To identify the spectator ions in the given chemical reaction, we need to analyze the full ionic equation and the net ionic equation. Spectator ions are ions that do not change during the reaction—they are present in both the reactants and the products in the same form.
Let's break down the given chemical reaction:
[tex]$
Pb(NO_3)_2(aq) + H_2SO_4(aq) \rightarrow PbSO_4(s) + 2HNO_3(aq)
$[/tex]
First, we write the full ionic equation showing all ions dissociated in aqueous solutions:
[tex]$
Pb^{2+}(aq) + 2NO_3^-(aq) + 2H^+(aq) + SO_4^{2-}(aq) \rightarrow PbSO_4(s) + 2H^+(aq) + 2NO_3^-(aq)
$[/tex]
Now, we identify which ions appear on both sides of the equation and do not participate in the precipitation reaction. These are the spectator ions.
Upon inspection:
- [tex]\(2NO_3^-\)[/tex] is present on both sides of the equation.
- [tex]\(2H^+\)[/tex] is also present on both sides of the equation.
These ions remain unchanged and are thus the spectator ions.
The [tex]\(Pb^{2+}\)[/tex] ion and the [tex]\(SO_4^{2-}\)[/tex] ion combine to form the precipitate [tex]\(PbSO_4(s)\)[/tex] and therefore are not spectator ions.
So, the spectator ions in this reaction are:
[tex]$
NO_3^-(aq), H^+(aq)
$[/tex]
Therefore, the correct answer is:
- [tex]\(NO_3^-(aq)\)[/tex]
- [tex]\(H^+(aq)\)[/tex]
