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I was curious if there is a specific reason why when writing mass balance reactions we always leave a $\ce{H+}$ on the left side of the equation as the professor did not explain so, or is it just convention?

For example, write the mass balance equation of $\ce{H3PO4}$:

$$\ce{H2O <=> H+ +OH-}$$ $$\ce{H3PO4 <=> H+ +H_2PO4^{-}(aq)}$$ $$\ce{H2PO4- <=>H+ + HPO4^{2-}(aq)}$$ $$\ce{HPO4^2- <=> H+ + PO4^{3-}(aq)}$$

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  • $\begingroup$ What do you mean by 'we always leave a $H^+$ on the left side of the equation' $\endgroup$
    – Nanoputian
    Commented Oct 29, 2015 at 10:39
  • $\begingroup$ It does not have to be done that way but that is how most people show it. $\endgroup$
    – paparazzo
    Commented Oct 29, 2015 at 10:53
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    $\begingroup$ Those are not mass balance equations. The mass balance equation is $[\ce{H3PO4}] + [\ce{H2PO3-}] + [\ce{HPO4^2-}] + [\ce{PO4^3-}] = c_{\ce{H3PO4}}$. $\endgroup$ Commented Oct 29, 2015 at 10:56
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    $\begingroup$ Do you mean why the proton is "always" written as the first product? I am not aware of such a convention. As long as all products are there with the correct stoichiometric coefficient, it does not matter if the proton is written first, last, or somewhere in-between. $\endgroup$
    – Yoda
    Commented Oct 29, 2015 at 12:22
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    $\begingroup$ Well, I guess you can call these "dissociations of $\ce{H3PO4}$". Anyway, as Anders and Frisbee have said, there's absolutely no requirement that the proton be the first product. $\endgroup$ Commented Oct 29, 2015 at 13:31

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There is no such convention explicitly telling what comes out first. Here, for a traditional formula of phosphoric(V) acid $\ce{H3PO4}$ writing a proton first serves only a didactic purpose. It's visually easier for students to keep a track on dissociation as the order of the elements both in formula and among the products is preserved.

On the other hand, shall one use a coordination formula of phosphoric(V) acid $\ce{[PO(OH)3]}$, it probably would make more sense to use a reversed order and put $\ce{H+}$ at the end:

$$ \begin{align} \ce{[PO(OH)3] &<=> [PO2(OH)2]- + H+}\\ \ce{[PO2(OH)2]- &<=> [PO3(OH)]^2- + H+}\\ \ce{[PO3(OH)]^2- &<=> [PO4]^{3-} + H+} \end{align} $$

One more minor thing. Equations are usually aligned about arrows, and a tabular array of products and reactants emerges when the same repeating compound ($\ce{H+}$) is shown first. Note how easier it is to grasp the information when more compounds are aligned:

$$ \begin{align} &\color{green}{\text{aligned}} & &\color{red}{\text{misaligned}}\\ \ce{H3PO4 &<=> H+ + H_2PO4^{-}(aq)} &\quad \ce{H3PO4 &<=> H_2PO4^{-}(aq) + H+} \\ \ce{H2PO4- &<=>H+ + HPO4^{2-}(aq)} &\quad \ce{H2PO4- &<=> HPO4^{2-}(aq) + H+} \\ \ce{HPO4^2- &<=> H+ + PO4^{3-}(aq)} &\quad \ce{HPO4^2- &<=> PO4^{3-}(aq) + H+} \end{align} $$

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