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I know as a fact that HCl is a stronger acid than H3O+, but I want to know the reason behind it. The definition of an acid is a substance which dissociates to give H3O+ ions. In that case H3O+ should be the strongest acid available because it does not even need to dissociate it give itself!!

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    $\begingroup$ chemguide.co.uk/physical/acidbaseeqia/theories.html $\endgroup$ – orthocresol Jul 15 '16 at 16:53
  • $\begingroup$ Does really answer my question.. $\endgroup$ – Mukul Goyal Jul 15 '16 at 16:58
  • $\begingroup$ Your definition is wrong and this mistake is unfortunately common. Sth acts as acid if protonates anything, not necessarily water. H3O+ is only a strongest acid which is present in significant concentrations in diluted aqueous solutions. $\endgroup$ – Mithoron Jul 15 '16 at 18:34
  • $\begingroup$ @Mithoron the point about H3O+ is kind of right, but in a very weird way; something that protonates anything would be the strongest acid possible, though. I think you mean "something acts as acid towards a compound if it protonates that compound". $\endgroup$ – Zubo Jul 15 '16 at 21:15
  • $\begingroup$ Yes, that's what I meant, I didn't say everything... I guess I should say "something", sorry for imperfection my English. $\endgroup$ – Mithoron Jul 15 '16 at 23:35
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Here is my opinion, which may be subject to errors and corrections are very welcome:

Using the definition of Brønsted, acidity is the tendency to give off protons. This, unlike the definition of Arrhenius, is not limited to aqueous solutions. However, if you do have an aqueous solution of an acid, something interesting happens: any acid HAc (or base B) stronger than H$_3$O$^+$ (or OH$^-$) completely dissociates via:

$$\ce{H2O + HAc <=> H3O+ + Ac-}$$ or $$\ce{H2O + B <=> OH- + BH+}$$

So the maximum acidic compound in water is the hydronium ion H$_3$O$^+$; you couldn't, for example, differentiate aqueous hydrochloric and perchloric acid solutions by pH. To rank very strong acids by their acid strength, non-aqueous solutions are used to determine the dissociation constants and the data obtained is then transferred approximately onto water as a solvent. Also note that we are talking about solutions of acids.

Specifically for HCl, we can denote the pairs of conjugated acids and bases:

$$\ce{H2O ~(~base~1) + HCl ~(~acid~2) <=> H3O+ ~(~acid~1) + Cl- ~(~base~2)}$$

So we can compare the strengths of the two acids: $\ce{HCl}$ and $\ce{H3O+}$. We know that this equilibrium is shifted to the right, so hydrochloric acid is stronger.

Very late edit to try and answer comment by MukulGoyal (unfortunately, he didn't elaborate on his questions), by number:

  1. Is the definition of acidity the tendency to give H+ ions, or H3O+ ions? That seems to matter here because in first case H3O+ will not even be acidic in water as the equilibrium H2O + H+ ---» H3O+ has a very high equilibrium constant.

  2. Similarly even if the solvent is NH3, the equilibrium constant of NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason, and hence it would actually be a very weak acid.

  3. Does your answer imply that any acid which has a Ka greater than 1 will be a stronger acid than H3O+?

First of all, please remember that equilibria are dynamic.

1) As I've mentioned, acidity, according to Brønsted, is the tendency to give off $\ce{H+}$. In $\ce{H2O}$, $\ce{H+}$ are solvated, on average, by 6 water molecules, giving rise to species like $\ce{H3O+(H2O)6}$ (for more information, check out Wikipedia and this).

1) and 2) you seem to be talking about the dissociation of $\ce{H3O+}$. However, you're using terms like "very high" and "very low", so that, in my opinion, becomes kind of a misleading discussion.

When you mention

$$\ce{H2O + H+ <=> H3O+}$$

you need to remember that the proton has to come from somewhere. In fact, what you are talking about there is

$$\ce{H2O + H2O <=> H3O+ + OH-}$$

which is the autoprotonation of water. Let's just calculate that equilibrium constant you talk about all the time:

$$K = \cfrac{c(\ce{H3O+})~ \cdot~ c(\ce{OH-}) }{c^2~(\ce{H2O})}$$

Well, guess what? The concentration of water (c = n/V = m / (V * M) = $\rho$ / M = 55,51 mol/L) remains about constant for pure water and diluted solutions (because not many water molecules form ions), so it is brought into the constant, yielding

$$K_w = c(\ce{H3O+})~ \cdot~ c(\ce{OH-})$$

$K_w$ depends on temperature and equals $10^{-14} $ mol$^2$ / L$^{2}$ at 25 °C. It remains constant in pure water and also in diluted solutions.

In pure water where $c(\ce{H3O+}) = c(\ce{OH-})$, we then get

$$K_w = c^2(\ce{H3O+})$$

and

$$c(\ce{H3O+}) = 1.0 * 10^{-7} ~mol~ / ~L~ $$

which essentialy quantifies what you, as I suspect, were talking about, in that it tells you the concentration of $\ce{H3O+}$ in the autoprotonation equilibrium for pure water. By the way, because pH is defined like so:

$$pH = -lg ~c~(\ce{H3O+})$$

at pH = 7 (neutral), $\ce{H3O+}$ and $\ce{OH-}$ are equal. In diluted solutions, when they are not equal, you can calculate one using the other.

Also, remember that in the Brønsted–Lowry theory acids and bases are defined by the way they react with each other, as mentioned in Wikipedia. So far, we have only looked at water and its autoprotonation, no solvents and other molecules, which have their own acidies and basicities.

2) As for the reaction you mentioned in ammonia as a solvent, well, first of all, $\ce{H3O+}$ needs to come from somewhere (let's imagine we drop some aqueous solution of hydrochloric acid in there). Second, I am not sure that constant would actually be as you describe it - I think that here, ammonia (which behaves ampholytic) would act as a base and pretty happily accept a proton. If anyone can provide insight into this, that would be great.

3) Acid strength is defined using $pK_a$, which is given by $pK_a$ = -lg $K_a$. From my experience, if $pK_a$ is defined via concentration, then $pK_a$ of hydronium ions is 0, and if it is defined via activity, it is -1.7 (-1.74, to be more exact; similarly, for $\ce{OH-}$, the analogous $pK_b$ is 14 via concentration and 15.7 via activity).

Coming back to your question about "Ka greater than 1", that seems about right, because -lg (1) = 0.

I refer you to an acid strength table, for instance, here. Also, I'd like to refer you to this excellent discussion on the matter, where, by the way, orthocresol lent a hand.

Also, just to add this, strong acids have $pK_a$ < 0, while weak acids have $pK_a$ > 0 (again, this is via concentration - for example, here in Wikipedia they use -1.74).

Everyone, corrections are very welcome!

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    $\begingroup$ This is an excellent answer. You might just want to clarify what happens to $\ce{HCl}$ in aqueous solution though to avoid any confusion. $\endgroup$ – bon Jul 15 '16 at 21:03
  • $\begingroup$ Thanks for the answer but I have 3 questions here: 1) Is the definition of acidity the tendency to give H+ ions, or H3O+ ions? That seems to matter here because in first case H3O+ will not even be acidic in water as the equilibrium H2O + H+ ---» H3O+ has a very high equilibrium constant. 2) Similarly even if the solvent is NH3, the equilibrium constant of NH3 + H3O+ ---» NH4+ + H2O would be very low due to the same reason, and hence it would actually be a very weak acid. 3) Does your answer imply that any acid which has a Ka greater than 1 will be a stronger acid than H3O+? $\endgroup$ – Mukul Goyal Jul 16 '16 at 1:36
  • $\begingroup$ 1) Read the theory on acids and bases. The link orthocresol gave is a good starting point. According to the Brønsted definition, acids are donors of H+. The equilibrium you noted doesn't make a lot of sense out of context, so please clarify what you mean - and why it should have a high constant. 2) What would be a weak acid? H3O+? How would that come to exist in ammonia? Why do you think this equilibrium would be shifted to the left? 3) H3O+ has a pKa of 0 (not Ka). Acids with higher Ka (lower pKa) are stronger (edit: wrote "lower" before, that's wrong, also had a wrong Ka definition, sorry). $\endgroup$ – Zubo Jul 16 '16 at 3:52
  • $\begingroup$ @MukulGoyal I'd like to improve the answer, but you need to clarify for me the things that I mentioned about your questions. $\endgroup$ – Zubo Jul 16 '16 at 12:25
  • $\begingroup$ @MukulGoyal I've attempted to answer your comment anyway - maybe this helps :) Everyone - I'd be very thankful for any corrections :) $\endgroup$ – Zubo Nov 3 '16 at 14:35

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