# Question 7bac6

Feb 22, 2017

${\text{Fe"_ ((aq))^(3+) + "PO"_ (4(aq))^(3-) -> "FePO}}_{4 \left(s\right)} \downarrow$

#### Explanation:

I will show you how to write the chemical equation that describes the reaction between aqueous solutions of sodium phosphate, ${\text{Na"_3"PO}}_{4}$, and iron(III) nitrate, "Fe"("NO"_3)_3.

SIDE NOTE If you're actually dealing with iron(II) nitrate, "Fe"("NO"_3)_2#, make sure to use this answer as an example of how to approach that particular reaction.

Both of these ionic compounds are soluble in water, which means that they dissociate completely in aqueous solution to form

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

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

When you mix these two solutions, the iron(III) cations and the phosphate anions combine to form iron(III) phosphate, ${\text{FePO}}_{4}$, an insoluble ionic compound that precipitates out of solution.

The second product is aqueous sodium nitrate, ${\text{NaNO}}_{3}$, a soluble ionic compound that will exist as ions in the resulting solution.

${\text{Na"_ 3"PO" _ (4(aq)) + "Fe"("NO"_ 3)_ (3(aq)) -> "FePO"_ (4(s)) darr + 3"NaNO}}_{3 \left(a q\right)}$

The complete ionic equation that describes this double-replacement reaction looks like this

$3 {\text{Na"_ ((aq))^(+) + "PO"_ (4(aq))^(3-) + "Fe"_ ((aq))^(3+) + 3"NO"_ (3(aq))^(-) -> "FePO"_ (4(s)) darr + 3"Na"_ ((aq))^(+) + 3"NO}}_{3 \left(a q\right)}^{-}$

To get the net ionic equation, eliminate the spectator ions, i.e. the ions that are present on both sides of the chemical equation

$\textcolor{red}{\cancel{\textcolor{b l a c k}{3 {\text{Na"_ ((aq))^(+)))) + "PO"_ (4(aq))^(3-) + "Fe"_ ((aq))^(3+) + color(red)(cancel(color(black)(3"NO"_ (3(aq))^(-)))) -> "FePO"_ (4(s)) darr + color(red)(cancel(color(black)(3"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(3"NO}}_{3 \left(a q\right)}^{-}}}}$

This will get you

${\text{Fe"_ ((aq))^(3+) + "PO"_ (4(aq))^(3-) -> "FePO}}_{4 \left(s\right)} \downarrow$