[Aurarice]

I've been taking a lab at University for this semester(Inorganic 1), and I feel like I'm incredibly behind.
I've also looked and flipped through the text, but I still can't seem to figure out some problems like how to see the formation of certain complexes.

I have a couple of questions that I'd like pointers on to figure out for my lab report(Which was due yesterday, so I already had to turn it in with a couple of blanks. Oh well, my fault.)
The experiment was basically the preparation of copper(I) iodide with copper metal and a few other compounds, and the post lab and my attempts are as follows(The lab was already due, but it's nagging me that I didn't know how to do these.):

Explain the effect of adding a drop of acetic acid: To remove copper oxides on the copper film.

Balanced eqn of CuO with acetic acid. Balanced eqn of CuCO₃.Cu(OH)₂ with the acid.

-2CH₃COOH + CuO -> (CH₃COO)2Cu+H₂O
-2CuCO₃.Cu(OH)₂+CH₃COOH->3Cu(CH₃COO)₂+2CO₂+4H₂O
The second equation I found online after a ton of searching, but I'm not sure how exactly they came up with it. Is it just known that the equation would form carbon dioxide and water, leaving behind a copper acetate? Is there a step-by-step process of elimination I can employ?

Balanced eqn between copper(II) acetate, sodium iodide, and copper
- Cu(CH₃COO)₂ + NaI + Cu(s) -> This is another issue I have. How do I know what happens? Considering that this includes three different compounds, what exactly can I do to know what will react?

Explain why no iodine escapes, nor condenses in the upper cold part of the tube while heating the mixture I₂-NaI-H₂O-Cu
- I'm assuming it's because iodine has reacted with the copper/rest of the solution, so gaseous iodine wouldn't escape while heating the mixture. I know that CuI is soluble in iodide, so would the sodium iodide react with the copper metal to make CuI and then that's what causes iodine to react and not escape?

Why is CuI soluble in concentrated iodide solutions?
- Because...it is? I'm not too sure about solubility in solvents besides water, unfortunately.

When the resulting 'diluted' NaI solution is concentrated and re-used for a further preparation of CuI, the yield eventually might go beyond 100%. Why?
- When I originally saw my yield go to high values I assumed it was just because of mechanical error and some of the excess washing materials used for the precipitated CuI got left over and weighed as well. Aside from that I'm not too sure what else would cause this.

Why are copper(II) and iodide incompatible in acid media and compatible in ammonia media.
- I wasn't aware they were incompatible. I thought in most any solution it'd just be Cu²⁺+2I^-  CuI₂ which decomposes to CuI and I₂.


Predict adding PPh₃ to suspension of CuI in CH₂Cl₂
- So I know PPh₃ is triphenylphosphine, CH₂Cl₂
 is dichloromethene. What would actually happen?
If I recall correctly, phosphines are easily oxidized, so would the triphenylphosphine act as an oxidizing agent for the copper?

[Borek]

2CuCO₃.Cu(OH)₂+CH₃COOH->3Cu(CH₃COO)₂+2CO₂+4H₂O
The second equation I found online after a ton of searching, but I'm not sure how exactly they came up with it. Is it just known that the equation would form carbon dioxide and water, leaving behind a copper acetate? Is there a step-by-step process of elimination I can employ?

All carbonates react with acids this way.

Balanced eqn between copper(II) acetate, sodium iodide, and copper
- Cu(CH₃COO)₂ + NaI + Cu(s) -> This is another issue I have. How do I know what happens? Considering that this includes three different compounds, what exactly can I do to know what will react?

This one is tricky. You can predict the outcome by taking into account solubility products and half reactions of copper and iodine.

Explain why no iodine escapes, nor condenses in the upper cold part of the tube while heating the mixture I₂-NaI-H₂O-Cu
- I'm assuming it's because iodine has reacted with the copper/rest of the solution, so gaseous iodine wouldn't escape while heating the mixture. I know that CuI is soluble in iodide, so would the sodium iodide react with the copper metal to make CuI and then that's what causes iodine to react and not escape?

What happens to iodine in the presence of iodides?

Why is CuI soluble in concentrated iodide solutions?
- Because...it is? I'm not too sure about solubility in solvents besides water, unfortunately.

It is not about iodide as a solvent, but about solutions containing iodide. Or, more likely, about solutions containing iodine - please check.

Once we get past these you should be able to answer some of the following questions.

[Aurarice]

All carbonates react with acids this way.

So all carbonate-acid reactions produce carbon dioxide & water, and then the metal bonded to the carbonate bonds with the acid?

This one is tricky. You can predict the outcome by taking into account solubility products and half reactions of copper and iodine.

NaI the copper solid would react to form CuI, correct? But I'm not sure what other products would result. Would the sodium cations react with acetate to make NaCH₃COO?

What happens to iodine in the presence of iodides?

Um. Iodine is the element, iodide is the anion. That's about all I know, though. Wouldn't the I₂ react with copper to form CuI₂ and degenerate to CuI and I₂ again?

It is not about iodide as a solvent, but about solutions containing iodide. Or, more likely, about solutions containing iodine - please check.

I'm lost.
It seems like my main problem is figuring out what happens to compounds containing iodide in solutions containing iodide. I'll see if I can research it in a textbook once I'm out of classes for the day, thanks!

However, I'm still fairly lost on the PPh₃ in CuI suspension in dichloromethane.
I asked my inorganic professor about it, and all she told me was that phosphine is usually oxidized, so I'd assume it reduces the copper...?

[Borek]

Na⁺ is just a spectator, you should think in terms of net ionic equations.

Iodide and iodine react in solutions, creating I₃^- anion.

[Aurarice]

I think I get it. Just went over a lab on Thursday about the formation of the triiodide ion. Thank you so much!