# Question #d60f9

Jul 12, 2015

A white precipitate will be formed.

#### Explanation:

Silver nitrate and sodium chloride are both soluble ionic compounds, which means that they actually exist as ions in aqueous solution.

$A g N {O}_{3 \left(a q\right)} \to A {g}_{\left(a q\right)}^{+} + N {O}_{3 \left(a q\right)}^{-}$

and

$N a C {l}_{\left(a q\right)} \to N {a}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

When these two substances are mixed, the silver cations, $A {g}^{+}$, and the chloride anions, $C {l}^{-}$, will bond together to form silver chloride, $A g C l$, an insoluble solid, which precipitates out of solution.

The other two ions, $N {a}^{+}$ and $N {O}_{3}^{-}$, will continue to exist as such in solution, since the potential compound sodium nitrate, $N a N {O}_{3}$, is soluble in aqueous solution.

So, the complete ionic equation looks like this

$A {g}_{\left(a q\right)}^{+} + N {O}_{3 \left(a q\right)}^{-} + N {a}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \to A g C {l}_{\left(s\right)} \downarrow + N {a}_{\left(a q\right)}^{+} + N {O}_{3 \left(a q\right)}^{-}$

The net ionic equation, which you get by eliminating spectator ions, i.e. the ions that are present on both sides of the eqaution, will be

$A {g}_{\left(a q\right)}^{+} + \cancel{N {O}_{3 \left(a q\right)}^{-}} + \cancel{N {a}_{\left(a q\right)}^{+}} + C {l}_{\left(a q\right)}^{-} \to A g C {l}_{\left(s\right)} \downarrow + \cancel{N {a}_{\left(a q\right)}^{+}} + \cancel{N {O}_{3 \left(a q\right)}^{-}}$

or

$A {g}_{\left(a q\right)}^{+} + C {l}_{\left(a q\right)}^{-} \to A g C {l}_{\left(s\right)} \downarrow$

So, as a conclusion, a double replacement reaction takes place in which the silver cations and chloride anions form an insoluble white solid called silver hcloride, $A g C l$.