joseph1990
Well-known member
- Joined
- May 9, 2017
- Messages
- 47
I am starting to question the whole process of gold thiosulfate refining.
There is a lengthy thread on gold thiosulfate with iron being the main reducer but it offers more criticisms than solutions.
Whenever you use thiosulfate as a leaching mechanism we have to use excess na2s2o3 since the s2o3 compound is used up at an accelerated rate in relation to ph...higher ph equates to the redevelopment of s2o3 but the near elimination of the electrodes. Once we figure out the leaching ratios we have to contend with reducing with na2s possibly with a higher ph.
My issue is with this formula
Na2S + Na2S2O3 + H20 = S + NaOH
if we were to use excess thiosulfate and perhaps we get some gold thiosulfate we know that the gold thiosulfate will in fact reduce to Au2S but it will also produce elemental sulfur.
We could then roast the colloids thereby getting some polysulfide derivative of gold sulfide but this would really cause smelly problems in the neighborhood. I recently bought 1kg NaBH4 and thought of using this instead of the na2s but I have no idea what the products will be but my hunch is that Na2S will be created. So if I successfully reduce the gold thiosulfate to Gold(0) and a side reaction were to occur with Na2S and water we would produce H2S thereby creating Au2S again on the surface of the Gold(0) sediment.
This is essentially a puzzle where the elimination of H2S and Na2S simply cannot be done chemically, the closest solution I currently have to this would be to add NaOH to the lixivant thereby creating Na2S and then reducing the gold thiosulfate to Au2s then adding NaBH4, since NaBH4 is extremely stable in basic solutions I can then add some acid to drop the gold particles but then this would create h2s and then I will need to accept that some gold sulfide is still in solution. Any input would be great.
There is a lengthy thread on gold thiosulfate with iron being the main reducer but it offers more criticisms than solutions.
Whenever you use thiosulfate as a leaching mechanism we have to use excess na2s2o3 since the s2o3 compound is used up at an accelerated rate in relation to ph...higher ph equates to the redevelopment of s2o3 but the near elimination of the electrodes. Once we figure out the leaching ratios we have to contend with reducing with na2s possibly with a higher ph.
My issue is with this formula
Na2S + Na2S2O3 + H20 = S + NaOH
if we were to use excess thiosulfate and perhaps we get some gold thiosulfate we know that the gold thiosulfate will in fact reduce to Au2S but it will also produce elemental sulfur.
We could then roast the colloids thereby getting some polysulfide derivative of gold sulfide but this would really cause smelly problems in the neighborhood. I recently bought 1kg NaBH4 and thought of using this instead of the na2s but I have no idea what the products will be but my hunch is that Na2S will be created. So if I successfully reduce the gold thiosulfate to Gold(0) and a side reaction were to occur with Na2S and water we would produce H2S thereby creating Au2S again on the surface of the Gold(0) sediment.
This is essentially a puzzle where the elimination of H2S and Na2S simply cannot be done chemically, the closest solution I currently have to this would be to add NaOH to the lixivant thereby creating Na2S and then reducing the gold thiosulfate to Au2s then adding NaBH4, since NaBH4 is extremely stable in basic solutions I can then add some acid to drop the gold particles but then this would create h2s and then I will need to accept that some gold sulfide is still in solution. Any input would be great.
