Does this look good so far?

[Josecuervohr]

453g of pins added 400 ml of clean strip muriatic acid, then added to that 100 ml nitric acid slowly. Reacted pretty good for about 15 mins then I was left with this.

 

[Yggdrasil]

453g of pins added 400 ml of clean strip muriatic acid, then added to that 100 ml nitric acid slowly. Reacted pretty good for about 15 mins then I was left with this. View attachment 67579View attachment 67580

Welcome to us.
I recommend studying more.

Aqua Regia are usually not recommended for pins.

100ml Nitric can dissolve as much as 100g or more Gold so you have used way too much Nitric.
Color seem odd but I guess the pins are Kovar based so then it makes sense.
Was there any metals left when all reactions had died down?

 

[Yggdrasil]

453g of pins added 400 ml of clean strip muriatic acid, then added to that 100 ml nitric acid slowly. Reacted pretty good for about 15 mins then I was left with this. View attachment 67579View attachment 67580

Here are some links to study.

We ask our new members to do 3 things.
1. Read C.M. Hokes book on refining jewelers scrap, it gives an easy introduction to the most important chemistry regarding refining.
It is free here on the forum: https://goldrefiningforum.com/phpBB3/viewtopic.php?f=54&t=19798
2. Then read the safety section of the forum: https://goldrefiningforum.com/threads/library-index-thread.22351/post-233795
3. And then read about "Dealing with waste" in the forum: https://goldrefiningforum.com/threads/library-index-thread.22351/post-233794

Suggested reading:
https://goldrefiningforum.com/forums/the-library.101/
https://goldrefiningforum.com/threads/when-in-doubt-cement-it-out.30236/
https://goldrefiningforum.com/threa...le-read-this-before-you-post-about-ore.35743/

Video section: https://goldrefiningforum.com/threa...our-visual-learning-members.35185/post-379931

List of abbreviations: https://goldrefiningforum.com/threads/a-glossary-of-common-terms.21242/post-219169

Forum rules is here.
https://goldrefiningforum.com/threads/gold-refining-forum-rules.31182/

 

[4metals]

Looks like you dissolved the metal and the gold and the excess copper in the original pins cemented out gold which is what you see on the bottom of the beaker. There also is likely some undissolved material from the pins.

check the liquid with stannous chloride to assure the liquid is not still holding gold as well.

You may have gotten lucky and dissolver all of the pins and cemented the gold all in one shot. But that would be very lucky. check the solution with stannous chloride and let us know the result.

 

[Josecuervohr]

Welcome to us.
I recommend studying more.

Aqua Regia are usually not recommended for pins.

100ml Nitric can dissolve as much as 100g or more Gold so you have used way too much Nitric.
Color seem odd but I guess the pins are Kovar based so then it makes sense.
Was there any metals left when all reactions had died down?

Yes there is. Took most of the plating off.

 

[Josecuervohr]

Looks like you dissolved the metal and the gold and the excess copper in the original pins cemented out gold which is what you see on the bottom of the beaker. There also is likely some undissolved material from the pins.

check the liquid with stannous chloride to assure the liquid is not still holding gold as well.

You may have gotten lucky and dissolver all of the pins and cemented the gold all in one shot. But that would be very lucky. check the solution with stannous chloride and let us know the result.

Stannous test
View attachment 1000003334.mp4

 

[Yggdrasil]

Yes there is. Took most of the plating off.View attachment 67581

Usually AR becomes green when dissolving Copper.
Take out all the pins that are stripped and wash them clean with a water spray and make surer there are no powder on them.
Spray the powder and liquid back into the beaker and heat it gently to 70-80 C and see if the powder and remnants dissolve.

 

[4metals]

The solution you are testing is very dark and difficult to see any reaction to the stannous. The better way to do the stannous test is to add a drop of the solution to the end of the paper test strip and allow it to spread up the paper.

The paper is acting like a chromatography column and hopefully filtering out the dark colors. Then a bit above the sample drop add a drop of stannous chloride. It too will migrate and where the 2 solutions touch, that is where you look for indication of gold.

This is useful with dark solutions.

 

[haveagojoe]

Also there could still be a lot of free nitric in the solution which might interfere with the stannous test, preventing the reduction of gold which produces the black colour, or immediately redissolving it.

 

[anarxi]

too little aqua regia.
to ensure dissolution, the ratio is 1 to 5.
for 100 grams of material 500 ml. aqua regia.

 

[Martijn]

Yes there is. Took most of the plating off.View attachment 67581

Then there will be hardly any precious metals in solution. The gold that was there, is now cemented out.
No need to test with stannous.

 

[Marcel]

Trying to process such large pins isn’t economical when the total quantity is so small. It’s fundamentally a scale problem: they have a very small surface area compared to their weight. Smaller, thinner pins can be about 100 times more productive per kilogram!

Furthermore, using Aqua Regia (AR) in this situation is not recommended because of cementation issues and the unfavorable copper-to-gold ratio. An electrolytic gold-stripping process with concentrated sulfuric acid would have been far more efficient.

Below is an example calculation to illustrate the difference between thick (5 mm) pins and thin (0.25 mm) pins, assuming a 0.05 µm gold plating. All numbers are approximate and assume ideal cylindrical geometry and perfect coverage:


ASSUMPTIONS:

• Copper density: 8.96 g/cm³
• Gold density: 19.3 g/cm³
• Gold plating thickness: 0.05 µm = 0.000005 cm

---

VARIANT A: THICK PIN (5 mm diameter, 10 mm length)

1. MASS OF ONE PINRadius = 2.5 mm = 0.25 cmHeight = 1.0 cmVolume = π × (0.25 cm)² × 1.0 cm ≈ 0.196 cm³Mass = 0.196 cm³ × 8.96 g/cm³ ≈ 1.76 g
2. PINS PER KILOGRAM1000 g / 1.76 g ≈ 568 pins/kg
3. SURFACE AREA OF ONE PINMantle area ≈ 1.57 cm²End faces combined ≈ 0.39 cm²Total ≈ 1.96 cm² per pin
4. SURFACE AREA PER KG568 pins × 1.96 cm² ≈ 1116 cm²/kg
5. GOLD MASS PER cm² (at 0.05 µm)19.3 g/cm³ × 0.000005 cm = 0.0000965 g/cm²
6. TOTAL GOLD IN 1 KG OF THICK PINS1116 cm² × 0.0000965 g/cm² ≈ 0.108 g of gold
7. GOLD IN 400 g OF THICK PINS0.4 × 0.108 g ≈ 0.043 g of gold

---

VARIANT B: THIN PIN (0.25 mm diameter, 10 mm length)

1. MASS OF ONE PINRadius = 0.125 mm = 0.0125 cmHeight = 1.0 cmVolume = π × (0.0125 cm)² × 1.0 cm ≈ 0.00049 cm³Mass = 0.00049 cm³ × 8.96 g/cm³ ≈ 0.0044 g
2. PINS PER KILOGRAM1000 g / 0.0044 g ≈ 227,000 pins/kg
3. SURFACE AREA OF ONE PINMantle area ≈ 0.0785 cm²End faces ≈ 0.00098 cm² (very small)Total ≈ 0.0795 cm² per pin
4. SURFACE AREA PER KG227,000 pins × 0.0795 cm² ≈ 18,080 cm²/kg
5. TOTAL GOLD IN 1 KG OF THIN PINS18,080 cm² × 0.0000965 g/cm² ≈ 1.74 g of gold
6. GOLD IN 400 g OF THIN PINS0.4 × 1.74 g ≈ 0.70 g of gold

---

COMPARISON (GOLD YIELD)

Side-by-Side Comparison​

Variant	Gold in 1 kg	Gold in 400 g
Thick Pin (5 mm)	~0.108 g of Au	~0.043 g of Au
Thin Pin (0.25 mm)	~1.74 g of Au	~0.70 g of Au

• At only 0.05 µm plating, 400 g of thick pins yields under 0.05 g of gold.
• The same mass of thin pins could yield almost 0.70 g of gold.

In real scenarios, other factors can lower the actual recovery (e.g., incomplete plating coverage, impurities, mechanical design, etc.). But this idealized comparison highlights just how important surface-to-mass ratio is – and underscores why it’s generally not economical to process large, heavy pins (with relatively tiny plating) using something like Aqua Regia. An electrolytic stripping process in concentrated sulfuric acid would typically be more suitable.

(Calculations done with a little help of my friend GPT)

 

[Alondro]

I use the thick heavy pins in my waste container to drag out any gold left in the solution. Then I filter out those foils along with the cemented gold when it's time to process the waste.

Also, it's better to get rid of the base metals BEFORE trying to dissolve the gold. HCl with a little copper chloride added to jump-start the reaction will get rid of most base metals over a few days (with perhaps a small amount of sodium persulfate added too, as a new member has discovered this can greatly speed up the process while still leaving the gold untouched) and leave you with gold foils to filter out, wash, and then purify further with one of several methods to dissolve the extremely thin gold. The cheapest method is HCl with a small amount of bleach (pure sodium hypochlorite solution ONLY!! No other additives!) added little by little, stirring briskly until the gold has dissolved). Then heat lightly for about 2 hours to drive off excess chlorine gas, cool, and drop the gold with SMB as usual.

 

[Josecuervohr]

Trying to process such large pins isn’t economical when the total quantity is so small. It’s fundamentally a scale problem: they have a very small surface area compared to their weight. Smaller, thinner pins can be about 100 times more productive per kilogram!

Furthermore, using Aqua Regia (AR) in this situation is not recommended because of cementation issues and the unfavorable copper-to-gold ratio. An electrolytic gold-stripping process with concentrated sulfuric acid would have been far more efficient.

Below is an example calculation to illustrate the difference between thick (5 mm) pins and thin (0.25 mm) pins, assuming a 0.05 µm gold plating. All numbers are approximate and assume ideal cylindrical geometry and perfect coverage:


ASSUMPTIONS:

• Copper density: 8.96 g/cm³
• Gold density: 19.3 g/cm³
• Gold plating thickness: 0.05 µm = 0.000005 cm

---

VARIANT A: THICK PIN (5 mm diameter, 10 mm length)

1. MASS OF ONE PINRadius = 2.5 mm = 0.25 cmHeight = 1.0 cmVolume = π × (0.25 cm)² × 1.0 cm ≈ 0.196 cm³Mass = 0.196 cm³ × 8.96 g/cm³ ≈ 1.76 g
2. PINS PER KILOGRAM1000 g / 1.76 g ≈ 568 pins/kg
3. SURFACE AREA OF ONE PINMantle area ≈ 1.57 cm²End faces combined ≈ 0.39 cm²Total ≈ 1.96 cm² per pin
4. SURFACE AREA PER KG568 pins × 1.96 cm² ≈ 1116 cm²/kg
5. GOLD MASS PER cm² (at 0.05 µm)19.3 g/cm³ × 0.000005 cm = 0.0000965 g/cm²
6. TOTAL GOLD IN 1 KG OF THICK PINS1116 cm² × 0.0000965 g/cm² ≈ 0.108 g of gold
7. GOLD IN 400 g OF THICK PINS0.4 × 0.108 g ≈ 0.043 g of gold

---

VARIANT B: THIN PIN (0.25 mm diameter, 10 mm length)

1. MASS OF ONE PINRadius = 0.125 mm = 0.0125 cmHeight = 1.0 cmVolume = π × (0.0125 cm)² × 1.0 cm ≈ 0.00049 cm³Mass = 0.00049 cm³ × 8.96 g/cm³ ≈ 0.0044 g
2. PINS PER KILOGRAM1000 g / 0.0044 g ≈ 227,000 pins/kg
3. SURFACE AREA OF ONE PINMantle area ≈ 0.0785 cm²End faces ≈ 0.00098 cm² (very small)Total ≈ 0.0795 cm² per pin
4. SURFACE AREA PER KG227,000 pins × 0.0795 cm² ≈ 18,080 cm²/kg
5. TOTAL GOLD IN 1 KG OF THIN PINS18,080 cm² × 0.0000965 g/cm² ≈ 1.74 g of gold
6. GOLD IN 400 g OF THIN PINS0.4 × 1.74 g ≈ 0.70 g of gold

---

COMPARISON (GOLD YIELD)

Side-by-Side Comparison​

Variant	Gold in 1 kg	Gold in 400 g
Thick Pin (5 mm)	~0.108 g of Au	~0.043 g of Au
Thin Pin (0.25 mm)	~1.74 g of Au	~0.70 g of Au

• At only 0.05 µm plating, 400 g of thick pins yields under 0.05 g of gold.
• The same mass of thin pins could yield almost 0.70 g of gold.

In real scenarios, other factors can lower the actual recovery (e.g., incomplete plating coverage, impurities, mechanical design, etc.). But this idealized comparison highlights just how important surface-to-mass ratio is – and underscores why it’s generally not economical to process large, heavy pins (with relatively tiny plating) using something like Aqua Regia. An electrolytic stripping process in concentrated sulfuric acid would typically be more suitable.

(Calculations done with a little help of my friend GPT)

I have about 75 pounds of these pins, and I didn't pay a whole lot for them. I'm just trying to find which process works best for me with small batches.