I am interested in preparing an anhydrous sample of sodium tungstate (Na2WO4), which is normally found as the dihydrate. Finding an example of this in literature is proving difficult, I am interested if someone can suggest a preparation method based on prior experience or difficulties I should encounter.

  • $\begingroup$ Is there a good reason you cannot dry dihydrate? Also, what are your starting materials and what equipment do you have access to? $\endgroup$
    – andselisk
    Dec 22 '20 at 16:41
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    $\begingroup$ I think I should be able to dry the dihydrate, previously I have normally found some literature before trying these procedures before. I have most standard lab equipment (heaters, vacuum pump etc) and IR which I think I can use to confirm it's no longer a hydrate. I am just unsure what conditions it would require. $\endgroup$ Dec 22 '20 at 17:07

Hoermann [1] prepared anhydrous sodium tungstate by melting the 1:1 mix of sodium carbonate and tungsten(VI) oxide (and, subsequently, growing single crystals); alternatively, he proposed prolonged drying of a dihydrate at 100 °C:

$\ce{Na2MoO3}$ und $\ce{Na2WO3}$ sind wasserfrei durch Zusammenschmelzen von 1 Mol $\ce{Na2CO3}$ mit 1 Mol $\ce{MoO3},$ bzw. $\ce{WO3},$ oder durch volliges Entwassern der Hydrate bei 100° zu erhalten. Die Salze zeigen ansgepragte Polymorphie.

Busey and Keller [2] obtained anhydrous sodium tungstate (reported water content 0.12%) by drying a dihydrate at 200 °C in vacuum. Purity has also been confirmed by Raman spectroscopy and powder x-ray diffraction:

The $\ce{Na2WO4}$ was prepared by heating the dihydrate at 200° in a vacuum.


  1. Hoermann, F. Beitrag zur Kenntnis der Molybdate und Wolframate. Die binären Systeme: $\ce{Li3MoO4-MoO3},$ $\ce{Na2MoO4-MoO3},$ $\ce{K3MoO4-MoO3},$ $\ce{Li2WO4-WO3},$ $\ce{Na2WO4-WO3},$ $\ce{K2WO4-WO3},$ $\ce{Li2MoO4-Na2MoO4},$ $\ce{Li2WO4-Na2WO4},$ $\ce{Li2MoO4-K2MoO4}.$ Z. Anorg. Allg. Chem. 1929, 177 (1), 145–186. DOI: 10.1002/zaac.19291770117.
  2. Busey, R. H.; Keller, O. L. Structure of the Aqueous Pertechnetate Ion by Raman and Infrared Spectroscopy. Raman and Infrared Spectra of Crystalline $\ce{KTcO4},$ $\ce{KReO4},$ $\ce{Na2MoO4},$ $\ce{Na2WO4},$ $\ce{Na2MoO4 · 2 H2O},$ and $\ce{Na2WO4 · 2 H2O}.$ The Journal of Chemical Physics 1964, 41 (1), 215–225. DOI: 10.1063/1.1725625.

Apart from the dehydration method mentioned by @andselisk, there are also 2 chemical methods where you can obtain the anhydrous salt:

  1. by the fusing together equivalent quantities of tungstic anhydride and sodium hydroxide or sodium carbonate, the resulting mass being taken up with water and allowed to crystallize. (@andselisk) Tungstic acid or hydrated tungsten oxide also works in this case.
  2. Fusion of alkali on mineral wolframite, $\ce{(Fe,Mn)WO4}$

For further information regarding the compound, see here


I have actually, to my surprise and that of others, successfully dissolved the filament from an incandescent oven bulb (see before and after pictures here, click to expand on the final picture), which is composed of, per a reference, to quote:

Incandescent light bulbs consist of an air-tight glass enclosure (the envelope, or bulb) with a filament of tungsten wire inside the bulb, through which an electric current is passed.

in a galvanic cell setting. The piece of tungsten metal serves as the anode and carbon (as graphite in my experiment) can act as the cathode. In the current case, I would recommend an electrolyte of NaOH plus 6% NaOCl (which relates to the so-called Bleach battery cell).

Heat in a microwave replenishing the electrolyte as needed. When dissolution is evident and Sodium hydroxide exhausted, remove electrodes and evaporate down the solution to effect salt separation.

As to mechanics, the NaOCl is expected to result in the anodic oxidation of the tungsten metal (as generally reported in the noted Bleach Battery reference) as follows:

$\ce{W + 3 NaOCl -> WO3 + 3 NaCl}$

Then, per Wikipedia, to quote:

$\ce{Na2WO4 + 2 HCl → WO3 + 2 NaCl + H2O}$

$\ce{Na2WO4 + 2 HCl → WO3.H2O + 2 NaCl}$

This reaction can be reversed using aqueous sodium hydroxide.

Hence, the presence of added Sodium hydroxide acting on $\ce{WO3}$ is expected to result in Sodium tungstate.

Now, why a microwave approach? To quote a 2013 paper: Microwave-Specific Enhancement of the Carbon−Carbon Dioxide (Boudouard) Reaction:

This changes the position of the equilibrium so that the temperature at which CO becomes the major product drops from 643 °C in the conventional thermal reaction to 213 °C in the microwave. The observed reduction in the apparent enthalpy of the microwave driven reaction, compared to what is determined for the thermal reaction from standard heats of formation, can be thought of as arising from additional energy being put into the carbon by the microwaves, effectively increasing its apparent standard enthalpy. Mechanistically, it is hypothesized that the enhanced reactivity arises from the interaction of CO2 with the steady-state concentration of electron−hole pairs that are present at the surface of the carbon due to the space-charge mechanism, by which microwaves are known to heat carbon. Such a mechanism is unique to microwave-induced heating...

Also, interesting cited chemistry on the microwave action on carbon reputedly creating some hydroxyl radicals (${.OH}$). Reference: see per this 2007 paper, Generation of hydroxyl radical in aqueous solution by microwave energy using activated carbon as catalyst and its potential in removal of persistent organic substances. Note: this is not unexpected, as per the prior reference especially in alkaline conditions, the action of an electron-hole on ${OH-}$ should produce ${.OH}$.

Note: The presence of radicals is expected to accelerate the oxidation of the tungsten metal.

Bottom line, this microwave-assisted electrosynthesis is relatively safe, effective, and avoids high temperatures. It requires only generally available inexpensive reagents.

Note, one can use alcohol to source an anhydrous Sodium tungstate as per Wikipedia, to quote: "insoluble in alcohol", but any Sodium chloride presence is sparingly soluble in methanol (14.9 g/L).

  • $\begingroup$ Sorry, but I fail to see what all of this has to do with the question. I find many of your answers pretty ingenious, but there is also a bunch that for some reason reminds me of Kissinger's shuttle diplomacy. This one leans towards the latter, I'm afraid. $\endgroup$
    – andselisk
    Dec 23 '20 at 17:21
  • $\begingroup$ Well, yes, I did forget to add the part of employing alcohol to create anhydrous crystals! However, given the general reluctance of old school chemists to ever consider a galvanic cell synthesis, albeit as my added pictures clearly demonstrate its successful feasibility here, I still expect significant dismissal. That, however, does not change pictures reflecting reality, a reality increasingly important for those without access to their labs! For those in want of a reagent, try expanding your horizon to encompass electrosynthesis and may I even further stray, with photochemical synthesis? $\endgroup$
    – AJKOER
    Dec 23 '20 at 17:34
  • $\begingroup$ At this point I'm no longer sure whether you are serious or just trolling. Neither your post nor any of the references don't have any mentions of anhydrous sodium tungstate, let alone a straightforward synthetic procedure which OP asked about. No one denies the credibility of the literature sources, it's that they are completely irrelevant for what's being asked. $\endgroup$
    – andselisk
    Dec 23 '20 at 18:17
  • $\begingroup$ To be completely honest, I did not separate out the sodium tungstate. I suspect using alcohol here would enable a salt separation leading to the anhydrous salt. Hence, a further edit to my answer. $\endgroup$
    – AJKOER
    Dec 23 '20 at 18:28
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    $\begingroup$ Again, there is no mentioning of sodium tungstate in your post or linked content whatsoever. Please post a clear and concise procedure of preparing anhydrous sodium tungstate $\ce{Na2WO4}$ backed up by relevant reputable sources and analytical data. Currently I don't see how any of this information makes any sense in the context of the question. $\endgroup$
    – andselisk
    Dec 23 '20 at 18:35

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