We've seen that although many ionic salts dissolve in water, those characterized by strong ionic bonds do not. We might wonder, then, what happens when we use two separate salts to introduce into solution a cation and anion that bond strongly to one another. Will the cation and anion stay in solution? Thermodynamically, it's clear that the formation of strong ionic bonds more than compensates for the loss of ion-dipole interactions around each ion (that's what we mean when we say that the ionic bonds are "strong"). As a result, ionic bonds will form spontaneously, and the ionic solid will come out of solution as the ions pair up. This process is the opposite of dissolution (literally its thermodynamic reverse), and it's called precipitation. Let's look briefly at an example of a precipitation reaction. Silver nitrate (AgNO3) is soluble in water—it's a nitrate, after all. Sodium chloride (NaCl) is of course also soluble in water. However, when an aqueous solution of silver nitrate is mixed with a solution of sodium chloride, a gray solid, or precipitate, comes out of the water. This solid is silver chloride, a salt which is insoluble in water itself. Note that the combination of a silver salt and a chloride salt in water allows the silver and chloride ions to pair up. They do so spontaneously because the energy released by ionic-bond formation is greater than the energy required to ruin solvation. Note also that the counterions to silver and chloride, sodium cation and nitrate anion, remain in solution. According to the solubility rules sodium nitrate is a soluble salt, so we should expect it to stay in solution. Predicting the Result of Mixing Two Salt SolutionsThe solubility rules are a powerful tool when it comes to precipitation reactions. Salts that are insoluble according to the rules will precipitate out of solution when the cation and anion are both present. Hence, to determine the results of a potential precipitation reaction, we need to follow two steps.
Consider the combination in aqueous solution of barium nitrate, Ba(NO3)2, and lithium sulfate, Li2SO4. To write the new salts that could form from these reactants, we simply trade either the cations or the anions while taking care to balance charges. The possible salts in this case are BaSO4 and LiNO3 (the ions involved are Ba2+, Li+, NO3–, and SO42–). Lithium nitrate is a salt of a group 1 cation with the eminently soluble nitrate anion, and it’s soluble in water. Barium sulfate, on the other hand, is insoluble (see rule 4). Thus, we can predict that barium sulfate will precipitate out of solution, while lithium cations and nitrate anions will remain in solution. Writing Precipitation EquationsThere are several different ways to write precipitation equations, each with their own advantages and disadvantages. The molecular equation shows ionic species as neutral, molecular entities in either aqueous or solid form. For example, when silver nitrate and potassium bromide are mixed, a precipitate of silver bromide forms. $$\mathrm{AgNO_3}(aq) + \mathrm{KBr}(aq) \rightarrow \mathrm{AgBr}(s) + \mathrm{KNO_3}(aq)$$ The ions that remain in solution are written together as an aqueous, ionic compound on the products side. The complete ionic equation shows all aqueous ions as separate species, but pairs up the ions that form a solid precipitate on the products side. $$\mathrm{Ag^+}(aq) + \mathrm{NO_3^-}(aq) + \mathrm{K^+}(aq) + \mathrm{Br^-}(aq) \rightarrow \mathrm{AgBr}(s) + \mathrm{K^+}(aq) + \mathrm{NO_3^-}(aq)$$ Note that the potassium and nitrate ions appear to do nothing during this process. These ions, which do not form a precipitate, are called spectator ions. The net ionic equation eliminates spectator ions to show only those ions involved in precipitate formation. The advantage of the net ionic equation is that it distills the precipitation process down to its essence: two or more ions coming together to form a precipitate. $$\mathrm{Ag^+}(aq) + \mathrm{Br^-}(aq) \rightarrow \mathrm{AgBr}(s)$$ Although the stoichiometric coefficients in all of these examples have all been equal to 1, they need not be in general. For example, lead(II) cations combine with two iodide anions to form insoluble lead(II) iodide. The net ionic equation for this process is shown below. $$\mathrm{Pb^{2+}}(aq) + 2 \: \mathrm{I^-}(aq) \rightarrow \mathrm{PbI_2}(s)$$ ← Back
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Chemical Reactions and Equations > Ask Alfie 
Silver nitrate reacts with sodium bromide
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Does a reaction occur when aqueous solutions of silver I nitrate and sodium bromide are combined?
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