# Question #73ac3

Jan 6, 2015

You are indeed dealing with a double replacement precipitation reaction, a reaction in which two soluble ionic compounds react to form an insoluble precipitate.

If you are familiar with the solubility rules (more here: http://www.csudh.edu/oliver/chemdata/solrules.htm), you'll recognize that both $N a O H$ (being a hydroxide salt of a group I element) and $A g N {O}_{3}$ (a common soluble salt of silver) are soluble in water. The general reaction looks like this

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

However, this reaction will not create siginificant amounts of $A g O H$ because of the more favorable energetics of this reaction:

$2 A g O H \to A {g}_{2} {O}_{\left(s\right)} + {H}_{2} {O}_{\left(l\right)}$

Silver oxide looks like this:

So, your general reaction actually looks like this:

$2 A g N {O}_{3 \left(a q\right)} + 2 N a O {H}_{\left(a q\right)} \to A {g}_{2} {O}_{\left(s\right)} + 2 N a N {O}_{3 \left(a q\right)} + {H}_{2} {O}_{\left(l\right)}$

Since we are dealing with ionic compounds, the complete ionic equation looks like this:

$2 N {a}_{\left(a q\right)}^{+} + 2 O {H}_{\left(a q\right)}^{-} + 2 A {g}_{\left(a q\right)}^{+} + 2 N {O}_{3 \left(a q\right)}^{-} \to A {g}_{2} {O}_{\left(s\right)} + 2 N {O}_{3 \left(a q\right)}^{-} + 2 N {a}_{\left(a q\right)}^{+} + {H}_{2} {O}_{\left(l\right)}$

The net ionic equation, which results from eliminating spectator ions, is:

$2 A {g}_{\left(a q\right)}^{+} + 2 O {H}_{\left(a q\right)}^{-} \to A {g}_{2} {O}_{\left(s\right)} + {H}_{2} {O}_{\left(l\right)}$

Here's a video of the reaction: