# Difficulty in understanding ionisation of sulphuric acid [closed]

$$\ce{H2SO4}$$ when added to water.

$$\ce{H2SO4 + 2 H2O -> SO4^2- + 2 H3O+}$$

My sir explained me these reaction as $$\ce{H}$$ will give $$\ce{2 H^+}$$ ions to the base $$\ce{H2O}$$.So why does it not become $$\ce{H4O^+}$$ ?

To try to make myself understand this , I checked my textbook.

$$\ce{H2SO4 + H2O <=> H3O+ + HSO4^−}$$

$$\ce{HSO4^− + H2O <=> H3O+ + SO4^2−}$$

I got even more confused that why is H2O is used twice in both the reactions?Why not use the products of 1st equation and then do the same reaction again.The one difference was that we substituted H20 instead of H3O there.

$$\ce{HSO4^− H3O+ <=>}$$ why not this ?

• Added proper formating and fixed 0 vs O errors. There is no compound like $\ce{H20}$ nor a ion $\ce{H30+}$. Dec 22 '20 at 9:55
• My sir explained me these reaction as H will give 2H+ ions to the base H2O It does not mean to a single molecule, but H2O as a compound. Dec 22 '20 at 10:00
• 0 instead of O again ? I have checked the editing history and no, I have written 2 times H2O instead of H3O. I have just replaced zeros by capital O. Or, I may have not got this your question correctly. Dec 22 '20 at 10:05
• Dec 22 '20 at 10:07
• not H4O2+. you have to edit it to $\ce{\underset{ }{H4O^2+}}$
– user99515
Dec 22 '20 at 10:48

Please read "The Aqueous Proton Is Hydrated by More Than One Water Molecule: Is the Hydronium Ion a Useful Conceit? by Todd P. Silverstein in the Journal of Chemical Education. J. Chem. Educ. 2014, 91, 4, 608–610. It is freely available from Google Scholar.

The latest thinking, past 2006, is that $$\ce{H3O+}$$ does not exist in any appreciable amount in aqueous systems. This ion exists in organic solvents. So consider all these protonated versions of water in your textbook as a very simplified version of a complex reality in water. The hydrated "proton" may look like this, where the blue part is the anion. Water chemistry is fascinating and yet to be fully explored.

Looking at the structure, where do you want to insert the second proton?

• This image looks awfully flat to me, and I would actually se quite a few places to stick another proton. Anyway, as I usually do with water posts: I link to the Grotthuss Mechanism, as that is always important to consider. Dec 23 '20 at 20:12
• Yes, Grotthuss mechanism is still remarkable despite its age. What students have to realize is the the H+ is mobile, so there is no rigid structure per se. Dec 23 '20 at 21:23

The simple answer why $$\ce{H4O^++}$$ (note that if another proton is added, the charge must be a double positive) doesn't form is that a water molecule has far more affinity for protons than the $$\ce{H3O^+}$$.

And there are also far more uncharged water molecules in the solution than products of the first protonation step. Don't forget this is taking place in a water solution so, although we don't usually include the solvent in the reaction equation, neutral water molecules are, by far, the most common molecules in the solution. So, even setting aside the fact that the second proton prefers to add to an uncharged water molecule, the likelihood of there being a water molecule available is far, far higher than there being a $$\ce{H3O^+}$$ andywhere.

Don't forget that reactions like this take place in a solution and that the molecules of the solution may participate in the reaction.

$$\ce{\underset{hydronium ion}{H3O+}}$$ may not further react because oxygen already has a formal positive charge and it maynot further bear a more positive charge to form $$\ce{\underset{ }{H4O^2+}}$$.

As there are approximately $$10^{21}$$ molecules of water in a drop and (you carry a test tube) so very large number of water molecules exist. $$\ce{\underset{bisulphate anion}{HSO4-}}$$ can react further with water(which is amphoteric in nature)

• While oxygen may bear a positive formal charge, it does not bear an actual positive charge. It also would not become significantly more positive adding another proton. After all $\ce{OH4^2+}$ is isoelectronic with methane. Some advice in non-chemical matters: Don't ask for justification of down-votes in comments; those who voted won't see it anyways and it usually puts people off who would have commented. Also you're only asking for identity, not why those people may have chosen to vote this way in the first place, that's certainly going to make people uneasy. Me, for example. Dec 23 '20 at 20:06