What Forms the Green-Blue Precipitate with Chromium (III) Ions and NaOH?

[TT0]

Homework Statement

Homework Equations

The Attempt at a Solution

I chose A as after googling, I learned it was green. My question is: what is the green-blue precipitate. I am inclined to say [Cr(H₂O)₆]²⁺ as it's solution is blue, but ions don't precipitate. I tried searching and could find nothing. Could someone tell me what it is?

Cheers!

 

[Borek]

Why not just a hydroxide?

 

[TT0]

I just did further googling and it seems that A is not green according to some and green according to others. I am lost. Chromium (III) is green I think, but in A, C and D, the oxidation number of chromium is also 3. Could you give me some advice?

Why not just a hydroxide?

Is this for the blue-green precipitate or the green solution?

Thanks!

 

[Borek]

Sorry, that was for the precipitate.

Cr is amphoteric, dissolution involves OH^-.

 

[TT0]

So I guess the precipitate is some kind of chromium hydroxide compound (not sure what is the oxidation of chromium is). So I guess that means that the ion causing the green colour has more OH anions than the precipitate. That means A, B and E is not possible. Since NaOH is a base, it doesn't seem the formation of H₃O+ is possible so I think the answer is C. My reasoning seems shaky so I could be very wrong. Is C the answer, if so, can you explain your logic so I can learn from it?

Cheers!

 

[Borek]

Oxidation doesn't change, there is no redox going on here.

As typical with most amphoteric elements, first hydroxide precipitates, then it dissolves in excess OH^-. This is awalys a series of reactions of the

Me(OH)_n^m + OH^- ↔ Me(OH)_n+1^m-1

type. One of them contains a neutral hydroxide, typically insoluble. Things get more complicated when you consider water ligands that are present as well (and replaced by OH^-), but as a first approximation they can be ignored.

 

[TT0]

So will the answer be C as it fits the formula you gave above? Thanks!

 

[CrazyNinja]

Point one I didnt get this statement from you.

So I guess that means that the ion causing the green colour has more OH anions than the precipitate.

Ions can't have anions. I think you had a moment of confusion and wrote something else other than what you intended.

Coming to the question, why don't we use the method of elimination? Since the NaOH is (aq) it can't be a strong enough oxidising agent to oxidise Cr(III) to anything higher, especially Cr(VI). That leaves out B and E as answers.

Now because we are using basic medium H₃0⁺ ions can't originate. that leaves out D as an answer.

Googling tells me that Cr[(H₂O)₆]⁺³ is bluish in color. Puff goes A as an answer.

Leaves us only with C. And this also makes sense, because in qualitative analysis we place Chromium along with best buddy Aluminium, which forms a similar compound on addition of excess of base.