# Will a double displacement reaction occur between copper (II) sulfate and hydrochloric acid? Why or why not?

Sep 18, 2015

For all intended purposes, no, it will not.

#### Explanation:

The idea here is that you're mixing two soluble compounds, copper(II) sulfate, ${\text{CuSO}}_{4}$, and hydrochloric acid, $\text{HCl}$, which are completely dissociate in aqueous solution.

In order for a double replacement reaction to take place, you need the reaction to produce an insoluble compound that precipitates out of solution.

However, this will not happen when you mix these wo substances because the products are also soluble in aqueous solution.

This reaction will produce copper chloride, ${\text{CuCl}}_{2}$, and sulfuric acid, ${\text{H"_2"SO}}_{4}$.

Theoretically, the overall reaction is

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

The complete ionic equation will thus be

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

All the ions are spectator ions, which means that they can be found on both sides of the reaction.

Keep in mind that this is what happens if you take sulfuric acid to dissociate completely in aqueous solution, i.e. if you take its second ionization to be complete as well.

If you don't do that, then you could write

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

The net ionic equation will be

${\text{SO"_text(4(aq])^(2-) + "H"_text((aq])^(+) rightleftharpoons "HSO}}_{\textrm{4 \left(a q\right]}}^{-}$

Anyway, I think that you were supposed to conside both of sulfuric acid's ionizations as complete, in which case the answer to the question is that this reaction does not take place because all species exist as ions in aqueous solution.

Sep 18, 2015

The change in colour results from the formation of complex ions ${\left[C u {\left({H}_{2} O\right)}_{6}\right]}^{2 +} \text{ and } {\left[C u C {l}_{4}\right]}^{2 -}$.

#### Explanation:

In addition to what he mentioned, I would like to add that the change in colour from light blue to green that could be observed is not resulting from a chemical reaction between $C u S {O}_{4}$ and $H C l$.
You should know that $C {u}^{2 +}$ once dissolved in water, it will form the complex ion ${\left[C u {\left({H}_{2} O\right)}_{6}\right]}^{2 +}$ which has a light blue colour, however, when $H C l$ is added, it will form the complex ion ${\left[C u C {l}_{4}\right]}^{2 -}$ which has a green colour, according to the following equilibrium:
${\left[C u {\left({H}_{2} O\right)}_{6}\right]}^{2 +} + 4 C {l}^{-} r i g h t \le f t h a r p \infty n s {\left[C u C {l}_{4}\right]}^{2 -} + 6 {H}_{2} O$