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Earlier I had a student come by wanting to know how to show the dissolution of $\ce{NH4OH}$ in water. I would think we would just write it: $$\ce{NH4OH(aq) <=> NH3(aq) + H2O(l)}$$ I say this as $$\ce{NH4OH(aq) + H2O(l) <=> NH3(aq) + H3O+(aq) + OH-(aq)}$$ I am struggling to understand when and why we put the $\ce{H2O}$ as a solvent above the equation arrows vs when we list it as a reactant. She seemed to believe the professor wanted the water to be expressed on the left as a reactant. Any assistance, please?

At first I believed perhaps this had something to do with whether the reactant is molecular or ionic and whether we are looking at ionization or disassociation but this does not appear to be the case as the book shows two examples of a molecular reactant in water differently. $$\ce{HNO3(aq) \overset{H2O}{->} H^+(aq) + NO3-(aq)} $$ and $$\ce{HCl(aq) + H2O(l) -> H3O^+(aq) + Cl-(aq)}$$

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  • $\begingroup$ What is NH4OH or NH4? H+ always exists as H3O+ . (H2O)n. H+ = proton, but is used as a shortcut for the above. For eventual writing and formatting of chemical or mathematical formulas or equations, see how to use MathJax $\endgroup$
    – Poutnik
    Commented Apr 9, 2021 at 19:41
  • $\begingroup$ I made a typing error and fixed it. It was supposed to be $NH_4OH$. My apologies. $\endgroup$
    – neurosciencecalc
    Commented Apr 9, 2021 at 19:48
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    $\begingroup$ $\ce{NH4OH}$ does not exist. It is just a solution of $\ce{NH3}$ in water. If a solution contains simultaneously $\ce{NH4+}$ and $\ce{OH-}$ ions, these ions will immediately react by proton transfer producing $\ce{NH3}$ and $\ce{H2O}$. Compare the pKa to be convinced. Don't speak of a solution of $\ce{NH4OH}$ $\endgroup$
    – Maurice
    Commented Apr 9, 2021 at 19:51
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    $\begingroup$ You still do not get it. There is no NH4OH in water. It is decades long taught nonsense. $$\ce{NH3(aq) + H2O <<=> NH4+(aq) + OH(aq)}$$ $\endgroup$
    – Poutnik
    Commented Apr 9, 2021 at 19:52
  • $\begingroup$ The professor must have made a typing error on the problem statement then. Thank you. $\endgroup$
    – neurosciencecalc
    Commented Apr 9, 2021 at 19:55

1 Answer 1

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There are few conceptual and factical points:

There is no such molecule in water solutions like $\ce{NH4OH}$. Nitrogen atom is not able to form 5 covalent bonds. There is either hydrated ammonia $\ce{NH3(aq)}$, either hydrated ammonium and hydroxide ions, following equilibrium:

$$\ce{NH3(aq) + H2O <<=> NH4+(aq) + OH-(aq)}$$ respectively $$\ce{NH3(aq) + H3O+(aq) <=>> NH4+(aq) + H2O}$$

$\ce{H2O}$ formula occurs in equations, if it explicitly takes part in the real or formal reaction and in the atom count enumeration. Otherwise, it occurs just implicitly, or explicitly as (aq) or formula above/below the arrow (see MathJax link in comments for coding it)

$$\ce{HSO4- <=> H+ + SO4^2-}$$ or $$\ce{HSO4- <=>[H2O] H+ + SO4^2-}$$ or $$\ce{HSO4-(aq)<=>H+(aq) + SO4^2-(aq)}$$ or $$\ce{HSO4- + H2O<=>H3O+ + SO4^2-}$$ or $$\ce{HSO4-(aq) + H2O<=>H3O+(aq) + SO4^2-(aq)}$$

  • Hydronium ion is often written as the shortcut $\ce{H+}$, or $\ce{H+(aq)}$, but this does not reflect reality. It would be a "naked" proton or deuteron which immediately attach first available electron pair, usually of an oxygen atom in a water molecule or hydroxide ion. More correctly would it be $\ce{H3O+(aq)}$ or explicitly $\ce{H3O+. (H2O)_n}$, where n depends mostly on solution molar concentration.
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