# Question ddea1

Jan 6, 2017

${\text{Al"("NO"_ 3) _ (3(aq)) + 3"NaOH"_ ((aq)) -> "Al"("OH")_ (3(s)) darr + 3"NaNO}}_{3 \left(a q\right)}$

#### Explanation:

Aluminium nitrate, "Al"("NO"_3)_3, and sodium hydroxide, $\text{NaOH}$, are soluble in water, which means that they dissociate completely in aqueous solution

${\text{Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO}}_{3 \left(a q\right)}^{-}$

${\text{NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH}}_{\left(a q\right)}^{-}$

When you mix these solutions, the aluminium cations, ${\text{Al}}^{3 +}$, and the hydroxide anions, ${\text{OH}}^{-}$, will combine to form aluminium hydroxide, "Al"("OH")_3, an insoluble ionic compound that will precipitate out of solution.

On the other hand, sodium nitrate, the other product of the reaction, is a soluble ionic compound, so it too will exist as ions in solution.

You will thus have

"Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + color(blue)(3) xx ["Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)] -> "Al"("OH")_ (3(s)) darr + color(blue)(3) xx ["Na"_ ((aq))^(+) + "NO"_ (3(aq))^(-)]

This is equivalent to

${\text{Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + 3"Na"_ ((aq))^(+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + 3"Na"_ ((aq))^(+) + 3"NO}}_{3 \left(a q\right)}^{-}$

That represents the complete ionic equation that describes this reaction.

Now, notice that some of the ions are present on both sides of the chemical equation. These ions are called spectator ions. You can remove the spectator ions from the chemical equation to get the net ionic equation

"Al"_ ((aq))^(3+) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-))))

You will thus have

"Al"_ ((aq))^(3+) + 3"OH"_ ((aq))^(-) -> "Al"("OH") _ (3(s)) darr#