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.