Hi, I'm just getting back into hobby gold refining after dabbling in it briefly in 2017. I've been reading the forum for a couple weeks now and working my way through Hoke.
I have a simple question that I can't seem to find an answer for in searching the forum, although it's no doubt been asked and answered numerous times. I know the preferred way to make AR is to add small amounts of HNO3 to HCl in stages until the gold is fully dissolved so as to minimize excess nitric at the end.
But I'm not sure what the chemical equation or the theoretical minimum amount of nitric is to dissolve a given amount of gold. I know that this would just be a theoretical minimum because AR produces chlorine, nitrosyl chloride, and varying amounts of nitrogen oxides which evolve from the solution, but I'm thinking it would be a good place to start with the first nitric addition, with more as needed after that first addition until the gold dissolves completely.
I could write something like Au + HNO3 + 4HCl --> HAuCl4 + NO + 2H2O. This would suggest that the minimum amount is just 1 mol HNO3 per 1 mol Au. This comes out to 63.012/196.967 = 0.320 g HNO3/g Au. Dividing by 68% and then by 1.41 g/mL, I get 0.334 mL 68% HNO3/g Au. But this seems far lower than is actually needed, far enough that it seems this reaction I wrote must be wrong by more than just a bit of loss from NOCl/NOx.
Does anybody know a more accurate amount for the minimum amount of nitric needed? And how much does it take in practice, in your experience?
I have a simple question that I can't seem to find an answer for in searching the forum, although it's no doubt been asked and answered numerous times. I know the preferred way to make AR is to add small amounts of HNO3 to HCl in stages until the gold is fully dissolved so as to minimize excess nitric at the end.
But I'm not sure what the chemical equation or the theoretical minimum amount of nitric is to dissolve a given amount of gold. I know that this would just be a theoretical minimum because AR produces chlorine, nitrosyl chloride, and varying amounts of nitrogen oxides which evolve from the solution, but I'm thinking it would be a good place to start with the first nitric addition, with more as needed after that first addition until the gold dissolves completely.
I could write something like Au + HNO3 + 4HCl --> HAuCl4 + NO + 2H2O. This would suggest that the minimum amount is just 1 mol HNO3 per 1 mol Au. This comes out to 63.012/196.967 = 0.320 g HNO3/g Au. Dividing by 68% and then by 1.41 g/mL, I get 0.334 mL 68% HNO3/g Au. But this seems far lower than is actually needed, far enough that it seems this reaction I wrote must be wrong by more than just a bit of loss from NOCl/NOx.
Does anybody know a more accurate amount for the minimum amount of nitric needed? And how much does it take in practice, in your experience?