amount of nitric acid needed
- Thread starter geonorts
- Start date
Help Support Gold Refining Forum:
machiavelli976
Well-known member
- Joined
- Mar 25, 2010
- Messages
- 263
3Ag + 4HNO3-> 3AgNO3+ NO+2H2O means for 3 moles Ag ( 323.61 grams) you need 4 moles of nitric(252 grams). for 110 grams of silver you need (252x110)/323.61=85.65 gr. HNO3, 100% . this is (85,65x100)/70=122.357 gr.HNO3 70% .for the proper volume of acid you should know the density of it witch i don't.
Close approximations:
- 7 pounds of silver per gallon of 70% nitric
- 100 tr.oz. per gallon
- 26.4 tr.oz per liter
- 822 grams per liter
- 1 gram per 1.22 ml
For your 110 grams of silver, therefore, it will take about 134 ml of 70% nitric.
For several reasons, these figures never come out to be 100% exact - due to reaction temperature, amount of dilution, actual strength of the concentrated nitric, etc. However, for planning purposes, they are very close. They assume that the dissolving vessel is open and not sealed in any way. They also assume that the 70% nitric has been diluted 50/50 (to 41%) with distilled water when dissolving the silver. When using this 50/50 solution, the volumes above, of course, are doubled. They also assume that the silver is pure to start with.
If the silver contains copper, such as in sterling or coin silver, the approximations will be slightly different, since:
- one gallon of 70% nitric will only dissolve about 2 pounds of copper
- about 4.3 ml of 70% nitric will dissolve 1 gram of copper
For 1 gram of Sterling silver - 92.5/7.5, Ag/Cu:
(1.22 x .925) + (4.3 x .075) = about 1.45 ml of 70% nitric or 2.9 ml of 50/50 nitric
For 1 gram of Coin silver - 90/10, Ag/Cu:
(1.22 x .90) + (4.3 x .10) = about 1.53 ml of 70% nitric or 3.06 ml of 50/50 nitric
Note: The amount of nitric needed to dissolve copper (1 gallon/2 pounds) also can be used as an approximation for some other base metals, such as nickel, zinc, or copper alloys, such as brass. This same figure, 1 gallon/2 pounds, can be approximated when dissolving some base metals, such as copper or iron, in 4 to 1 aqua regia.
Here again, except under very stringent controls, there can be no exact figures given for the amount of acid needed. For planning purposes, however, the above amounts have been proven in practice to be very good approximations. You can use these figures as a starting point and then make adjustments, one way or another, if needed. This is not rocket science.
- 7 pounds of silver per gallon of 70% nitric
- 100 tr.oz. per gallon
- 26.4 tr.oz per liter
- 822 grams per liter
- 1 gram per 1.22 ml
For your 110 grams of silver, therefore, it will take about 134 ml of 70% nitric.
For several reasons, these figures never come out to be 100% exact - due to reaction temperature, amount of dilution, actual strength of the concentrated nitric, etc. However, for planning purposes, they are very close. They assume that the dissolving vessel is open and not sealed in any way. They also assume that the 70% nitric has been diluted 50/50 (to 41%) with distilled water when dissolving the silver. When using this 50/50 solution, the volumes above, of course, are doubled. They also assume that the silver is pure to start with.
If the silver contains copper, such as in sterling or coin silver, the approximations will be slightly different, since:
- one gallon of 70% nitric will only dissolve about 2 pounds of copper
- about 4.3 ml of 70% nitric will dissolve 1 gram of copper
For 1 gram of Sterling silver - 92.5/7.5, Ag/Cu:
(1.22 x .925) + (4.3 x .075) = about 1.45 ml of 70% nitric or 2.9 ml of 50/50 nitric
For 1 gram of Coin silver - 90/10, Ag/Cu:
(1.22 x .90) + (4.3 x .10) = about 1.53 ml of 70% nitric or 3.06 ml of 50/50 nitric
Note: The amount of nitric needed to dissolve copper (1 gallon/2 pounds) also can be used as an approximation for some other base metals, such as nickel, zinc, or copper alloys, such as brass. This same figure, 1 gallon/2 pounds, can be approximated when dissolving some base metals, such as copper or iron, in 4 to 1 aqua regia.
Here again, except under very stringent controls, there can be no exact figures given for the amount of acid needed. For planning purposes, however, the above amounts have been proven in practice to be very good approximations. You can use these figures as a starting point and then make adjustments, one way or another, if needed. This is not rocket science.
machiavelli976
Well-known member
- Joined
- Mar 25, 2010
- Messages
- 263
makes me feel embarrassed for the years spent in school ! great post GSP and thanks from me too.
machiavelli976
Well-known member
- Joined
- Mar 25, 2010
- Messages
- 263
i was alking about grams and GSP about ml. mass against volume!? it is not the same thing chuckle !
machiavelli976
Well-known member
- Joined
- Mar 25, 2010
- Messages
- 263
my final result is 122.357 grams of HNO3 70% and GSP's final calculation is 134 ml . where did you get " about 4 times too much" ? there's a weird math' in your head lol !
machiavelli976,
Actually, the SG of 70% is 1.42 or, about 11.8#/gal
If you search the internet, you will find at least 4 different equations given for the dissolving of silver with nitric acid:
(1) 3Ag + 4HNO3 = 3AgNO3 + NO + 2H2O
(2) 4Ag + 6HNO3 = 4AgNO3 + NO + NO2 + 3H2O
(3) 2Ag + 3HNO3 = 2AgNO3 + HNO2 + H2O
(4) Ag + 2HNO3 = AgNO3 + NO2 + H2O
(1) was given by machiavelli976. I don't know his source.
(2), I found here: http://www.saltlakemetals.com/Solubility_Of_Silver_Nitrate.htm
(3) and (4), here: http://answers.yahoo.com/question/index?qid=20071218133113AAa89Mt
He gives a BASF site as his reference, although I couldn't find it. (4) is when using strong nitric and (3) for weak nitric. He doesn't say how weak, but I think I have seen other references that, to satisfy this equation, the nitric would have to be very weak.
In (1), the Molar ratio of HNO3/Ag is 4:3. In (2) and (3), it is 3:2. In (4), it is 2:1. Big differences in nitric usage.
The 134 ml of nitric I gave to dissolve 110 grams of silver is about a 2:1 Molar ratio - equation (4). The 134 ml of nitric is about 2.1 Moles (.134 x 15.6) and the 110 grams of silver is about 1.02 Moles (110/107.87), for a Molar ratio of 2.06:1
The equation (or combination of equations) that actually fits the bill is dependent on the conditions - mainly temperature, nitric acid strength, and whether or not the gases produced are condensed and re-enter the solution. Also, on the forum, there has been discussion of adding H2O2 to continually convert the gases back to HNO3. In this case, it is theoretically possible to get close to a HNO3/Ag Molar ratio of 1:1 and, therefore, only require half as much nitric. Ag + HNO3 + ? = AgNO3 + ? Call this Equation (5).
I have made up solutions for 30 gallon silver cells many times. For the silver (200 oz), I used pure silver and calculated the nitric needed based on a 2:1 Molar ratio (Eq. 4) or, 7 pounds of silver per gallon of nitric. I shotted the Ag and divided it equally in 2 buckets. I then diluted the nitric with distilled water, 50/50, and added it to the buckets (in increments, at first, to prevent it from foaming over - it also got quite hot). The next morning, the reaction was complete and there was always 1 or 2 oz of Ag undissolved. This was probably due to the fact that I was using tech grade HNO3, which was only about 67%. Having a little undissolved silver was ideal, since I didn't want a bunch of excess nitric in the cell solution.
For the copper, I used clean copper electrical wire and based the nitric usage on 2# of copper per gallon. Same result. I ended up with a little undissolved copper in the morning.
All of this proved to me that Eq (4), with a 2:1 Molar ratio was right on, UNDER THE CONDITIONS I WAS USING - hot; open container; strong acid solution, etc. It also proved to me that, under these common conditions, 1 gallon of nitric would dissolve either 7 pounds of silver or 2 pounds of copper. I love these numbers, since they are very easy to remember.
Actually, the SG of 70% is 1.42 or, about 11.8#/gal
If you search the internet, you will find at least 4 different equations given for the dissolving of silver with nitric acid:
(1) 3Ag + 4HNO3 = 3AgNO3 + NO + 2H2O
(2) 4Ag + 6HNO3 = 4AgNO3 + NO + NO2 + 3H2O
(3) 2Ag + 3HNO3 = 2AgNO3 + HNO2 + H2O
(4) Ag + 2HNO3 = AgNO3 + NO2 + H2O
(1) was given by machiavelli976. I don't know his source.
(2), I found here: http://www.saltlakemetals.com/Solubility_Of_Silver_Nitrate.htm
(3) and (4), here: http://answers.yahoo.com/question/index?qid=20071218133113AAa89Mt
He gives a BASF site as his reference, although I couldn't find it. (4) is when using strong nitric and (3) for weak nitric. He doesn't say how weak, but I think I have seen other references that, to satisfy this equation, the nitric would have to be very weak.
In (1), the Molar ratio of HNO3/Ag is 4:3. In (2) and (3), it is 3:2. In (4), it is 2:1. Big differences in nitric usage.
The 134 ml of nitric I gave to dissolve 110 grams of silver is about a 2:1 Molar ratio - equation (4). The 134 ml of nitric is about 2.1 Moles (.134 x 15.6) and the 110 grams of silver is about 1.02 Moles (110/107.87), for a Molar ratio of 2.06:1
The equation (or combination of equations) that actually fits the bill is dependent on the conditions - mainly temperature, nitric acid strength, and whether or not the gases produced are condensed and re-enter the solution. Also, on the forum, there has been discussion of adding H2O2 to continually convert the gases back to HNO3. In this case, it is theoretically possible to get close to a HNO3/Ag Molar ratio of 1:1 and, therefore, only require half as much nitric. Ag + HNO3 + ? = AgNO3 + ? Call this Equation (5).
I have made up solutions for 30 gallon silver cells many times. For the silver (200 oz), I used pure silver and calculated the nitric needed based on a 2:1 Molar ratio (Eq. 4) or, 7 pounds of silver per gallon of nitric. I shotted the Ag and divided it equally in 2 buckets. I then diluted the nitric with distilled water, 50/50, and added it to the buckets (in increments, at first, to prevent it from foaming over - it also got quite hot). The next morning, the reaction was complete and there was always 1 or 2 oz of Ag undissolved. This was probably due to the fact that I was using tech grade HNO3, which was only about 67%. Having a little undissolved silver was ideal, since I didn't want a bunch of excess nitric in the cell solution.
For the copper, I used clean copper electrical wire and based the nitric usage on 2# of copper per gallon. Same result. I ended up with a little undissolved copper in the morning.
All of this proved to me that Eq (4), with a 2:1 Molar ratio was right on, UNDER THE CONDITIONS I WAS USING - hot; open container; strong acid solution, etc. It also proved to me that, under these common conditions, 1 gallon of nitric would dissolve either 7 pounds of silver or 2 pounds of copper. I love these numbers, since they are very easy to remember.
kalay,
With a density of 9#/gallon, the nitric concentration, by weight, would only be about 16%.
With a density of 9#/gallon, the nitric concentration, by weight, would only be about 16%.
machiavelli976
Well-known member
- Joined
- Mar 25, 2010
- Messages
- 263
i have told you GSP makes me feel embarrassed for the years spent in school !! you should feel the same too. anyway thanks again guys !
I appreciate the compliment but, don't say that! I do have 10 units of College Chemistry but I doubt if any of it covered precious metals. How much time did you spend in college studying silver chemistry? Probably very little. PM chemistry is specialized and is very lacking in all but specialized schools. I'm sure you could run rings around me in most anything else. I've just been doing this same stuff for 40+ years and, I guess, something had to soak in. You do something for a long time and you just develop a feel for it.
You weren't off in your calculations. You just picked the wrong equation or a reference that claimed it was the right one. The only reason I know that the 2:1 ratio works in practice, in most cases, is because I've done it a jillion times.
goldsilverpro said:
Very quickly - Im on a lunch break
Ideally :
2Ag + 2HNO3 +H2O2 -> 2AgNO3 + 2H2O
ToDo:
1) tests if and how much emission of NOx is reduced/eliminated
2) calculations to find out if it makes economical sense (It would make perfect sense from health point of view - no NOx fumes)
There have been several discussions on the forum about using H2O2 to make better use of the acid and to reduce the fuming. I think Noxx did some work on it. It's also covered in some patents. The problem is that the H2O2 breaks down quickly in the hot solution and it takes a lot of additions to maintain it. Some patents that use H2O2 add a little ethylene glycol (antifreeze), which I think acts as a preservative for the H2O2. Since I haven't tried this, I won't comment on it. Maybe Noxx can jump in.
Barren Realms 007
In Remembrance
goldsilverpro said:
It can become a run away reaction with the heat and it seems the the glycol leavess an oily residue form what I have experienced. I used it a few times and then quit using it. IMHO
goldenchild
Well-known member
- Joined
- Aug 14, 2009
- Messages
- 1,809
Someone made a video experimenting with ethylene glycol, nitric and h202.
etack
Well-known member
I thought that was you goldenchild
Eric
Eric
