Question

Question #73ac3

Answer 1

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 `NaOH` (being a hydroxide salt of a group I element) and `AgNO_3` (a common soluble salt of silver) are soluble in water. The general reaction looks like this

`NaOH_((aq)) + AgNO_(3(aq)) -> AgOH_((s)) + NaNO_(3(aq))`

However, this reaction will not create siginificant amounts of `AgOH` because of the more favorable energetics of this reaction:

`2AgOH -> Ag_2O_((s)) + H_2O_((l))`

Silver oxide looks like this:

So, your general reaction actually looks like this:

`2AgNO_(3(aq)) + 2NaOH_((aq)) -> Ag_2O_((s)) + 2NaNO_(3(aq)) + H_2O_((l))`

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

`2Na_((aq))^(+) + 2OH_((aq))^(-) + 2Ag_((aq))^(+) + 2NO_(3(aq))^(-) -> Ag_2O_((s)) + 2NO_(3(aq))^(-) + 2Na_((aq))^(+) + H_2O_((l))`

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

`2Ag_((aq))^(+) + 2OH_((aq))^(-) -> Ag_2O_((s)) + H_2O_((l))`

Here's a video of the reaction: