What are the products of #K_3PO_4 + AlCl_3#?

1 Answer

Answer:

Aluminium phosphate and aqueous potassium chloride.

Explanation:

Potassium phosphate, #"K"_3"PO"_4#, and aluminium chloride, #"AlCl"_3#, are soluble ionic compounds, which means that they exist as ions in aqueous solution.

#"K"_3"PO"_text(4(aq]) -> 3"K"_text((aq])^(+) + "PO"_text(4(aq])^(3-)#

#"AlCl"_text(3(aq]) -> "Al"_text((aq])^(3+) + 3"Cl"_text((aq])^(-)#

When these two solutions are mixed, the aluminium cations, #"Al"^(3+)#, and phosphate anions, #"PO"_4^(3-)#, will react to form aluminium phosphate, #"AlPO"_4#, an Insoluble solid that precipitates out of solution.

The potassium cations, #"K"^(+)#, and chloride anions, #"Cl"^(-)#, will form potassium chloride, #"KCl"#. However, this salt is soluble in aqueous solution, so it will actually exist as cations and anions.

This means that you have

#"K"_3"PO"_text(4(aq]) + "AlCl"_text(3(aq]) -> "AlPO"_text(4(s]) darr + 3"KCl"_text((aq])#

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

#3"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"_text((aq])^(-)#

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

#color(red)(cancel(color(black)(3"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"_text((aq])^(-))))#

which looks like this

#"Al"_text((aq])^(3+) + "PO"_text(4(aq])^(3-) -> "AlPO"_text(4(s]) darr#