# Question 941e1

Aug 27, 2016

${\text{FeBr"_ (3(aq)) + "K"_ 3"PO"_ (4(aq)) -> "FePO"_ (4(s)) darr + 3"KBr}}_{\left(a q\right)}$

#### Explanation:

Double replacement reactions are all about the solubility rules that govern ionic compounds in aqueous solution.

Iron(III) bromide and potassium phosphate will react to form iron(III) phosphate and aqueous potassium bromide

${\text{FeBr"_ (3(aq)) + "K"_ 3"PO"_ (4(aq)) -> "FePO"_ (4(s)) darr + 3"KBr}}_{\left(a q\right)}$

Iron(III) bromide, ${\text{FeBr}}_{3}$, and potassium phosphate, ${\text{K"_3"PO}}_{4}$, are both soluble in water, which means that they dissociate completely to produce their respective cations and anions

${\text{FeBr"_ (3(aq)) -> "Fe"_ ((aq))^(3+) + 3"Br}}_{\left(a q\right)}^{-}$

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

Now, when you mix these two solutions, the iron(III) cations will combine with the phosphate anions to form the insoluble iron(III) phosphate, ${\text{FePO}}_{4}$, which precipitates out of solution.

On the other hand, the other possible product, which would be potassium bromide, $\text{KBr}$, is soluble in aqueous solution, which means that the potassium cations and the bromide anion will not combine with each other, they will remain as ions in solution.

You will thus have

${\text{Fe"_ ((aq))^(3+) + 3"Br"_ ((aq))^(-) + 3"K"_ ((aq))^(+) + "PO"_ (4(aq))^(3-) -> "FePO"_ (4(s)) darr + 3"K"_ ((aq))^(+) + 3"Br}}_{\left(a q\right)}^{-}$

That is the complete ionic equation, which features all the ions that exist in solution. To get the net ionic equation, remove the spectator ions, which are those ions that are present on both sides of the equation

"Fe"_ ((aq))^(3+) + color(red)(cancel(color(black)(3"Br"_ ((aq))^(-)))) + color(red)(cancel(color(black)(3"K"_ ((aq))^(+)))) + "PO"_ (4(aq))^(3-) -> "FePO"_ (4(s)) darr + color(red)(cancel(color(black)(3"K"_ ((aq))^(+)))) + color(red)(cancel(color(black)(3"Br"_ ((aq))^(-))))

The net ionic equation will thus be

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

Now, when aqueous solutions of calcium nitrate, "Ca"("NO"_3)_2 and aluminium bromide, ${\text{AlBr}}_{3}$, are mixed, a precipitate does not form.

That is the case because both possible products are soluble in aqueous solution. The calcium cations, ${\text{Ca}}^{2 +}$, will not combine with the bromide anions, ${\text{Br}}^{-}$, because calcium bromide is soluble.

Likewise, the aluminium cations, ${\text{Al}}^{3 +}$, will not combine with the nitrate anions, ${\text{NO}}_{3}^{2 -}$, because aluminium nitrate is soluble.

Therefore, you can say that when those two solutions are mixed, no reaction will take place

"Ca"("NO"_ 3)_ (2(aq)) + "AlBr"_ (3(aq)) -> color(red)("N. R.")#