# Sodium chloride reacts with copper sulfate to produce sodium sulfate and copper chloride. 2NaCI (aq) + CuSO_4 (aq) -> Na_2SO_4 (aq) + CuCl_2 (s). What type of reaction is this?

Jul 7, 2016

No reaction occurs here.

#### Explanation:

Aqueous sodium chloride, $\text{NaCl}$, will not react with aqueous copper(II) sulfate, ${\text{CuSO}}_{4}$, because the two potential products are soluble in aqueous solution.

The chemical equation given to you is actually incorrect because copper(II) chloride, ${\text{CuCl}}_{2}$, is not insoluble in aqueous solution. In fact, it is quite soluble.

This means that the reaction does not produce an insoluble solid that precipitates out of solution.

Sodium chloride and copper(II) sulfate are both soluble ionic compounds that dissociate completely in aqueous solution

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

${\text{CuSO"_ (4(aq)) -> "Cu"_ ((aq))^(2+) + "SO}}_{4 \left(a q\right)}^{2 -}$

Sodium sulfate, ${\text{Na"_2"SO}}_{4}$, and copper(II) chloride, ${\text{CuCl}}_{2}$, are soluble in aqueous solution as well, which means that they exist as ions in solution

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

${\text{CuCl"_ (2(aq)) -> "Cu"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

This means that the chemical equation given to you can be written as

$2 \overbrace{\left[{\text{Na"_ ((aq))^(+) + "Cl"_ ((aq))^(-)])^(color(blue)("NaCl"_ ((aq)))) + overbrace("Cu"_ ((aq))^(2+) + "SO"_ (4(aq))^(2-))^(color(red)("CuSO"_ (4(aq)))) -> overbrace(2"Na"_ ((aq))^(+) + "SO"_ (4(aq))^(2-))^(color(purple)("Na"_ 2"SO"_ (4(aq)))) + overbrace("Cu"_ ((aq))^(2+) + 2"Cl"_ ((aq))^(-))^(color(darkgreen)("CuCl}}_{2 \left(a q\right)}\right)}$

This is equivalent to

$\textcolor{b l u e}{\cancel{\textcolor{b l a c k}{2 {\text{Na"_ ((aq))^(+)))) + color(blue)(cancel(color(black)(2"Cl"_ ((aq))^(-)))) + color(red)(cancel(color(black)("Cu"_ ((aq))^(2+)))) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) -> color(blue)(cancel(color(black)(2"Na"_ ((aq))^(+)))) + color(red)(cancel(color(black)("SO"_ (4(aq))^(2-)))) + color(red)(cancel(color(black)("Cu"_ ((aq))^(2+)))) + color(blue)(cancel(color(black)(2"Cl}}_{\left(a q\right)}^{-}}}}$

As you can see, all the ions present on the reactants' side are also present on the products' side. These ions are called spectator ions because they essentially don't take part in the reaction.

In your case, all the compounds exist as ions in solution, which is why no reaction occurs here.

2"NaCl"_ ((aq)) + "CuSO"_ (4(aq)) -> color(red)("N.R.")