Question #ad8d5

1 Answer
Mar 13, 2015

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

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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.