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Apologies for this extremely basic question, I'm just beginning with Chemistry so please don't be too harsh on me.

My book says that sulfuric acid, $\ce{H2SO4}$, dissociates in its ions following this reaction: $$\ce{H2SO4 -> H2^+ + SO4^{2-}}$$

My question is, why can't the dissociation reaction happen like this: $$\ce{H2SO4 -> 2H^+ +SO4^{2-}}$$

I know hydrogen is a diatomic gas, but here I don't know if H will dissociate as a gas or as a liquid (since $\ce{H2SO4}$ is a liquid, not a gas).

I'm trying to learn, thank you for your understanding and your time.

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    $\begingroup$ It can and does happen as you suggested. Your book is wrong. Hydrogen the diatomic gas is simply not here. $\endgroup$
    – Ivan Neretin
    Commented Jun 17, 2018 at 12:51
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    $\begingroup$ Thanks, but then how do I know when I will have $H_2^+$ and when $2H^+$? $\endgroup$
    – Jose Lopez Garcia
    Commented Jun 17, 2018 at 13:00
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    $\begingroup$ (The book was written by my teacher, I suppose he made a mistake in this exercise) $\endgroup$
    – Jose Lopez Garcia
    Commented Jun 17, 2018 at 13:03
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    $\begingroup$ I would agree that $\ce{H2^+}$ is not present. The overall reaction is the dissociation of both hydrogen ions, but I'd suggest that the dissociations happen one at a time. Both dissociations would be very fast, but not instantaneous. $\endgroup$
    – MaxW
    Commented Jun 17, 2018 at 14:03
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    $\begingroup$ @Jose On your current level of theory, this is pretty simple: you always have $\ce{2H+}$ and never $\ce{H2+}$. You might want to ask this question again, say, after a year. $\endgroup$
    – Ivan Neretin
    Commented Jun 17, 2018 at 15:48

1 Answer 1

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$\ce{H2SO4}$ is one of common strong acids, meaning that $\ce{K_{a(1)}}$ is large and that its dissociation even in moderately concentrated aqueous solutions is almost complete.

Arrhenius dissociation:

$$\ce{H2SO4 <=> H+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

Brønsted-Lowry Dissociation:

$$\ce{H2SO4 + H2O <=> H3O+ + HSO4-}~~~~~~~~~~\ce{K_{a(1)}}=\ce{large}$$

This accounts for the vast majority of protons donated by the acid. However, since it is diprotic, you may want to take into account the second dissociation, which is technically weak but has a larger $\ce{K_a}$ than many weak acids.

Arrhenius 2nd Dissociation:

$$\ce{HSO4- <=> H+ + {SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}=1.2\times10^{-2}$$

Brønsted-Lowry 2nd Dissociation:

$$\ce{HSO4- + H2O <=> H3O+ +{SO_4}^2-}~~~~~~~~~~\ce{K_{a(2)}}= 1.2\times10^{-2}$$

This second dissociation may need to be taken into account for some calculations, but it is negligible in concentrated solutions.

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    $\begingroup$ There might be only 6 strong acids mentioned in your book, but it's by no means total number. Also this Also this Arrhenius/Bronsted division is kinda silly IMO. Both H+ and H3O+ are only symbolical and don't truly reflect hydration of proton. $\endgroup$
    – Mithoron
    Commented Jun 22, 2018 at 22:29
  • $\begingroup$ @Mithoron My teacher defined “strong” acids as those with a “large” Ka (as in too big to be measured). Do you know of a list of the rest? $\endgroup$
    – Shady Puck
    Commented Jun 22, 2018 at 22:33
  • $\begingroup$ There is no list as their number is limitless. Add -SO3H group to one of millions organic groups and you have strong acid, voila! $\endgroup$
    – Mithoron
    Commented Jun 22, 2018 at 22:40
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    $\begingroup$ @Mithoron Good to know! I have not taken organic chemistry yet, so I was not aware of this. In the future, you should try to find a better way of critiquing than a downvote and a reprimand. Perhaps an edit to the post in question and a comment explaining it? Just a thought — and I will edit this post to reflect your insight. $\endgroup$
    – Shady Puck
    Commented Jun 22, 2018 at 23:00
  • $\begingroup$ Sulfonic acids are just an example. There's also a lot of inorganic acids, just less known, and their number is also probably limitless. $\endgroup$
    – Mithoron
    Commented Jun 22, 2018 at 23:11

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