Sciencemadness Discussion Board - Lime green solution...

Lime green solution...

blogfast25 - 23-11-2010 at 10:37

Below is a photo of a few hundred ml of lime green waste solution, by-product of making (NH4)2SnCl6 from pewter metal dissolved in a mixture of strong (15 – 22 %) HCl and 38 % HNO3 and adding about a stoichiometric amount of NH4Cl. The solution is then basically simmered down to almost nothing (during which the green colour clearly intensifies) and then cooled and iced. The hexachlorostannate drops out with yields now of up to 92 % of theo. Replace the NH4Cl with KCl mole for mole and obtain K2SnCl6, also with the same green by-product.

Using pure tin metal (99.9 % as advertised) no green is obtained.

The pewter should contain some antimony (but I’ve not been able to demonstrate that), no copper (verified) and no appreciable amounts of lead (also verified). The HCl does contain small amounts of iron even though switching to a cleaner grade didn’t stop the lime green from appearing.

The colour is quite striking and unusual in inorganic chemistry. Its intensity indicates appreciable concentration…

watson.fawkes - 23-11-2010 at 11:59

The easiest OTC source, to my thinking, for a bit of antimony would be in hardened lead, often used for wheel balancing weights. A local tire shop might spare a few. In the US, bullet casting alloy is readily available in many places and generally has some antimony in it. Along this line, perhaps harder to find but also possible is "linotype" alloy, which is lead-antimony-tin.

blogfast25 - 23-11-2010 at 12:26

Oh but I can get 500 g of > 99 % Sb for 8 quid (plus PP) from a British eBayer (and I probably will) but I’m not sure how this would solve this mini mystery: I know of no lime green antimony compounds, do you? Antimony from the pewter will almost certainly be found in the (NH4)2SnCl6 (IV) as NH4SbCl6 (V), unless it is very much more soluble than the Sn homologue. If the solubility is about the same, Sb may stay in the mother liquor as a minority constituent… There’s a paper freely available on the Net that describes the separation of Sn and Sb using the hexachlorostannates, a malonated Dowex-1 equivalent exchange resin and malonic acid solution as eluant… Sn and Sb are remarkably similar, chemically.

DJF90 - 23-11-2010 at 12:32

How about nickel...? I seem to recall that colour from my school days, [NiCl4]2- IIRC.

[Edited on 23-11-2010 by DJF90]

watson.fawkes - 23-11-2010 at 14:06

Quote: Originally posted by blogfast25

Oh but I can get 500 g of > 99 % Sb for 8 quid (plus PP) from a British eBayer (and I probably will) but I’m not sure how this would solve this mini mystery: I know of no lime green antimony compounds, do you? Antimony from the pewter will almost certainly be found in the (NH4)2SnCl6 (IV) as NH4SbCl6 (V), unless it is very much more soluble than the Sn homologue.

I was thinking that you might try making some NH4SbCl6 directly and see if it's green; nothing particularly sophisticated. If you were in the US I'd recommend Rotometals. They sell Sb in both ingot and shot, but I'm not sure that it's economical overseas.

The nickel suggestion might be it too. I was looking at babbitting metals (for bearings) today. The tin-based babbitts are very much like pewter, and one of them listed had 0.15% Monel in it, which of the Monel alloys not listed, but they're all high-nickel alloys, with varying amounts of copper, iron, manganese, and others.

blogfast25 - 23-11-2010 at 14:17

Yes, NH4SbCl6 can almost certainly be made as above, substituting Sb for Sn, the former also dissolves in aqua regia.

But if that’s right and the pewter (it’s a food grade –from a drinking tankard) contains, say nominally 5 % Sb, then the hexachloroantimonate would have to be mightily coloured for the lime colour to be accounted for. It’s not very likely: it has basically the same electronic configuration as the stannate equivalent: a full d subshell coordinating the chloride ligands..

Still, worth a ponder… The hexachloroantimonate has (or had?) use because the caesium salt is very insoluble.

Nickel is something I'll test for but the colour just seems so ntense...

[Edited on 23-11-2010 by blogfast25]

MrHomeScientist - 23-11-2010 at 14:48

The only thing I've run across in my own experimentation that was similar to that color was copper (II) chloride in HCl - it turns a similar shade of green when the solution is high on Cl-.

But since you tested negative for copper, my post was useless

Just thought I'd mention it in case that sparked an idea for you.

watson.fawkes - 23-11-2010 at 15:56

Quote: Originally posted by blogfast25

Nickel is something I'll test for but the colour just seems so intense...

When you put it that way, you spark a different question: Is the color monochromatic? You don't need a calibrated spectroscope to answer that, just a basic one. A simple prism or diffraction grating will suffice. If you tilt the optic around, you should see only one color-image of the bottle, not more than one and not a streaked-across image.

Arthur Dent - 23-11-2010 at 17:47

Indeed the green color of your solution is intense. I've made some Nickel Chloride by dropping a few early 70's canadian 5 cent pieces (99.9% Ni) in conc. HCl and almost immedioately, the acid turned light green, and the finish of the nickels wasn't even affected yet. Now the pieces are completely dissolved and the solution is the deepest dark emerald green you've ever seen... really beautiful.

So I think it doesn't take very much nickel to give an acid solution a green tinge.

Robert

woelen - 24-11-2010 at 04:04

The color of the solution is not really like the color of nickel(II) in solution. That is really green and not yellowish green. The complex NiCl4(2-) is not very stable and even a fairly strong solution of NiCl2 in HCl has a more green color.

I can imagine it contains nickel, but its not the only contamination. Molybdenum may also be one of the constituents. Molybdenum can give intense yellow/green solutions.

Can you add some solution of Na2SO3 to the yellow solution and boil? When molybdenum is present, then you get a blue color.

blogfast25 - 24-11-2010 at 07:21

@Watson: I do have a primitive spectroscope but more to the point I also have a DVD-R: perfect for this purpose. Will test later on. To the naked eye this doesn’t look monochromatic to me. I’ve always suspected some complex and they don’t usually generate monochromatic colours…

@woelen: molybdenum, eh? What the hell would that be doing in pewter?? I will certainly test this but with metabisulphate, that also generates SO2 which I’m guessing is the purpose of the sulphite?

************

Well, the green is certainly not monochromatic, rather it’s a wide green streak in my spectroscope.

Here’s some test tube tests with the liquid:

1: control; solution as such.

2: with small amount of sodium metabisulphite after boiling. Slight intensification of colour due to boiling in. No blue.

3: with a larger amount of sodium metabisulphite after boiling. Slight intensification of colour due to boiling in. No blue.

4: with weak NH3 solution: Sn(OH)4 drops out, no blue Cu2+/NH3 complex (as expected).

5: with 5 M NaOH: Sn(OH)4 drops out but the green hue is clearly maintained; so it can exist in alkaline conditions too. That rather rules out Ni2+ too.

The only conclusion so far is that the 'waste' solution still contains appreciable amounts of hexachlorostannate: the ammonium salt appears very soluble, even in the cold...

[Edited on 24-11-2010 by blogfast25]

kmno4 - 24-11-2010 at 09:02

Take sol.1 and make it acidic with HNO3, next add H2O2, boil it - any changes ?
Take sol.1 make it strongly alkaline with NaOH, filter out Sn salts and to clear solution add H2O2 and boil it - any changes ?

blogfast25 - 24-11-2010 at 11:01

Quote: Originally posted by kmno4

Take sol.1 and make it acidic with HNO3, next add H2O2, boil it - any changes ?
Take sol.1 make it strongly alkaline with NaOH, filter out Sn salts and to clear solution add H2O2 and boil it - any changes ?

@kmno4: done!

The series above is with addition of reagents but BEFORE boiling:

1: control; solution as such

6: control plus HNO3 plus H2O2. Solution darkens. Possibly typical of Fe3+ or Fe.NO (2+).

7. Test tube 5 filtered and H2O2 added. Same colour as 6.

Above, AFTER boiling 6 and 7:

6’: after boiling 6 the green re-appears. I’ve seen this many times when making the hexachlorostannates (NH4 or K) from pewter: at some point during the simmering process the colour turns lime green immediately…

7’: the alkaline solution does not change colour upon boiling of 7…

I really suspect iron but what compound?

[Edited on 24-11-2010 by blogfast25]

kmno4 - 24-11-2010 at 11:22

I think that the green something could be some (probably chloride) complex of Cr(III).
Such complexes are intesively green colored.
If 7 really contains chromates - additional tests are needed.

[Edited on 24-11-2010 by kmno4]

not_important - 24-11-2010 at 12:26

I'm going for iron, too. Hard to say for certain, almost certain a chlorine containing complex but it may have other things in it. Doesn't take too much Fe to give colours, and it's everywhere. A complex might interfere with some of the simple tests for iron, and Fe(III) hydrated oxide goes colloidal fairly easily when the Fe is not too concentrated so simple filtering might not removed it. Indeed, when purifying compounds that might be contaminated with iron, a common method involves oxidising the iron to Fe(III), then using a low solubility oxide/hydroxide/carbonate of the cation metal to push the pH alkaline and precipitate the Fe(III) out on the surface of that so as to be more easily removed.

Cr would seem to be low probability but a check might be in order.

BTW - at lower concentrations Cu and Ni look similar when aq NH3 is added - the blue colour from each is similar.

DougTheMapper - 24-11-2010 at 17:40

It's probably iron. I have obtained similar results while reacting shavings of Mg with HCl to yield MgCl2. An identical brilliant lime green color formed which I had assumed was dissolved chlorine. I was surprised, however, when it remained after the solution was boiled so I suspected contamination of some kind - probably from the steel cutting tools used to machine the Mg. The only other known elements in the Mg alloy were trace amounts of aluminum and manganese.

kmno4 - 24-11-2010 at 21:59

Quote: Originally posted by DougTheMapper

It's probably iron. I have obtained similar results while reacting shavings of Mg with HCl to yield MgCl2

Very often Mg shavings are coated with Cr (to prevent oxidation).
Besides - appropriate tests for Fe (II/III) are commonly known and very easy to perform.

woelen - 25-11-2010 at 00:43

If you want to confirm that the color is due to iron, then add a solution of K3Fe(CN)6 or K4Fe(CN)6 to a mix of the green and brown solution. A deep green or blue color confirms iron. This test is very sensitive! The color is due to formation of prussian blue.

Another very sensitive test is to add the brown solution to a solution of a thiocyanate. This will lead to formation of a bloodred complex, thiocyanatoferrate(III).

kmno4 - 25-11-2010 at 00:55

Yes, but one must remember that tested solutions should be acidic if you want to get reliable results.

blogfast25 - 25-11-2010 at 06:07

Thanks ALL for your responses.

Tests with KSCN had been carried out before (Fe was always my prime suspect) but were negative, to my surprise. I will now repeat a thiocyanate test on solution 6. Bear in mind that the FeSCN(2+) complex is a fairly weak one: anything else complexing the Fe more strongly may obscure it.

I’ve unfortunately no ferro or ferri hexacyanides.

Recapping a bit:

Sources of Fe are: HCl grade used, contaminate Fe in pewter, cutting tool (a hacksaw).

Green colour appears only with pewter, not with 99.9 % tin metal. With the latter a slightly yellow solution is obtained, pointing to naked Fe3+.

Switching from a fairly iron rich HCl grade to a much purer one made no difference.

Although this is essentially still nothing more than a hypothesis, it would seem to point to a complex of Fe3+ in the presence of Sb (V). I’ve ordered some 99.2 % antimony (of course that too maybe Fe contaminated) for some tests.

One way to almost completely eliminate Fe (while also avoiding Sb) would be to make some (NH4)2SnCl6 from pure tin and to recrystallise it carefully, then treat the dried product with acetone to leach out any FeCl3. The Fe, Sb free hexachlorostannate could then be reduced with Al/HCl and reoxidised with HNO3 (or H2O2) to check if the lime gree colour appears or not.

Quick update:

Adding 1 M KSCN to test tube 1 (the lime green control) yields a very strong positive for Fe3+, unmistakably so. So there is definitely quite a bit of Fe (III) in the solution. Question is, what Fe (III) compound or complex has a lime green colour? I still suspect antimony has something to do with this and will run some tests when I get my consignment of Sb...

[Edited on 25-11-2010 by blogfast25]

kmno4 - 25-11-2010 at 10:49

Fe does not have to be the only metal present in this solution.
It also does not explain why solution after precipitation Sn salts is still gren and why this green colour turns to yellow after H2O2 treatment.
I would strongly recommend to do test for Cr (via CrO5) in solution number 7. All tests you performed tell me that it must be Cr.
Besiedes, if you have some fluoride (KF, NaF...) you can mask Fe(III) directly in solution 1. Then you can see if green colur still remains. Fluoride gives more stable complex with Fe(III) than thiocyanate.
After masking sol.1 with F(-) and adding larger excess of SCN(-), it is quite possible that you will get (after warming) dark violet-bule Cr(III)-SCN(-) complex.
Of course, I can be wrong but experiment is a king.

blogfast25 - 25-11-2010 at 12:32

@KMnO4: yes, chromium is still in the race, I’d say. I have no soluble fluorides right now though. Not sure by what you mean by ‘via CrO5’? You mean precipitate the Cr as an insoluble chromate (lead for instance)... Please explain… And Cr (III) forms a thiocyanate complex? News to me…

Edit: oh, I see: CrO5: Cr (VI) peroxide. That was kind of tested for in tube 2, wasn't it?

Carrying on:

So, about 6 g (theor.) of (NH4)2SnCl6 were prepared from tin (99.9 %), HCl 15 % and HNO3 38 %, as per usual. The solution assumes a light yellow colour right from the start and after simmering and collection of the product a yellowish supernatant solution was obtained and tested for iron (III).

Before addition of KSCN 1 M:

8: supernatant liquor from an (NH4)2SnCl6 preparation from tin (99.9 %) metal: slightly yellow with a greenish tinge…

9: HCl 15 w% grade used in the preparation: clear and colourless.

After addition of KSCN 1 M:

8’: tube 8 reacting strongly positive for Fe3+

9’: tube 9 reacting weakly positive for Fe3+

In short, there’s Fe3+ in there too. Due to the iron contamination of the HCl used it’s impossible to tell what the main source is but the HCl is a clear suspect…

The (NH4)2SnCl6 was rinsed thoroughly with iced DIW and turned snow white. I saw no good reason to recrystallise it as it appears to be iron free. A KSCN test confirmed that.

When I’ve made some Sb compound, I should be able to test that the complex may form only in the presence of both Sn and Sb.

[Edited on 25-11-2010 by blogfast25]

woelen - 25-11-2010 at 23:36

Chromium does form a complex with thiocyanate, but it is not red. It has a deep purple color, but one needs quite some chromium and thiocyanate at high concentration to see this complex. It hardly interferes with the very sensitive test for iron(III).

'Naked' iron(III) is not yellow, but it is (nearly) colorless. Only at very high concentration, iron(III) has a weak brown/pinkish color. Only when chloride is added, a yellow color appears and due to hydrolysis in aqueous solution the color turns more like brown.

http://woelen.homescience.net/science/chem/solutions/fe.html

If you add either phosphate or fluoride to your yellow solutions, then you'll completely mask the iron(III). Both phosphate and fluoride form a rather stable and colorless complex. If you have phosphoric acid or some soluble phosphate, just add some of this to your solutions. The color of the iron(III) then is masked and if you still have some color after addition of that, then this is due to some other metal.

blogfast25 - 26-11-2010 at 06:20

@ woelen: sorry, I did mean FeCl3 which does have a distinct, almost glowing, yellow colour. All my solutions here are born in a combination of HCl and HNO3 and are naturally high in Cl-. Solid FeCl3.6H2O is yellow too.

As regards kmno4’s idea of testing for Cr as CrO5 (VI – peroxide), I doubt very much if Cr (III) – that is the most likely oxidation state of Cr if there really is any present – can be oxidised in aqueous solution with HNO3/H2O2 in acid conditions as he suggested. A couple of quick tests yesterday seemed to suggest that.

It maybe possible to oxidise chromite – Cr (III) in alkaline conditions – with H2O2 and heat to Cr (VI), I’ll check that this afternoon. But industrially chromates are the result of oxidising Cr (III) with alkali fusion and in the presence of an oxidiser (although air oxygen can suffice).

Another reason why I now believe Cr is far less likely than an Fe complex as the cause of the colour lays in test tube 5: control plus strong alkali gives white precipitate of Sn(OH)4 and lime green supernatant liquid. While ‘in theory’ it’s possible that the suspected Cr (III) had gone into solution as chromite, in reality that’s not very likely: the pH in solution wasn’t measured but it can’t be that high otherwise the Sn(OH)4 would have redissolved as stannate (Sn(OH)6 (2-)) too. By rights the Cr (III) should have dropped out as Cr(OH)3.nH2O.

So I think it’s probably possible to eliminate Cr from the list of suspects with some more tests along those lines…

[Edited on 26-11-2010 by blogfast25]

woelen - 26-11-2010 at 08:15

The CrO5 test is not easily performed here. CrO5 (actually, CrO(O2)2, the diperoxo complex of hexavalent chromium) is not easily made in your solutions. H2O2 does not oxidize chromium(III), not even in the presence of a strong oxidizer like HNO3. The only oxidizer which can do the job is peroxodisulfate, S2O8(2-), with silver ions as a catalyst. This combo quickly oxidizes chromium(III) to chromium(VI) in acidic solution. Next, you have to be sure that all peroxodisulfate is destroyed before any H2O2 is added, otherwise the peroxodisulfate also oxidizes H2O2, giving sulfate and oxygen.

It indeed is surprising to see such a strange lime green color. The only iron(III) complex I know which has such a lime green color is the trisoxalato complex. I have some of that material, the ammonium salt:

http://woelen.homescience.net/science/chem/compounds/ferric_...

It can be made by dissolving iron(III) hydroxide in a solution of excess oxalic acid and adding ammonia to neutralize excess oxalic acid.

kmno4 - 26-11-2010 at 09:18

"The CrO5 test is not easily performed here. CrO5 (actually, CrO(O2)2, the diperoxo complex of hexavalent chromium) is not easily made in your solutions. H2O2 does not oxidize chromium(III), not even in the presence of a strong oxidizer like HNO3"

That is why I suggest to use solution no. 7
It is alkaline (strongly I suppose) and in these conditions
Cr is in Cr(OH)4 (-) form. This hydroxo comlex is easily attacked by H2O2 to give CrO4(2-).
Next step is to make it acidic (HNO3) and again add H2O2.
If Cr is present, under these conditions it will give CrO5.
It is unstable and this test should be made on cold.
Additionally, you should add (and shake with) some extrahent : diethyl ether, isoamyl or buthyl alcohol etc.. to extract CrO5.
Organic layer will be deeply blue colored if Cr is present.
Optionally you can use persulfates instead of H2O2.

woelen - 26-11-2010 at 10:43

The latter is not true. Peroxodisulfate does NOT give peroxo complexes. It reacts with hydrogen peroxide, giving oxygen, where hydrogen peroxide acts as reductor and the peroxodisulfate acts as oxidizer.

not_important - 26-11-2010 at 10:51

Dump some Al foil bits into some of the solution, to reduce the transition metals down to lower oxidation states. Add aqueous ammonia to make alkaline, then evaporate everything and finally heat it hot enough to drive off ammonium salts. Then check the residue for transition metals, sans interference from excess ammonium and chloride ions.

Note the Al will give hydroxide/oxide that serves as a carrier for trace metal ppts. As example if the heated solids, after cooling, were to be treated with a small amount of H2SO4 to dissolve the solids, then strong aq NH3 added, Ni and Cu will go into solution while Fe and other metals that don't form NH3 complexes will remain with the Al hydroxide/oxide.

[Edited on 26-11-2010 by not_important]

blogfast25 - 26-11-2010 at 13:22

Well, I carried out two tests on Cr (III) solutions, one an old Cr alum solution, one a freshly prepared Cr3+ solution (from dichromate + alcohol), both free of iron. I got some strange results. I alkalised both strongly to dissolve as chromite. Then I added 3 % H2O2 and heated. From the first solution Cr(OH)3 precipitated but the supernatant solution was yellowish.

The second one got the same treatment and turned… brown! To both was then added some conc. sulphuric acid and the first solution then also turned brown:

I transferred a bit of the left hand side solution to a test tube and added a solution of Mohr’s Salt (ferrous ammonium sulphate), the solution turns green (photo doesn't do justice to colour, the green is lighter to the naked eye):

This strongly suggests some oxidation took place during the alkaline H2O2 treatment: Cr (VI) formed oxidises Fe (II) to Fe (III) and is reduced back to Cr (III)…

The problem with not_important’s idea is that ammonia will precipitate also the Sn, it would be nice to just gather all the D-blocks without the Sn… But that might not matter much…

[Edited on 26-11-2010 by blogfast25]

blogfast25 - 28-11-2010 at 09:15

Well, it looks like I’ve proven my own hypothesis regarding the cause of the lime green colour, namely an unknown complex involving Fe (III), Sn (IV) and Sb, fairly conclusively wrong. Here’s what I did:

1. 0.9 g of Sn (99.9 %) and 0.1 g of Sb (99.2 %) together were given the usual HCl/HNO3 mix + NH4Cl treatment to simulate making the hexachlorostannate from an Sb-rich pewter. Iron was supplied from the 22 % HCl. No green was ever observed, instead the solution was a strong yellow (see below, tube 10).

2. an attempt to make NH4SbCl6 resulted in an unknown but water soluble Sb compound. Here, 5 g of Sb (99.2 %, a coarse powder) was given the same treatment as I would pure tin but the result was different: no well formed crystals appeared and at the end of the simmering a yellow syrupy resulted. Overnight that solidified into a hard, opaque mass, clearly looking like a hydrate, very different from (NH4)2SnCl6. Is this a (III) or a (V) compound of Sb? Must look into that…

3. Some of the hard mass of 2. was combined with an more or less equal amount of iron free, earlier prepared (NH4)2SnCl6 and dissolved (with heat) in about 20 ml of iron bearing HCl 22 %. It dissolves into a yellow perfectly clear liquid. Which was then reduced to just the few ml seen below (tube 11).

The photo below summarises the results:

10: lime green control

10 and 11: solutions obtained from experiments 1. and 3.

I might corroborate this by reducing the solutions with Al and reoxidising with HNO3…

Assuming my results here are correct then it’s reasonably to assume that something else in the tankard pewter is causing the lime green colour.

Further future experiments should include working with completly Fe free HCl and adding a small amount of a Cr (III) compound to the experiments above...

[Edited on 28-11-2010 by blogfast25]

not_important - 28-11-2010 at 10:48

Cold be both Fe and Cu, the 1st being a common contaminate and occasionally used in base metal alloys, the 2nd definitely being used in some such alloys. The high concentrations of other ions might be messing up the tests being used.

Re my suggestion - the heating to drive off ammonium salts would likely reduce the solubility of the hydrated tin oxides, a leach with cold dilute H2SO4 may just selectively pull ot the transition metals while leaving most Sn (and Sb) behind.

Like how you are trying to figure this one out, theory and experiment. While modern equipment might quickly give the answer, the old methods can do the job while helping improve lab techniques too; useful for those with limited funds.

blogfast25 - 28-11-2010 at 12:35

Cu had already been eliminated basically on another thread: the pewter doesn’t appear to contain any significant amounts (traces cannot be excluded, of course).

Regarding modern equipment, trust me that if I had some ‘CSI Miami style’ stuff I’d have peeked by now! It’s only because the colour is really unusual in inorganic chemistry that I’m still pursuing it…

The objection I have to reduction with Al followed by precipitation with strong ammonia is that you don’t really separate anything from anything: ammonia isn’t strong enough to dissolve, at least AFAIK, Sn, Al and Cr (assuming present) as anions. So basically you’ll get a mass of everything plus alumina that should be OK for XRF. Unless selective leaching as you suggest works. Instead I’ll reduce with Al and precipitate with strong NaOH: at least Sn, Al and any Cr will be separated from the Fe.

The antimony compound I prepared is possibly slightly hydrolysed HSbCl6, a strong acid that, acc. Holleman, can be crystallised as HSbCl6.4.5 H2O. This is a very strong acid, probably highly soluble in water (as consistent with observation) as indicated by the existence of a hydrate. I’ve been preparing a second batch now, using a slightly modified procedure…

Arthur Dent - 29-11-2010 at 04:59

blogfast25 said: "trust me that if I had some ‘CSI Miami style’ stuff I’d have peeked by now!"

http://orbitingfrog.com/blog/2008/07/02/make-your-own-spectr...

and with that crude apparatus, maybe a flame test could give you info on the trace elements of your solution.

I have some spare platinum wire if you want to make a little flame test wand...

Robert

blogfast25 - 29-11-2010 at 06:44

@ Arthur:

I’ve got one of those (but mine is better – reflector diffractors give better results than ‘see through’ diffractors).

The problem is that trace elements, even when flamed, often only emit very weak lines that are hard to see. For that you often need very hot flames, a fairly high feed of solution into the flame or even better (but harder to do), electrical excitation (electrical arcing) of the trace element atoms. Also you need to be able to precisely measure the wavelength of any observed lines to be able to tag them to a specific element. I’ve got blueprints for a home made spectrogoniometer (allows determination of wavelength) in my drawer for months now but haven’t gotten round building the damn thing yet. And the instrument doesn’t solve the problem of sufficient atom excitation either.

So, dear Watson, it’s not an open and shut case…

BTW, if you want to have a go at the spectroscope you linked to, don’t use ‘an old CD’ (that’s yesterday’s homemade spectroscope technology – LOL!), instead use an R-DVD: the datalines on R-DVDs are much closer together leading to much better resolution (about 1 nm for a good build). Mine uses an R-DVD reflector diffractor…

Arthur Dent - 29-11-2010 at 07:07

Thanks for the tip about DVDs, I didn't know that. I do know that the very small plastic-encased lens right in front of the laser diode in some CD/DVD drives (like the PHR803) is in fact a diffraction grating. Unfortunately, it's about 1.5mm x 1.5mm, too small for this particular application. This site:
http://www.rainbowsymphonystore.com/difgratfilsh.html
sells diffraction gratings in flexible sheets, and the price is right!

But definitely, I'll try to use an old DVD for this project.

One of my "other" hobbies is playing around with lasers and laser optics, still hoping to get my hands on a nice (cheap) multiline argon laser some day. :cool:

Robert

[Edited on 29-11-2010 by Arthur Dent]

blogfast25 - 29-11-2010 at 09:20

Well, well, well: chromium is back on the top of the list of suspects with a simple experiment, similar to the one above. Tin filings (a ‘pinch’) and Cr filings (from aluminothermic chromium – a smaller ‘pinch’) were combined with 15 ml of HCl 22 % and 5 ml of HNO3 38 %. Reaction starts immediately and a lime green colour appears.

The immediate appearance is a slight difference with the green appearing when dissolving pewter in ‘aqua regia’: there it appears much later on. But the physical form of the Cr in this test is also very different.

The solution (appears a little darker than to the naked eye):

A small amount of the solution was then mixed with yellow HCl 22 % and it was possible to obtain a perfect colour match: see 1 (control) and 12 (matched solution):

In itself this still doesn’t prove anything conclusively and only proving the pewter actually contains Cr will do. But it does quite fit the facts: due to the concentrating of pewter AR solutions to crystallise out the hexachlorostannate, as well as the concentrating of an iron bearing HCl solution during the same step, relatively little Cr would be needed for the yellow FeCl3 and green CrCl3 to combine to a lime green colour.

O/T:

It’s that cold here that despite some electrical heating a saturated solution of oxalic acid on my shelves started to crystallise out! California it ain’t…

blogfast25 - 29-11-2010 at 09:30

Quote: Originally posted by Arthur Dent

Thanks for the tip about DVDs, I didn't know that. I do know that the very small plastic-encased lens right in front of the laser diode in some CD/DVD drives (like the PHR803) is in fact a diffraction grating. Unfortunately, it's about 1.5mm x 1.5mm, too small for this particular application. This site:
http://www.rainbowsymphonystore.com/difgratfilsh.html
sells diffraction gratings in flexible sheets, and the price is right!

Robert

[Edited on 29-11-2010 by Arthur Dent]

Robert:

The place to be for homemade spectroscopy is here (click around, it’s full of useful info for a beginning spectroscopist!):

http://astro.u-strasbg.fr/~koppen/spectro/mk2be.html

That’s the model (well, three of them!) I built using an R-DVD. It resolves the famous sodium D doublet (on a good day!) If set up and used properly emission lines are very clear and narrow. But for serious spectroscopy you need good excitation and a goniometer to measure angles and thus calculate wavelengths. My planned instrument will measure angels down to 0.1 (360) degree. That’s about 1 nm. Commercial instruments like that should not cost much more than about 200 bucks I’m guessing.

Arthur Dent - 30-11-2010 at 10:02

Shweeet!

I guess I have my work cut out this weekend, really cool project! I happen to have a nice black cardboard box just right for the job.

Glad the mystery trace impurity got identified!

- Robert

not_important - 30-11-2010 at 10:26

If you can co-display, as two bands, the sample spectrum and that of a neon light - cost around a half Euro, you can do a decent job of wavelength identification. Neon hass a number of lines throughout much of the visible and so makes a decent low cost reference.

blogfast25 - 30-11-2010 at 13:10

@not_important:

You mean a baby room night light or something like that? I've used the Na D line, Li scarlet line and a 617 nm red laser as references. Never thought of Ne though...

Back to the job at hand: Cr in the lime green solution. Despite a rather elaborate effort I've still not been able to conclusively prove the green is that of Cr3+.

50 ml of the solution was neutralised with NaOH, to as close as possible as pH = 7. This solution is a light sky blue. Filtered and dried the precipitate (mainly SnO2) till it was a paste. About 1 g of it was mixed with an excess of Na2CO3.10H2O and quite a bit of NaNO3. The crystal water was first removed by gently heating the mix in a ceramic crucible and the dry was mix then fired at max heat for 30 mins (propane Bunsen). The bottom of the crucible was glowing bright red-orange.

After cooling the frit showed a yellow tinge at the bottom and was removed with small aliquots of 50 % H2SO4 (quite a bit of NO2 was surprisingly generated at that stage…) The slurry (the SnO2 remains as solid) was then filtered.

But the filtrate didn’t test positive for Cr: with H2O2 (3 %) the colour changes to yellow, adding lead nitrate to it yields a white precipitate, not yellow lead chromate. Adding lead nitrate to the solution containing peroxide also gave a white precipitate.

Tomorrow a test with NaOH – KClO3… and a higher quantity of the precipitate.

not_important - 30-11-2010 at 14:17

Neon bulb, plus a series resistor. Bulb like this : http://www.weisd.com/store2/EIKA1A.php

sample spectra including Ne here http://www.amateurspectroscopy.com/color-spectra-of-chemical... notce that there are strong lines for most regions of the visible.

Someone else mentions the concept here http://chemcom.tripod.com/cTakingSpectraPictures.htm

I'd be surprised if there's Cr in that alloy, as that doesn't seem to be common, but if the tests indicate it is there then I'll accept that. Perhaps the Cr got in through salvaged metal, a small amount of Cr such as from plating on metal or plastic could dissolve in the alloy and as it's not lead get past normal tests.

Does the white precipitate with lead nitrate match the characteristics of the chloride or sulfate? (check solubility in hot water) Also boil the ppt with Na2CO3 solution, filter off lead, check solution for other complex anions.

kmno4 - 30-11-2010 at 15:34

Quote: Originally posted by blogfast25

After cooling the frit showed a yellow tinge at the bottom and was removed with small aliquots of 50 % H2SO4 (quite a bit of NO2 was surprisingly generated at that stage…)

In this way you probably reduced all, or almost all hypotetical chromates.

Quote:

But the filtrate didn’t test positive for Cr: with H2O2 (3 %) the colour changes to yellow

So what was the colour of the filtrate ?
blogfast25 - I think you should exercise converting Cr(III) to Cr(VI) via NaOH/H2O2 methode and convertig it further to CrO5.
I did this several times and it always works if done properly.
If you have already Cr in solution then all "hot methodes" are simply waste of time.
If content of Cr is small - you possibly will not see any blue colour of CrO5 in such diluted solution. Then you must use extrahent for CrO5: to acidic chromate (max 0,1 M) at 10 C (or less) add ~1cm3 of diethyl ether or butanol (as far I remember butyl acetate will not give good results) and next add H2O2 (excess is better than too small amount). Shake it at once - if organic layer is blue, then Cr is present. If not - most probably it is not present.
If there is no colour then you can add small amount of any di/chromate (soluble) and see if blue colour will really appear. If yes, then conditions are proper. If not.... something is wrong.

blogfast25 - 1-12-2010 at 07:31

@ kmno4:

OK, back to the drawing board.

The control solution in a test tube was first alkalised with 5 M NaOH: SnO2 precipitates from it, then redissolves to weak green solution (stannate and chromite). Weak H2O2 added: colour change to yellow-orange is very marked, this clearly indicates chromate formation. Then re-acidified with HNO3 38 %: no notable change (pH is now < 1). Added Pb(NO3)2, alas no yellow lead chromate forms, instead some PbCl2 slowly dropped out.

To half of the acidified solution more H2O2 was then added, no change occurred though…

I believe to see PbCrO4, I’d have to convert all to nitrates, washing out any chlorides, then re-apply the procedure…

There’s another elephant in the room that everyone seems to have missed, even kmno4 (;-)): MnO4(-) according the reduction potential series should be capable of comfortably oxidising Cr (III) to (VI). Titration of a test tube of the lime green solution using 0.1 N KMnO4 confirms that: the MnO4(-) reacts away swiftly and the green colour of Cr3+ disappears… Could even be the basis of a volumetric determination of the Cr...

So it must be chromium alright. I’m 100 % sure, as the early tests confirm, that it’s part of the alloy and didn’t come flying in from a cutting tool or post-digestion. Weird…

I’ve another piece of pewter, an old style hip flask (ideal for the drink-driving ‘hip flask defence’! – ‘I just took a big swig from my steeping alcohol filled hip flask when getting out of the car, officer, you’ve got nothing on me!’), I’ll see if that contains any soon…

[Edited on 1-12-2010 by blogfast25]

blogfast25 - 1-12-2010 at 08:02

@ not_important:

Like the neon bulb but its wattage is seriously low. Thanks for the interesting links…

The one about digital photography of spectra is very useful: these things are really seriously difficult to capture well. Here’s my best shot of a saver bulb spectrum with an R-DVD based spectroscope with an 8 MPixel camera:

To the naked eye the lines are much finer.

Where the link goes a bit off the rails is where he writes about the ‘calibration scale’. I tried several times something similar with very conflicting results until thick as two short planks here remembered an experiment carried out a few weeks before for the determination of the wavelength of a pointer-style red laser. Below is shown the principle of the method: Top: a diffraction grating (a CD and subsequently also an R-DVD). Left: path of incident ray with laser in left hand bottom corner, right path of reflected/diffracted ray for order n=0 and n=1 (this is a real life situation):

The wavelength was determined from (a set of measurements with varying beta) the angles alfa and beta, in turn calculated with Pythagoras from the lengths O’O, O’B and O’C. It doesn’t take much to understand that the scale obtained is highly non-linear. With photos the situation is even more complicated because the camera plane and the reflector/diffractor plane are unlikely to be exactly parallel…

The only real way to measure the angles easily is by means of a Bunsen-Kirchoff style ‘swivel’ spectrometer but with a diffractor instead of a prism (prisms still work too of course…).

So the author’s points on calibration seems a mix of wishful thinking and confirmation bias to me…

[Edited on 1-12-2010 by blogfast25]

watson.fawkes - 1-12-2010 at 12:42

Quote: Originally posted by blogfast25

With photos the situation is even more complicated because the camera plane and the reflector/diffractor plane are unlikely to be exactly parallel…[...]
So the author’s points on calibration seems a mix of wishful thinking and confirmation bias to me…

I don't see any fundamental reason why it can't be made to work, with appropriate caution. I should note that I didn't see on that page anywhere he was trying to do calculations of wavelength, nor did he make any claiming of linearity with the picture of the ruler. I think you may be reading something into the narrative, which seemed to me just about image capture.

What I think is quite clever about the idea is that you can photograph quite weak intensities with this scheme by means of long exposure times. Getting a picture of the scale itself can be done with a short flash of a white LED at some point during the exposure. By manipulating these two times, you can get both adequate signal and adequate contrast.

As you point out, the scale will be nonlinear and require calibration. Shooting an NE-2 lamp spectrum, though, can provide that each time the device is used. You need to know the angle from the principal axis to the scale (canonically 90°

and the distance from the optic to the scale. Three lines of known wavelength suffice. If you don't know the spectrum grating spacing, you'll need a fourth line. You might want to measure it anyway, since it has a non-zero temperature coefficient. The law of cosines gives slightly easier arithmetic, FWIW, although you've likely more-or-less already derived it already.

As for the camera's orientation to the scale, that won't matter if you're correlating the spectral lines to the rulings on the scale directly. Coincidence there is what you're looking for. If you feel the need to interpolate, you might have to determine distance and orientation from camera to scale, but the easiest way to deal with this is to mechanically lock the scale at right angles to the axis of the optic, look only at the foot of the perpendicular, ensure adequate distance, and then rely on the sinθ ~ θ small-angle approximation. In other words, by constructing the apparatus properly, you can assume that your image is linear to the scale. And this is just for interpolation, recall, so any error introduced here is second-order.

As to the quality of your images, I suspect vibration. Human vision compensates for vibration quite well. Mechanical systems need care. It's one of the reasons for fancy and expensive optical tables.

Lastly: Traditional spectroscopy was always done in very dark rooms.

blogfast25 - 1-12-2010 at 13:38

@ watson:

Yes, I’ve always set mine to the longest exposure times my camera allowed.

As regards the calculation you suggest, it’s principally exactly what I used for the measurement of the wavelength of the laser.

Re. the non-zero temperature coefficient, I didn’t think of that: R-DVDs and CDs are made of some plastic or other, so quite susceptible to dilation by temperature.

I used a tripod and set timer (the camera takes a specified number of shots without operator intervention) to avoid all camera shake: I think the main source of line widening may simple be due to limited resolution of the camera.

Dark rooms? Yes of course: not so easy to achieve when you have a de facto light source in there…

A few facts:

The actual wavelength, using the definitions used in the photo above is calculated from:

n λ = D ( sin β - sin α )

with n the order of the spectrum and D the distance between adjacent lines on the diffraction grating.
For ordinary CD ROMs, D = 1.6 μm (micrometer) and for DVD-R (single layer, 4.7 GByte storage capacity) it is 0.74 μm.

The laser light measured was 659 nm (with a sample standard deviation of 9 nm).

[Edited on 1-12-2010 by blogfast25]

watson.fawkes - 1-12-2010 at 20:36

Quote: Originally posted by blogfast25

I used a tripod and set timer (the camera takes a specified number of shots without operator intervention) to avoid all camera shake: I think the main source of line widening may simple be due to limited resolution of the camera.

There's a remarkable amount of vibration in an ordinary room unless you take special precautions.

There's also, I see now, the possibility that there's a diffractive spreading going on having to do with one of the restricted pupils in the camera. What kind of camera is it?

The imaging chip is also a candidate for strange monkey business, like internal side scattering that's ordinarily never a factor in snapshots, where there aren't huge order-of-magnitude jumps in intensity from pixel to pixel.

blogfast25 - 2-12-2010 at 07:32

Going back to the lead chromate, I now believe it didn’t form because the conditions weren’t favourable at such low pH. The solubility product of PbCrO4 is Ks = 2.8 x E-13, so not massively small (a bit unusual for something that occurs as a mineral) and low pH pushes the CrO4(2-) to HCrO4(-) and Cr2O7(2-): there probably wasn’t enough CrO4(2-) at pH < 1 to exceed Ks…

Re the camera, it’s a box standard Canon PowerShot A560, zoom 4X, 7.1 MPixels. You can choose from a number of preset settings, such as exposure time (on a relative scale). Decent but not professional and certainly not designed for shooting spectra! Basically an ‘instant’ digital camera with some extra options. I use it for all pix on this forum…

What’s even stranger is that I started off using a proper non-digital camera of very high quality: a semi-professional Zeiss Ikon, about 60 years old but in perfect condition). It’s one of those where the viewfinder looks right through the lens, until you snap and the mirror quickly veers away briefly, exposing the analog film. Just about everything can be set as required by the operator.

Well, I set up my camera, could see the spectra perfectly (as to the naked eye), took some twenty shots varying the settings and when the film was developed… nothing, nada, zilch! Not a single line… I used box standard Kodak middle of the range ASA film for that. Explain that to me!

I was also wondering what’s the lower limit of D for a VIS diffraction grating, the R-DVD is 740 nm. There’s a point where D is so small that light waves basically ‘wash’ over the lines, like sea waves over a small object or light waves over an atom or simple molecule… But before that point is reached, smaller D leads to better resolution, see CD v. R-DVD…

watson.fawkes - 2-12-2010 at 10:59

Quote: Originally posted by blogfast25

What’s even stranger is that I started off using a proper non-digital camera of very high quality: a semi-professional Zeiss Ikon, about 60 years old but in perfect condition). [...] Well, I set up my camera, could see the spectra perfectly (as to the naked eye), took some twenty shots varying the settings and when the film was developed… nothing, nada, zilch! Not a single line… I used box standard Kodak middle of the range ASA film for that. Explain that to me!

The dynamic range of the eye is about 90 dB, so it's certainly possible that the film remained underexposed. What was the longest exposure time? My immediate intuition is that you'd need an external exposure timer, but that's only a guess.

blogfast25 - 2-12-2010 at 12:24

Longest exposure time a few seconds. Yes, to an exposure timer. Film with higher light sensitivity should also help. Still, digital has such tremendous advantages…

woelen - 3-12-2010 at 02:16

Do not get surprised if you need very long exposure times. I have done quite a lot of old-fashioned photography and for me, an exposure of 1 minute is nothing special when using standard 100 ASA films. At twilight, exposure times already go up to half a minute with aperture settings of f/8 or so and when it is dark, required exposure times may well be multiple minutes.

blogfast25 - 3-12-2010 at 08:49

Thanks Woelen, but I'll try and improve things 'the digital way'!

not_important - 3-12-2010 at 11:11

Quote: Originally posted by watson.fawkes

...
Lastly: Traditional spectroscopy was always done in very dark rooms.

And because of that not uncommonly used for courses in advanced snogging....

More seriously it can take quite long exposures, doing a series with big increases in exposure time can be used to bracket the needed time, followed by a series with lesser steps in increasing times to get the needed time, is usually needed when first using a particular setup.

The difficulties with using lower end digital cameras is when some folks have gone with cobbling together a system with the CCD and support chips, but their own controller and software. May digital cameras switch to lower resolution modes, including ones that enable adjacent pixel summing, for low light levels or fast exposures; this can really drop the resolution. Doing your own controller means you can do long exposures at full resolution, some even add TEC cooling to reduce dark current and noise.

For elemental analysis, an arc/spark between iron electrodes was a common source of reference lines, Fe having a a large number of lines in the visible and UV makes it useful for identifying lines in a sample.

blogfast25 - 4-12-2010 at 07:51

not_important: Yes, I’m aware that there are people messing about with the optical part of digicams and using their own controller software but that’s not for simpletons like me. I’m just gonna plow on to my Bunsen-Kirchoff – grating/swivel instrument for best identification of single lines… Thanks for the tip on Fe arcing: makes a lot of sense!