[chrome72]

I am a watch maker and want to replicate an old method that the likes of rolex used.

They would plate brass with zinc then apply a hexavalant black chromate to turn the zinc black. Over time some of these dials when exposed to UV turned various shades of brown.

My understanding is the chromate contained silver which is photosensitive.

I want to make some watches that mimic this and could either be black or over time turn brown with UV exposure. I know I could tinker with the silver content in a black chromate solution.

I was looking through old parents and came across a recipe for hexavalent chromate :

50 grams chromic acid and 3 ml sulfuric acid per 1 liter of water.  I would then add silver ions to the mixture in varying %s.

Is making my own black chromate as simple as buying the ingredient and simply mixing them?  I fully understand that hexavalent chromate is hazardous and would be limiting exposure and having appropriate disposal.

For silver ions would silver nitrate be a good option or finding “silver ions” from a lab supply a better option ?

Thanks for any advice!

[Borek]

apply a hexavalant black chromate

Hexavalent chromate compounds - at least these soluble - are yellow.

My understanding is the chromate contained silver

Chromate per se doesn't contain silver. Perhaps what you means is silver chromate. If you add silver salt (like silver nitrate) to sodium chromate solution (technically any soluble chromate will do, just sodium and potassium are most common in the lab practice), silver chromate will precipitate. It not black though, more like red-brown.

which is photosensitive.

Silver is not photosensitive. Some silver compounds are (halogens and pseudohalogens for sure, no idea about chromate).

50 grams chromic acid and 3 ml sulfuric acid per 1 liter of water.

That's probably more like a recipe for dichromate solution - chromate converts to dichromate in low pH, and adding sulfuric acid makes the solution acidic. Dichromate solutions are orange (compared to the yellow chromate).

Is making my own black chromate as simple as buying the ingredient and simply mixing them?

Assuming "black chromate" is just a slurry of silver chromate - mostly yes. It won't be easy though, you need to control three parameters - pH (if too low chromate converts to dichromate) and concentrations of chromate and silver. A bit like guessing PIN to the dial padlock found lying on the ground

For silver ions would silver nitrate be a good option or finding “silver ions” from a lab supply a better option ?

No such thing as "silver ions" separated from counterions, you can only make/buy salts. Yes, silver nitrate is the most common and the most useful source of silver in inorganic synthesis.

[Corribus]

which is photosensitive.

Silver is not photosensitive. Some silver compounds are (halogens and pseudohalogens for sure, no idea about chromate).

Complex topic. Silver ions are light sensitive and photoreduction is a route to make silver nanoparticles. Generally a sensitizer is needed to make it efficient enough for synthesis.

See, e.g., this paper that used light to generate metallic silver particles from Ag chromate.

Harada et al. Langmuir 2010, 26, 23, 17896–17905 https://pubs.acs.org/doi/10.1021/la102705h

Silver is a complicated metal

[chrome72]

Got any advice then how I should proceed with this then?

[Corribus]

I am no expert on metal plating, sorry.

But silver chromate seems fairly simple to make provided you are willing to hazard working with chromate solutions.

I found this report from 1927(!): https://pubs.acs.org/doi/10.1021/j150276a012

The color is described as brick red and/or lustrous depending on how the reactants were added. I guess you can find more modern procedures.

Ag chromate is a semiconductor, like AgCl, so I guess it will have some photoactivity. Probably shorter wavelength (UV) light will be better, but visible light seems to be effective as well.

https://www.tandfonline.com/doi/full/10.1080/17458080.2015.1110624

Couldn't say what the result will be in your application.

[chrome72]

Black chromate looks to be yellow:orange chromate with the addition of soluble silver. Is that the same thing as silver chromate?

[Borek]

Please elaborate, it is not clear what your question is.

Chromate anion in solution is yellow, dichromate anion in solution is orange. Precipitated salt - silver chromate - is basically red, or dark red. No idea about color of silver dichromate.

[chrome72]

Please elaborate, it is not clear what your question is.

Chromate anion in solution is yellow, dichromate anion in solution is orange. Precipitated salt - silver chromate - is basically red, or dark red. No idea about color of silver dichromate.

I think I know more what I am looking to do.

First I need to make 50g chromic acid.

Chromic acid is sodium dichromate + water + sulfuric acid.

I am using the weights of each ingredient to help me understand % weights.  1 gram of sodium dichromate = 1 g.  1ml of water = 1 g.  1 ml of sulfuric acid =1.84g.

I see a standard recipe for chromic acid is: 
200g sodium dichromate (aka 200 g) this is 6.5% of total weight
100ml water (aka 100 g) this is 3.3% of total weight
1500 ml sulfuric acid (aka 2,760g)  this is 90.2% of total weight

Scaling that to make 50 grams of chromic acid, i really only need to have:
200g sodium dichromate * 6.5% = 3.2g sodium dichromate
100g water * 3.3% = 1.65g water
2,760g sulfuric acid * 90.2% = 45.09g sulfuric acid.
That totals 50 grams of chromic acid. 

Is this the right way to think about it?  Trying to figure out how much to buy.

[Borek]

Numbers are more or less correct, you keep the right proportions, but:

First: that's not a way of making 50 g of dichromic acid, that's a way of making 50 g of a solution containing about 2.7 g of the dichromic acid (at least that's what the stoichiometry says).

Second: while you are right about sulfuric acid mass, you will be measuring it by volume, don't forget to take into account its high density.

[chrome72]

Well shoot.

How do I ge to 50g chromic  acid then?

[Borek]

This is all in a basic stoichiometry, you should start with the balanced reaction reaction, calculate how much sodium dichromate is needed and scale everything to that number. But I will give you a hint: original recipe produced about 166 grams.

Note: you won't be able to isolate the acid itself, so you will be working with the solution. Why do you think you need 50 g of the acid?

[chrome72]

Thanks so much. I haven’t had a chemistry class since high school.

Sounds like I need to do 166/50=30% of the recipe.

I want 50g so I can make a liter of black chromate. The original patent calls for 50 grams chromic acid and 3 ml sulfuric acid per 1 liter of water.

1 liter will give me enough chromate after I add silver nitrate for the blackening.

[chrome72]

Another  question. How would just chromic acid crystals be made so it’s not in a liquid form?

Trying to figure out if this is worthwhile making or should I buy chromic acid crystals online. I know you will probably say just buy but the researcher in me is curious the how

[Borek]

As I already hinted at: there is no such thing as a pure dichromic acid (or at least not in a standard lab conditions).

No idea what they mean by "50 g of the acid". Definitely not the compound itself. Perhaps exactly the mixture you posted recipe for, but that's just a random guess.