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.

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.