Earlier I had a student come by wanting to know how to show the dissolution of $NH_4OH$ in water. I would think we would just write it: $$NH_4OH(aq) <=> NH_3(aq) + H_2O(l)$$ I say this as $$NH_4OH(aq) + H_2O(l) <=> NH_3(aq) +H_3O^+(aq) + OH^-(aq)$$ I am struggling to understand when and why we put the $H_2O$ 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. $$HNO_3(aq) \xrightarrow{H_2O} H^+(aq) + NO_3^-(aq) $$ and $$HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$$

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 $$\ce{HCl(aq) + H2O(l) -> H3O^+(aq) + Cl-(aq)}$$

Earlier I had a student come by wanting to know how to show the dissolution of $NH_4OH$ in water. I would think we would just write it: $$NH_4OH(aq) <=> NH_3(aq) + H_2O(l)$$ I say this as $$NH_4(aq) + H_2O(l) <=> NH_3(aq) +H_3O^+(aq) + OH^-(aq)$$ I am struggling to understand when and why we put the $H_2O$ 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. $$HNO_3(aq) \xrightarrow{H_2O} H^+(aq) + NO_3^-(aq) $$ and $$HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$$

Earlier I had a student come by wanting to know how to show the dissolution of $NH_4OH$ in water. I would think we would just write it: $$NH_4OH(aq) <=> NH_3(aq) + H_2O(l)$$ I say this as $$NH_4OH(aq) + H_2O(l) <=> NH_3(aq) +H_3O^+(aq) + OH^-(aq)$$ I am struggling to understand when and why we put the $H_2O$ 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. $$HNO_3(aq) \xrightarrow{H_2O} H^+(aq) + NO_3^-(aq) $$ and $$HCl(aq) + H_2O(l) \rightarrow H_3O^+(aq) + Cl^-(aq)$$