The molarity for pyrogallol in the standard stock solution of PMK is (100g/1L)x(1mole/126.11g) = 0.79M. What is the molarity when we have 6 ml of the same stock solution in 1518 ml of water? I keep coming up with .003, but that seems too low. Thanks.

Where does the 100 g/L come from? Is this your weight of pyrogallol in the stock solution? As for the rest, using the numbers you have above, I also get 0.003 M.

Dilution rate I agree on 0.8 mol/l stock solution. Isn't PMK 1:2:100 dilution, so it would be 1/103 of 0.8 which is about 0.008 mol/l working solution. 6 ml in 1518 would be 1/253 which sounds like 1:2:250 to me. My experience with PMK is quite limited as I used it some years ago on 135 and liked its sharpness but it was pretty grainy, so that grainsize limited resolution. These days I play around with Pyrocat and other Pyrocatechine developers similar at dilution rates of 1:1:100. Typical developing times are 15 mins at 20°C. So I'm afraid that 1:2:250 would take like for ever or even might not be strong enough to get a reasonable dmax. Before adding any film to this thin soup (as thick as japanese Miso soup) take a 135 film leader etc. and develop it for 45-60 mins in this soup. That should give you an impression of dmax available by this soup. (I use this to test Xtol prior to use, if dmax is too low I can save the test film, if dmax is fine I use a test film and if that is still ok I use it for the real films) Let us know if want to use PMK at 1:2:250 or just made a mistake in the maths. Kind regards, Wolfram :confused:

Thank you for your answers. The standard stock solution is made up at a rate of 50g/500ml, i.e., 100g/L. I am looking only at pyrogallol in this analysis, hence M = 0.79 for the stock solution. The standard working solution comprises 10ml A + 20ml B + g/L + 970ml water for a total of 1L (not 1030ml), and for pyrogallol M=0.079. I am sure I have made no error in these calculations. If we have 6ml A in a solution totaling 1518ml, M=.003 for pyrogallol. My reasoning is that 6ml of a 10% solution = 0.6g; (0.6g/1.58L) x (1 M/126.11g) = 0.003. Alternatively, 0.6g/1.518L = 0.39g/L, and (0.39g/1L) x (1M/126.11g) = 0.003. Finally, where M1 is the molarity of the stock solution; V1 is the volume of the stock solution used; M2 is the molarity of the solution made; and V2 is the volume of the working solution: M1 x V1 = M2 x V2. In rearranged form: M1 x V1 / V2 = M2. Plugging in the values, we see that (0.79)(6)/1.518 = .003. It still seems surprisingly low, but is evidently correct. So why do I care? Because I notice staining but no relief image when using the diluted working solution described above, i.e., the one containing 6ml of A in 1.518L of water. That solution has a pH between 9.5 and 10. I know that stain image formation and relief image formation are separate mechanisms, that conditions favoring one are unfavorable to the other, and that when used at a rate of 0.05M, pyrogallol gives a relief image with or without supplemental oxidation between pH 9.5 and 12.5. It therefore occurred to me to calculate the molarity of pyrogallol in the diluted solution to see whether my observation was consistent with the chemistry, or whether I needed new glasses. The answer is .003, meaning that my observation probably is correct, i.e., no relief image is produced at this concentration.

Here's a quick way: http://www.hometrainingtools.com/articles/chemistry-molar-solutions-science-teaching-tip.html http://www.graphpad.com/quickcalcs/Molarityform.cfm