# Spectator ions are those which are in solution but do not react. Are there any spectator ions for the reaction of sodium phosphate with calcium bromide?

Sep 29, 2016

Yes, there are.

#### Explanation:

In order to see whether or not you'll have spectator ions when this reaction takes place, examine the solubility of the chemical species involved in the reaction.

To do that, use the solubility rules for ionic compounds.

Sodium phosphate, ${\text{Na"_3"PO}}_{3}$, is a soluble ionic compound, which means that it exists as ions in aqueous solution

${\text{Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO}}_{4 \left(a q\right)}^{3 -}$

The same can be said of calcium bromide, ${\text{CaBr}}_{2}$.

${\text{CaBr"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Br}}_{\left(a q\right)}^{-}$

Now, when you mix these two solutions, you will end up with calcium phosphate, "Ca"_3("PO"_4)_2, and sodium bromide, $\text{NaBr}$.

As you can see from the table, calcium phosphate is insoluble in aqueous solution. This means that when the calcium cations combine with the phosphate anions, a solid will precipitate out of solution.

On the other hand, sodium bromide is soluble in aqueous solution, which implies that it exists as ions in solution.

This means that the sodium cations, $\textcolor{b l u e}{{\text{Na}}^{+}}$, and the bromide anions, $\textcolor{p u r p \le}{{\text{Br}}^{-}}$, will be spectator ions because they exist as ions on both sides of the chemical equation.

2 xx [3"Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-)] + 3 xx ["Ca"_ ((aq))^(2+) + 2"Br"_ ((aq))^(-)] -> "Ca"_ 3("PO"_ 4)_ (2(s)) darr + 6"Na"_ ((aq))^(+) + 6"Br"_ ((aq))^(-)

This is equivalent to

$\textcolor{b l u e}{6 {\text{Na"_ ((aq))^(+)) + 2"PO"_ (4(aq))^(3-) + 3"Ca"_ ((aq))^(2+) + color(purple)(6"Br"_ ((aq))^(-)) -> "Ca"_ 3("PO"_ 4)_ (2(s)) darr + color(blue)(6"Na"_ ((aq))^(+)) + color(purple)(6"Br}}_{\left(a q\right)}^{-}}$

Removing the spectator ions will get you the net ionic equation

3"Ca"_ ((aq))^(2+) + 2"PO"_ (4(aq))^(3-) -> "Ca"_ 3("PO"_ 4)_ (2(s)) darr