Chemistry of Photography. The light sensitive emulsion The latent image Developing the image Fixing the image. The emulsion. AgNO 3 + KBr = AgBr + KNO 3 in gelatin AgBr precipitates (WHY??) and remain in the gelatin to form minute grains.
The light sensitive emulsion
The latent image
Developing the image
Fixing the image
Br - + light Br + e –
The electron then migrates to a shallow “trap” (called a sensitivity site).
Ag + + e - Ag
Species produced include: Ag2+, Ag2o, Ag3+, Ag3o, Ag4+, Ag4o
Why doesn’t it go the other way? i.e. why is it stable?
Ag + + e Ag
(i.e. the reduction of silver ion to metallic silver by a developing solution), proceeds much more easily for an exposed silver halide grain than for an unexposed grain. The “gain” can be ~109.
The reduction potential of the developer must be such that it will develop those exposed silver halide grains, but not large enough to develop them all. (A “fogging” developer…)
What actually happens?
C6H4(OH)2 + Na2SO3 + 2AgBr +NaOH C6H3(OH)2SO3Na +2NaBr+H2O +2Ag
Hydroquinone sodium sulphite silver bromide sodium hydroxide hydroquinone sulphonate sodium bromide water SILVER!
stabilizer ya gotta do something for the bromine! (plus it adjusts the pH)
Chemical velocity: ΔT = 1o C Δvchem = 10%.
The biggest problem after the invention of photography in the 1830’s was the lack of permanency. You have to get rid of that remaining bromide, or eventually the photograph will go black. There are no true solvents of AgBr. When sugar is dissolved in water, and then evaporated, the sugar is recovered. This never happens with AgBr. The residue left behind is always a transformed salt. So what we need to do is make sure the transformed salt is soluble, so it can be washed away.
AgBr + Na2S2O3 = AgNaS2O3 + NaBr (only slightly soluble)
But if we have a more liberal solution of sodium thiosulphate:
2AgBr + 3 Na2S2O3 = Ag2Na4(S2O3)3 + 2 NaBr (bingo!)
Does anything else work? KCN. We won’t go there….