adam_mizer
Well-known member
- Joined
- Nov 4, 2010
- Messages
- 176
On a post written by GoldenChild in another thread:
Now trying to use the approach above with the silver bars I have.
The bars are 94.5% silver. EDIT(base them at 94%)
If I take 1 of those bars weighing 54 grams that = 50.76 grams silver + 3.24 grams copper (lets use copper but I'm not sure).
Now I have distilled and condensed my nitric and its aweful red fuming, I estimated 95% but as the above post states 90% would be a safe guess-timate.
Now if it takes 1.17ml (edit) 70% nitric per gram of silver and 3 times more per gram of copper then should I use XXml of the distilled/condensed nitric.
(50.76gr silver x 1.17ml (70%) nitric + 3.24gr copper x (3x1.17ml) = 3.51x1.17ml=11.37ml (70%) nitric) = 70.76ml (70%) nitric for that bar.
It appears I need 70.76ml 70% = so how many XXml's of 90% would equal the 70.76ml of 70%?
Whats the math ratio?
If a person has different amounts of ml's as not stated in the above quote, I need help with the math basing it all on the 90% value.
Somehow it seems simple enough but I just can't figure it out right now.
(I added a picture maybe we could make a comparison on the percentage of nitric?)
This is just over a 1 liter bottle.
Lots of edits and another: I know the contaminent is nitrogen dioxide what I don't really know is can I just release this fuming contaminent or does it even matter much?
Changed the picture, added some more home brew. Found out with my setup (bunson burner) it can process much faster with smaller amounts (100ml sulphuric) rather than trying to process several 100ml's at a time.
Is there a weight measurement that can be done to determine the percent strength of the nitric? Might not matter and I'll probably just use only a few ml's extra.
To answer my own question on the amount I can just break the 125ml above down into increments and figure exactly what I need, maybe by doing this I'll get the ratio math answered.
Another piece to the puzzle. When making poor mans AR....
If you distill your own nitric it takes 255 grams of high purity (98-99%) sodium nitrate to produce approximately 125ml of red fuming nitric. Distillation should in theory produce 100% pure nitric acid. In reality the home refiner will only be able to produce 92-96% pure nitric. Let's use 90% to be safe.
If you take your 125ml of 90% nitric and cut it with 36ml of distilled water you will now have 161ml of 70% nitric.
((90/70) * 125) - 125
Thanks to GSP for helping me figure that one out.
Using these numbers you can now more accurately guess how much nitric acid you are putting into the HCL to prevent excess nitric additions.
Now trying to use the approach above with the silver bars I have.
The bars are 94.5% silver. EDIT(base them at 94%)
If I take 1 of those bars weighing 54 grams that = 50.76 grams silver + 3.24 grams copper (lets use copper but I'm not sure).
Now I have distilled and condensed my nitric and its aweful red fuming, I estimated 95% but as the above post states 90% would be a safe guess-timate.
Now if it takes 1.17ml (edit) 70% nitric per gram of silver and 3 times more per gram of copper then should I use XXml of the distilled/condensed nitric.
(50.76gr silver x 1.17ml (70%) nitric + 3.24gr copper x (3x1.17ml) = 3.51x1.17ml=11.37ml (70%) nitric) = 70.76ml (70%) nitric for that bar.
It appears I need 70.76ml 70% = so how many XXml's of 90% would equal the 70.76ml of 70%?
Whats the math ratio?
If a person has different amounts of ml's as not stated in the above quote, I need help with the math basing it all on the 90% value.
Somehow it seems simple enough but I just can't figure it out right now.
(I added a picture maybe we could make a comparison on the percentage of nitric?)
This is just over a 1 liter bottle.
Lots of edits and another: I know the contaminent is nitrogen dioxide what I don't really know is can I just release this fuming contaminent or does it even matter much?
Changed the picture, added some more home brew. Found out with my setup (bunson burner) it can process much faster with smaller amounts (100ml sulphuric) rather than trying to process several 100ml's at a time.
Is there a weight measurement that can be done to determine the percent strength of the nitric? Might not matter and I'll probably just use only a few ml's extra.
To answer my own question on the amount I can just break the 125ml above down into increments and figure exactly what I need, maybe by doing this I'll get the ratio math answered.