Step 1:
In the first reaction, $\mathrm{HI}(a q)+\mathrm{H}_{2} \mathrm{O}(l) \longrightarrow \mathrm{H}_{3} \mathrm{O}^{+}(a q)+\mathrm{I}^{-}(a q)$, we can see that $\mathrm{HI}$ donates a proton to $\mathrm{H}_{2} \mathrm{O}$, forming $\mathrm{H}_{3}
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