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Old 13th September 2015, 10:21 AM
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Default Simple fixer experiment

I'm doing a simple experiment about Sodium Thiosulfate as the only component in the fixer for an entry I'm writing in my blog.

Reading the Photography Theory and Practice book, from L.P.Clerc, second edition, 1947 reprint, found in page 276 a graph that find strange (I reproduced that graph as my first attached graph). The graph as I understand it means that, if you increase the Sodium Thiosulfate percentage up to 40%, the fixing time decreases, but from that percentage on the fixing time increases.

As I find no logical in this, I reproduced the experiment with two film in sheets: Fomapan 100 and Efke Aura IR 820. I used anhydrous Sodium Thiosulfate which I diluted in demineralized water. Fixer temperature has been always constant at 27C. Times I obtained were with continuous fixer agitation. My times are placed in second attached graph - I always stopped the clock when the milky layer's film disappeared.

Why, do you think, the two graphs are so different? What changed from 1937 (second edition first reprint of the book)? Or did I misunderstood something?

One more appreciation: as the Sodium Thiosulfate concentration increases, base film colour become more green (Foma) and more blue (Efke). At 20% concentration, all two film bases are pale green and you can't differentiate one from the other.



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Old 13th September 2015, 11:34 AM
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Purely as an outsider the first graph appears to make sense in that the fixing time becomes longer the more dilute the fixer.

The second graph seems a reverse/contradiction of the first which pretty much goes against most common theories and practices.

If I were to jump to a conclusion, I would say that that there must be some kind of human error.
What were the precise quantities you used in order to get the 20-60% dilutions?

Sorry to question your methods Domingo but the second graph looks fundamentally flawed proving the opposite of the norm/standard that higher dilutions lead to longer processing times.

I'm curious what others think and make of your results and observations.
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Old 13th September 2015, 11:39 AM
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Has it something to do with when the concentration is increased you are nearing the saturation point and the liquid (water) carrying the solution cannot be absorbed into the film sensitive layer as easily or as quickly
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Old 13th September 2015, 11:52 AM
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There is always an optimum for any process, so I can imagine that the dip in the first graph shows that at higher concentrations the reaction becomes inhibited.
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Old 13th September 2015, 11:59 AM
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Also, I'm not too sure if the 2 graphs show the same experiment..

Surely fixing has to do with the silver halides and not so much about the base+fog, base tint etc.?

Just trying to think along with you here to figure out what's going on.
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Old 13th September 2015, 12:12 PM
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Quote:
Originally Posted by MartyNL View Post
[...]
What were the precise quantities you used in order to get the 20-60% dilutions?
[...]
300 grams of Sodium Thiosulfate in 500 ml demineralized water as concentration A.
Then
  1. 50 ml A + 100 ml water to get 150 ml at 20%
  2. 50 ml A + 50 ml water to get 100 ml at 30%
  3. 100 ml A + 50 ml water to get 150 ml at 40%
  4. 100 ml A to get 100 ml at 60%

Quote:
Sorry to question your methods Domingo
[...]
no need to apologize, of course, because I want to know if I did right.
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Old 13th September 2015, 12:42 PM
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Now based on the info. you've provided Domingo I understand it differently;
1. 1+2
2. 1+1
3. 2+1
4. Neat stock

Your "A" dilution is your neat stock and I don't see how the concentrations you've provided bring you to your respective % dilutions. But hey, I'm not a mathematician!
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Old 13th September 2015, 03:05 PM
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Domingo if I've understood your dilution ratios correctly then your concentrations become increasingly more concentrated and NOT dilute.

This will account for your graph which is the same as the first but in reverse.

But maybe something is lost in translation?
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Old 14th September 2015, 07:25 AM
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Quote:
Originally Posted by MartyNL View Post
[...]
But maybe something is lost in translation?
maybe, English is not my first (nor my second) language, so it's likely I have not explained my idea well.

Dilutions come from the formula

V1xC1 = V2xC2

being
  • V1 = Volume 1
  • C1 = Percentile 1
  • V2 = Volume 2
  • C2 = Percentile 2

So you can apply

50 ml x 60% = V2 x 20%

And have

(50x60)/20 = 150

These 150 are the total millilitres you need. Since you have 50 ml (from the neat stock) you only need to fill with 100 ml plain water. Hence the

50 ml A + 100 ml water to get 150 ml at 20%

I think I'm doing well up to this point. Or maybe not.
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Old 14th September 2015, 10:35 AM
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You've lost me already
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