Identify the 2 metal cations through reactions

[TT0]

Homework Statement

A coloured solution, known to contain two metal ions, was treated with excess cold sodium hydroxide solution. When filtered a whitish solid, slowly changing to brown, was retained on the filter paper and a colourless solution collected as the filtrate. Dropwise addition of hydrochloric acid to the filtrate produced a white precipitate which dissolved in excess acid. Treatment of the residue from the filter paper with a solution of a strong oxidiser produced a reddish-violet solution.

Indicate any pairs of ions which on testing as above leads to the observed changes.

1. A Zn2+ and Mn2+ ions
2. B Mg2+ and Zn2+ ions
3. C Mn2+ and Mg2+ ions
4. D Fe2+ and Zn2+ ions
5. E Mn2+ and Fe2+ ions

Homework Equations

The Attempt at a Solution

"Treatment of the residue from the filter paper with a solution of a strong oxidiser produced a reddish-violet solution." I am pretty sure this indicates Mn2+ is present as I think it turns purple. This cancels out B and D. This means that the second cation when treated with excess NaOH becomes a whitish solid, slowly changing to brown. I think this is Fe as when exposed to air it rusts and turns brown. I am pretty sure this logic is wrong though. Could someone give me some advice?

Thanks!

 

[Borek]

Please reread the experiment description, as you are ignoring a very important part of the information given (plus, you are confused about which cation changes color during the experiment).

Sadly, the description is not clear and I see two fitting answers. Matter of difference between reddish-violet and purple-violet solution.

 

[TT0]

I think the information I am ignoring is that "a colourless solution collected as the filtrate". This probably means that the cation left in the filtrate is not a transition metal as I think they generally don't give colourless solutions. There are 4 cations that they give us: Zn2+, Mn2+, Mg2+, Fe2+. So I guess Mg2+ is the second cation.

Since excess NaOH is added in the first step, that means a zinc precipitate won't form so this leaves us with Fe2+ and Mg2+ as the first cation. Since the whitish solid turns brown, I am guessing it is Fe2+ as when iron oxidises, it rusts which is brown. There is no Fe2+ and Mg2+ option though so I guess I am wrong. This leaves us with Mn2+ and Mg2+ which is option C. Is this correct? If it is, could you give me your thought process as my logic seems terrible.

Cheers!

 

[Borek]

Think about the filtrate: why on addition of an acid does the white precipitate appear first, then disappears?

 

[TT0]

The filtrate contains the metal cation, hydroxide anions, and sodium cations. When a bit of HCl is added, I can't see any combination that will precipitate other than the metal hydroxide. However, the metal hydroxide is clearly soluble as it was part of the filtrate. That means when HCl is added the metal hydroxide becomes insoluble probably because of solubility products. I have no clue though. I have tried googling but can't find anything relevant. Could you please explain this to me?

 

[Borek]

Again, you are forgetting amphotericity.

 

[TT0]

I have googled about amphotericity. From the knowledge that the most common hydroxides that are amphoteric are zinc and aluminium hydroxide. So I guess the reaction before the addition of HCl, after addition of a bit of HCl and after addition of HCl is:

Before HCl: Zn²⁺ + 4OH^- ↔ Zn(OH)₄^2-
After dropwise HCl: Zn(OH)₄^2- + 2H⁺ ↔ Zn(OH)₂ + 2H_Cl2O
After excess HCl: Zn(OH)₂ + 2H⁺ ↔ Zn²⁺ + 2H_Cl2O

Is this the correct reaction equations?

Since first cation is Mn2+ and second is Zn2+ is A the answer?

Thanks a lot!