5
$\begingroup$

I have read that the concept of Brønsted and Lowry is based on the fact that acid-base reactions involve the transfer of $\ce{H+}$ ions from one substance to another, and that an acid is a substance that donates a proton to another substance, but:

can a substance donate a proton without the proton getting accepted by another substance? So for example: $$\ce{AH -> A- + H+}$$ without reacting with another substance?

If yes: Is this a Brønsted-Lowry acid?

Improve this question
$\endgroup$
2

1 Answer 1

Reset to default
3
$\begingroup$

The closest thing I can find to this involves a most unlikely proton donor. Given an appropriate base, sodium hydride donates a proton for encapsulation in the base (rather than forming a localised covalent bond), leaving the valence electrons on sodium (!) in inverse sodium hydride. The proton is not free (a completely free proton is possible only in plasma), but analogous to complexed metal ion or solvated electron.

Improve this answer
$\endgroup$
7
  • $\begingroup$ "Encapsulation in the base"? Does it mean that the base receives/accepts the proton? $\endgroup$
    – Tan Yong Boon
    Jan 1, 2018 at 14:23
  • 1
    $\begingroup$ Yes, some bonding interaction (in this case between the protons and lone pairs on nitrogen atoms in the base) to drive the "reversed" oxidation states. But you avoid the usual localized covalent bonding, the encapsulated species may be fairly called ironically bonded $\ce{H^+}$. $\endgroup$
    – Oscar Lanzi
    Jan 1, 2018 at 14:26
  • $\begingroup$ I don’t think this qualifies as the proton not getting accepted by anything since it does get accepted by the encapsulation host. $\endgroup$
    – Jan
    Jan 1, 2018 at 14:58
  • 1
    $\begingroup$ Hence, "the closest thing I can find". $\endgroup$
    – Oscar Lanzi
    Jan 1, 2018 at 15:02
  • $\begingroup$ It's not closest by any means and no idea why you put such example. BTW it would be good to use links instead of bare urls. $\endgroup$
    – Mithoron
    Jan 1, 2018 at 16:35

Not the answer you're looking for? Browse other questions tagged or ask your own question.