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A chemical engineer doing chemistry is always an interesting thing, but sometimes we bring something to the table with our simple ways of looking at chemistry.
For me it's been fascinating rationalizing the positions that the proponents of the various precipitating agents take on this site. We have proponents of SO2, of sodium sulfite, and of sodium metabisulfite (not to mention copperas, but that's a different issue).
Here's my 2 cents on the issue:
1. There will be excess HCl in our AR solutions after we've dissolved the gold. Both sodium sulfite and SMB are going to react with that HCl to make SO2:
Na2SO3 + 2HCl = 2 NaCl + H2O + SO2
Na2S2O5 + 2HCl = 2NaCl + H2O + 2SO2
This is how they tie into the "it's the same as using SO2" discussion.
2. SO2 will, obviously, precipitate gold.
3SO2 + 2AuCl3 + 3H2O = 2Au + 3SO3 + 6HCl
two interesting issues here:
- we've now made more HCl (to react with the sulfites)
- the SO3 will immediately react with water to make sulfuric acid: SO3 + H2O = H2SO4
so the reaction can probably be better written like this:
3 SO2 + 2 AuCl3 + 6 H2O = 2 Au + 3 H2SO4 + 6 HCl
3. We can write similar equations for sodium sulfite going through SO2
3 Na2SO3 + 6 HCl = 6 NaCl + 3 H2O + 3 SO2
and
3 SO2 + 2 AuCl3 + 6 H2O = 2 Au + 3 H2SO4 + 6HCl
summing those reactions:
3 Na2SO3 + 2 AuCl3 + 3 H2O = 2 Au + 3 H2SO4 + 6 NaCl
4. And for SMB, through an SO2 intermediate
3 Na2S2O5 + 6 HCl = 6 NaCl + 3 H2O + 6 SO2
6 SO2 + 4 AuCl3 + 4 H2O = 4 Au + 6 SO3 + 12 HCl
summing the reactions (and including the H2SO4 step)
3 Na2S2O3 + 4 AuCl3 + 9 H2O = 4 Au + 6 NaCl + 6 HCl + 6 H2SO4
If one calculates the stoichiometry of the reactions, to obtain 1 g gold requires = 0.49 g SO2, or 0.96 g sodium sulfite, or 0.72 g SMB
Anyway, my 2 cents... (assuming I didn't mistype from my notes) :mrgreen:
For me it's been fascinating rationalizing the positions that the proponents of the various precipitating agents take on this site. We have proponents of SO2, of sodium sulfite, and of sodium metabisulfite (not to mention copperas, but that's a different issue).
Here's my 2 cents on the issue:
1. There will be excess HCl in our AR solutions after we've dissolved the gold. Both sodium sulfite and SMB are going to react with that HCl to make SO2:
Na2SO3 + 2HCl = 2 NaCl + H2O + SO2
Na2S2O5 + 2HCl = 2NaCl + H2O + 2SO2
This is how they tie into the "it's the same as using SO2" discussion.
2. SO2 will, obviously, precipitate gold.
3SO2 + 2AuCl3 + 3H2O = 2Au + 3SO3 + 6HCl
two interesting issues here:
- we've now made more HCl (to react with the sulfites)
- the SO3 will immediately react with water to make sulfuric acid: SO3 + H2O = H2SO4
so the reaction can probably be better written like this:
3 SO2 + 2 AuCl3 + 6 H2O = 2 Au + 3 H2SO4 + 6 HCl
3. We can write similar equations for sodium sulfite going through SO2
3 Na2SO3 + 6 HCl = 6 NaCl + 3 H2O + 3 SO2
and
3 SO2 + 2 AuCl3 + 6 H2O = 2 Au + 3 H2SO4 + 6HCl
summing those reactions:
3 Na2SO3 + 2 AuCl3 + 3 H2O = 2 Au + 3 H2SO4 + 6 NaCl
4. And for SMB, through an SO2 intermediate
3 Na2S2O5 + 6 HCl = 6 NaCl + 3 H2O + 6 SO2
6 SO2 + 4 AuCl3 + 4 H2O = 4 Au + 6 SO3 + 12 HCl
summing the reactions (and including the H2SO4 step)
3 Na2S2O3 + 4 AuCl3 + 9 H2O = 4 Au + 6 NaCl + 6 HCl + 6 H2SO4
If one calculates the stoichiometry of the reactions, to obtain 1 g gold requires = 0.49 g SO2, or 0.96 g sodium sulfite, or 0.72 g SMB
Anyway, my 2 cents... (assuming I didn't mistype from my notes) :mrgreen:
