Equation question

[Manda55]

Im confused about how H2SO4 is produced from the following equation:

CuSO4 + 2HCl --> CuCl2 + H2SO4

[jdurg]

What are you confused about? It's a fairly standard reaction equation.

[Tetrahedrite]

If you don't mind Jdurg, could you explain it to me as well. Both CuSO4 and CuCl2 are exceedingly soluble, how is the H2SO4 extracted without getting HCl contamination????

[woelen]

This reaction eqation is not what happens. You can only describe this reaction in terms of ions and then you'll see that only very small amounts of H2SO4 will be formed.

 

CuSO4 dissolves in hydrochloric acid as follows (and even this is strongly simplified):

 

CuSO4(s) + 4Cl(-) + H(+) ---> CuCl4(2-) + HSO4(-)

 

From this you will see many equilibrium reactions, leading to small quantities of H2SO4, SO4(2-), HCuCl4(-)

 

The complex CuCl4(2-) is green.

[Tetrahedrite]

This reaction eqation is not what happens. You can only describe this reaction in terms of ions and then you'll see that only very small amounts of H2SO4 will be formed.

 

CuSO4 dissolves in hydrochloric acid as follows (and even this is strongly simplified):

 

CuSO4(s) + 4Cl(-) + H(+) ---> CuCl4(2-) + HSO4(-)

 

From this you will see many equilibrium reactions, leading to small quantities of H2SO4, SO4(2-), HCuCl4(-)

 

The complex CuCl4(2-) is green.

This was exactly my point above, I didn't think the reaction was " a fairly standard reaction equation" at all. Just a note though, the quantity for CuCl4(2-) and bisulfate that is formed is highly dependent on the concentration of the initial HCl(aq). Low concentrations will yield CuCl(-) and CuCl2(0) rather than the tetrachloride species.

[woelen]

Low concentrations will yield CuCl(-) and CuCl2(0) rather than the tetrachloride species.

As I said, what I wrote down still is highly simplified. You are right, but not entirely 100%.

 

What happens in reality is that water-ligands are replaced by chloride ligands. So, you will get ions [ce]Cu(H_2O)_3Cl^{+}[/ce], [ce]Cu(H_2O)_2Cl_2[/ce], [ce]Cu(H_2O)Cl_3^{-}[/ce] and of course [ce]CuCl_4^{2-}[/ce].

 

At low pH, also H(+) ions can be coordinated to the complex, making the situation even more complicated.

[jdurg]

If you don't mind Jdurg, could you explain it to me as well. Both CuSO4 and CuCl2 are exceedingly soluble, how is the H2SO4 extracted without getting HCl contamination????

 

I'm not stating that the reaction itself is simple, but the equation itself is a simple double replacement reaction. I was a bit confused as to what was actually being asked.

[annalisa]

I have a very big difficulty in forming the equation between Sodium Sulphite and Iodine? and Sodium Metabisulphite with Iodine?

[woelen]

The dry chemicals do not react, there is no way that the sulfite (or metabisulfite) can be oxidized by iodine.

 

In the presence of water, a reaction can occur. This is a very common and typical high school redox reaction:

 

SO3(2-) + I2 + H2O --> SO4(2-) + 2H(+) + 2I(-)

 

A solution of sodium sulfite in water will become acidic in the presence of iodine.

 

When (meta)bisulfite is oxidized by iodine, a very similar reaction occurs, the only difference is that things are even more acidic:

 

S2O5(2-) + H2O <--->>> 2HSO3(-)

 

HSO3(-) + I2 + H2O ---> SO4(2-) + 3H(+) + 2I(-)

 

=======================================

 

What is not known widely, but which is visible very well, is that excess sulphur dioxide reacts with iodide ion, to form a deep yellow complex.

 

So, what high school text books teach is that sulfite and sulphur dioxide in the presence of water react with iodine to form iodide, sulfate ion and acid. This is not surprising. What high school text books do not teach is that with excess sulphur dioxide, the liquid becomes deep yellow, and a complex is formed. This complex can be written as [i.nSO2](-), with n equal to 1, 2, 3 or 4.

For the interested one, try it. Take some KI and dissolve that in some acid (e.g. dilute HCl or dilute H2SO4). Take some Na2SO3 or Na2S2O5 and dissolve that in some acid as well. Now you have two colorless solutions, one having a smell of SO2. Now mix the two solutions. You will obtain a deep yellow liquid, with a smell of SO2, and in solution the iodide/sulphur dioxide complex.