# Schwerters recipe



## Slochteren (Sep 11, 2015)

Hi i found this recipe posted by Butcher:

Schwerters solution, which is made from potassium dichromate, and dilute nitric acid solution.
One recipe:
1/4 Oz. (7.9ml) distilled H2O
15grams K2Cr2O7 (potassium dichromate) (bright orange salt)
3/4 Oz. (22ml) HNO3 

The question i have is how strong is the HNO3 in this recipe? If it is 70% i have to use more HNO3 because mine is 53%.

Thanks Paul


----------



## MarcoP (Sep 11, 2015)

Yes, 70% nitric as a common rule in this board is to refer to an acid as full strength. Otherwise "dilute" or ratios are used. Eg.: simply calling Sulfuric Acid it means 98%, Nitric 70% and so on.

I believe you should add less water too, last time I did mine following butcher's recipe I just used 30ml 52% Nitric, close enough, and it works great. My instruments marks 1ml, smaller measurements had to be approximated.

Marco


----------



## Slochteren (Sep 11, 2015)

Thanks,

also the right recipe with 53%nitric is:

3.5 ml Distilled Wtaer
29 ml 53% nitric
15 Gram K2Cr2O7

Right?


----------



## MarcoP (Sep 11, 2015)

Hum, I get

29ml 53% nitric
0.9ml distilled water
15gr potassium dichromate

Edit: usually I'm wrong, better wait seniors 8)


----------



## Slochteren (Sep 11, 2015)

Thanks anyway.

Sunday i have time to make it also i have some time left for searching/advise.

Paul


----------



## g_axelsson (Sep 11, 2015)

I've seen several different recipes of Schwerters solution, so I made a page about it.
http://goldrefiningwiki.com/mediawiki/index.php/Schwerter's_solution

Göran


----------



## goldsilverpro (Sep 11, 2015)

Slochteren said:


> Thanks,
> also the right recipe with 53%nitric is:
> 3.5 ml Distilled Wtaer
> 29 ml 53% nitric
> ...


Wrong



MarcoP said:


> Hum, I get
> 29ml 53% nitric
> 0.9ml distilled water
> 15gr potassium dichromate
> Edit: usually I'm wrong, better wait seniors 8)


Wrong

The question here is the relative strengths of the 53% solution alone and the 70% solution after diluting 25%. By relative strength, I mean how many grams of HNO3 are in, say, one liter (1000ml) of solution. To do this, you MUST find a chart that compares %HNO3, by weight, to the specific gravity (S.G.). From a chart, you can compute the grams per liter of HNO3. 
NOTE: I have searched several times and have found no simple formula for determining the S.G. from the weight %, except for the iffy ball park one I concocted on the 2nd page of the threaded linked below - Specific Gravity = (.006 x % nitric, by weight) + 1. I prefer using a chart for any problem of this type, because it's more accurate. Check this thread for for more info on this subject. 
http://goldrefiningforum.com/phpBB3/viewtopic.php?f=48&t=10457&hilit=graph+specific+gravity&start=0

Here's the chart I use. I use the 20C (68F) column, since it is the closest to normal room temperature:
http://www.handymath.com/cgi-bin/nitrictble2.cgi?submit=Entry

The weight of HNO3 in one liter of 70% nitric acid solution, before and after dilution.
(1a) *Before dilution of 70%*. Find 70% concentration on the left. Move over to the 20C column and read the S.G. = 1.4134 kg/l or, since there are 1000g in a kg, 1413.4 g/l. A liter of 70% nitric weighs 1413.4 grams. However, only 70% of that weight is HNO3 - the rest is water.
Therefore, total HNO3/liter of 70% HNO3 = 70% of 1413.4 = .70 X 1413.4 = 989 grams.

(1b) *After dilution of 70%*. The formula is 3/4, 70% nitric + 1/4, water. Therefore, one liter contains 3/4 (75%) as much HNO3, per liter, as it did before dilution. Therefore, after dilution, there is 989 X .75 = 741g HNO3/liter

The weight of HNO3 in one liter of 53% HNO3
(2) On the chart, the S.G. of 53% HNO3 is 1.3278. Therefore, a liter would weigh 1327.8 grams. Since this 1327.8g is only 53%, the total HNO3 is 1327.8 X .53 = 704g/l

Conclusion:
Straight 53% contains 704g/l of HNO3/liter
Diluted 70% contains 741g/l of HNO3/liter

Straight 53% is weaker than the diluted 70% HNO3 called for in the Schwerter formula. Therefore, there can be no addition of water - more water will make it even weaker. Use it straight. The 53% is a little weaker than it calls for, but, for this particular solution, anything in the ball park will probably work fine

For the math people: If you had both 53% and 70%, how many ml of 70% would you add to a liter of 53% to make it conform to the Schwerter formula?

I might note that, when you dilute 70% nitric with an equal amount of water, the result is 40.1% HNO3, by weight, not 35%.


----------



## MarcoP (Sep 11, 2015)

At the same percentage and at the same volume we end up having different amount on nitric available. A step further, thank you!


----------



## MarcoP (Sep 11, 2015)

If Schwerter formula requires 741gr/l (straight ~55%) then I should add 36.6ml of 70%?


----------



## goldsilverpro (Sep 11, 2015)

MarcoP said:


> If Schwerter formula requires 741gr/l (straight ~55%) then I should add 36.6ml of 70%?



If you add 36.6ml of 70% to 1 liter of 53%, you would then have 1036.6ml


----------



## MarcoP (Sep 11, 2015)

goldsilverpro said:


> MarcoP said:
> 
> 
> > If Schwerter formula requires 741gr/l (straight ~55%) then I should add 36.6ml of 70%?
> ...


That's right, I should need 150ml 70% for each liter of 53% nitric to reach a concentration of 741gr/l.

150ml of 70% contains 148.35gr nitric, plus 704gr already present in the 53% equals 852gr.
1 liter 53%, plus 150ml equals 1150ml.
852/1150=741/1000

Marco


----------



## goldsilverpro (Sep 11, 2015)

MarcoP said:


> goldsilverpro said:
> 
> 
> > MarcoP said:
> ...



Very Good!! I think you're the 1st person to really get the hang of this. Did you guess 150ml or did you calculate it? If you calculated it, what were the steps?


----------



## MarcoP (Sep 11, 2015)

After your very well explained and detailed posts I should have smacked my self for not doing it right. Thank you for your time, patience and knowledge shared! Sure, charts helps but you got us to understand them!

Marco


----------



## MarcoP (Sep 11, 2015)

At 2 am after a continuously bad week I had to guess it. I kept adding till I had the desired concentration and I'm not sure if I could come up with a fixed formula.

Marco


----------



## goldsilverpro (Sep 11, 2015)

I guessed too. When math things are difficult or confusing or too time consuming to set up, zeroing in on the answer with multiple trial and error guesses and a calculator often works the fastest. Don't some computer programs work that way?


----------



## MarcoP (Sep 12, 2015)

goldsilverpro said:


> I guessed too. When math things are difficult or confusing or too time consuming to set up, zeroing in on the answer with multiple trial and error guesses and a calculator often works the fastest. Don't some computer programs work that way?


A software could scroll trough the data and come up with a close match, or range. That's what we roughly do and a guesstimator could be written.

Since this morning I've been trying to find a [stt]formula[/stt] function for a one liter solution but still no go. I recently come up with the idea that I "only" need two common factors (one for each acid) that multiplied by each acid gr/ml will give me the amount of each acid to use. Those factors have something to do with 741gr/L, or 0.741gr/mL.

Marco


----------



## goldsilverpro (Sep 13, 2015)

MarcoP said:


> goldsilverpro said:
> 
> 
> > I guessed too. When math things are difficult or confusing or too time consuming to set up, zeroing in on the answer with multiple trial and error guesses and a calculator often works the fastest. Don't some computer programs work that way?
> ...


Good ol' Algebra

[53%] = .703 g/ml of HNO3
[70%] = .989 g/ml of HNO3
[desired mix] = .741 g/ml of HNO3

X is the number of ml of 70% added to 1000ml of 53%

(.703)(1000) + (.989)(X) = (.741)(1000 + X)
703 + .989X = 741 + .741X
.989X - .741X = 741 - 703
.248X = 38
X = 38/.248 = 153ml


----------



## MarcoP (Sep 13, 2015)

goldsilverpro said:


> Good ol' Algebra
> 
> [53%] = .703 g/ml of HNO3
> [70%] = .989 g/ml of HNO3
> ...



Great work! I, instead, wanted to go directly to the 1 Liter solution, or just change 1000 with the desired volume. Have been a while since my last algebra problem

0.703 · x + 0.989 · y = 0.741 · 1000

I don't know if you have ever noticed but handymath's site have few calculators to make or change concentration using same acid with different concentrations http://www.handymath.com/calculators.html

Anyhow, your function does the job and no need to guess anymore.

Marco


----------



## kadriver (Sep 15, 2015)

Slochteren said:


> Thanks,
> 
> also the right recipe with 53%nitric is:
> 
> ...



This don't look right. I made some with 1 gram potassium dichromate, 10ml 70% nitric and 10ml distilled water.

If I remember (I made it a long time ago) when I added the water and acid in a graduated cylinder to the potassium dichromate, it turned hard as a rock.

I think that if the two liquids are gently heated and the potassium dichromate is added a little at a time the result would have been much better.

Seems like 15 grams mixed with 29ml nitric and 3.5 ml DH2O would make a slurry, not even liquid.

After about 3 years I had to add some more potassium dichromate because it started to weaken.

kadriver


----------



## goldsilverpro (Sep 15, 2015)

kadriver said:


> Slochteren said:
> 
> 
> 
> ...


Just about any combination of the 2 will work, as you've proven. Weaker solutions will probably work slower. All you need is nitric strong enough to dissolve some silver and, if plated, some of the metal underneath the plating and you want this to happen fairly quickly. At the same time, you need a supply of dichromate ions, to form with the dissolved silver the bright-red silver dichromate.


----------



## MarcoP (Sep 15, 2015)

If any amount of Potassium Dichromate could do the job, then each of us could use a concentration that could be used in most circumstances.

Solubility in water:
4.9gr/100mL @ 0C°
11.3gr/100mL @ 20C°
26.3gr/100mL @ 40C°

Since most recipes wants a 30mL solution and Nitric acid @ 741gr/L (~55.2%) weights ~1350gr, we should have about 609gr of water, rounded to 600mL to also account 0C°. In 30mL of 55% Nitric we will have about 18mL of water and able to dissolve, and keep in solution during the whole year, 0.8gr of K2Cr2O7 @ 0C° or 2gr @ 20C°.

In my area temperature rarely falls under 10C° I might dilute mine to 1.5gr/30mL, in fact as kadriver note my mix was also dense and hard to dissolve all of it, daily temperature has always ranged from 30 and 40 C°, in winter I'm sure the solution will be so dense to make it useless.

New to this kind of math, hopefully I didn't make any big mistake.

Kadriver, my math says you'r in the ball mark, the recipe isn't.

Marco


----------



## goldsilverpro (Sep 16, 2015)

I once bought an neat acid testing kit, with 4 glass dropper bottles, 3 for 10, 14, 18K gold test acids and 1 for Schwerter's solution. And, of course, a touchstone. The kit came in a wooden box and there was a little vial containing 20g of potassium dichromate. The directions said to dissolve all the dichromate salt in 3/4 oz of nitric acid + 1/4 fl.oz. of water. If I remember right, there was some salts that didn't dissolve. I let it settle and poured it off into another container. Didn't seem to affect the testing, but I didn't like it.

I made it weaker after that, so all the salts dissolved, but I don't remember the numbers.


----------

