Question #73ac3

1 Answer
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:, 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: