Mark - you've probably answered this before, but how are you exposing the test strips?
This is just a simple theory based on lack of knowledge, but could it be that the precipitate is oxidized ascorbate?So I learned that PG + sodium metaborate + ascorbic acid can crystallize. The Dimezone S seems to encourage and/or cause it though. So I think something other than hydrolysis is happening.
I would like to know to which extent variations in composition have a visible effect on characteristic courve, grain, sharpness, etc. Since you have obviously done lots of experiments before you came up with your final result, could you comment on that? Would a 10% variation make a big difference or does it take much more before you saw anything different?And the density-curves comparing it with XTOL:
I've never seen such close graphs. The grain looks the same as XTOL under close exam on the light-table with 22X loupes.
We know that Phenidone/Dimezone S catalyzes oxidation of ascorbate,
But in the end it has the effect of a catalyst. Ascorbate, which would reduce silver very slowly on its own if at all (especially at pH of 8), does it rapidly in the presence of trace amounts of Phenidone or Dimezone S. During regular film development, a lot more Ascorbate is oxidized than Phenidone. Curious why this wouldn't count as a catalytic reaction.This is just not true you misunderstand the meaning of catalysis. Ascorbate ion reduces oxidized Phenidone/Dimezone back to the parent compound. However this is a moleculae for molecule process not the massive oxidation of the ascorbate that your statement implies.
NO, in a catalytic readtion the catalyst remains unchanged. In the catalytic converter of a car the platinum catalyst can be extracted unchanged when the unit is discarded. It is not used up. Super additivity is NOT a catalytic reaction, the Phenidone or Dimezone does get used up. The regeneration of the Phenidone by the ascorbate is not 100% effective.But in the end it has the effect of a catalyst.
Even if some of the Phenidone gets used up in the reaction, this is not a key feature of the reaction, and in fact we would all be a lot happier if the Phenidone would not get used up at all. Fact is that lots of silver ions get reduced per active molecule of Phenidone and fact is also that platinum catalysts in cars don't last forever.Catalysis is the change in rate of a chemical reaction due to the participation of a substance called a catalyst. Unlike other reagents that participate in the chemical reaction, a catalyst is not consumed by the reaction itself.
Fact is that lots of silver ions get reduced per active molecule of Phenidone
if Jerry kindly provided me with a link to his exchange with Pat, we might save this forum from a pointless reiteration of arguments that have already been beaten to death
Your precipitate may be glycol-borate complex ,we will never know.I don't think esters are formed at the mixing temperatures you use:
http://www.chemicalforums.com/index.php?topic=41216.0
For other PC developer PC-TEA when selecting development times for different films I have used a simple multiplier of Xtol time,is there any reason not to use this with one of the concentrates?
Exciting stuff! And just what i need. Wonder how it works on delta 3200?
Mark, once you get a formula with 99% tea, i think i will jump aboard and start doing some testing of my own. I just dumped over 3 liters of dead xtol.
I would like to know to which extent variations in composition have a visible effect on characteristic courve, grain, sharpness, etc. Since you have obviously done lots of experiments before you came up with your final result, could you comment on that? Would a 10% variation make a big difference or does it take much more before you saw anything different?
Measuring viscous liquids by volume becomes tricky! Best to use a syringe from a pet store or vet, or use weight. PE
Has anyone seen this happen?
This is the same beaker of developer, developing test-strips for 9.5 and then 10.25 minutes:
View attachment 55406
What's odd is the left half of the curves are identical, despite the time-difference. The right half of the chart behaves as we'd expect. Any idea why?
Mark Overton
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