How much AR is needed?

[bswartzwelder]

Hi all. I've been away from the board for quite a while and much of what I had been doing is exactly where I left it. I have done a search of the board, but I must be doing something wrong. I am looking to see how much AR is needed to dissolve 1 gram of gold. Specifically, how much Nitric acid does it take to dissolve 1 gram of gold? I can always start out with a slight excess if HCl, but I didn't want to waste any of the Nitric acid. I don't remember seeing this previously, but I know this can't be the first time this question has been asked. Maybe I phrasing it wrong. Thanks.

 

[Palladium]

3 to 1 ratio
Some use 4 to 1 ratio.

3 ml of hcl (31%) to 1 ml of nitric (67-69%) for gram of gold.

Give or take.

 

[Palladium]

Here's a good post GSP made:

The ratios of HCl and HNO3 in the equations you find are molar ratios and not volume ratios. For example, the equation given by T.K. Rose in his "Metallurgy of Gold" is:
Au + HNO3 + 4HCl = 2H20 + NO + HAuCl4
As you can see, the molar ratio of HCl:HNO3 in this equation is 4:1

You can also find an equation that gives a 3:1 molar ratio, but this is for combining HCl and HNO3 to make aqua regia only and doesn't consider the dissolving of gold or anything else.
HNO3 + 3HCl = NOCl + Cl2 + 2H2O
The NOCl then breaks down:
2NOCl = 2NO + Cl2

To compute the volume ratios, I first must determine the number of moles/liter (that ratio is called the Molarity) in the various strengths of the acids in question. I did this for 31% and 38% HCl and for 69% HNO3.
69% HNO3 = 15.43 Molar
31% HCl = 9.81 Molar
38% HCl = 12.4 Molar

From this, you can compute the volumes of each needed to give a 3:1 and a 4:1 molar ratio. For the HCl, I used 31% because that is the common HCl strength of Muriatic acid and 38% because it is the common strength of lab grade HCl.

For 3:1 molar ratio
31% HCl and 69% HNO3 = 4.7:1 ratio by volume
38% HCl and 69% HNO3 = 3.7:1 ratio by volume

For 4:1 molar ratio
31% HCl and 69% HNO3 = 6.3:1 ratio by volume
38% HCl and 69% HNO3 = 5:1 ratio by volume

So, what volume ratio should you use? Most of you know that, when dissolving 1 ounce of gold, it takes about 125ml of muriatic acid and about 25ml of nitric. I have proven these numbers to myself, with about a +/- 10% variation, many times. Thus, the practical ratio, by volume, seems to be about 5:1, which is slightly more than the 3:1 molar ratio given above. In practice, though, since I prefer an excess of HCl, I usually start with about 150ml of muriatic acid per ounce and end up with about a 6:1, by volume, ratio.

There is much erroneous information about this on the internet. The Wikipedia article on aqua regia, for example, says to use a 3:1, by volume, ratio.

Chris

 

[nickvc]

Use GSPs method of an excess of the Hcl needed and slowly add nitric until all reaction stops with your beaker been on heat and with a watch glass on, this method should avoid an excess of nitric making precipitation easier and saving nitric.

 

[bswartzwelder]

Thank you for the replies. Right now, I think I have about 40 grams of gold to process. I didn't want to mix up a gallon of AR when a few ounces might be enough. I also didn't want to just put it in a little bit of AR and keep having to add more and more to get it to all dissolve, either. Last time I looked, it was still January here in Southern Maryland, so I won't be outside working on this stuff for a while, yet I did want to be ready once the nice weather does finally arrive. Thanks again.

 

[goldsilverpro]

Thank you for the replies. Right now, I think I have about 40 grams of gold to process. I didn't want to mix up a gallon of AR when a few ounces might be enough. I also didn't want to just put it in a little bit of AR and keep having to add more and more to get it to all dissolve, either. Last time I looked, it was still January here in Southern Maryland, so I won't be outside working on this stuff for a while, yet I did want to be ready once the nice weather does finally arrive. Thanks again.

It's really not something you can pinpoint exactly. I usually say that a gallon will dissolve about 2 pounds of gold or iron or copper or zinc or nickel, etc. That would be for a 4 to 1 or 5 to 1, HCl to HNO3, solution. That's about 4.0 to 4.5ml of the AR mix per gram of metal.

 

[Chivingtonj]

I have a problem.
My gold is not dissolving in the aqua Regia solution.
I'm 99% sure I mixed it properly.

Any suggestions??? Thank you!

 

[kernels]

I have a problem.
My gold is not dissolving in the aqua Regia solution.
I'm 99% sure I mixed it properly.

Any suggestions??? Thank you!

You should not be doing any mixing, a quick read of the forum would have taught you that you want to cover your Gold with HCl and slowly add HNO3 until all Gold is dissolved. This avoids an excess of HNO3 which has to be removed before Gold can be precipitated.

Now, as to why your Gold is not dissolving. . .

1. What form is the Gold in ? Jewelry, E-waste . . . ?
2. How did you mix the AR ?
3. What temperature are you trying to dissolve at ?

 

[Palladium]

Need more information?
What karat is the gold?
Is it in sold form or grains and cornflakes?
Did you inquart it?
What strength are the acids?
Heat or no heat?

Better descriptions mean we can give more targeted replies.

 

[Palladium]

You beat me to it!

 

[EMRE]

Here's a good post GSP made:

The ratios of HCl and HNO3 in the equations you find are molar ratios and not volume ratios. For example, the equation given by T.K. Rose in his "Metallurgy of Gold" is:
Au + HNO3 + 4HCl = 2H20 + NO + HAuCl4
As you can see, the molar ratio of HCl:HNO3 in this equation is 4:1

You can also find an equation that gives a 3:1 molar ratio, but this is for combining HCl and HNO3 to make aqua regia only and doesn't consider the dissolving of gold or anything else.
HNO3 + 3HCl = NOCl + Cl2 + 2H2O
The NOCl then breaks down:
2NOCl = 2NO + Cl2

To compute the volume ratios, I first must determine the number of moles/liter (that ratio is called the Molarity) in the various strengths of the acids in question. I did this for 31% and 38% HCl and for 69% HNO3.
69% HNO3 = 15.43 Molar
31% HCl = 9.81 Molar
38% HCl = 12.4 Molar

From this, you can compute the volumes of each needed to give a 3:1 and a 4:1 molar ratio. For the HCl, I used 31% because that is the common HCl strength of Muriatic acid and 38% because it is the common strength of lab grade HCl.

For 3:1 molar ratio
31% HCl and 69% HNO3 = 4.7:1 ratio by volume
38% HCl and 69% HNO3 = 3.7:1 ratio by volume

For 4:1 molar ratio
31% HCl and 69% HNO3 = 6.3:1 ratio by volume
38% HCl and 69% HNO3 = 5:1 ratio by volume

So, what volume ratio should you use? Most of you know that, when dissolving 1 ounce of gold, it takes about 125ml of muriatic acid and about 25ml of nitric. I have proven these numbers to myself, with about a +/- 10% variation, many times. Thus, the practical ratio, by volume, seems to be about 5:1, which is slightly more than the 3:1 molar ratio given above. In practice, though, since I prefer an excess of HCl, I usually start with about 150ml of muriatic acid per ounce and end up with about a 6:1, by volume, ratio.

There is much erroneous information about this on the internet. The Wikipedia article on aqua regia, for example, says to use a 3:1, by volume, ratio.

Chris

Chris,
I came across your old post about the volumes needed to dissolve gold. Those are fine numbers but how would you factor in all other metals that are in karat jewelry when stone removal is conducted. Have you ever come up with milliliter values for Hydrochloric and nitric acid for karat gold digestion?
Thanks

 

[goldsilverpro]

Ralph,

I posted this earlier in the thread. It takes about the same ball park amount of AR, per gram, to dissolve those base metals as it does gold.

"It's really not something you can pinpoint exactly. I usually say that a gallon (of AR) will dissolve about 2 pounds of gold or iron or copper or zinc or nickel, etc. That would be for a 4 to 1 or 5 to 1, HCl to HNO3, solution. That's about 4.0 to 4.5ml of the AR mix per gram of metal.

 

[Platinum]

But you make it look so concise and eloquent the other way!
My post of your post was meant as more of a citation to your post!

 

[goldsilverpro]

Sorry EMRE. I somehow read that incorrectly and thought that your post was from Ralph (Platinum, Palladium, and Rhodium).

 

[EMRE]

Sorry EMRE. I somehow read that incorrectly and thought that your post was from Ralph (Platinum, Palladium, and Rhodium).

Np)))

 

[Wonka]

Ralph,

I posted this earlier in the thread. It takes about the same ball park amount of AR, per gram, to dissolve those base metals as it does gold.

"It's really not something you can pinpoint exactly. I usually say that a gallon (of AR) will dissolve about 2 pounds of gold or iron or copper or zinc or nickel, etc. That would be for a 4 to 1 or 5 to 1, HCl to HNO3, solution. That's about 4.0 to 4.5ml of the AR mix per gram of metal.

Does this apply to platinum aswell?

 

[FrugalRefiner]

If you're asking goldsilverpro to respond, I'm afraid he passed away several years ago. If you move your cursor over a member's name or avatar, you'll see when they last visited the forum.

Dave

 

[Wonka]

Oh, sorry. Didn’t know.

 

[FrugalRefiner]

No problem. It's all a learning opportunity.

Dave