Question 470ad

Jun 21, 2016

Here's my take on this.

Explanation:

The problem wants you to predict the precipitate produced by mixing two solutions, one that contains aluminium nitrate, "Al"("NO"_3)_3, a soluble ionic compound, and one that contains sodium hydroxide, $\text{NaOH}$, another soluble ionic compound.

The reaction takes place in aqueous solution, so right from the start you should be aware that you're dealing with ions.

Soluble ionic compounds dissociate completely in aqueous solution to form cations, which are positively charged ions, and anions, which are negatively charged ions.

The two solutions can thus be written as

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

It's absolutely crucial to make sure that you add the charges of the ions to the chemical equation.

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

Now, you should be familiar with the solubility rules for aqueous solutions. The aluminium cations, ${\text{Al}}^{3 +}$, will combine with the hydroxide anions, ${\text{OH}}^{-}$, to form the insoluble aluminium hydroxide, which precipitates out of solution.

Notice that the aluminium cations have a $3 +$ charge and the hydroxide anions have a $1 -$ charge. This means that you're going to need $3$ hydroxide anions in order to balance the positive charge of the cation.

Therefore, the chemical formula for aluminium hydroxide is "Al"("OH")_3, .

The other product of the reaction will be aqueous sodium nitrate. The sodium cations have a $1 +$ charge and the nitrate anions have a $1 -$ charge, so the chemical formula for sodium nitrate will be ${\text{NaNO}}_{3}$.

The balanced chemical equation will thus looks like this

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

The complete ionic equation for this reaction looks like this

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

To get the net ionic equation, eliminate spectator ions, which are those ions that are found on both sides of the 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))^(-))))

This will get you

color(green)(|bar(ul(color(white)(a/a)color(black)("Al"_ ((aq))^(3+) + 3"OH"_ ((aq))^(-) -> "Al"("OH")_ (3(s)) darr)color(white)(a/a)|)))

Now, I'm not really sure if the math formatting is to blame for some of the equations you wrote, but you're missing subscripts and charges in all of them.

Take the first equation, for example. "Al"("NO"_3) is actually "Al"("NO"_3)_3#.

I'm assuming that $\text{Al"^3"OH}$ is actually ${\text{Al"^(3+)"OH}}^{-}$. If that is the case, then the correct version would be

${\text{Al"^(3+)("OH"^(-))_3 implies "Al"^(3+) + 3"OH}}^{-}$

There's no such thing as ${\text{Na}}_{3}$. Sodium cannot form molecules, it can exist either as a solid, $\text{Na}$, or a cation, ${\text{Na}}^{+}$.

In your case, the sodium cation was present in solution, so ${\text{Na}}_{3}$ is actually $3 {\text{Na}}^{+}$.

Once again, remember to always add charges when dealing with ions. An ion without an added charge is not actually an ion.

For example, there's no such thing as ${\text{NO}}_{3}$. Instead, add the $1 -$ charge that belongs to the ions to get ${\text{NO}}_{3}^{-}$ anion.

Also, keep in mind that a coefficient add to a soluble ionic compound gets distributed to all the ions that are produced by said compound in solution.

For example,

$\textcolor{red}{3} {\text{NaOH" = color(red)(3)("Na"^(+) + "OH"^(-)) = color(red)(3)"Na"^(+) + color(red)(3)"OH}}^{-}$

$\textcolor{b l u e}{2} {\text{Al"("NO"_3)_3 = color(blue)(2)("Al"^(3+) + 3"NO"_3^(-)) = color(blue)(2)"Al"^(3+) + 6"NO}}_{3}^{-}$

Notice that the charges must remain balanced at all times.

All in all, you should review ionic compounds before diving into complete and net ionic equations.