Question

What are the products of `K_3PO_4 + AlCl_3`?

Answer 1

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`