Question

The following chemical reaction takes place in aqueous solution:

ZnBr2(aq)+2 NaOH(aq) --Zn(OH),(s)+2 NaBr(aq)
Write the net ionic equation for this reaction.

Answer 1

`{\rm Zn^(2+)}\, (aq) + {\rm 2\; OH^(-)}\, (aq) \to {\rm Zn(OH)_2}\, (s)`.

Step-by-step 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
`{\rm ZnBr_2}\, (aq)` and
`{\rm NaBr}\, (aq)` are salts. The state symbol "
`(aq)`" suggests that both of these salts are soluble. Hence, both of these salts exist as ions and should be rewritten:

• Each
  `{\rm ZnBr_2}\, (aq)` formula unit would exist as one
  `{\rm {Zn}^(2+)}\, (aq)` and
  `2\; {\rm Br^(-)}\, (aq)`. Notice how there are twice as many
  `{\rm Br^(-)}` ions as
  `{\rm {Zn}^(2+)}` ions.
• Each
  `{\rm NaBr}\, (aq)` formula unit would exist as one
  `{\rm Na^(+)}\, (aq)` and one
  `{\rm Br^(-)}\, (aq)`.

Similarly, the state symbol "
`(aq)`" suggests that the base
`\rm NaOH` is also soluble:

• Each
  `{\rm NaOH}\, (aq)` formula unit would exist as one
  `{\rm Na^(+)}\, (aq)` and one
  `{\rm OH^(-)}\, (aq)`.

On the other hand, the state symbol "
`(s)`" suggests that the base
`{\rm Zn(OH)_2}\, (s)` is a precipitate and is not soluble. Rather, the bonds within
`{\rm Zn(OH)_2}` stay mostly intact, and this species would not exist as ions. Hence, do not rewrite
`{\rm Zn(OH)_2}\, (s)\!` when deriving the ionic equation for this reaction.

Hence, the ionic equation for this reaction would be:

`\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}`.

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:

`{\rm Zn^(2+)}\, (aq) + {\rm 2\; OH^(-)}\, (aq) \to {\rm Zn(OH)_2}\, (s)`.