Answer
Sulfate ion ($S{O_4}^{2-}$) and Sulfuric acid ($H_2SO_4$) are the conjugate partners of $HS{O_4}^-$.
Work Step by Step
1. Identify if the substance is an acid or a base (or both):
- $HS{O_4}^-$ can act as an acid or as a base.
When it acts as a base:
2. Add one proton to it:
- Add one hydrogen and sum one to the charge:
$H{SO_4}^- -- \gt H_2SO_4$
- This is the conjugate acid of $HS{O_4}^-$.
3. Name it:
Sulfur ("S") has 2 oxyacids:
$H_2SO_3$ and $H_2SO_4$.
So, since $H_2SO_4$ is the one where "S" has the highest oxydation number, it receives an "ic" on the name:
$H_2SO_4$ = "Sulfuric acid."
When it acts as an acid:
4. Remove one proton from it:
- Remove one hydrogen and subtract one from the charge:
$HS{O_4}^- -- \gt S{O_4}^{2-}$
- This is the conjugate base of $HS{O_4}^-$.
5. Name it:
$H_2SO_4$ = Sulfuric acid.
"ic" acids $->$ "ate" ions.
$S{O_4}^{2-}$ = Sulfate ion.
