Answer:
[tex]{\rm Zn^{2+}}\, (aq) + {\rm 2\; OH^{-}}\, (aq) \to {\rm Zn(OH)_2}\, (s)[/tex].
Explanation:
Ionic Equation for this reaction
Rewrite only the species that exist as ions. Those species typically include:
- soluble salts,
- strong acids, and
- soluble bases.
In this reaction, both [tex]{\rm ZnBr_2}\, (aq)[/tex] and [tex]{\rm NaBr}\, (aq)[/tex] are salts. The state symbol "[tex](aq)[/tex]" suggests that both of these salts are soluble. Hence, both of these salts exist as ions and should be rewritten:
- Each [tex]{\rm ZnBr_2}\, (aq)[/tex] formula unit would exist as one [tex]{\rm {Zn}^{2+}}\, (aq)[/tex] and [tex]2\; {\rm Br^{-}}\, (aq)[/tex]. Notice how there are twice as many [tex]{\rm Br^{-}}[/tex] ions as [tex]{\rm {Zn}^{2+}}[/tex] ions.
- Each [tex]{\rm NaBr}\, (aq)[/tex] formula unit would exist as one [tex]{\rm Na^{+}}\, (aq)[/tex] and one [tex]{\rm Br^{-}}\, (aq)[/tex].
Similarly, the state symbol "[tex](aq)[/tex]" suggests that the base [tex]\rm NaOH[/tex] is also soluble:
- Each [tex]{\rm NaOH}\, (aq)[/tex] formula unit would exist as one [tex]{\rm Na^{+}}\, (aq)[/tex] and one [tex]{\rm OH^{-}}\, (aq)[/tex].
On the other hand, the state symbol "[tex](s)[/tex]" suggests that the base [tex]{\rm Zn(OH)_2}\, (s)[/tex] is a precipitate and is not soluble. Rather, the bonds within [tex]{\rm Zn(OH)_2}[/tex] stay mostly intact, and this species would not exist as ions. Hence, do not rewrite [tex]{\rm Zn(OH)_2}\, (s)\![/tex] when deriving the ionic equation for this reaction.
Hence, the ionic equation for this reaction would be:
[tex]\begin{aligned}&\underbrace{{\rm Zn^{2+}}\, (aq) + 2\, {\rm Br^{-}}\, (aq)}_{\text{from ${\rm ZnBr_2}\, (aq)$}} + \underbrace{2\, {\rm Na^{+}}\, (aq) + 2\, {\rm OH^{-}}\, (aq)}_{\text{from $2\, {\rm NaOH}\, (aq)$}} \\ & \to \underbrace{{\rm Zn(OH)_2}\, (s)}_{\text{precipitate}} + \underbrace{2\, {\rm Na^{+}}\, (aq) + 2\, {\rm Br^{-}}\, (aq)}_{\text{from $2\, {\rm NaBr}\, (aq)$}}\end{aligned}[/tex].
Net Ionic Equation for this reaction
Eliminate species that are present on both sides of the ionic equation to obtain the net ionic equation. A species should be eliminated if only if an equal number of this species are found on both sides of the ionic equation. Otherwise, subtract from the side with a larger number of that species.
For this reaction, the net ionic equation would be:
[tex]{\rm Zn^{2+}}\, (aq) + {\rm 2\; OH^{-}}\, (aq) \to {\rm Zn(OH)_2}\, (s)[/tex].
