An acid is:
A molecule or other species which can donate a proton
So when hydrogen appears by itself on one side of a dissociation reaction e.g. $\ce{HOCl -> H+ + OCl-}$ is the hydrogen an acid? As the hydrogen can donate itself?
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asked Oct 15, 2016 at 19:42
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1$\begingroup$ Yes in such a dissociation the $\ce{H^+}$ ion is an acid. // A naked $\ce{H^+}$ ion doesn't really float around in aqueous solution but it is rather the $\ce{H3O^+}$ ion. The $\ce{H3O^+}$ ion is used so much that is has a special name hydronium. $\endgroup$– MaxWCommented Oct 15, 2016 at 19:54
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$\begingroup$ Just as an afterthought to previous answers: while the H+ ion, which you use in your example, is an acid, hydrogen gas (H2) is not. $\endgroup$– AleksanderCommented Oct 15, 2016 at 21:42
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2$\begingroup$ H2 has (very weak) acidic properties. $\endgroup$– MithoronCommented Oct 15, 2016 at 23:49
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1 Answer
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Well, you can never have H+ by itself. In aqueous media it's always associated with the lone pairs of water molecules and exists as H3O+. But yes, theoretically speaking H+ is a Brønsted–Lowry acid as the definition of such is to donate a proton and it donates itself to water in the case. It's also a Lewis acid as by donating itself it accepts electrons in its 1s emptry orbital.
But be aware that in water, you never have H+, it's always H3O+ which is the strongest acid that can exist in water solutions, all acids are limited by the strength of H3O+ when dissolved in water. Therefore, a species is an acid in water if it generates H3O+. In other solvents, for example liquid NH3, the corresponding species is NH4+ and an acid in ammonia is defined as species that generates NH4+. This whole paragraph relates to the solvent definition of acids and bases.
I hope all that answers your question, please comment if you want anything clarified.
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$\begingroup$ Nope that explains it perfectly :) It was written in my book two ways 1) $\ce{HOCl (aq)⟶H+ +OCl-}$ and 2) $\ce{HOCl (aq) +H2O (l)⟶H3O+ (aq) + OCl- (aq)}$ but I didn't connect the two before reading your answer. $\endgroup$ Commented Oct 15, 2016 at 20:00