Question

Question #5c053

Answer 1

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

`"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

`"2Na"_3"PO"_4"(aq)" + "3CuCl"_2"(aq)" → "6NaCl(aq)" + "Cu"_3"(PO"_4)_2"(s)"`

The precipitate will be Cu₃(PO₄)₂.