# Dissolving the Copper in Sterling (nitric Qty?)



## cejohnsonsr (Oct 7, 2020)

I'm trying to determine how much nitric I need. I've read that about 2ml will dissolve 1gr of silver. But how much more to account for the copper? For example, if I had enough sterling to yield 10ozt silver, I'd also have about 1ozt copper. According to my calculations I'd need about 660ml to dissolve the silver. How much more for the copper? Thank you for any help.


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## butcher (Oct 7, 2020)

Approx:
there are many factors such as alloy makeup, surface oils, dilution, temperature, hydrogen peroxide all can change the reactions and how much metal is oxidized into solution as ions, these are rough figures but should get you started.

One liter (1000ml) 35% HNO3 will dissolve about 410grams of silver or about 120grams of copper.

thus about 32 ml of 35% HNO3 (1/3 liter) will dissolve about 13.2 grams silver or about 3.9 grams of copper.

so about 2.4 ml 35% HNO3 will dissolve around a gram of silver, and about 8.2ml 35% HNO3 to dissolve a gram of copper.

AgNO3 is around nine times more soluble in hot water than cold.

not figuring in an excess of nitric you can figure about 0.25 grams of copper will cement about a gram of silver from the AgNO3 solution, or about an ounce of copper will cement about 3.4 ounces of silver.

The more NO2 and NO gas you can keep in solution the more silver it will dissolve, using a cover glass sort or refluxing, adding more water to begin with and some peroxide can help to keep from wasting the nitric solution, letting the heat of the reaction go until it slows and then only adding heat to increase reaction as necessary, then as it seems to stop reacting you can raise the heat again as needed concentrating the solution which will dissolve even more metal into solution...

P.S. I put up a 500' TV/radio tower in Joplin many years ago.


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## kurtak (Oct 9, 2020)

cejohnsonsr said:


> I'm trying to determine how much nitric I need.



Hard to say - Why ? - because in the first place we have no idea what the concentration of the nitric you are working with is in the first place --- is it 30% - 50% - 70% or ????

also - as butcher pointed out --- there are MANY factors/conditions that can/will determine the actual amount nitric to dissolve a given amount of silver or copper

Roughly speaking - if you are starting with 70% nitric it takes (about) 1 ml of nitric to dissolve 1 gram silver --- that is 1ml 70% nitric diluted with 1 ml distilled water --- so -------



> I've read that about 2ml will dissolve 1gr of silver.



IF (big if) you are starting with 70% nitric - that number is (about) twice "more" nitric needed to dissolve a gram of silver - don't know where you got that number - but it's wrong & you would be wasting nitric

That said - here are a couple of examples of how factors/conditions can/will change the amount of nitric needed

(1) when I make my electrolyte for my silver cell - I use 68% nitric & I can dissolve 500 grams of silver crystals (harvested from the cell) with as little as 350 ml but no more then 450 ml nitric --- so less then 1 ml nitric per gram silver (I have successfully dissolve 500 grams silver with as little as 300 ml 68% nitric)

(2) on the other hand - if I tried to dissolve a solid 500 gram bar of silver - it would likely take between 1.2 to 1.5 ml of 68 % nitric to dissolve that same 500 grams of silver --- so more then 1 ml per gram

concerning dissolving copper with nitric - again - depending on factors/conditions - it takes (about) 4 times more nitric to dissolve 1 gram copper then it takes to dissolve 1 gram silver 

Bottom line --- before we can even come close to giving you an answer we need to know (1) what is the percentage of the nitric you are starting with & (2) what is the method/conditions you are using to do the job

Kurt


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## Palladium (Oct 9, 2020)

HOW TO CALCULATE THE AMOUNT OF NITRIC ACID
NEEDED FOR YOUR SILVER CELL
It takes 1.2 ml of 67-70% nitric acid to dissolve 1 gram of pure silver.
It takes 1.45 ml of 67-70% nitric acid to dissolve 1 gram of sterling silver.
It takes 1.53ml of 67-70% nitric acid to dissolve 1 gram of coin silver.
It takes 4.15 ml of 67-70% nitric acid to dissolve 1 gram of pure copper.

HOW TO CALCULATE THE AMOUNT OF NITRIC ACID
NEEDED FOR YOUR STERLING SILVER
Example: I have 2148 grams of sterling silver I want to process.
2148 grams X 1.45 ml of nitric acid = 3114.6 ml of nitric acid and to that you will add
3114.6 ml of distilled water.

How to figure any kind of silver and the amount of nitric needed.
Let say we have something that we know is 40% silver. Then that means the rest is
either copper or some other base metal. Say you material weighs 100 grams. We know
that the silver content of the item at 40% silver should be 40grams of silver and 60
grams of the other base metals. So looking at the chart above for nitric acid consumption
we can figure: 40 grams of pure silver X 1.2 ml per gram = 48 ml of nitric for the silver portion.
For the base metals we figure 60 grams base metal (Copper) X 4.15 ml = 249 ml nitric.
So to refine this 100 grams of 40% silver we would add 249ml + 48ml = 297 ml nitric.
To that we would add 297ml of distilled water.

Note: 1 gram of copper will precipitate 3.4 grams of silver from solution.


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## kurtak (Oct 11, 2020)

Palladium said:


> HOW TO CALCULATE THE AMOUNT OF NITRIC ACID
> NEEDED FOR YOUR SILVER CELL
> It takes 1.2 ml of 67-70% nitric acid to dissolve 1 gram of pure silver.
> It takes 1.45 ml of 67-70% nitric acid to dissolve 1 gram of sterling silver.
> ...



The problem I have with that is the words "NEEDED" and "It takes"

When I first joined the forum & was just getting started with refining - member "stihl88" had formulas for dissolving silver & gold in his signature line - which is what I used to calculate the amount of acid I would "need" to use in my refining processes --- (note; - this was when I was still learning)

The formulas in stihl88 sig. line can be seen here

:arrow: Stainless Steel Silver Cell

However - the more I dove into refining (silver in particular) I "soon" found out that the 1.17 ml 70% nitric to dissolve 1 gram "pure" silver was "in fact" NOT an exact true calculation --- using that calculation I would sometimes end up with free nitric remaining (all the silver was dissolved but the nitric was not all used up) - other times all the nitric was used up but not all the silver was dissolved

At that time - I was getting my nitric in 2.2 liter bottles of 70% "lab grade" nitric

Making electrolyte for my silver cell is the best example I can give

The very first time I made my electrolyte I went & bought some 999 silver coins (5 ozt.) - calculated the 50/50 nitric/water (with a little extra water for evaporation) - put the coins in a beaker along with "all' the acid/water & put it on the hot plate - there came a point where there was still about (plus/minus) one ozt silver left - but also clearly no more reaction taking place on the remaining silver --- I "needed" to add more nitric (& water) then the "so called" calculated amount to finish dissolving the silver

After that first time making my electrolyte I started using the "pure" silver crystals from my silver cell to make my electrolyte - using that same calculation to make the electrolyte --- & what I found was that sometimes that calculation would not dissolve all the silver - sometimes it would be very little remaining silver but other times there would be quite a fair amount of remaining silver 

Then there were other times when using that same calculation it would dissolve "all" the silver - so I would add some more silver crystals to make sure the electrolyte was absolutely pregnant with silver - sometimes it took very little silver to be sure all the nitric was used up - but - other times I was amazed at how much more silver I had to add

Now why is that ? (that sometimes it takes more then the calculated acid & other times it takes less)

It's because it depends on the conditions under which you are trying to dissolve the silver (or copper) with the nitric- & there are 2 main conditions that come into play here

(1) the surface area of the metal you are trying to dissolve - as in - trying to dissolve a solid piece of silver (say a solid 500 gram bar) - or - trying to dissolve 500 grams of say silver crystals from your silver cell (which has a far greater surface area) --- I can assure you that it will take less acid to dissolve the crystals then to dissolve the solid bar

(2) How much of the NOx you keep in - or loose from your reaction vessel when dissolving the metal - I can assure you that it will take more nitric if the NOx is allowed to escape the reaction vessel & take less nitric if you can keep the NOx in your reaction vessel

Taking those two conditions into consideration I can assure you that you can dissolve more then a gram of silver with less then 1 ml 70% nitric - & on the other hand it can (& likely will) take more then the calculated 1.2 ml 70% nitric to dissolve 1 gram silver

Just as an example of using less acid to dissolve more metal --- when I make the electrolyte for my silver cell I always make it with 500 grams of crystals from my cell (so lots of surface area) I can consistently dissolve that 500 grams silver with (about) 400 ml of 68% nitric - so about .8 ml nitric per gram silver

I say "about" because sometimes the 400 ml nitric will dissolve all 500 grams silver & I will have to add some silver to make sure all the nitric is used up - other times there will be around ten to thirty grams undissolved silver

The reason it is never an exact amount of nitric per grams silver is the two above conditions (1) how dense &/or how fine the crystals I am dissolve are - &/or - (2) how big the pour spout is on the beaker I am using is (some of my beakers have a small pour spout - others have a large pour spout) - so it depends on the amount of NOx escaping (after putting the watch glass on the beaker) small spout retains more NOx - large spout more NOx escapes

I did a post about this back in 2015

:arrow: Silver from Silver solder 40%

On the other hand I have dissolved several large batches of "gold plated" 999 silver ozt coins (to recover the gold plating) & I can assure you that it will take more then the 1.2 ml nitric to dissolve a gram of silver

That's because of the larger surface area - so the reaction is slower - which in turn means less NOx is being produced - which though it may not make a lot of sense - actually allows more of the NOx to escape the spout --- that's because more of the NOx is escaping over the longer time of the slower reaction

Bottom line - there is no such thing as a calculation to determine exactly how much nitric it will take to dissolve exactly how much silver (&/or copper)

Exactly how much will very much depend on the conditions

The best advice I can give is to start with figuring 1 ml 67 - 70 % nitric per gram silver (&/or figure 4 ml nitric per gram copper) - put the silver(or copper) in your beaker - cover it with the calculated amount of distilled water (plus some for evaporation) - put the calculated amount of nitric in another beaker - pour about 50% (half) that nitric in the beaker of the metal/water - let it react until done reacting (or at least until "near" done reacting) - depending on the amount silver (or copper) that has not dissolved add some more nitric - at some point (depending on conditions) you will end up with ether - all the silver dissolved but still have unused nitric in your nitic beaker - or all the nitric used with undissolved silver left so a little more nitric needed --- rarely if ever will you end up using the exact amount nitric calculated per gram metal - because of "the fact" that depending on the conditions - no mater how exact you try to be in your calculation - you will likely end up using ether less nitric then actually needed - or not have enough nitric for what you figured

Opps - Edit to add - there is a third condition (3) heat can/does also play a roll in the conditions as heat plays a roll in the speed of the reaction - which in turn plays a roll in the production of the NOx

Kurt


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## g_axelsson (Oct 11, 2020)

Kurt, that is how I do it too, to put the expected amount of acid in a beaker or a measuring cylinder and add it bit by bit.

The amount of nitric needed is hard to calculate as there are more than one way the reaction can go depending on concentration and temperature. Even pressure can affect the amount needed, 4metals have described high pressure digestion chambers that can use all nitric and don't produce any NOx.

See this video from Nurdrage. https://www.youtube.com/watch?v=U8v2zbUYBgw

The amount needed is as much as it takes to finish the job. Any calculation can only get you in the ballpark.

Göran


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