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.
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.
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
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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
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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
