# Question #072db

Oct 20, 2015

Here's what I got.

#### Explanation:

You know that your solution contains metal cations.

You also know that this solution will produce a precipitate, which is an insoluble solid, when mixed with a solution of sodium chloride, $\text{NaCl}$.

Sodium chloride is a soluble ionic compound, which means that it will dissociate completely to form sodium cations, ${\text{Na}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$.

This means that the pricipitate will be a halide, or, more specifically, a chloride.

If you take a look at the solubility rules for chlorides, you'll notice that three common metal ions will produce insoluble solids with the chloride anion

• Lead(II) cation $\to$ ${\text{Pb}}^{2 +}$
• Mercury(I) cation $\to$ ${\text{Hg}}_{2}^{2 +}$
• Silver(I) cation $\to$ ${\text{Ag}}^{+}$

So, the three ionic compounds that can precipitate out of solution are

• Silver chloride, $\text{AgCl}$

• Mercury(I) chloride, ${\text{Hg"_2"Cl}}_{2}$

• Lead(II) chloride, ${\text{PbCl}}_{2}$