Question #ddea1
1 Answer
Explanation:
Aluminium nitrate,
#"Al"("NO"_ 3)_ (3(aq)) -> "Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-)#
#"NaOH"_ ((aq)) -> "Na"_ ((aq))^(+) + "OH"_ ((aq))^(-)#
When you mix these solutions, the aluminium cations,
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
#"Al"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) + 3"Na"_ ((aq))^(+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr + 3"Na"_ ((aq))^(+) + 3"NO"_ (3(aq))^(-)#
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#