Step 1: Recall the Bronsted-Lowry acid-base theory.
In the Bronsted-Lowry framework, an acid is a proton ($ H^+ $) donor and a base is a proton acceptor. When an acid donates a proton, the species formed is its conjugate base - it is the same species minus one proton.
Step 2: Write the conjugate base relationship.
\[ \text{Acid} \rightleftharpoons \text{Conjugate base} + H^+ \] The conjugate base has one fewer proton than the acid and one more negative charge (or one less positive charge).
Step 3: Apply to H3O+.
Hydronium ion $ H_3O^+ $ can act as an acid by donating one proton: \[ H_3O^+ \rightleftharpoons H_2O + H^+ \] The species formed after proton donation is $ H_2O $. Therefore, the conjugate base of $ H_3O^+ $ is $ H_2O $.
Step 4: Verify by checking the charge relationship.
$ H_3O^+ $ has charge $ +1 $. After losing $ H^+ $ (charge $ +1 $), the residual species has charge $ +1 - 1 = 0 $. $ H_2O $ has charge 0. Consistent.
Step 5: Check why other options are wrong.
$ OH^- $ is the conjugate base of $ H_2O $, not of $ H_3O^+ $. $ H^+ $ is what is donated, not the conjugate base. $ H^- $ is a hydride ion, completely unrelated here.
Step 6: Final answer.
\[ \boxed{H_2O} \]