Question #bde65

Hydrogen sulfate, or bisulfate ion, as you'll sometimes see it called, cannot act as a conjugate base in solution because sulfuric acid, `H_2SO_4`, is a strong acid that dissociates completely.

In other words, in the first dissociation of sulfuric acid, the equilibrium that forms hydronium ions, `H_3O^(+)` and hydrogen sulfate lies so far to the right that, for all intended purposes, you can consider that no sulfuric acid molecules remain undissociated.

`H_2SO_(4(aq)) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + HSO_(4(aq))^(-)`, `K_(a1) = 2.4 * 10^6`

Notice the very large value of the acid dissociation constant, `K_a`. In theory, `HSO_4^(-)`, being the conjugate base of sulfuric acid, can pick up a proton to reform sulfuric acid. However, for practical purposes, the dissociation of sulfuric acid can be seen as going to completetion because of the very large `K_a`.

`H_2SO_(4(aq)) + H_2O_((l)) -> H_3O_((aq))^(+) + HSO_(4(aq))^(-)`

So, when you place sulfuric acid in water (never do the other way around!), all the sulfuric acid molecules will lose their proton, which is then picked up by water molecules `->` hydronium is formed.

![http://webgenchem.com/gc2600012.htm]()

Once this happens, you can consider that the solution no longer contains `H_2SO_4` molecules. Hydrogen sulfate goes on to act as a weak acid, donating its proton to form bisulfate ions, `SO_4^(2-)`.

However, hydrogen sulfate is a weak acid and thus cannot dissociate completely.

`HSO_(4(aq))^(-) + H_2O_((l)) rightleftharpoons H_3O_((aq))^(+) + SO_(4(aq))^(2-)`, `K_(a2) = 1.0 * 10^(-2)`

This means that your solution will contain both hydrogen sulfate and bisulfate ions.