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Traveller
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Leaching Gold with Lugol's Soultion and Recycling Lugol's Solution
Hello
I am fairly new to this forum and possess a limited knowledge of chemistry; hence, my appeal to you for advice.
I am attempting to recover ultra-fine gold particles from the top surface of a deposit of clay. It is a fairly rich deposit of gold a few millimeters
thick on the upper surface of the clay.
The method I am attempting is leaching with Lugol's Solution. Reputedly, it is a selective leach and will only dissolve gold, leaving the base metals.
Lugol's Solution (iodine, sodium iodide, water) is not the most popular method of leaching, due to the high cost of iodine, but it has been well known
as a leach since the late 1800's. Many prospectors carried a bottle of Lugol's with them to assay samples in the field.
As the process has been described to me, one takes finely ground ore, places it in a container and covers it with Lugol's Solution. A tight fitting
lid is placed on the container and the container shaken regularly for 4-6 hours, at which time any free gold should be in solution. Maximum particle
size should be about 80 mesh. The solution is filtered and the ph raised by the addition of sodium hydroxide. Raising the ph should cause the gold to
come out of solution as a fine powder. The solution is again filtered to separate gold and solution. The iodine is recovered as elemental iodine by
the addition of hydrochloric acid and an oxidizer such as hydrogen peroxide or sodium hypochlorite. Lugol's Solution is re-made by adding sodium
hydroxide to the iodine to make sodium iodide, then adding iodine and distilled water to this mix.
Two questions I have:
If all I am attempting to do is raise the ph to precipitate gold, can I use something besides sodium hydroxide, such as sodium carbonate? NaOH is a
little dangerous to use.
Once I have raised the ph and dropped the gold from solution, do I not still have Lugol's Solution, just at a higher ph? Using a ph meter, could I not
just slowly add HCl until I brought it back to the original ph of Lugol's Solution? Would this not be normal Lugol's Solution again? What is the
normal ph of Lugol's Solution?
Oops, that was more than two questions. Hope someone can help me.
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weiming1998
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Quote: Originally posted by Traveller  | Hello
I am fairly new to this forum and possess a limited knowledge of chemistry; hence, my appeal to you for advice.
I am attempting to recover ultra-fine gold particles from the top surface of a deposit of clay. It is a fairly rich deposit of gold a few millimeters
thick on the upper surface of the clay.
The method I am attempting is leaching with Lugol's Solution. Reputedly, it is a selective leach and will only dissolve gold, leaving the base metals.
Lugol's Solution (iodine, sodium iodide, water) is not the most popular method of leaching, due to the high cost of iodine, but it has been well known
as a leach since the late 1800's. Many prospectors carried a bottle of Lugol's with them to assay samples in the field.
As the process has been described to me, one takes finely ground ore, places it in a container and covers it with Lugol's Solution. A tight fitting
lid is placed on the container and the container shaken regularly for 4-6 hours, at which time any free gold should be in solution. Maximum particle
size should be about 80 mesh. The solution is filtered and the ph raised by the addition of sodium hydroxide. Raising the ph should cause the gold to
come out of solution as a fine powder. The solution is again filtered to separate gold and solution. The iodine is recovered as elemental iodine by
the addition of hydrochloric acid and an oxidizer such as hydrogen peroxide or sodium hypochlorite. Lugol's Solution is re-made by adding sodium
hydroxide to the iodine to make sodium iodide, then adding iodine and distilled water to this mix.
Two questions I have:
If all I am attempting to do is raise the ph to precipitate gold, can I use something besides sodium hydroxide, such as sodium carbonate? NaOH is a
little dangerous to use.
Once I have raised the ph and dropped the gold from solution, do I not still have Lugol's Solution, just at a higher ph? Using a ph meter, could I not
just slowly add HCl until I brought it back to the original ph of Lugol's Solution? Would this not be normal Lugol's Solution again? What is the
normal ph of Lugol's Solution?
Oops, that was more than two questions. Hope someone can help me. |
1, I'm not sure about that. It depends on whether the gold exists in solution as gold (III)/(I) iodide/triiodide or as a complex. It also depends on
what pH the gold "un-complexes" and precipitates out. I have not found a mechanism on the action of Lugol's Solution on gold. It's best to use NaOH.
You don't even have to use concentrated solutions of it. A 1 molar solution can bring the pH up very quickly. Just put on a pair of gloves and eye
protection and you're pretty much safe.
2, Once you raised the pH, you don't have Lugol's solution any more, but a mix of IO3- and I- ions. That's because Lugol's solution is a solution of
triiodide and sodium/potassium ions. The I3-, (probably) disproportionates like I2 to I- and IO3- ions when basic, since it is I2 dissolved in I-.
Look at your solution afterwards. If it is colourless, the I3- ions are gone. Thus, it is not Lugol's Solution anymore. Adding HCl will simply result
in an acidified iodide/iodate mix.
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Traveller
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Weiming
Thank you for your reply.
Uh, not quite sure what IO3- is. Is this iodate?
Any ideas on regenerating the solution, following the addition of NaOH and the precipitation of gold, back to the original mixture of I2 and NaI found
in Lugol's Solution? I imagine the elevated ph has to be dealt with at some point.
Perhaps the only way of recycling is taking everything back to elemental iodine and starting over, as I read.
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weiming1998
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| Quote: Originally posted by Traveller | Weiming
Thank you for your reply.
Uh, not quite sure what IO3- is. Is this iodate?
Any ideas on regenerating the solution, following the addition of NaOH and the precipitation of gold, back to the original mixture of I2 and NaI found
in Lugol's Solution? I imagine the elevated ph has to be dealt with at some point.
Perhaps the only way of recycling is taking everything back to elemental iodine and starting over, as I read. |
Take the solution and heat/evaporate until the water is gone. Then take the solid and heat in a crucible. This will decompose the iodate to iodide.
After that, dissolve some iodide in water (not all of it) and react with H2SO4 or acidified H2O2. This will produce iodine. Dissolve your remaining
iodide in water and start adding iodine until no more can dissolve, or until all of it is used up. This will regenerate your Lugol's solution,
although I would just buy a new bottle.
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Traveller
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Weimer
Thanks again for the reply. You have an ability for making a complicated matter simple which, for the simple minded like me, is a blessing. 
In the method I read about, they reduced the solution to elemental iodine by adding HCl and an oxidizer. They then made sodium iodide by adding NaOH.
However, I have read, on this forum, that this will give you sodium iodide AND sodium iodate.
For my purpose of making Lugol's solution for leaching gold, will the presence of sodium iodate interfere with the process or would it need to be
driven off by heat?
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weiming1998
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| Quote: Originally posted by Traveller | Weimer
Thanks again for the reply. You have an ability for making a complicated matter simple which, for the simple minded like me, is a blessing.
In the method I read about, they reduced the solution to elemental iodine by adding HCl and an oxidizer. They then made sodium iodide by adding NaOH.
However, I have read, on this forum, that this will give you sodium iodide AND sodium iodate.
For my purpose of making Lugol's solution for leaching gold, will the presence of sodium iodate interfere with the process or would it need to be
driven off by heat?
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According to Wikipedia, iodates will react with iodides under acidic conditions,(http://en.wikipedia.org/wiki/Iodine_clock_reaction) to form iodine, so you might be right in acidifying the spent solution to regenerate the
Lugol's Solution, but I'm not sure if that's true or if HCl can provide the H+ needed.
I'm also not sure if the iodate will interfere with the leaching. A precipitate of gold (III)/(I) iodate might form and thus interfere with leaching.
I would get rid of the iodate, just in case, by heating or adding a reducing agent (like sulfite).
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hissingnoise
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If the clay particles are small could you not use the physical method of weight-separation?
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AJKOER
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Alternately, here is a description of the less expensive commercial cyanide-process for Gold extraction being essentially the use of air and a dilute
potassium-cyanide solution followed by treatment with a large excess of Zinc. To quote (source: http://gold.atomistry.com/extraction.html ):
"Access of air is essential to solution of gold in potassium cyanide, the process being attended by evolution of hydrogen. Lead, bismuth, antimony,
cadmium, silver, and mercury also dissolve in presence of air; but copper, iron, aluminium, nickel, cobalt, and zinc dissolve in absence of air. Gold
and silver are distinguished by the fact that their maximum solubility corresponds with a very low concentration of the potassium-cyanide solution, a
phenomenon probably due to the slight solubility of air in concentrated solutions of this salt. The solution of gold in the cyanide solution is
accompanied by the intermediate formation of hydrogen peroxide, and the process is accelerated by addition of this substance:
2Au+4KCN+2H2O+O2 = 2KAu(CN)2+2KOH+H2O2;
2Au+4KCN+H2O2 = 2KAu(CN)2+2KOH.
A similar accelerating effect is exerted by other substances, such as potassium ferricyanide, potassium permanganate, potassium chromate, sodium
peroxide, barium peroxide, cyanogen bromide, cyanogen chloride, persulphates, and certain organic compounds. The best method of reducing the
proportion of the other metals is to maintain the cyanide solution dilute.
In precipitating the gold by zinc, the proportion required is about seven times that indicated by the equation
Zn+2Au•=2Au + Zn••,
the discrepancy being due to solution of part of the zinc in the cyanide solution, with evolution of hydrogen. Purity of the zinc is an important
factor in counteracting this loss."
Other Gold extraction processes are discussed, but in less detail.
[Edited on 13-10-2012 by AJKOER]
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Traveller
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No, the particle size of the gold involved makes it impractical for removal with gravity methods.
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Traveller
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| Quote: Originally posted by AJKOER | Alternately, here is a description of the less expensive commercial cyanide-process for Gold extraction being essentially the use of air and a dilute
potassium-cyanide solution followed by treatment with a large excess of Zinc. To quote (source: http://gold.atomistry.com/extraction.html ):
"Access of air is essential to solution of gold in potassium cyanide, the process being attended by evolution of hydrogen. Lead, bismuth, antimony,
cadmium, silver, and mercury also dissolve in presence of air; but copper, iron, aluminium, nickel, cobalt, and zinc dissolve in absence of air. Gold
and silver are distinguished by the fact that their maximum solubility corresponds with a very low concentration of the potassium-cyanide solution, a
phenomenon probably due to the slight solubility of air in concentrated solutions of this salt. The solution of gold in the cyanide solution is
accompanied by the intermediate formation of hydrogen peroxide, and the process is accelerated by addition of this substance:
2Au+4KCN+2H2O+O2 = 2KAu(CN)2+2KOH+H2O2;
2Au+4KCN+H2O2 = 2KAu(CN)2+2KOH.
A similar accelerating effect is exerted by other substances, such as potassium ferricyanide, potassium permanganate, potassium chromate, sodium
peroxide, barium peroxide, cyanogen bromide, cyanogen chloride, persulphates, and certain organic compounds. The best method of reducing the
proportion of the other metals is to maintain the cyanide solution dilute.
In precipitating the gold by zinc, the proportion required is about seven times that indicated by the equation
Zn+2Au•=2Au + Zn••,
the discrepancy being due to solution of part of the zinc in the cyanide solution, with evolution of hydrogen. Purity of the zinc is an important
factor in counteracting this loss."
Other Gold extraction processes are discussed, but in less detail.
[Edited on 13-10-2012 by AJKOER] |
Hello AKJOER
I have read extensively about the use of potassium or sodium cyanide for dissolving gold and wish that I could use it, mostly because it can be
prepared to only dissolve gold and ignore base metals. It is an effective and economical method of gold extraction; evidenced by the fact that upwards
of 90% of gold extracted annually is extracted with cyanide.
The down side of cyanide is its tendency to form toxic gases at lower ph's, its environmentally detrimental byproducts and the fact that an
individual, such as myself, cannot legally purchase it in Canada.I know it is possible to take the necessary safety precautions and it is possible to
neutralize cyanide solutions with sodium thiosulphate but, is it possible to purchase cyanide on the open market?
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Eddygp
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If you plan to take a piece of clay to your lab, I would suggest mercury, because it is quite efficient. However, it is very toxic and difficult to
find these days.
there may be bugs in gfind
[ˌɛdidʒiˈpiː] IPA pronunciation for my Username |
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zed
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Seems to me Oxone plus a Halide salt may be used for some metal leaching processes.
[Edited on 13-10-2012 by zed]
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hissingnoise
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| Quote: | | I would suggest mercury, because it is quite efficient. However, it is very toxic and difficult to find these days. |
I used this this seller fairly recently and cost and delivery was reasonable . . .
He specifies collection, now, but that may change!
[Edited on 13-10-2012 by hissingnoise]
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Traveller
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I don't mean to be critical but, I did not ask for advice on mercury, gravity separation or cyanide. I have looked into all of these methods and can
find no way to apply them to my situation.
I am, however, eager to learn anything you have to offer about iodine/iodide leaching of gold and recycling of this leach.
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hissingnoise
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I see! Your entire focus is on acquisition of I<sub>2</sub>?
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Traveller
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No, my focus is on recovering fine gold using Lugol's Solution and being able to recycle that solution, due to its high cost.
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hissingnoise
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You already have the required Lugol's Solution then, I take it?
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Traveller
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I have a small supply I am using to experiment with. There is a chemical supply house in Vancouver that tells me they will supply me with iodine and
sodium iodide once I am ready to scale up to an operational volume.
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Traveller
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I have been informed that a second thread I began about Lugol's Solution should have been part of this thread. I will transfer it's contents here plus
the one reply to it thus far.
Thread content:
Is this guy making Lugol's Solution in a kind of backwards way?
http://www.youtube.com/watch?v=CLhwkFKLdPA
Reply from Vargouille:
Short answer, no.
Lugol's solution is a mixture of iodine and an iodide salt. This process is a simple oxidation of iodide by acidified hydrogen peroxide, which
converts almost all of the iodide into iodine, with little to no production of triiodide. One could argue that because of the lack of precise
measurements on the amount of reagents used, some triiodide may be produced if the amount of iodide salt is in excess, however, the main product is
free iodine, which precipitates and is removed. The dark color of the solution remaining is, I believe, from the small amount of iodine that is
soluble in water, rather than a concentrated solution of potassium triiodide.
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Traveller
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So, if the solution he ended up with is free iodine in water, what would happen if he added more potassium iodide powder to this solution?
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hissingnoise
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| Quote: | Now the ore is in the jar. Pour in the Iodine solution so that the ore is well covered. Shake it occasionally for some time, maybe ½ to 1 hour. Now
you must remove the ore. You can let it settle and pour off the Iodine containing the gold or you can filter it with a small funnel and some coffee
filter paper. The idea is to get the solution as clear as possible. <b>Now you add a small glob of mercury and shake.</b> At some point
the solution will lose it's red-brown Iodine color and become a (usually) clear yellowish liquid with sediment of heavy <b>"floured
mercury"</b> in the bottom of the jar/test tube. Allow this to settle for a few minutes and then carefully pour off the liquid. Add some water
to the mercury, shake, allow it to settle, and pour off. Don’t throw the liquids away. Now you have your precious metals amalgamated in the "floured
mercury". Now you simply add some nitric acid (not more than 50/50 with water) and dissolve the mercury. When the mercury is all dissolved you will be
left with a black or brown material that you cannot dissolve. This is your precious metal. Don’t throw the nitric solution away. Now take a good
look at the black sediment. Try to get a feel for how much there is so that you will have a comparative idea of how much you have. You can buy, from
chemical supply companies, a graduated, conical test tube. If you run the assay in this you can simply read off the quantity of precious metals on the
scale engraved on the tube. It’s a nicety that the old timers didn’t have but it is convenient.
The reason to save all the liquids is simple, we are going to recover all your Iodine and mercury. The solution that contained Iodine is treated with
a few drops of chlorox. The Iodine will settle to the bottom. The liquid is poured off and then you add a little lye water until the solution becomes
clear and colorless. Your iodine is in solution ready to use again . . .
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"Floured mercury" is the metal in the form of small globules which do not agglomerate to the liquid form!
It must be carefully handled due to its toxicity . . .
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Traveller
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Thank you for this information. However, I am really not interested in using mercury. One of the reasons I became interested in the iodine leaching
process is that I do not wish to introduce toxic compounds into the environment I live in.
Remember this guy?
[Edited on 14-10-2012 by Traveller]
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Vargouille
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| Quote: Originally posted by Traveller | | So, if the solution he ended up with is free iodine in water, what would happen if he added more potassium iodide powder to this solution?
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In that case, I believe that it would form a yellowish solution of dilute potassium triiodide.
CRC states that iodine has a solubility of 0.03 g/100 g H2O at 20C. This lab from the University of Delaware states that dilute solutions of triiodide are yellowish, while more concentrated ones are brown. That,
incidentally, is why starch is used during titrations of iodine, because it forms an intensely blue-black solution even at very low concentrations of
triiodide. It was probably warmer than 20C when the video was taken, but let's assume that the solubility wasn't altered too drastically. That
corresponds to a 0.001 M solution of iodine. Now, the concentration of triiodide that constitutes a dilute solution isn't defined, but it's safe to
say that 0.001 M is dilute enough.
I would not, mind you, call this "Lugol's Solution". Lugol's solution is significantly more concentrated based on iodine, either about 0.2 M (for 5%)
or 0.08 M (for 2%).
[Edited on 14-10-2012 by Vargouille]
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hissingnoise
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Milliner's disease resulted from prolonged exposure to mercury vapour!
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Traveller
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There were lots of prospectors in Canada that went crazy or died from inhaling mercury fumes while retorting gold/mercury amalgam. As I said, I am not
interested in using it.
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