Question

Question #47c74

Answer 1

See explanation.

Explanation:

Silver nitrate, `"AgNO"_3`, will react with sodium sulfate, `"Na"_2"SO"_4`, to produce silver sulfate, `"Ag"_2"SO"_4`, an ionic compound that is considered insoluble in aqueous solution, and aqueous sodium nitrate.

`2"AgNO"_ (3(aq)) + "Na"_ 2"SO"_ (4(aq)) -> "Ag"_ 2"SO"_ (4(s)) darr + 2"NaNO"_ (3(aq))`

As you can see by inspecting the solubility rules, silver cations, `"Ag"^(+)`, form an insoluble solid when paired with sulfate anions, `"SO"_4^(2-)`.

The nitrate anions, `"NO"_3^(-)`, and the sodium cations, `"Na"^(+)`, are spectator ions, which means that you have

`2"Ag"_ ((aq))^(+) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-)))) + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + "SO"_ (4(aq))^(2-) -> "Ag"_ 2"SO"_ (4(s)) darr + color(red)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)(2"NO"_ (3(aq))^(-))))`

The net ionic equation that describes this double replacement reaction looks like this

`2"Ag"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) -> "Ag"_ 2"SO"_ (4(s)) darr`

The silver sulfate will precipitate out of solution, provided that the concentrations of the silver nitrate solution and of the sodium sulfate solution are high enough.

The solubility product constant for silver sulfate is listed as

`K_(sp) = 6.0 * 10^(-5)`

http://www.wiredchemist.com/chemistry/data/solubility-product-constants

In order for a precipitate to form, you must have

`["Ag"^(+)]^2 * ["SO"_ 4^(2-)] > K_(sp)`

`["Ag"^(+)]^2 * ["SO"_ 4^(2-)] > 6.0 * 10^(-5)`