# Question 5c053

Mar 13, 2015

It will occur, since one of the products, "Cu"_3"(PO"_4)_2#, is insoluble.

#### Explanation:

This looks like a possible double replacement reaction.

So the likely equation for your reaction is

$\text{2Na"_3"PO"_4"(aq)" + "3CuCl"_2"(aq)" → "6NaCl(?)" + "Cu"_3"(PO"_4)_2"(?)}$

To decide if a precipitate forms, you have to review the "Solubility Rules". The two important ones for this question are:

1. All compounds of Group 1 elements are soluble.
2. All sulfites (SO₃²⁻), carbonates (CO₃²⁻), chromates (CrO₄²⁻), and phosphates (PO₄³⁻) are insoluble except for those of NH₄⁺ and Group 1 metals.

NaCl is soluble (Rule 1).

Cu₃(PO₄)₂ is insoluble (Rule 2).

So the balanced equation is

$\text{2Na"_3"PO"_4"(aq)" + "3CuCl"_2"(aq)" → "6NaCl(aq)" + "Cu"_3"(PO"_4)_2"(s)}$

The precipitate will be Cu₃(PO₄)₂.