[fran008]

In the lab we were provided with an inorganic compound labelled S. The following tests were carried out and the following results obtained :

To a little aqueous solution of S about 4cm3 of aqueous NaOH were added and boiled. A greyish grenn precipitate is formed which i concluded is probably Cr(OH)3

With dilute sulfuric acid followed by 3 drops of potassium  manganate(VII) the purple solution rapidly turns colourless suggesting a reducing agent is present.

Now where I do not know how to interpret any longer

A white precipitate was obtained when a little aqueous S was added to dilute HCl and aqueous Barium chloride

Also a caramel precipitate was obtained after 20 drops of concentrated nitric acid followed by excess aqueous NaOh were added.

The most confusing test is the one of S with aquoeous ammonia. It gave a blue ppt . I would have reasoned out that the precipitate is Cu(OH02 but at the end of the session our lab eacher told us that S was a chromium transition metal!

Could any one help?

[nodyl]

The only answer I can think of is that the white precipatate that was obtained when a little aqueous S was added to dilute HCl and aqueous Barium chloride indicates the presence of sulfate ions (barium sulfate as the white precipitate).

And as you said, it is quite confusing the fact that a blue precipitate was observed when s was added to ammonia solution. I would have expected the same colour when s was added to NaOH. Both reactions (and thus precipitats) must be Cr(OH)3.

Since your lab teacher told you that the inorganic compond was a chromium compound, Salt s is Chromium(III) sulfate - Cr2(SO4)3

Hope that this was a little bit of help for you although it is quite confusing the reaction between chromium and ammonia solution.

[Alberto_Kravina]

Was the precipitate that you obtained with ammonia soluble in excess NH₃?

[fran008]

Yes it was soluable, giving a bluish solution. I would have concluded it was Copper, but from the other tests it clearly was not!

I've askedpermission to go to the lab and check out the experiment again!

Will keep u informed if i had any error!

[fran008]

From the retake of my experiment it seems I had a mistake. this time, a greyish green precipitate insoluable in excess was formed showing that the ion is Cr 3+

Hence the nature of S is Cr2(SO)4. But could anyone explain to me what happens when concentrated nitric acid followed by NaOH dilute was added to s? didnt quite get that part!

Thanks!

[plu]

I believe your "caramel" precipitate would be Cr(OH)₃, the Cr³⁺ ions resulting from oxidation of the original Cr²⁺ ions in compound S by the concentrated nitric acid.

[Alberto_Kravina]

the Cr3+ ions resulting from oxidation of the original Cr2+ ions in compound S by the concentrated nitric acid.

I don't think that the original sample contained Cr²⁺