Help with some basic chemistry (theory) on developers

[Rudeofus]

There was a lot of very active and very important research in B&W processing during late sixties and early seventies, with a completely revised view on development kinetics, induction time. There may be value in books which predate this period, but a book written by a credible source after 1975 is IMHO essential reading material. I do not have the third edition, but did buy the fourth edition. I did find it now for quite moderate prices at abebooks, although their search engine is a bit odd and sometimes only yields very expensive offers.

 

[elerion]

When these old books say "some alkali % concentration is pH x" (like Mees's book, p391), is it % weight or % molar ?

For example, pH of "Na2SO3 (sodium sulfite) 10%" is 9.9 (Developing, the negative technique, C.I and R.E Jacboson, p90)
0.1 molar solution?
or
100g per liter?

Doing my own pH calculations by hand, I get very close results if I consider the concentrations to be molar (mol/L), not if considering % weight.

 

[Photo Engineer]

If it is per cent, then it is likely g/l total or ml / l total. Moles are generally expressed as such.

It is always good to measure pH with a meter though, as the variations in local water supply (even distilled) and age of chemicals can give some variation in final pH.

PE

 

[GLS]

If a text talks about an X% solution of a dissolved solid, it is almost invariably talking about weight%. Molar% is only ever used in the context of comparing reagent stoichiometries; most typically with catalysts.

So, a 10% solution of something in water would mean 100 g/L.

 

[Photo Engineer]

Well, it would mean 100 grams in a total of 1 liter of solution. This varies with the solvent and the solid. It is not precisely 100 g/l. You generally start with 800 ml and add the 100 g of solid and then slowly dilute to a total of 1L while stirring and the solid dissolves.

PE

 

[GLS]

Yes, that is what is meant by 100 g/L: 100 g of solute per litre of resulting solution. However the weight% does not vary with the solute dissolved, only with the solvent used (due to differing solvent densities). However, if we limit the discussion to aqueous solutions, then a 10 weight% solution of anything is 100 g in a total solution volume of one litre.

 

[elerion]

Following with the developer oxidation state restoration... I found this:

[...]developing agent due to reduction of silver halide is decomposed on inactive substances. Sodium sulfite can react with those by-products of the reaction and recover them back into the substances with the properties of the developing agent. Most by-products of the developing agents of Kendall's type has possibility to be regenerated by sodium sulfite.

Which reinforces attachment post #17
Reaction is:
benzoquinone + Na2SO3 + H2O = Na2SO4 + HQ
HQ: hydroquinone
benzoquinone: oxidated HQ
Na2SO4: sodium sulfate

"Most by-products of the developing agents of Kendall's type has possibility to be regenerated"
Seems to be only those with two OH groups? That, doesn't mention it.

 

[Rudeofus]

@elerion: Nope, unfortunately it doesn't work like this. Which source did you find this in? The correct reaction would be benzoquinone + Na2SO3 <==> hydroquinonemonosulfonate sodium salt.

 

[Anon Ymous]

@elerion: Nope, unfortunately it doesn't work like this. Which source did you find this in? The correct reaction would be benzoquinone + Na2SO3 <==> hydroquinonemonosulfonate sodium salt.

IIRC, the products of this reaction are pH dependant and at some point benzoquinone is converted to hydroquinone when reacting with sodium sulfite. I'm not at home at the moment, I'll try to find some details later.

 

[elerion]

Which source did you find this in?

http://lapskojs.klingt.org/lab0d0ble/pdf/film_development.pdf

 

[Rudeofus]

IIRC, the products of this reaction are pH dependant and at some point benzoquinone is converted to hydroquinone when reacting with sodium sulfite. I'm not at home at the moment, I'll try to find some details later.

AFAIK this reaction happens at low pH, i.e. if Quinone is reacted with Sulfur Dioxide. If you react Quinone with Sodium Sulfite at pH 9, you should produce mostly HQMS. Just think of it, most photographic developers could be formulated with minimal amounts of HQ and high amounts of cheap Sodium Sulfite, if Sodium Sulfite could restore oxidized Hydroquinone in meaningful amounts at pH typical for photographic developers.

 

[Rudeofus]

http://lapskojs.klingt.org/lab0d0ble/pdf/film_development.pdf

This is by no means a scientic publication on photo chemistry. Some amateur with interest in Caffenol type developers hacked together a few dozen pages of stuff about developers, most of which contain serious inaccuracies or outright errors.

This publication contains such gems as:

The presence of two strong oxidants like a sodium sulfite and developing agent in the
same solution increases activeness of their oxidation. The stronger oxidant react with
oxygen intensively in the presence of other oxidant than alone. In most cases sodium
sulfite behaves as a stronger oxidant.

and

Each of the developing agents has own limited effectiveness and can
develop the film during certain time defined by Redox potential of the developing agents.

Feel free to use the recipes from this PDF, but please don't bother with the technical back ground presented in it.

 

[Anon Ymous]

AFAIK this reaction happens at low pH, i.e. if Quinone is reacted with Sulfur Dioxide. If you react Quinone with Sodium Sulfite at pH 9, you should produce mostly HQMS. Just think of it, most photographic developers could be formulated with minimal amounts of HQ and high amounts of cheap Sodium Sulfite, if Sodium Sulfite could restore oxidized Hydroquinone in meaningful amounts at pH typical for photographic developers.

Ah, yes, I just checked it. pH needs to be below 4,5 and even then, sodium hydroquinone monosulfonate is also produced to some varying percentage. At the typical developer pH range no hydroquinone is formed.

 

[elerion]

Rudeofus, I missed that about the strong oxidants, so thanks for pointing that out.
The other gem,... yes, I realized too. Just wasn't sure if it was me who had the thing wrong. Found no reference to that in any books. Superadditive action in mainly due to one agent being "close" to the halides, and the other one "refilling" electrons of the former. Meanwhile, I found that redox potential of the source of electrons agent has to be lower than that of the energetic agent. Well, common sense, really, but I didn't realized until then.

Ah, yes, I just checked it. pH needs to be below 4,5 and even then, sodium hydroquinone monosulfonate is also produced to some varying percentage. At the typical developer pH range no hydroquinone is formed.

Thanks for the clarification!

 

[Alan Johnson]

Here's LFA Mason Photographic Processing Chemistry p74 to give hopefully the correct reaction for posts 32:and 17:
" By comparing the reaction products of benzoquinone and sulphite with those of hydroquinone, sulphite and an oxidising agent , [Luvalle] found that over the alkaline pH range the latter gave quantitative yields of hydroquinone monosulfonate, whereas the former gave low yields of monosulphonate together with much of [a] blue-green compound. From these results he concluded that the sulphonate is formed by the reaction of sulphite with the semiquinone of hydroquinone or its dimer rather than with the quinone. The further oxidation of this reaction product to hydroquinone sulphonate is presumeably effected by another molecule of semiquinone"
Appears that if Mason is correct, your conclusions from the attachment to post 17 are not correct and that benzoquinone plays very little part in the process.
btw it is helpful to know what are semiquinones and free radicals as they are often mentioned in this organic chemistry branch.

 

[Rudeofus]

To contribute to our confusion, T. H. James's "Theory of the Photographic Process" adds yet another reaction: HQ + O2 + 2 Na2SO3 <===> HQMS-Na + Na2SO4 + NaOH

All in all I'd conclude, that the overall reaction scheme is a lot more complex than what can be expressed by a simple reaction, and the complete process may depend on concentration of compounds, pH, oxidation rate, presence of trace metals, and others. Some Sulfite ions will end up as Sulfonate, others will be turned into Sulfate. However, no credible source I have seen has claimed, that Quinone is restored to HQ by Sulfite, so I'd suggest we drop that one from our list of possible reactions.

 

[Alan Johnson]

Rudi, the same equation is given in Mason p73 and relates to the case of .aerial oxidation. He says "a similar sequence of reactions will obtain when oxidation is effected by exposed silver halide..The oveall reaction under these conditions then becomes:
HQ + 2AgBr +Na2SO3 --> HQMS + 2Ag + NaBr +HBr "
This leads to the quote from post 40, "the sulphonate is formed by the reaction of sulphite with the semiquinone of hydroquinone or its dimer rather than with the quinone"
IMO there is no confusion but these overall reaction equations leave out the semiquinone step.

 

[Rudeofus]

Many times people here on APUG/Photrio have tried to synthesize HQMS, and yield was sometimes better, sometimes worse, but never 100%, and always an excess of Sulfite had to be used. This tells me, that there is not "one reaction path for oxidation through silver ions and one distinct reaction path for oxidation through aerial oxygen", but that there is a mixture of reaction schemes which will yield variable amounts of Sulfate and HQMS.

And I guess we both agree, that most definitely none of the reaction schemes seen in photographic developers convert nontrivial amounts of Quinone and Sulfite into HQ and Sulfate.

PS: Synthesis descriptions for HQMS state quite clearly, that Quinone is unsuitable for making HQMS, and that Quinone + Sulfite form all kinds of colorful products but not much HQMS. This also supports your statement, that the reactions in photographic developers goes directly from semiquinone to HQMS and do not produce Quinone as intermediary.

 

[Photo Engineer]

I have made HQMS using a method from the internet. I found the concentration to be low requiring recrystallization or the use of more concentrated solutions. This latter becomes very dangerous and so I abandoned it. I find that the reaction to be quite correct, and is what leads to a small change in activity of any HQ developer in the first day or so of activity. However, you do have to be quite careful to be able to detect this change.

PE

 

[elerion]

I'd suggest we drop that one from our list of possible reactions.

Totally agree.

The recent discussion about reactions is a little bit obscure for me right now.

But, meanwhile, I found this, written by Scott Williams, PH.D., which suggests that there're optimun values for the pH, from developer activity point of view, which I didn't know about.

"...developing agents are most active in their anionic state (developers comprising a hydroxyl group) and in a neutrally charged state (for the amine class).
The ionization sweet spot corresponds to the developer’s ionization constant.
For hydroquinone, there will be two pH values whereby a dramatic increase in activity results (one for each hydroxyl group)"

AFAIK, OH-()-OH developers need somewhat higher pH than OH-()-Nxx.
But also, that the activity increases monotonically.

In Glafkides, there's this equation for the oxidation potential of a reducer:
E_red = E0 + 0.058/n + log((Ox)/(Red)) - 0.058*pH

Where it is obvious that a higher pH helps, as we need a E_red below a certain point, stated as 0.12 V as a reasonable value for AgBr film.
From here, we do not see any sweet spot. The higher, the more the molecules are prone to donate e-, i.e. more activity, more grain, more constrast.

Now, I think this is the trick:
If we maintain the pH fixed (good buffering) then, changes to E_red are due to:

E0 + 0.058/n + log((Ox)/(Red))

The relation between Ox and Red concentration in

(Quinone)/(HQ)

(Quinone) = Ka1*Ka2 + Ka1(H+) + (H+)^2
(HQ) = Ka1 * Ka2

E_red decreases with a slope of 0.058 for pH <pKa1
then, the slope softens, to aprox 0.025~0.03, for pKa1<pH<pKa2
then restores to 0.058.

I wouldn't describe this as sweet spot, but the effect is there.
Mees states these transitions at pH 9.8 and 12.3
I suppose the sweet spot is pH 9.8, as the slope get less favorable for us to decrease redox potential whit increasing pH.
But "dramatic increase in activity" ?

 

[Rudeofus]

As you raise pH, HO-()-OH type developers will deprotonize, while R1R2N-()-NR3R4 type developers will give up their extra protons, and as you already described, both will become more active until there is only completely deprotonized species. Once that pH is reached, further increases in pH will no longer change developer activity.

Note: this "sweet spot" strongly depends on the exact kind of substituents: CD-3 needs pH 12 to get going, while CD-4 is perfectly happy at pH 10. Butyl-HQ is more active at lower pH than Methyl-HQ. You really need to look at each developer individually to make that call. There are some rules, but lots of room for surprises, too.

Note: in many cases you do not want to run developers at their "sweet spot". Running them at lower pH can give better image properties such as lower granularity, and most people are perfectly happy with a developer which takes 10+ minutes to develop but gives superior image results.

 

[elerion]

As you raise pH, HO-()-OH type developers will deprotonize

I think I missed something.
Does it mean that as you raise pH, HO-()-OH ends up being O-()-O ?
If so, how does it help activity?

 

[Rudeofus]

I think I missed something.
Does it mean that as you raise pH, HO-()-OH ends up being O-()-O ?
If so, how does it help activity?

As you raise pH, you increase OH- in solution. The more OH- you have, the further the reaction 2 OH- + HO-()-OH <===> 2 H2O + -O-()-O- is shifted to the right. Since -O-()-O- is actually the active species, the concentration of -O-()-O- determines the activity of the developer.

With R1R2N-()-NR3R4 it's the exact same thing: 2 OH- + R1R2H+N-()-NH+R3R4 <===> 2 H2O + R1R2N-()-NR3R4, R1R2N-()-NR3R4 being the active species.

Obviously at some sufficiently high pH all HO-()-OH or R1R2H+N-()-NH+R3R4 will be converted to -O-()-O- or R1R2N-()-NR3R4, respectively, and further increase in pH will and can not convert any more.

 

[Photo Engineer]

All well and good, but to what purpose? I have designed many B&W and color developers without using any of this, but merely the practical knowledge of chemistry. Without that, you need the above and it becomes very technical and difficult. And so, I have answered my own question. Without a few years of the study of chemistry, design or even understanding developers becomes difficult and in the lab it becomes a tedious process. With the knowledge, you have a myriad of good starting points.

Bottom line? Study chemistry. Start with the basics and then do some Organic and Physical chemistry.

PE

 

[elerion]

Thanks to Rudeofus I've realized there's a mistake in my long post:
It is not HQ, but HQ anion (in the relation (Quinone)/(HQ))
That is, C6H4O2--

Also, even though it has been stated that quinone does not convert back to HQ by sulfite action in developers, it does seem to be partially converted back to HQ when pH > 10

Mees p479:
"It is known that at high pH a very rapid disproportionation of quinone into hydroquinone and hydroxyquinone occurs
2 C6H4O2 + H2O -> C6H4(OH)2 + C6H3(OH)O2

which may be followed by the analogous formation of hydroxyhydroquinone and dihydroxyquinone,
2 C6H3(OH)O2 + HO2 -> C6H3(OH)3 + C6H2(OH)O2"

PE, what purpose? Well, I'm not paying attention to these very specific topics from a practical point of view (e.g. design my own developer), but because it helps me to guide my progress, and face it highly motivated.
I'm studying. This obviously takes time. I do have a very basic orgo base, which I'm reviewing and improving gradually, but it isn't trivial to step up from basic theory to something very specific and practical. All your answers help me a lot, and definitely guide me. Thanks a lot for that