Kodalith Type Developer Formula

sanking · Feb 21, 2009

I need to develop some lith film to very high contrast for a project but I don't have any kodalith type developer on hand. Could someone send me a formula that would work more or less like Kodalith? I have a fairly good supply of chemicals and would likely have on hand what I need to mix the developer.

Sandy

bill williams · Feb 21, 2009

Kodak D-85

Water 500 mliters
Sodium Sulfite (A) 30 grams
Paraformaldehyde 7.5 grams
Sodium Bisulfite 2.2 grams
Boric Acid crystals 7.5 grams
Hydroquinone 22.5 grams
Potassium Bromide 1.6 grams

WTM 1 Liter

Nicholas Lindan · Feb 21, 2009

sanking said:
I need to develop some lith film to very high contrast for a project but I don't have any kodalith type developer on hand

Ansco 81 isn't bad. It doesn't have the rapid 'infectious' action of Kodalith AB, but it doesn't require any formeldehyde or lye, either. It keeps for months.

Per liter:

35. gm Hydroquinone
55. gm S. Sulfite
80. gm S. Carbonate
5.5 gm Citric Acid
10. gm P. Bromide

Use full strength, developing time not longer than 3 minutes.

sanking · Feb 21, 2009

Hi Bill and Nicholas,

Thanks for the formulas. I have everything but the paraformaldehyde so will probably go with Nicholas's formula. What does that do?

Sandy

Ian Grant · Feb 21, 2009

You can substitute Formalin/formaldehyde solution for the Paraformaldehyde 7.5ml of 40%, then make up a Part B of 5% Sodium Hydroxide in water, mix 1+1 for use. This works very well with very sharp edges.

It's not a scientific substitution but one based on experience, we used this commercially for about 12 years in the mid 70' & 80's.

Ian

nworth · Feb 21, 2009

Kodak D-8 and D-9 are a pair you can choose from. D-9 is a classic A-B formula for infectious development of litho film. D-8 is a normal very high contrast developer.

Kodak D-8 high contrast film developer

Water 750 ml
Soduim sulfite (anh) 90 g
Hydroquinone 45 g
Sodium hydroxide 37.5 g
Potassium bromide 30 g
WTM 1 l

For litho type films
Dilute 2:1 for use. Develop about 2 minutes at 20C. The diluted solution does not keep.
Variation: for low temperature processing (to 0C) use undiluted, to -20C add 25% ethylene glycol.

Kodak D-9 hydroquinone caustic film developer
For extreme contrast with process films

Solution A
Water (53C) 500 ml
Sodium bisulfite 22.5 g
Hydroquinone 22.5 g
Potassium bromide 22.5 g
WTM 1 l

Solution B
Water 1 l
Sodium hydroxide 52.5 g

Use equal parts of A and B. Wash negative thoroughly after development and before fixing to avaoid stains. Use at 18C.

sanking · Feb 21, 2009

Ian,

Just to clarify, I substitute 7.5ml of 40% formalin for the 7.5 g of parafomaldehyde in Part A. Then I mix a separate Part B solution of 5% sodium hydroxide. Then I dilute the Part A with Part B 1:1 for use?

Sandy

Ian Grant said:
You can substitute Formalin/formaldehyde solution for the Paraformaldehyde 7.5ml of 40%, then make up a Part B of 5% Sodium Hydroxide in water, mix 1+1 for use. This works very well with very sharp edges.

It's not a scientific substitution but one based on experience, we used this commercially for about 12 years in the mid 70' & 80's.

Ian

Ian Grant · Feb 21, 2009

Yes, that's it.

Back in the 70's none of my chemical suppliers could find para-Formadehyde anywhere, and no references to tell them what it was so I adder Formalin/Formaldehyde solution. When I found the developer didn't work I realised there was no alkali So I tried it 1:1 with the Part B from my Ilford ID-13, and it worked perfectly.

I now know thanks to the internet that parformaldehyde dissolves to form Sodium Hydroxide & Formaldehyde. It's a US name and only crops up in US designed formulae from Kodak & Agfa Ansco. I've never bothered to check it's more common International name.

Made up the way I suggest it has an extremely long shelf life until mixed.

Ian

bill williams · Feb 21, 2009

Ian,

Thanks very much for the info about the sodium hydroxide! I'd made up the D85 with formalin instead of the paraformaldehyde and didn't get the expected results. Reverted back to my LD20. Now I can try the D85 again for lith printing!

Photo Engineer · Feb 21, 2009

Ian Grant said:
I now know thanks to the internet that parformaldehyde dissolves to form Sodium Hydroxide & Formaldehyde. It's a US name and only crops up in US designed formulae from Kodak & Agfa Ansco. I've never bothered to check it's more common International name.

It is also known as Tri oxy methylene IIRC. (CH2O)3 in a 6 membered ring that falls apart in alkali.

PE

2F/2F · Feb 21, 2009

Does it need to be totally halftone, or just "very high contrast"? Not that these other formulas are not superior, but plain Dektol does quite a job of giving high contrast results on that film.

sanking · Feb 21, 2009

For my project I need totally halftone.

Sandy King

2F/2F said:
Does it need to be totally halftone, or just "very high contrast"? Not that these other formulas are not superior, but plain Dektol does quite a job of giving high contrast results on that film.

Photo Engineer · Feb 21, 2009

sanking said:
For my project I need totally halftone.

Sandy King

Sandy;

To get true Kodalith contrast, you actually need both a Kodalith film and a Kodalith developer. The film itself was specially made to have higher than normal contrast for a film material, and to be especially sensitive to the Kodalith developer.

PE

Kirk Keyes · Feb 22, 2009

Ian Grant said:
I now know thanks to the internet that parformaldehyde dissolves to form Sodium Hydroxide & Formaldehyde.

This is incorrect. I guess you can't trust everything you read on the internet. That said...

Paraformaldehyde is the condensatoin product of formaldehyde, that is, formaldehyde can polymerize with itself and form a solid material. If you let a bottle of formaldehyde solution sit around long enough, you may find that it's got a layer of white goo on the bottom of the bottle. That's from the formaldehyde in the formalin condensing and making paraformaldehyde.

And as PE showed, paraformaldehyde has no sodium in it, so there's no way it can dissolve in water and form sodium hydroxide and formaldehyde. PE mentions that it forms 6-membered rings, but it can also forms a monomer containing up to around 100 formaldehyde units.

It's soluble in dilute acid and base solutions, as they both hydrolyze the paraformaldhyde back into formaldehyde.

Sandy, you asked what the paraformaldehyde did in the D-85 formula - I suspect it's there to harden the emulsion. Perhaps PE can address this.

Anyway, I've used the D-9 formula given below with Kodak LDP-4 direct reversal copy line film with good results.

drazak · Feb 22, 2009

Kirk: When paraformaldehyde dissolves in water it creates a basic solution, I think that's what Ian was refering to. As I'm sure you know, most of our developing agents are base catalyized, thus if there was no other agent to turn the developer basic, it was near inactive.

Ian Grant · Feb 22, 2009

Kirk, I don't disagree about Paraformldehyde, (polyoxymethylene), but the reference I saw was to how it formed formaldehyde & sodium hydroxide in D85, there's plenty of Sodium ions in the solution, and in the Agfa Ansco equivalent AN79 Potassium ions as well.

After I found D85 made up with 7.5ml Formaldehyde solution in place of Paraformaldehyde barely even tarted to develop lith & line films I measured the pH of. It wasn't alkaline, it's 30 years ago so I can't remember the exact pH meter reading now

Kodak now use Sodium Formaldehyde Bisulphite instead of Para-formaldehyde in Kodalith developers and a part B of Sodium/Potassium Carbonate and usually Sodium/Potassium Hydroxide.

Paraformaldehyde in water is neutral so where else is the alkali needed for this lith developer to work coming from ? The developer needs a working pH of around 10. There's a reaction somewhere between the Paraformaldehyde and most likely the Sulphite or (Meta)bisulphite that's generating the hydroxide ions.

Sandy, this Reprolith developer (D85/AN79) gives excellent edge effects on all the lith/line films I've used, and was far superior to any of the non formaldehyde types of developer, which is it's why it's still in production today although the formula has evolved.

Ian

dancqu · Feb 22, 2009

Sandy, search Google for, lith formulas . I've used
a variation on Wall's Normal Hydroquinone with some
success with paper. As with most lith formulas it was
designed for film. Very simple. No guarantee. Dan

Ian Grant · Feb 22, 2009

Kirk Keyes said:
This is incorrect. I guess you can't trust everything you read on the internet. That said...

Perhaps I can almost trust what I read after all

You'd be right to say the statement was simplistic though as it doesn't say what happens with the formaldehyde.

A quick Google search shows the Formaldehyde-Sulphite Clock reaction, and this can take an essentially acidic sulphite/(meta)bisuphite & formaldehyde solution from a pH of less than 7 & turning it to pH between 9-11. The reaction uses H+ leaving an excess of the hydroxide ion -OH,

This is quite obviously the explanation for how D85/AN79 works, it's quite possible if I'd added substantially more than 7.5ml of 40% Formaldehyde and waited sometime for the clock reaction to occur I wouldn't have needed to add a Part B

So yes, the Paraformaldehyde dissolves, forms Formaldehyde, which in turn reacts with the Sulphite/(meta)bisulphite and liberates the hydroxide ion, which in turn forms Sodium or Potassium Hydroxide.

Ian

Photo Engineer · Feb 22, 2009

Actually, any aldehyde reacts with sodium sulfite to form an aldehyde.sulfite adduct. (actually, I believe it is the bisulfite adduct to be more precise). This product is a solid white powder that dissolves in water and can be used for timed release of aldehyde. (see my patent or Kodak pre-bleach formulas for examples)

Formaldehyde also reacts with itself to form a variety of cyclic and polymeric structures of the type stated above, and these can be dissolved in either acid or base to release the original formalin.

Aldehydes also are powerful reducing agents, formalin being at the top of the list. (reducing agents can be developers).....

Therefore, use of formalin or paraformaldehyde in alkali will give you a stronger developer with lith like properties due to the formalin and the increase in pH. The reaction (CH2O)n -> CH2O is electronically neutral to appearances, but is going from a "neutral" species to a reducing species.

http://www.chemindustry.com/chemicals/725857.html or here http://en.wikipedia.org/wiki/Paraformaldehyde

Formaldehyde is a developing agent or reducing agent with low discriminating powers. It fogs film when used as a hardener and has been abandoned in the industry for this purpose. It is very toxic and is suspected as being a carcinogen.

The original formula may be incomplete in that it may not contain enough alkali to do the entire conversion job. I have seen some formulas with NaOH in them. I have also seen (IIRC) formulas with acetone in them. Someone should chime in here on that one as acetone is much less toxic!

PE

Ian Grant · Feb 22, 2009

Photo Engineer said:
The original formula may be incomplete in that it may not contain enough alkali to do the entire conversion job. I have seen some formulas with NaOH in them. I have also seen (IIRC) formulas with acetone in them. Someone should chime in here on that one as acetone is much less toxic!

PE

The whole formaldehyde-sulphite clock reaction starts without any alkali, so that's why D85 seems incomplete at a first glance, It seems that it's the balance of Formaldehyde, sulphite and (meta)bisulphite that's critical to when the pH suddenly switches from mildly acidic to highly alkali. It doesn't happen immediately either. Kodak & Agfa Ansco were obviously using this specific reaction.

Kirk may well be able to tell us more.

Ian

Photo Engineer · Feb 22, 2009

You have it correct Ian.

The FORMATION of the bisulfite adduct releases base, but the alkali catalyses release by the paraformaldehyde with no overt pH change. Thank you.

I have not done the Formaldehyde-Bisulfite reaction in the lab for nearly 50 years.

PE

Kirk Keyes · Feb 22, 2009

Ian Grant said:
This is quite obviously the explanation for how D85/AN79 works, it's quite possible if I'd added substantially more than 7.5ml of 40% Formaldehyde and waited sometime for the clock reaction to occur I wouldn't have needed to add a Part B

So yes, the Paraformaldehyde dissolves, forms Formaldehyde, which in turn reacts with the Sulphite/(meta)bisulphite and liberates the hydroxide ion, which in turn forms Sodium or Potassium Hydroxide.

I suspect you are right that simply adding sufficient formalin solution would have worked. It should have only needed 2.5 times of 7.5 mls to do it. (Paraformaldehyde is essentially 100% formaldehyde, so 100%/40% = 2.5 times needed.)

I used to do formaldhyde analysis, and I could never get reliable results using formalin solutions as my source of formaldehyde. I switched to sodium formaldehyde bisulfite, and I was able to get check samples to be within 10% of the calculated value. The other nice thing about Na Formaldehyde bisulfite is that is dissolves easily in water.

bill williams · Feb 22, 2009

Ok, from the Ted Pella. INC. web site http://www.tedpella.com comes this statement: "10% formalin and 4% formaldehyde on the one hand, and 4% paraformaldehyde on the other, are essentially all the same and differ only in the presence of a small amount of methanol in the former." So, apparently using 2.5 times the 7.5 ml of formalin would equal the 7.5 grams of paraformaldehyde.

Ian Grant · Feb 22, 2009

bill williams said:
Ok, from the Ted Pella. INC. web site http://www.tedpella.com comes this statement: "10% formalin and 4% formaldehyde on the one hand, and 4% paraformaldehyde on the other, are essentially all the same and differ only in the presence of a small amount of methanol in the former." So, apparently using 2.5 times the 7.5 ml of formalin would equal the 7.5 grams of paraformaldehyde.

Which is what Kirk has already said

Photo Engineer · Feb 22, 2009

Formalin and Formaldehyde are synonyms actually.

Formaldehyde is a gas that dissolves in water. The water solution is called Formalin or colloquially 37% Formaldehyde. Chemically then Formalin is the water solution of a gas. The maxium concentration normally found, of Formaldehyde gas in water is 37% which is why the terms have become synonymous.

So, 4% formaldehyde and 4% paraformaldehyde are the same but are not equal to 10% formalin since formlain is formaldehyde. Sorry.

PE

Kodalith Type Developer Formula

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