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  1. #1

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    Sep 2002
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    I thought I'd try 6% in Sodium Sulfite in Rodinal just to see what happens. So I got 1 lb of Lauder SF ("photo grade"), mainly because it was $6.50 and the Kodak 1lb bottle was $8.50. Yeah, I'm cheap! The Kodak one spells out that it's Anhydrous, but the Lauder one doesn't say. If it's monohydrous, does that mean it's half water by weight, or that there's a 1:1 SSF-to-H2O molecular ratio and the weight ratio is in relation to their molecular weights? (Sorry if this sounds silly, but I haven't spent a single attosecond thinking about chemistry since high school 20 years ago.) I would guess the SSF is a much larger molecule and many times heavier than water. Basically, are they the same thing, or should I use twice as much of the Lauder for the same W:V ratio, or something inbetween? Also, I don't have a scale, is there a way to convert SSF weight to volume? E.g. if it's 1.5x the weight of water, I'd use 6%/1.5 by volume which would be pretty easily measured by displacement. I do understand that the measurement isn't critical, but I'd like to have a method to the madness.

  2. #2
    Ole
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    Monohydrate meant one molecule of water to each molecule of sodium sulfite: Na2SO3+H2O. But the hydrated sulfite is often a heptahydrate, Na2SO3+7(H2O), which weighs about twice what the anhydrous does.
    -- Ole Tjugen, Luddite Elitist
    Norway

  3. #3

    Join Date
    Sep 2002
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    Heptahydrated makes sense, in that I suspect it's half sulfite to H2O, so simply use twice as much as anhydrous. Probably preferable stock, too, if it's supposed to go into solution in water since the anhydrous will presumably degrade quicker. So not knowing whether I bought anhydrous or hydrous, I now have a 2X error margin. I guess I'll go back and get the Kodak anhydrous.

    But... I did a search on photo.net, and found someone posted using 4g/l with Rodinal 1+75 -- that's 0.4%!!! Someone else agreed, but they used 5g/l! That's got to be a teaspoon or so. I hope they were simply mistyping, meaning 40 or 50g? I would have expected half a percent to have a nonexistent effect? Now I'm totally confused.



 

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