How come mixing sodium sulfate and sodium chloride doesn't make them react?

Dec 21, 2015

Here's why that is the case.

Explanation:

As you know, sodium sulfate ,${\text{Na"_2"SO}}_{4}$, and sodium chloride, $\text{NaCl}$, are both ionic compounds that are soluble in aqueous solution.

This means that in a solution of sodium sulfate, the salt will dissociate completely to form sodium cations, ${\text{Na}}^{+}$, and sulfate anions, ${\text{SO}}_{4}^{2 -}$.

${\text{Na"_2"SO"_text(4(aq]) -> 2"Na"_text((aq])^(+) + "SO}}_{\textrm{4 \left(a q\right]}}^{-}$

Similarly, sodium chloride will dissociate completely in aqueous solution to form sodium cations and chloride anions, ${\text{Cl}}^{-}$

${\text{NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

Since both ionic compounds exist as cations and anions in solution, mixing two solutions will not result in a reaction because, well, all ions will continue to exist as ions in solution.

Simply put, there is no compound that can be formed by mixing these two solutions. All the ions present on the reactants' side will also be present as ions on the products' side.

2"Na"_text((aq])^(+) + "SO"_text(4(aq])^(-) + "Na"_text((aq])^(+) + "Cl"_text((aq])^(-) -> color(red)("N". "R".)

In order for a double replacement reaction to take place here, you would need two of the ions to form an insoluble solid that precipitates out of solution.

As discussed above, this is not the case here.