Diluted Nitric Acid Recovery
- Thread starter jimbo12
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Well read the dealing with waste section and act accordingly then.
Martijn
Well-known member
21 + 20 +13 = 54 oz, not 51.
This makes me think your "inquartation" was actually 21/54= 38.88%
I assume with these numbers the 21oz of gold is carat gold, and not the pure percentage of the lot.
So just checking: was the ratio 21 oz of pure gold to 20oz of copper and 13oz of silver? your gold content with inquarting can only be 54/4=13.5 oz, if your gold was 13.5/21= 15.4K Gold.
Or to ask another way: did you make the gold 25% of the total weight?
if not, there could still be silver or copper in the gold sponge.
Your assumption is partly correct 21 oz was K and Not Pure the total K was approx.15.5K prior to inquarting..Upon Cementing out my silver I was able to recover 411 grams of silver...I am not quite sure how i am going to recover nitric without a lot of work unless i just denox solution for disposal
You either deNOx or dispose, if you deNOx you can't recover.
How did you create your Silver Chloride?
Salt or HCl?
kurtak
Well-known member
In his OP he said ---------
So based on the two numbers he has posted I assume he means his starting material was 21 oz with expected recovery of about 8 oz (not sure if he is using 28 gram ounce or 31.1 troy ounce)
Anyway - that would put his starting material at "about" 10 carat
Edit to add; - opps - he posted just about the same time I did
Kurt
kurtak
Well-known member
Per the bold print (20 oz copper) - why did you choose copper?
Though copper will certainly work silver is by far a better choice
First of all - 67% - 70% nitric (diluted 50/50 with D-water) will dissolve 4 times more silver then copper
To put that in perspective - (& roughly speaking) 1 gallon nitric will dissolve about 8 pounds of silver were as it will only dissolve about 2 pounds of copper (or about 1 ml nitric to dissolve 1 gram silver compared to 4 mil nitric to dissolve 1 gram copper) --- so using copper to inquart is a good way to waste nitric - especially if you pay a high price for nitric
Second - if done right - it is much easier to recover the silver (then copper) from the nitric & end up with nitric that you can re-use to dissolve silver again --- so your nitric even goes further
Third - you never loose your silver so you always have your silver for inquarting again
Forth - it's easier to calculate the amount of nitric needed when dissolving one metal (silver) then dissolving two metals (silver/copper)
I mean - I get it - you likely didn't have enough silver to do all the gold in one batch - BUT - you could have done it in 2 or 3 batches - by reusing the silver from/for each batch AND used WAY less nitric
Also - when you say oz --- are you talking about a 28 gram (regular) ounce - or a 31.1gram Troy ounce
Generally speaking - when talking about refining - we use Troy ounces - which is ozt - not oz
Kurt
I appreciate the information and you are correct I did not have enough silver and yes I was working in Troy Oz.
...Based those calculations I should have used approx 1/3 US Gal. when in fact I used approx. 1 US Gal. Nitric @ approx. 70%
Silver Chloride was Pulled by HCL and Then Sodium Hydroxide and Sugar
...Based those calculations I should have used approx 1/3 US Gal. when in fact I used approx. 1 US Gal. Nitric @ approx. 70%
Silver Chloride was Pulled by HCL and Then Sodium Hydroxide and Sugar
So you did not precipitate more than 70% of the Silver?
Is that a guess or fact?
If fact, why not all?
I’m just puzzled so I’d would love to hear the reasoning.
Do not edit your posts without making a note that it has been edited.
Thanks for the explanation.
kurtak
Well-known member
Current copper spot price (Kitco) is "about" $4.07
He says he used 20 oz copper for his inquart
Assuming he is talking about a regular ounce (oz) & not a troy ounce (ozt) --- 20 oz copper = about 1.25 pounds copper
1.25 pounds copper times $4.07 = $5.08 --- not 60-70 cents (0.6-0.7 dollar U.S.)
That is why I asked him he means ozt when he says oz --- it makes a difference
If he means ozt when he says oz - then it would actually be "a bit" more copper then 1.25 pounds (about 60 grams more)
Kurt
Sorry for the confusion , everything I do is in Grams but the conversion for Ounces was Troy ozt.
My Kitco says not quite 9 USD per Kilo which give 5.4 USD.
So the divider has been on a trip again
kurtak
Well-known member
Per the bold print - nope - your calculation is wrong
1/3 gallon of nitric would not even be enough nitric to dissolve the amount of copper you used to inquart the karat gold let alone the silver used to inquart plus the Ag/Cu in the karat gold
20 ozt copper X 31.1 = 622 grams copper
it takes "about" 4 ml 70% nitric to dissolve 1 gram copper
622 X 4 = 2,488 --- so you needed right close to 2.5 liters of nitric just to dissolve the copper you used for inquarting
13 ozt silver X 31.1 = 404 grams silver
it takes "about" 1 ml 70% nitric to dissolve 1 gram silver --- (in other words - it takes 4 time more nitric to dissolve copper then it takes to dissolve silver --- which is why we use silver to inquart rather then copper)
So - 2,488 ml nitric to dissolve the copper + 404 ml nitric to dissolve the silver = 2,892 ml nitric (or 2.89 liters)
A gallon is 3.78 liters minus the 2.89 need to dissolve just the metal used for inquarting = .89 liter (or 890 ml) left to dissolve the Ag/Cu that is in the karat scrap
you said the karat gold was 15.5K ---- 15.5/24 = .645 (64.5%) gold
21 ozt X .645 = 13.5 ozt gold & 7.5 ozt Ag/Cu
7.5 X 31.1 = 233.25 grams Ag/Cu
Because we don't know how much copper & how much silver is actually in the karat gold - figure you need to dissolve 2/3 copper & 1/3 silver just to be on the safe side
So "about" 155 grams copper & about 77.5 grams silver
155 grams copper X 4 ml nitric = 620 ml nitric to dissolve that copper + another 78 ml nitric to dissolve the silver (in the original karat scrap) = 698 ml nitric
so - add the nitric needed to dissolve the metal used to inquart plus the nitric needed to dissolve the Ag/Cu in the karat scrap & you are very close to 1 gallon nitric
2,893 ml + 698 = 3,591 ml (or 3.59 liters)
3.78 (a gallon) - 3.59 = .19 liter (or 190 ml) unused nitric
Now - it is also important to understand that when dissolving metals - these are rough numbers
Depending on the conditions employed when dissolving metal - it can take more or less acid to dissolve the same amount of metal
That means that - depending on conditions - our 1 gallon of nitric may have been more then enough - just enough - or NOT enough - to do the job --- because your 1 gallon was very close to what you needed
Side note; - without doing any actual calculating - if you had used only silver to inquart - you could have probably got away with only using (plus/minus) 1.5 liters nitric
Kurt
Is 60 degrees C ideal for evaporation of water without loss of nitric? How did you come up with this figure? I plan on evaporating quite a bit of solution soon, and I don't want to lose nitric, but I couldn't come up with an ideal temperature for evaporation... Thanks!
Well, Based on my calculations the Gold that was left (which I have Not Done AR on as of yet) weighs pretty close to what it should be .However The Silver That I added and Recovered Was Spot On . so, I missed some Silver Recovery or there was No Silver at all in the gold that was processed which is Impossible... At this point I will Run My Gold 1 more time in 50/50 Nitric / Distilled Water just to be sure.
You will always lose some nitric when evaporating. Nitric is a gas dissolved in water. Even at room temperature, some nitric will gas off as water evaporates. The more dilute the solution, the greater the percentage of water that will evaporate compared to the nitric, until it reaches its azeotrope. The azeotrope is the concentration at which a solution cannot be further concentrated by simple evaporation or distillation. For nitric, that concentration is about 68%. So if your solution is less than 68% nitric, evaporation will cause the concentration to increase, but some nitric is always going to gas off in the process.
Dave
