Question

Question #d60f9

Answer 1

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.

`AgNO_(3(aq)) -> Ag_((aq))^(+) + NO_(3(aq))^(-)`

and

`NaCl_((aq)) -> Na_((aq))^(+) + Cl_((aq))^(-)`

When these two substances are mixed, the silver cations, `Ag^(+)`, and the chloride anions, `Cl^(-)`, will bond together to form silver chloride, `AgCl`, an insoluble solid, which precipitates out of solution.

The other two ions, `Na^(+)` and `NO_3^(-)`, will continue to exist as such in solution, since the potential compound sodium nitrate, `NaNO_3`, is soluble in aqueous solution.

So, the complete ionic equation looks like this

`Ag_((aq))^(+) + NO_(3(aq))^(-) + Na_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s)) darr + Na_((aq))^(+) + NO_(3(aq))^(-)`

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

`Ag_((aq))^(+) + cancel(NO_(3(aq))^(-)) + cancel(Na_((aq))^(+)) + Cl_((aq))^(-) -> AgCl_((s)) darr + cancel(Na_((aq))^(+)) + cancel(NO_(3(aq))^(-))`

or

`Ag_((aq))^(+) + Cl_((aq))^(-) -> AgCl_((s)) darr`

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, `AgCl`.