Question

Write the net ionic equation for the precipitation of zinc hydroxide from aqueous solution: ?

Answer 1

`Zn^(2+) + 2OH^(-) -> Zn(OH)_2`

Explanation:

A net ionic equation shows us the ions that are actually participating in the reaction (versus the remaining ions that do not participate, called spectator ions). In your case the zinc and hydroxide ions were the only ions participating, as they were the only ions that formed a new substance; a precipitate. The other ions remained ions in solution.

This website: http://www.occc.edu/kmbailey/chem1115tutorials/Net_Ionic_Eqns.htm does an excellent job explaining it.

Therefore we have `Zn^(2+) + 2OH^(-) -> Zn(OH)_2`

Answer 2

Here's what I get.

Explanation:

The precipitate may have come from a reaction like

`"Zn"("NO"_3)_2"(aq)" + "2NaOH(aq)" → "Zn"("OH")_2"(s)" + "2NaNO"_3"(aq)"`

This equation is the molecular equation.

Step 1. Ionic equation

We separate all the soluble ionic compounds into their ions to get the ionic equation.

`"Zn"^"2+""(aq)" + "2NO"_3^"-""(aq)" + "2Na"^"+""(aq)" + "2OH"^"-""(aq)" → "Zn"("OH")_2"(s)" + "2Na"^"+""(aq)" + "2NO"_3^"-""(aq)"`

Step 2. Net ionic equation

We cancel all spectator ions to get the net ionic equation.

`"Zn"^"2+""(aq)" + color(red)(cancel(color(black)("2NO"_3^"-""(aq)"))) + color(red)(cancel(color(black)("2Na"^"+""(aq)"))) + "2OH"^"-""(aq)" → "Zn"("OH")_2"(s)" + color(red)(cancel(color(black)("2Na"^"+""(aq)"))) + color(red)(cancel(color(black)("2NO"_3^"-""(aq)")))`

The net ionic equation is

`color(blue)("Zn"^"2+""(aq)"+ "2OH"^"-""(aq)" → "Zn"("OH")_2"(s)")`