What is the conjugate acid and base of #HSO_4^-#?

The thing to remember about conjugate acids is that they are the chemical species that is formed when a Bronsted-Lowry base accepts one proton, #"H"^(+)#.

This means that in order to find the conjugate acid of a substance that can act as a Bronsted-Lowry base, all you have to do is add a proton to it.

But keep in mind that a proton carries a #1+# charge, so make sure that you take this into account.

In your case, the hydrogen sulfate anion, #"HSO"_4^(-)#, can act as a Bronsted-Lowry base and accept a proton to form sulfuric acid, #"H"_2"SO"_4#

#overbrace("HSO"_ (4(aq))^(-))^(color(blue)("base")) + "H"_ ((aq))^(+) -> overbrace("H"_ 2"SO" _(4(aq)))^(color(darkgreen)("conjugate acid"))#

Notice that because you're adding a #1+# charge to a compound that has a #1-# charge, the resulting compound will have a zero net charge.

Similarly, conjugate bases are chemical species that are formed when a Bronsted-Lowry acid donates one proton.

This means that you can find the conjugate base of a Bronsted-Lowry acid by removing a proton from it.

In your case, the hydrogen sulfate anion can act as a Bronsted-Lowry acid and donate a proton to form the sulfate anion, #"SO"_4^(2-)#.

#overbrace("HSO"_ (4(aq))^(-))^(color(darkgreen)("acid")) -> "H"_ ((aq))^(+) + overbrace("SO"_ (4(aq))^(2-))^(color(blue)("conjugate base"))#

Once again, notice that the charge is balanced because removing a #1+# charge from a compound that has a #1-# charge will give you a compound that has a #2-# net charge.