Modern Rodinal Substitutes Part II

Photo Engineer · Sep 2, 2009

I'm sorry I was unclear about the sulfuric acid. Kirk is right of course. The reason I have a gallon of 37% Sulfuric Acid is that for a molar equivalent of 1 L of 50% NaOH, you will need about 1 L of 37% Sulfuric Acid, which is the strongest that can be bought on the open market here. You do not dump the acid onto the hydroxide solution. You must have it in a dilute form. I dilute it to about 3.7% so I have plenty of water ready to flush things with. I also like to have a little extra acid to be diluted kept around the lab to get the splashes, if any.

Sodium Hydroxide is far more harmful and quick acting to cause injury than Sulfuric Acid. That is not to say that they are not each dangerous, but the harm from Hydroxide is virtually instantaneous, especially to the eyes, whereas, you have a few seconds to rinse off the acid.

Also, this combination will generate heat but few fumes are given off. This often depends on the floor material.

This combination is also selected to help preserve my floor just in case.

Take care. All is well in my lab this AM. I left the still warm 1 L bottle in its containment bath of cold water overnight just in case. The lid was just a bit loose. So, if the bottle broke the hydroxide would be contained in a diluting water bath and that was in my stainless steel sink. The loose cap would prevent any potential explosive condition in case contaminants had gotten into the hydroxide somehow. It would also protect against a vacuum being created over the hydroxide as air and water vapor condensed as it cooled further.

Rodinal is perfectly safe albeit the solutions are very very alkaline and they must be handled as such. It is the mixers that have to take extra precautions.

PE

gainer · Sep 2, 2009

We do not need a 50% solution of either NaOH or KOH. The KOH I got from The Chemistry Store was in the flake form. Less than 100 grams per liter are needed for Rodinal. If the flakes are poured into 750 ml of water at room temperature while stirring, there is no sticking to the bottom and the heat generated does not cause boiling. Add the sulfite next. The p-aminophenol base should dissolve, and you have produced a Rodinal expedient.

Can anyone tell me the solubility of potassium 4-methyl paraminophenolate in water, or where to find it?

Ian Grant · Sep 2, 2009

No idea Patrick but I do have some formulae for High Concentration Metol devs, and I'm doing some sorting at the moment.

Ian

Kirk Keyes · Sep 2, 2009

Ian Grant said:
They don't have that free SO3--

Something's different in that early "Rodinal type" developer and it is very high dilution.

I'd say most all developers regardless of whether they start with sulfite, bisulfite, or metabisulfite have free SO3--. That's sulfite in solution, after all. Nothig special there.

Did you mean free SO2? (And not "free SO3--?) If so, once you've raised the pH so that the solution to where it is pretty much not acidic, then you don't have an free SO2 either. The SO2 that's dissolved will become SO3--. So there's no free SO2 in Rodinal.

So what did you mean?

Photo Engineer · Sep 2, 2009

Several of the existing formulas use 25 - 50% NaOH. That is the reason I mention what I did. Many require the use of solid NaOH which is just as bad.

PE

Ian Grant · Sep 2, 2009

It is but we have an excess, that's the nature of Metabisulphite it's seems more than the Molecular weight, and it acts that way in use.

I think the whole Bisulphite/Metabisuphite properties are poorly defined and explained, there is lot of free SO2 when you buy fresh Metabisulphite, not included in the MW.

All the work I can find is into how Metabisulphite is used in practice, and the small difference in the US grade "Bisulphite". I know from practical experience there's a difference.

Ian

gainer · Sep 2, 2009

With regard to the EZ-Rodinal, It would probable be best to dissolve the sulfite and hydroxide at about the same time. IIRC, the sulfite is endothermic. Certainly, the KOH is potentially dangerous. Worse yet is adding water to the dry hydroxide, either K or Na.

I have made KOH upon occasion. I had the potassium carbonate, and can get calcium hydroxide at a garden shop (they call it "slaked lime" or "hydrated lime") . Mix one molecular weight of each in water and you get a precipitate of calcium carbonate and 2 molecular weights of KOH dissolved in the amount of water you used. You'll be hard pressed to recover all the solution by filtering, but the amount that can be decanted after the limestone has settled will be the same strength as what is trapped in the limestone. Just get the precipitate out of the container before it dries if you want to get it out without some acid.

Go ahead. It's OK to laugh.

Kirk Keyes · Sep 2, 2009

Ian Grant said:
It is but we have an excess, that's the nature of Metabisulphite it's seems more than the Molecular weight, and it acts that way in use.

I think the whole Bisulphite/Metabisuphite properties are poorly defined and explained, there is lot of free SO2 when you buy fresh Metabisulphite, not included in the MW.

There's more than it "seems" as the metabisulfite uses some of the water it's dissolved in to make the SO3--.

But there's not going to be more than there actually is...

gainer · Sep 2, 2009

Almost forgot. The reason I asked about the metolate is that I tried the following:
KOH.....................28 g.
K2SO3..................222 g, 45% w/w K2SO3 (100 g dry.)
Metol....................41.8 g.
Water to 500 ml.
If all dissolves, the solution will contain 28.8 grams K2SO4. It did not all dissolve. I wasted a lot of time trying to dissolve it by bringing the pH up to 14.5 (stupid meter didn't know any better). I added water to make 750 ml and it became clear in more ways than one. It still is a potent stock solution.
The K2SO4 is not much in the working solution, but is a fair proportion of the stock in terms of 4methyl paraminolphenol. In any case, something is not all dissolving that I thought ought to.

Photo Engineer · Sep 2, 2009

Patrick;

I won't laugh, but one operation I will not do is filter a concentrated KOH solution. Too much chance of an accident, and the KOH usually ends up dissolving the filter paper. The CaCO3 is colloidal and can go through the filter paper.

Best use a sintered filter for this one, if it even works.

PE

Kirk Keyes · Sep 2, 2009

gainer said:
If all dissolves, the solution will contain 28.8 grams K2SO4.

Why is there 28.8 grams K2SO4 in there? I get 21.2 g by my calcs.
The MW of metol is slightly less (172.19) vs that of K2SO4 (174.3), but there is only 1/2 mole of sulfate for every mole of metol free base, so you should get just a bit more than half of what you started with. Metol, the source of the sulfate ion, is the limiting reagent, and there is an excess of KOH.

Kirk Keyes · Sep 2, 2009

By the way, Cole-Parmer sells 45% w/w potassium hydroxide for $44 for a 1 L bottle. And that's AR grade, really pure stuff! That will have 655 grams KOH per liter. 500 g of ACS Certified KOH pellets (really pure in solid form) will cost you $76.

It's worth buying it in liquid form in my book. And a lot safer...

gainer · Sep 2, 2009

The MW of Metol is listed as 344.39, which includes 1 MW of H2SO4, 98.08. I used 1/8 MW of Metol, so 0.125 MW H2SO4 went in and were neutralized by 0.25 MW of KOH leaving 0.125 MW of K2SO4 and some water. I calculate now 22.01 g K2SO4. I must have slipped a cog sometime about 2 AM. I'm close to you, but not for the same reasons. There are TWO molecules of the Metol base for each molecule of H2SO4, so each MW of Metol requires 4 MW of KOH to completely free two MW of the base, which is (344.39 - 98.08)/2 or 123.16. Since I'm using 1/8 MW of Metol, I need 28 g KOH.

Kirk Keyes · Sep 3, 2009

gainer said:
There are TWO molecules of the Metol base for each molecule of H2SO4

Ha! There's one half sulfuric acid molecule complexed with every N-methyl-p-aminophenol! :^)

By the way, 41.8/344.38= 0.121 moles, and not 0.125 moles. So you don't actually have a full 1/8th mole.

Even then, 0.125 moles of K2SO4 (MW = 174.26) should equal 21.78 grams.

But we're splitting hairs.

gainer · Sep 3, 2009

I wasn't splitting my hair. I just used the round numbers for most of them. I can't afford to lose too many hairs.

Kirk Keyes · Sep 3, 2009

gainer said:
I wasted a lot of time trying to dissolve it by bringing the pH up to 14.5 (stupid meter didn't know any better).

I think I mentioned earlier in this thread or in Part I that pH meters just are not up to measuring above about pH 12.5 unless you have special electrodes and suitable standards. You should really take any pH meter reading above that level with a really big grain of salt unless you konw what you are doing and have the right equipment...

By the way, why not stop at pH 14?

Kirk Keyes · Sep 3, 2009

gainer said:
I wasn't splitting my hair. I just used the round numbers for most of them. I can't afford to lose too many hairs.

Looking at your photo, it looks like you have more than me!

Kirk Keyes · Sep 3, 2009

gainer said:
In any case, something is not all dissolving that I thought ought to.

Have you collected some and done some spot tests to see what it is? Does it burn after drying? What's the pH if you dissolve it in some more water? If you can dissolve some in water, does it form a precipitate with calcium ions?

There are 3 simple tests to get you headed in the right direction for identifying the precipitate.

Photo Engineer · Sep 3, 2009

Many pH meters come with an electrode correction table for temperature and pH values that deviate too far from the norm. I used to use that when I made up high pH solutions or when I ran processes above 75F.

PE

Photo Engineer · Sep 3, 2009

An important note regarding old and new emulsion types has been bothering me regarding the evolution of the Rodinal developers and several other formulations.

Old style emulsions were washed to a given conductivity, and this was done after the treatment with sulfur. Later emulsions were still washed to a given conductivity, but some had sulfur + gold added, the sulfur being in the form of thiocyanate.

Modern formulas are untreated with sulfur, washed to a given vAg, treated to balance the final salt content at a fixed figure, and then sulfur + gold finished in a manner in which the sulfur may be in the form of some quite exotic compounds. They are then surface treated for spectral sensitivity with some dyes quite different than those used 50 years ago and are treated with agents to preserve the emulsion that are quite different than the early chemistry.

So, today's emulsions are vastly different in precipitation but also in workup prior to coating, and in such a way that the silver solubility is quite different.

Just some thoughts from someone who has worked on each side of the silver curtain.

PE

Kirk Keyes · Sep 3, 2009

Photo Engineer said:
Old style developers were washed to a given conductivity, and this was done after the treatment with sulfur.

You probably mean "Old style emulsions" and not "Old style developers".

Ray Rogers · Sep 3, 2009

Kirk Keyes said:
You probably mean "Old style emulsions" and not "Old style developers".

without a doubt...

Photo Engineer · Sep 3, 2009

Fixed. Thanks.

PE

Ray Rogers · Sep 3, 2009

Photo Engineer said:
An important note regarding old and new emulsion types has been bothering me regarding the evolution of the Rodinal developers and several other formulations.
<cut>
PE

I really don't think I can totally agree with everything said in your orignal post; I can see how the silver halide solubility could be potentially different, due to new grain structures and halide profiles, possibly even affecting the deposted silver and its solubilty, but your other claims of "exotic" sulfur compounds and vastly differing color senstizers really need more comment...

Ray

Photo Engineer · Sep 3, 2009

Ray;

Just converting from conductivity to vAg will change the excess halide available to diffuse into the developer at the start of the process. Using an organic antifoggant or antioxidant instead of Sodium Sulfite in an overcoat will change emulsion properties and use of Thiosulfate as sensitizer instead of Thiocyanate will change the properties.

Modern sensitizing dyes are quite different than the early dyes and we both know that. Examples are modern cyanines and merocyanines and earlier ones include many ortho dyes instead of pan dyes. Erythrosine comes to mind here.

So, I'm sure with some thought, we all could come up with a long long list of simple differences that are quite well known having been posted here before and posted elsewhere.

PE

Modern Rodinal Substitutes Part II

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