What are the products from the Reaction of water and sulphuric acid?

2 Answers
Mar 21, 2018

Hydrogen sulfide anion #(HSO_4^-)# and hydronium cation #(H_3O^+)#.

Explanation:

Well, I would say that two reactions will occur.

The first reaction is the dissociation of sulfuric acid. The equation is:

#H_2SO_4(aq)+H_2O(l)->H_3O^+(aq)+HSO_4^(-)(aq)#

This creates a hydrogen sulfate anion and hydronium cations.

Then, the hydrogen sulfate anion dissociates further into sulfate ions and even more hydronium ions. The equation for that is:

#HSO_4^(-)(aq)+H_2O(l)rightleftharpoonsH_3O^+(aq)+SO_4^(2-)(aq)#

Note that this reaction is reversible since the hydrogen sulfate anion does not ionize completely, and so there is a chance of recombination, but sulfuric acid is a really strong acid, and so will ionize completely with little chance of reversibility.

Well, hydronium ion, and sulfate dianion #(H_3O^+, SO_4^(2-))#....

Explanation:

#H_2SO_4(l) +2H_2O(l)rarr2H_3O^(+)+SO_4^(2-)#

And note that this is stepwise reaction....the first protonolysis proceeds quantitatively...

#H_2SO_4(aq) + H_2O(l) rarr HSO_4^(-) + H_3O^+#

...the second protonolysis is not as complete, however, we often treat #H_2SO_4# as a quantitative diacid...

#HSO_4^(-) + H_2O(l) rightleftharpoons SO_4^(2-) + H_3O^+#

The overall reaction is given by the first equation...

#H_3O^+# is rather a conceptual entity....the which we could (and do) write as #H^+#... as an acid, sulfuric acid ACTS by protonating the solvent molecule to increase concentrations of the characteristic cation of the solvent, here water, to give protonated water....i.e. #H_3O^+#...