complete dissociation" that the carborane superacid in water, ns don"t recognize what would.

The acidity range goes more still than even carborane superacids. There are a few known acids which will practically always totally dissociate, no issue what. The representatives of this class are rather exotic points such as $\ceH_3^+$, $\ceHN2^+$, $\ceHNe^+$, $\ceHHe^+$, and so forth. These acids have the right to only be detected in the gas phase, together they completely dissociate by protonating basically anything lock come into contact with (including any anion, i m sorry is why they cannot be isolated as a neutral compound).

The last means to ensure complete dissociation the an acid in a equipment is to change the solvent in which the is dissolved. All points considered, water is a pretty weak base. If, because that example, ammonia (am) were provided as the solvent, the relevant equilibrium because that $\ceHCl$ would be:

$$\ceHCl(am)->H+(am) + Cl-(am) \quad \quad \quad \mathrmK_a(am)=\fraca_H^+(am)a_Cl^-(am)a_HCl(am)=10^-pK_a(am)$$

If i were to provide a rudely guess, I"d imagine that the $\mathrmpK_a$ because that $\ceHCl$ in fluid ammonia would very roughly it is in somewhere about -15; the is, dissociation that $\ceHCl$ in liquid ammonia is around a billion time easier compared to water. This is due to the fact that ammonia is a more powerful base 보다 water.

You know just how we to be talking around hydronium salts together if they were some unusual thing? Well, the equivalent concept in ammonia would be the development of ammonium salts. And indeed, $\ceHCl$ does easily kind a salt through ammonia in a exact 1:1 molecule ratio, specific ammonium chloride, $\ceNH4Cl$. Not precisely an inexplicable compound.

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There"s nothing preventing you from using an extremely strong base as a solvent. Because that example, in pure liquid phosphazene superbases, even really weak acids may be fully dissociated in approximately conditions, even in the most rigorous sense.