Using equations, show the formation of hydroxonium ions in an aqeous solution of tetraoxosulphate(VI) acid
I assume you meant sulfuric acid, H2SO4. The weird name you wrote is not an IUPAC approved name for H2SO4.
H2SO4 + H2O ==> 2[H3O]^+ + [SO4]^2-
To explain the formation of hydroxonium ions in an aqueous solution of tetraoxosulfate(VI) acid (also known as sulfuric acid), we need to write the relevant chemical equation.
The molecular formula for sulfuric acid is H2SO4. When it is dissolved in water, it undergoes ionization, which means it breaks up into ions.
The first step in the ionization process is the separation of the hydrogen ions (H+) from the rest of the molecule. In this case, two hydrogen ions are released from each sulfuric acid molecule.
H2SO4 → 2H+ + SO4^2-
Two hydrogen ions combine with water molecules (H2O) in the solution to form hydroxonium ions (H3O+):
2H+ + 2H2O → 2H3O+
So, the final equation showing the formation of hydroxonium ions in an aqueous solution of tetraoxosulfate(VI) acid would be:
H2SO4 + 2H2O → 2H3O+ + SO4^2-
It's important to note that hydroxonium ions (H3O+) are essentially the same as hydronium ions (H+). They both represent the presence of a single proton (H+) and are used interchangeably to describe the acidity of a solution.