Back to the topic of difference between with plain water and acidified water in “development” I thought I will demonstrate the role of equilibria as elucidated by Mike Ware here. In case of plain water (neutral to slightly alkaline) development, the idea is that some of the ferrous ions produced from photo-reduction of ferric ammonium citrate do not react with the ferricyanide either in the paper or in the solution because of thermodynamic impediment and are lost to the developing effluent as a result. In this case, the effluent should still contain these ferrous ions and given the right condition should form Prussian blue.
Figure 1 is the plain water developing solution as it came after ½ minute with an exposed strip of sensitized paper – it is clear, slightly yellow as would be expected from unreacted compounds from the sensitizer.
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However, as soon as I added a little bit of citric acid and gave it a stir – the solution turned blue instantly (Figure 2.)
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So there was still a lot of ferrous ions in the solution and since it did not get a chance to form Prussian blue on the paper, that much density is lost. And that’s the reason we get lower D
max (as well as clearer whites and higher contrast) in a cyanotype developed in
plain water as opposed to one developed in
acidic water, even though the former looks clear and the latter is blue post-development.