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This question already has an answer here:

For what reason is it universally agreed upon that we use $$\ce{H+ + OH-}$$ ions in balancing water on either side of the reaction, but not the $$\ce{H3O+}$$ ions?

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marked as duplicate by Mathew Mahindaratne, Karsten Theis, Nilay Ghosh, Jon Custer, Tyberius Oct 21 at 15:30

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    $\begingroup$ Who said hydronium isn't used? You can use either $\ce{H+}$ or $\ce{H3O+}$ or even $\ce{H9O4+}$ — it just adds complexity in terms of bloated stoichiometry. So, if you don't want to focus on what business does a proton have in aqueous solution, in most cases you get away with $\ce{H+}$ just fine. $\endgroup$ – andselisk Oct 20 at 17:49
  • $\begingroup$ @andselisk so basically it's because of the accuracy and complexity of which we want to refer to when doing redox reactions? $\endgroup$ – user243851 Oct 20 at 17:57
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    $\begingroup$ "on average, each hydronium ion is solvated by 6 water molecules which are unable to solvate other solute molecules." en.wikipedia.org/wiki/Hydronium#Solvation $\endgroup$ – Poutnik Oct 20 at 18:00
  • $\begingroup$ @user243851 Yes, I suppose you can put it this way. $\endgroup$ – andselisk Oct 20 at 18:01
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It is by far not universally agreed upon. My year 10 chemistry teacher taught me to use $\ce{H3O+}$ in balancing. The result is a larger number of water molecules on the other side of the equation but everything else is the same.

Many chemists will prefer $\ce{H+}$ because it is shorter and simpler, that’s all.

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Because what we really need to know here is the concentration of protons. The concentration of water is considered to be constant in most of the calculations involving the water autodissociation. Thus, you may hydrate the said proton in your equation however you want — or not at all — but $K_\mathrm{w} = [\ce{H+}][\ce{OH-}]$ would still be equal to $K_\mathrm{w} = [\ce{H9O4+}][\ce{OH-}]$ and, well, who needs extra complications?

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