# What are the products of K_3PO_4 + AlCl_3?

Jan 6, 2016

Aluminium phosphate and aqueous potassium chloride.

#### Explanation:

Potassium phosphate, ${\text{K"_3"PO}}_{4}$, and aluminium chloride, ${\text{AlCl}}_{3}$, are soluble ionic compounds, which means that they exist as ions in aqueous solution.

${\text{K"_3"PO"_text(4(aq]) -> 3"K"_text((aq])^(+) + "PO}}_{\textrm{4 \left(a q\right]}}^{3 -}$

${\text{AlCl"_text(3(aq]) -> "Al"_text((aq])^(3+) + 3"Cl}}_{\textrm{\left(a q\right]}}^{-}$

When these two solutions are mixed, the aluminium cations, ${\text{Al}}^{3 +}$, and phosphate anions, ${\text{PO}}_{4}^{3 -}$, will react to form aluminium phosphate, ${\text{AlPO}}_{4}$, an Insoluble solid that precipitates out of solution.

The potassium cations, ${\text{K}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$, will form potassium chloride, $\text{KCl}$. However, this salt is soluble in aqueous solution, so it will actually exist as cations and anions.

This means that you have

${\text{K"_3"PO"_text(4(aq]) + "AlCl"_text(3(aq]) -> "AlPO"_text(4(s]) darr + 3"KCl}}_{\textrm{\left(a q\right]}}$

The complete ionic equation, which features all the ions present in solution, will be

$3 {\text{K"_text((aq])^(+) + "PO"_text(4(aq])^(3-) + "Al"_text((aq])^(3+) + 3"Cl"_text((aq])^(-) -> "AlPO"_text(4(s]) darr + 3"K"_text((aq])^(+) + 3"Cl}}_{\textrm{\left(a q\right]}}^{-}$

If you remove the spectator ions, which are ions present on both sides of the equation, you'll get the net ionic equation

$\textcolor{red}{\cancel{\textcolor{b l a c k}{3 {\text{K"_text((aq])^(+)))) + "PO"_text(4(aq])^(3-) + "Al"_text((aq])^(3+) + color(red)(cancel(color(black)(3"Cl"_text((aq])^(-)))) -> "AlPO"_text(4(s]) darr + color(red)(cancel(color(black)(3"K"_text((aq])^(+)))) + color(red)(cancel(color(black)(3"Cl}}_{\textrm{\left(a q\right]}}^{-}}}}$

which looks like this

${\text{Al"_text((aq])^(3+) + "PO"_text(4(aq])^(3-) -> "AlPO}}_{\textrm{4 \left(s\right]}} \downarrow$