# Does it matter if for a dissolution reaction, we write "H"_2"O"(l) above the reaction arrow or as a reactant, since it becomes a spectator particle?

Jun 18, 2015

A lot of chemists alter notation for convenience. For example, we have the atomic unit for the Bohr radius, ${a}_{0}$. Why not use meters of some sort? Because no one wants to keep writing out $\frac{4 \pi {\epsilon}_{0} {\left(\frac{h}{2 \pi}\right)}^{2}}{{m}_{e} {e}^{2}}$. We just want to write ${a}_{0}$ and leave it at that. :)

Anyways, writing it with water above the reaction arrow or not, it doesn't matter, so long as it's balanced. If we had an ionic compound dissociate in water, the following two are equivalent:

$\text{NaCl" (s) + "H"_2"O"(l) -> "Na"^(+)(aq) + "Cl"^(-)(aq) + "H"_2"O} \left(l\right)$

${\text{NaCl" (s) stackrel(color(red)("H"_2"O"(l))" ")(->) "Na"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

It's just more compact to write it the second way, but either way is technically fine. Just make sure you put what solvent it is.

Or, if we compare hydronium with hydrogen:

$\text{NH"_3(aq) + "H"_3"O"^(+)(aq) rightleftharpoons "NH"_4^(+)(aq) + "H"_2"O} \left(l\right)$

vs...

${\text{NH"_3(aq) + "H"^+(aq) rightleftharpoons "NH}}_{4}^{+} \left(a q\right)$

We get this from:

"NH"_3(aq) + "H"^(+)cancel(("HOH")^(0))(aq) rightleftharpoons "NH"_4^(+)(aq) + cancel("HOH")^0(l)

It's like a net ionic equation with water present on both sides of the written reaction. Hydronium is really just protonated water.