Question

Question #ad8d5

Answer 1

To have a better understanding of what happens when two compounds are mixed together in aqueous solution, you should become familiar with the solubility rules

![http://highered.mheducation.com/olcweb/cgi/

Usually, you are asked to determine whether or not a certain reaction will lead to the formation of a precipitate. In this case, you must work backwards - determine what reactants could be mixed starting from the product(s).

So, you're dealing with a double replacement reaction in which your cations and anions will exchange partners. You know that one of the products must be calcium carbonate, `CaCO_3`.

You can use the solubility rules to help you pick a compound containing the `Ca^(2+)` cation, and another compound containing the `CO_3^(2-)` anion - keeping in mind that both the reactans must be soluble.

So, a soluble compound containing `Ca^(2+)` could be calcium nitrate, or `Ca(NO_3)_2`, because all nitrates are soluble. Likewise, a soluble compound containing `CO_3^(-)` could be sodium carbonate, or `Na_2CO_3`, because carbonates are insoluble with the exception of those formed with alkali metal cations.

`Ca(NO_3)_(2(aq)) + Na_2CO_(3(aq)) -> CaCO_(3(s)) + 2NaNO_(3(aq))`

When in aqueous solution, the complete ionic equation will be

`Ca_((aq))^(2+) + 2NO_(3(aq))^(-) + 2Na_((aq))^(+) + CO_(3(aq))^(2-) -> CaCO_(3(s)) + 2Na_((aq))^(+) + 2NO_(3(aq))^(-)`

If you remove spectator ions, i.e. the ions present on both sides of the equation, you'll get the net ionic equation

`Ca_((aq))^(2+) + CO_(3(aq))^(2-) -> CaCO_(3(s))`

You can use the same approach to find other compounds soluble in aqueous solution that form a precipitate when they react, the reaction above is just one example.