Sulfuric acid \(\mathrm{H}_2\mathrm{SO}_4\) is renowned for its strength as an acid in chemical reactions and solutions. However, its dissociation in water occurs in two distinct steps, as it is a diprotic acid.
In the first dissociation step, sulfuric acid loses one proton, forming hydrogen sulfate ions \(\mathrm{HSO}_4^-\). This step is complete because sulfuric acid is a strong acid:
\[\mathrm{H}_2\mathrm{SO}_4 \rightarrow \mathrm{H}^+ + \mathrm{HSO}_4^-\]
In the second step, \(\mathrm{HSO}_4^-\) ions can lose another proton to form sulfate ions \(\mathrm{SO}_4^{2-}\):
\[\mathrm{HSO}_4^- \rightleftharpoons \mathrm{H}^+ + \mathrm{SO}_4^{2-}\]
This second dissociation is not complete because \(\mathrm{HSO}_4^-\) acts as a weak acid. Due to this partial dissociation, there are more \(\mathrm{HSO}_4^-\) ions than \(\mathrm{SO}_4^{2-}\) ions in the solution.
- Sulfuric acid's two-step dissociation is key to its behavior in solutions.
- This process results in a greater concentration of \(\mathrm{HSO}_4^-\) ions over \(\mathrm{SO}_4^{2-}\) ions.
Grasping this concept helps in understanding the unique role that sulfuric acid plays across various chemical reactions.