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 $\ce{H3PO4}$:

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

  • $\begingroup$ 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
  • 2
    $\begingroup$ 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
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

1 Answer 1


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:

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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.