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 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, $C a C {O}_{3}$.

You can use the solubility rules to help you pick a compound containing the $C {a}^{2 +}$ cation, and another compound containing the $C {O}_{3}^{2 -}$ anion - keeping in mind that both the reactans must be soluble.

So, a soluble compound containing $C {a}^{2 +}$ could be calcium nitrate, or $C a {\left(N {O}_{3}\right)}_{2}$, because all nitrates are soluble. Likewise, a soluble compound containing $C {O}_{3}^{-}$ could be sodium carbonate, or $N {a}_{2} C {O}_{3}$, because carbonates are insoluble with the exception of those formed with alkali metal cations.

$C a {\left(N {O}_{3}\right)}_{2 \left(a q\right)} + N {a}_{2} C {O}_{3 \left(a q\right)} \to C a C {O}_{3 \left(s\right)} + 2 N a N {O}_{3 \left(a q\right)}$

When in aqueous solution, the complete ionic equation will be

$C {a}_{\left(a q\right)}^{2 +} + 2 N {O}_{3 \left(a q\right)}^{-} + 2 N {a}_{\left(a q\right)}^{+} + C {O}_{3 \left(a q\right)}^{2 -} \to C a C {O}_{3 \left(s\right)} + 2 N {a}_{\left(a q\right)}^{+} + 2 N {O}_{3 \left(a q\right)}^{-}$

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

$C {a}_{\left(a q\right)}^{2 +} + C {O}_{3 \left(a q\right)}^{2 -} \to C a C {O}_{3 \left(s\right)}$

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.