Inquarttion vs. silver dilution

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RickRag

Active member
Joined
Sep 13, 2018
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26
Location
Dallas, TX
Hi guys,

I have some yellow dental scrap that I would classify as first-grade gold scrap because it has no non-metallic impurities and is likely 14-18K material. Most dental casting golds contain some percentage of silver up to approximatley 25% along with palladium and some platinum. According to Hoke, in recovering alloys containing gold and PM's as noted in Chapter XII, one of the preparatory steps is to do a nitric leach to remove base metals, the drawback being that any palladium would have to be recoverd as well as whatever platinum leached along with the silver. Knowing that my sample likely contained PM's, per her instructions, I granulated it and attempted an AR dissolution on a small sample, not knowing how much silver was present. Some of the sample did dissolve but the rest did not. In the event that the acids were slow to act it is likely that the alloy contains more than 12% silver and she suggests melting some other metal, copper, brass, gold or zinc, in with the scrap to dilute the silver to less than 10%. She does not suggest inquartation, however others in this forum might choose to inquart the alloy with additional silver. It would seem to me that dilution of the silver would be more efficient and use less nitric than inquartation. If the above is a correct interpretation, which metal would be best used to dilute the silver. I have no more gold to do so, only base metals?
 
Gram for gram it takes twice as much nitric to dissolve copper, zinc, or brass which is a combination of zinc and copper than it does silver. Silver is your friend!
 
Silver is a good choice. Keep in mind the Palladium is also nitric soluble. Parting will give you gold as an insoluble. Rarely see platinum in dental. The silver and Palladium will be in the nitric acid. Drop the silver with hydrochloric acid to get silver chloride which you can reduce to silver metal. Now Palladium is in solution and copper will cement it out leaving the acid with only base metals ready for waste treatment.
 
Were the pieces of metal attacked at all or are they still intact?

If you have tried to digest gold alloy with silver present, you could have a crust of silver chloride. This silver chloride is riddled with fine particle gold and residual dissolved gold. Melting with this crust present can result in gold losses, though I'm not sure what kind of quantities you are dealing with.

If you go ahead with melting, heat slowly with excess soda ash in the flux to help reduce losses.



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For 1 gram of copper about 4.3 ml of nitric.
For 1 gram of pure silver about 1.2 ml of nitric.
1 gallon of nitric will dissolve about 2 lbs copper or 7 lbs of silver.
 
Thanks for the replies. Just for grins and giggles, I decided to run some numbers. Please tell me if I've made any errors in the following calculations.

Lets assume a 100 gm. sample is 16K gold (66.6 gms.), and 18% silver (18 gms.), the rest being base metal (15.34 gms). I realize Pd and some Pt might be present but for now these are left out of the calculation for the amount of HNO3 required.

As a silver dilution example, in a 100 gram sample of which 18 gms is silver, in order to dilute the silver to less than or equal to 10% would require a total weight of 180 gms., thus an additional 80 gms. of diluent metal, copper, zinc, etc. wolud be required.

Inquartation calculates thusly, (66.66 gms. gold/.25) = 266.6 total weight - 100 gms. original sample weight = 166.66 gms. total silver needed less the 18gms. silver already present = 148.64 gms. silver added for a net 6K sample weighing 266.6 gms.

Because there are disparate total weights to be processed, 180 gms. vs. 266 gms., the total amount of silver, 18 gms in the dilution example vs. 166.66 gms in the quartation example and the differing amount of base metals likely to be present (assuming no Pd or Pt), 15.34gms. + 80gms. added diluent = 95.34 gms base metal in the dilution example vs. 15.34 gms in the inquartation example, the following HNO3 requirements can be determined.

Using Platinum's above reference of 4.3 ml vs. 1.2 ml HNO3 to dissolve copper vs. silver, respectively, it would thus require 166.66 gms. silver * 1.2 ml HNO3/gm = 192 ml for the inquartation leach vs. 95.34 gms. base metal * 4.3 ml HNO3/gm. of copper/zinc/brass = 409.9 ml HNO3 for the dilution method.

Question #1

It's not quite clear what Hoke recommends in diluting the silver to less than 10% and subsequent dissolution. Her suggestion to dissolve a small sample of a high silver content sample in AR to determine if dilution is necessary does not address the method of subsequent dissolution after dilution, ie., whether one uses AR to dissolve the entire mass, or whether one uses a nitric leach on the diluted silver/high gold content sample to dissolve the base metals and AG similar to the leach used after inquartation. I'm unsure (and frankly doubt) if a high gold/low silver content sample would dissolve the base metals and silver using HNO3 alone. My interpretation is that AR is used after dilution. If that is the case, does the calculation for the dissolving the base metal above require 409.9 ml of HNO3 as a component of AR in order to dissolve that amount of base metal? If AR is needed using the amount of HNO3 calculated above, that would require almost 2L of HCl, and subsequent reclamation of the values from AR would necessitate a lot of HNO3 to be driven off before one could drop the gold. Please clarify this for me. As for inquartation, the gold is left behind and only minimal amounts of HNO3 would need to be dealt with. So far, it seems to me to be a no brainer.

AR would also be required using inquartation in the approximate amount of 5.8-7.4 ml AR/gm gold which calculates (using the larger amount of AR per gram) to be 66.66 gms gold * 7.4 ml ml AR/gm. = 492 ml AR. Using a common ratio of AR, HCl (4 parts), H2O (2 parts), and HNO3 (1 part), then 14.3% of the AR is HNO3. 14.3% of 492 ml needed is 70.3 ml HNO3, for a total of 192 ml. for inquartation leach + 70.35 ml. for AR = 262.35 ml HNO3 needed for inquartation and subsequent gold dissolution vs. 409.9ml HNO3 for base metal (and +70.35 ml needed??) for gold in the dilution method.

Question #2

Does the presence of HCl (AR) lower the required amount of HNO3 when dissolving base metals?? Would in fact 409.9 ml of HNO3 in an AR solution be needed??

In summary,

Inquartation with Ag requires 192 ml for leaching plus 70.35 ml for the AR gold dissolution.
Total HNO3 required using silver for inquartation = 262.35

Dilution with base metal(s) requires 409.9 ml for AR/leaching (plus 70.35 for AR gold dissolution, if only a leach is used to dissolve base metals)
Total HNO3 required using base metals for dilution = 409.9 - 480.2 ml

If my calculations are correct, (if I have made any mistakes or improper assumptions, please correct me) then the answer is obvious. Inquartaion is definitely more efficient than dilution regardless of how one dissolved the diluted/low silver content sample.
 
Sorry i'm on the road now so i don't have time to run the numbers.
Here's a worksheet :arrow:
 

Attachments

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snoman701 said:
Were the pieces of metal attacked at all or are they still intact?

If you have tried to digest gold alloy with silver present, you could have a crust of silver chloride. This silver chloride is riddled with fine particle gold and residual dissolved gold. Melting with this crust present can result in gold losses, though I'm not sure what kind of quantities you are dealing with.

Hi snoman701,

Because this is a small sample of a larger batch (approx. 500 gms.), I'm trying to determine if I should inquart the rest or not.

I returned the vessel to heat for an additional 4 hrs. or so to see if any further dissolution would take place. I then decanted the gold/AR solution off and examined the remaining solids. They remainder, (of which there was about 30% of the original volume) resembled the original granulated scrap but was very friable. It was not powder. I attempted to crush what remained with a glass stir rod (I did not pulverize it) to remove/disoldge any encrusted AgCl and covered it again with the AR/Gold solution hoping to dissolve any remaining gold and using up more HNO3, if any. I heated this to boiling for another 2-3 hrs. but did not discern any further reaction. Solids still remain, although I haven't weighed it. I'm confident there is AgCl in there but not certain what, if any, values remain.

Should I add a little more HNO3 in case all has been consumed in the original AR? If that doesn't reveal anything, I thought about dissolving the crud in ammonia to see what solids remain or possibly adding fresh AR to the detritus after pulverizing in a mortar and pestle to see if any further dissolution occurs.

Is it likely this is AgCl with entrained precious metals? How should I determine if it is or not? If it does have values entrained, what is the best way to recover them? If not, should I proceed with the entire lot as above or inquart the remaining scrap?

It appears from the other posts that inquartation would be a good idea for the remainder. It certainly looks like it would make things easier and more clear cut given the above. Any help would be appreciated.
 
If your solution seems to have stopped working I’d advise adding more hcl first as this causes no problems unlike too much nitric.
The fact you dissolved 70% of the metals I think points to the fact you didn’t mix the melt enough and some of it has passivated, with what you have left try the ammonia and then flatten as well as you can the pieces left as a larger surface area will allow the acids to penetrate better.
 
I added some newly mixed AR to the remains and your suspicion is confirmed Nick. I hadn't added enough reagent to totally dissolve the mass. I'll let this reaction go as far as it can. I think it's just a matter of time and enough AR. I didn't seem to find a great deal of AgCl that precipitated out, just undissolved metal which may have been passivated. I'll crush and/or mash what's left until AR no longer acts on it. I have a feeling that it will ultimately dissolve completely. If this turns out to be the case, should I inquart the rest or just hit it with AR? There may not be enough silver to worry about. Thanks for your help.
 
My advice would be to put the scrap into your beaker add enough Hcl to cover and heat then slowly add nitric in small increments , if after an addition of nitric you get no reaction add more Hcl and proceed until your scrap has dissolved.
 
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