Answer:
a. [tex]Al(OH)_3(s)+3H^\rightarrow Al^{3+}+3H_2O(l)\\[/tex]
b. [tex]Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)[/tex]
Explanation:
Hello,
a. In this case, the overall reaction is:
[tex]Al(OH)_3(s)+3HBr(aq)\rightarrow AlBr_3(aq)+3H_2O[/tex]
Nevertheless, the ionic version is:
[tex]Al(OH)_3(s)+3H^++3Br^-(aq)\rightarrow Al^{3+}+3Br^-(aq)+3H_2O(l)\\[/tex]
Since the base is insoluble, thereby, the balanced net ionic equation turns out:
[tex]Al(OH)_3(s)+3H^\rightarrow Al^{3+}+3H_2O(l)\\[/tex]
Since bromide ions become spectator ions.
b) In this case, the overall reaction is:
[tex]Pb(NO_3)_2(aq)+2LiI(aq)\rightarrow PbI_2(s)+2LiNO_3(aq)[/tex]
Nevertheless, the ionic version is:
[tex]Pb^{2+}(aq)+2(NO_3)^-(aq)+2Li^+(aq)+2I^-(aq)\rightarrow PbI_2(s)+2Li^+(aq)+2(NO_3)^-(aq)[/tex]
Since lead (II) iodide is insoluble whereas lithium nitrate does not, thereby, the net ionic equation turns out:
[tex]Pb^{2+}(aq)+2I^-(aq)\rightarrow PbI_2(s)[/tex]
Since lithium and nitrate ions become spectator ions.
Regards.
