Question

How do you identify the spectator ions in the neutralization of lithium hydroxide, LioH, with hydrobnomic acid, HBr?

Answer 1

In this case, the hydroxide `OH^-1` and hydride `H^+` ions from the `LiOH` and `HBr` respectively remain in solution.

Explanation:

"Spectator ions" are simply those ions in a reaction solution that do not themselves react. They remain in solution as ions.

In this case, the hydroxide `OH^-1` and hydride `H^+` ions from the `LiOH` and `HBr` respectively remain in solution. If you consider their possible combination into water they might not be considered "spectators". But if the reaction takes place in a water solution, that is hard to define unless there is a noticeable change in the pH of the solution.

Answer 2

How else but by writing the stoichiometric equation?

Explanation:

We can write the stoichiometric in this way:

`LiOH(aq) + HBr(aq) rarr H_2O(l) + LiBr(aq)`

Of course, when we write `LiBr(aq)` we mean the aquated species, which at a micro-level we might represent as `[Li(OH_2)_(4-6)]^+` or `[Br(H_2O)_(4-6)]^-`; these are coordination complexes if you like of the `Li^+` or `Br^-` ions with water ligands. Because water is present in such abundance, such solvation often tends to be ignored.

In the neutralization reaction, bond formation has EXPLICITLY occurred in the reaction to form water:

`H_3O^+ +HO^(-) rarr 2H_2O(l)`

In the ionization reaction, to form the aquated ions from the salt, bond-formation has only implicitly occurred, and we have the idea that lithium and bromide ions are present as their aquated ions. Of course we might, say, add ethanol, to the solution and precipitate out the ionic species, `LiBr`, as a solid.