Searching for Lithium and Rubidium in Lepidolite

Lepidolite, empirically KLi₂AlSi₄O₁₀F(OH) (MM = 388.3 g/mol), is a mica type mineral that often contains rubidium (Rb, presumably through substitution of the K) and sometimes caesium (Cs), besides the valuable lithium (Li).

Lepidolite pieces, slightly over-exposed:



This is an attempt to show the presence of lithium and rubidium (and possibly also caesium) by means of simple spectrometry.

Lepidolite is usually fused with an excess of crushed limestone at 1200 – 1500 C and until most of the limestone has been converted to lime (CaO). As this is outside my temperature envelope I tried to fuse 1 part crushed Lepidolite with 2 parts of slaked lime (Ca(OH)₂ at high Bunsen heat for 1 hour but on cooling it soon became apparent no reaction had taken place at all: it takes a much higher temperature to digest this alumino silicate in a solid-solid reaction.

I decided then to resort to alkaline fusion with NaOH, in the knowledge that most of the sodium would be separable because NaCl is poorly soluble in concentrated HCl.

So to that effect 11.3 g of crushed Lepidolite was mixed with 22.6 NaOH small prills and fused for 20 min in a nickel crucible with lid, on high Bunsen heat.

It has to be said that this mineral is hard to grind because of its lamellar structure, which causes these lamellae to glide over each other with minimal friction. With a combination of pounding and grinding action, something was obtained that I deemed sufficiently rich in surface area, although by no means a powder in the narrow sense of the word.

After fusion, the contents were transferred into a beaker, not without considerable effort: this frit was very hard. Softening up with small aliquots of hot water and creative use of a chisel did the trick. No unreacted Lepidolite was observed. Grinding the frit also proved hard.

To it was added about 50 ml of HCl 37 % and after simmering this was diluted to about 200 ml. An insoluble, off white residue remained and was filtered off. At this point I assumed all silica and fluorine (as sodium hexafluoroaluminate) to be in the residue and all Al, Na, K, Li, Rb and Cs to be in the clear filtrate.

The filtrate was carefully alkalised with NH3 33 w% to pH about 9 and the aluminium precipitated as Al(OH)3:



Note that this causes ammonium chloride to be introduced into the mix of alkali metal chlorides.

The filtrate was then boiled in, in an Erlenmeyer, until crystals started to appear and bumping became intolerable. On cooling the crystals were filtered off and about 36 ml of clear filtrate was obtained. 72 ml of HCl 37 w% was added and another batch of NaCl dropped out:



It’s possible that these crystals contain also KCl, as it too is insoluble in strong HCl.

These crystals were filtered off on a ceramic filter of medium porosity with vacuum. The filtrate is now supposed to contain all LiCl, all RbCl, all CsCl, most KCl, some NaCl and the NH₄Cl.

That is about as far as I got to date.

The plan is now to split the alkali metal chlorides into two groups, according to a scheme which is elaborated here (I have the free, full version on paper):

http://www.worldcat.org/title/radioactivity-of-the-rubidium-...

(The Radioactivity of the rubidium extracted from the lepidolite and zinnwaldite of Japan, by S. Iimori and J. Yoshimura Lepidolite from Nagatori, Chikuzen Province, and the lithium content of Japanese mica, by S. Iimori and J. Yoshimura.)

In it, the authors make use of the insolubility of the hexachlorostannates of K and Rb in conc. HCl. Adding H₂SnCl₆ to the last filtrate should precipitate NH4, K, Rb (and presumably but not mentioned in that paper, also Cs) as M₂SnCl₆. Filtering then creates a separation between (Na, Li) and (NH4, K, Rb, Cs).

I have in the past independently verified the insolubility of both ammonium and potassium hexachlorostannates in conc. HCl.

For K and Rb the paper lists solubility values of these hexachlorostannates in d = 1.185 HCl at 20 C as 0.0689 (as KCl) g/100 ml for K and 0.0093 (as RbCl) g/100 ml for Rb. No values for NH4 or Cs are given.

A solution of H₂SnCl₆ (from HCl, SnCl₂ and H₂O₂ has been prepared and will be allowed to stand overnight, to allow full oxidation of the Sn(II) to Sn(IV).

[Edited on 2-2-2014 by blogfast25]

Quote: Originally posted by elementcollector1

Interesting! Although, rubidium is present in lepidolite in *extremely* low concentrations, so I'm interested to see your yield.

@UC:

Wunderbar!

A few points.

Coffee grinder: should have thought of that! But my small, thin lamellae dissolved easily into the molten NaOH.

Crucible: I've never looked back since I bought a nickel one. Indestructable and oblivious to molten alkali or hot conc. H2SO4. Drop it and it bounces back! Every home lab should have one, mandatory!

Ca separation: precipitate as carbonate, the sulphate is still slightly soluble.

Do you plan any type of separation of Li, K, Rb and if so, how?

Personally I think 'fusion' (not literally) with CaO would have a lot going for it. This way, you 'block' all Si, Al and F immediately. Then leach with water for Group I hydroxides. But temperature needed is very high.

Thanks for the links! The minsoscam *.pdf is interesting to me because of the spectrum. I'll check these values against NIST's vast spectral database.


[Edited on 4-2-2014 by blogfast25]