# Question #a8a12

Jun 21, 2017

No.

#### Explanation:

The trick here is to recognize that both products are soluble in water, so they will exist as ions in aqueous solution.

In other words, none of the ions will combine to form an insoluble compound, so you can say that no reaction will take place when you mix these two solutions.

$\text{NaNO"_ (3(aq)) + "Li"_ 2"SO"_ (4(aq)) -> "N.R.}$

You can see that this is the case by writing the complete ionic equation. The possible products are aqueous lithium nitrate, ${\text{LiNO}}_{3}$, and aqueous sodium sulfate, ${\text{Na"_2"SO}}_{4}$.

$2 {\text{Na"_ ((aq))^(+) + 2"NO"_ (3(aq))^(-) + 2"Li"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) -> 2"Na"_ ((aq))^(+) + "SO"_ (4(aq))^(2-) + 2"Li"_ ((aq))^(+) + 2"NO}}_{3 \left(a q\right)}^{-}$

As you can see, all ions are essentially spectator ions because they are present on both sides of the chemical equation.

So instead of writing

$2 {\text{NaNO"_ (3(aq)) + "Li"_ 2"SO"_ (4(aq)) -> "Na"_ 2"SO"_ (4(aq)) + 2"LiNO}}_{3 \left(a q\right)}$

you must say that

$\text{NaNO"_ (3(aq)) + "Li"_ 2"SO"_ (4(aq)) -> "N.R.}$