# How much nitric...



## pinman (Feb 3, 2010)

How much nitric is required to dissolve 1oz of gold? Sorry if this question is redundant.


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## Harold_V (Feb 3, 2010)

pinman said:


> How much nitric is required to dissolve 1oz of gold? Sorry if this question is redundant.


Generally a fluid ounce, along with no fewer than 4 ounces of HCl, will dissolve a troy ounce of gold. 

Harold


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## pinman (Feb 3, 2010)

That being the case I may consider having some shipped. Thank you.


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## Oz (Nov 16, 2010)

I ran across this old thread a few days ago and it is a frequent question. As Harold said typically it is stated that it takes 4 oz HCl + 1 oz HNO3 or 118.2941ml HCl + 29.57353ml HNO3 to digest a troy ounce of fine gold.

Well I had some time on my hands tonight just babysitting solutions so I decided to find out for myself. The materials used were as follows;

131.8 grain .9995 fine gold button
Technical grade HCl
68-70% ACS grade HNO3

Since I wanted reasonably accurate results I did fluid measurements to the drop, and mass to better than a tenth of a grain. First I used a class A 100ml graduated cylinder to find out down to the drop (from “my” dropper) how many drops of HNO3 it takes to equal 1ml. To get a good average I worked my way up the cylinder counting how many drops per 5ml and dividing. It worked out to exactly 200 drops per 5ml all six times working up the cylinder. So for “my” dropper, 40 drops of HNO3 equals 1milliliter.

The classic 4:1 ratio is in excess of HCl and I needed the volume to cover the button, so I started with 30 milliliters HCl then added 1 milliliter (40 drops) of HNO3 and put on medium heat. I added water as necessary to make up for evaporation until the end when I raised the heat to just start boiling until the solution was down to 20milliliters and all action had stopped. 

The fine gold was force dried before weighing;
Starting weight was 131.8grains
Finishing weight is 112.3grains
Digested weight is 19.5grains 
So 1milliliter of HNO3 with excess HCl will digest 19.5grains of gold.

It takes 29.57353 milliliters to equal 1 fluid US ounce, so 29.57353 X 19.5grains = 576.683835 grains of fine gold will be digested per fluid ounce of HNO3. There are 480 grains to the troy ounce so 576.683835 grains divided by 480 = 1.201425 Troy ounces of fine gold that 1 fluid ounce of HNO3 will digest. 

In summery;
1 milliliter of nitric will digest 19.5grains or 0.8125 pennyweight or 0.040625 Troy ounces of fine gold.
1 fluid ounce US of nitric will digest 576.68grains or 24.0285 pennyweight or 1.201425 Troy ounces of fine gold.

Said another way;
To digest 1 gram of fine gold requires 0.79 milliliters of nitric
To digest 1 pennyweight of fine gold requires 1.23 milliliters of nitric
To digest 1 Troy ounce of fine gold requires 24.61 milliliters of nitric

If you choose to try replicating this you need to calibrate how many drops it requires from your dropper it takes to make 1 milliliter of HNO3. Be sure to use the actual acid for this as different liquids will have different size drops. It goes without saying that the nitric you use matters. I was using 68-70% ACS Grade nitric made by Mallinckrodt that has impurities in the parts per million range. I would guess that most here are using a lesser grade.

Someone should save me from myself as now I am tempted to try this with silver, copper, palladium, and the list goes on.


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## 4metals (Nov 16, 2010)

Thanks for working that out OZ! I have always used the rule of thumb7.5 ounces of karat gold will dissolve in 1 liter of aqua regia. 

Your number for the nitric, rounded up to a whole number is 25 ml per ounce, add 4 times that as HCl and we have 125 ml per ounce.

1000 /7.5 = 133 ml Aqua regia per ounce. Pretty Close !!!!!

I use that figure for any high grade karat gold, different proportions of copper, nickel etc. cause the in practice number to vary a bit but I was happy to see your evening getting intimate with an eye dropper produced very similar results to observations of large lot refining of karat material.

Thanks!!


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## Oz (Nov 16, 2010)

I am somewhat surprised that those numbers hold up that well for karat gold vs. the fine gold I was using. This is why I was tempted to perform similar tests with the typical alloying elements used in karat scrap. Taken together it would allow for more accurate acid use when dealing with known alloys or similar materials. 

I do not expect it to be all that useful in my refining on the scale I process now, but down the road on assayed bars showing the other elements it could be useful. Hopefully it will help those that use WAY too much nitric to see how little is actually required. That could be a substantial cost savings for those that buy expensive nitric by the liter. 

If those nitric consumption numbers also work for high grade karat gold (say 18K) as well as fine gold, then 1 liter of nitric could digest 40.63 Troy ounces of 18K scrap for a theoretical yield (if plumb) of 30.47 Troy ounces of fine gold with a market value of $40,726.20 at the current spot price. Kinda makes nitric cost seem trivial in that light.


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## goldsilverpro (Nov 17, 2010)

As a close approximation, I've been saying all along that 1 gallon of concentrated nitric (diluted to 2 gallons with water, of course) *OR*, 1 gallon of aqua regia, will dissolve 2 pounds of most base metals (copper, nickel, etc.). That figures to 129 ml/tr.oz. of base metals, which is about the same as the volume of AR needed to dissolve an oz of pure gold. Therefore, it takes the same amount of AR to dissolve pure Au or karat Au.


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## gold4mike (Nov 17, 2010)

Thanks, OZ, for shining a very bright light on the subject. I was feeling bad about the 6 liters of Nitric that I paid to have shipped to me. I see now that using it for karat gold will not be anywhere near as costly as I previously thought.

I'll continue to use homemade nitric or HCl alone for dissolving pins and similar items. It also tells me that the glassware I already own will be sufficient for me for the foreseeable future. No need for 4 liter beakers for 2 or 3 ounces of gold.

Great post!


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## FrugalEE (Nov 17, 2010)

Oz,

Thanks for that post. I learned several things ie calibrating dropper, partially dissolving & measuring the difference, big excess HCL not a factor. When you say "all action had stopped" isn't there some residual very slow action that could continue and even a tiny bit more of gold could be dissolved?

I tried my first AR yesterday using it to clean up a purple stained dishpan and found out that water dilution (to better cover the stained area and reduce fumes) really slowed down the action. Actually, it seemed to stop it.

FrugalEE


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## JATSLO (Jul 20, 2022)

Oz said:


> I ran across this old thread a few days ago and it is a frequent question. As Harold said typically it is stated that it takes 4 oz HCl + 1 oz HNO3 or 118.2941ml HCl + 29.57353ml HNO3 to digest a troy ounce of fine gold.
> 
> Well I had some time on my hands tonight just babysitting solutions so I decided to find out for myself. The materials used were as follows;
> 
> ...


Sodium nitrate is a good substitute when combined with HCL. The reaction produces sodium chloride and nitric acid. The salt, the HCL & nitric acid can be separated from each other producing three chemical compounds. A lot of people waste chemicals rather than refine them.


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