I'm an idiot when it comes to math and need help with this -- even if I had a calculator, I wouldn't know what buttons to push to figure it out: Johnathan Bailey's formula for gold toning solution calls for part B created by dissolving 100g of Sodium Thiocyanate into 1250ml of water. Eyeballing, it that looks like an 8% solution -- correct? But suppose you've got thiocyante in 51% solution form, how do you get to 8% solution? My guess is that mixing 51% solution 1:2.5 with water would get me really close, but what do I know? The goal is to get 1 liter of the stuff and, frankly, the biggest help would be for someone to just say, "Put X ml of 51% in a graduate, fill to 1000ml with water and there you have it -- idiot!"

Working Solutions You are not alone Poco. I finally found the definitive answer to this common problem in Dr. Tim Rudman's excellent book "the Photographer's Master Printing Course". At the back of the book are several pages of tables and formulas for solving lots of these little problems. To get a working strength dilution from a higher percentage solution: 51-8=43 so, 43 parts water with 8 parts of 51% solution will yield an 8% solution. I hope this is helpful. Tim

Think of it in a simple way. A 51% solution has 51 ml of active ingredient and 49 ml of water. To make an 8% solution you need 92 ml of water and 8 ml of active ingredient. (or 920 ml and 80 ml to make a liter). Now you need to put in the graduate 80 ml of active ingredient. Since 100 ml contain only 51 ml of ingredient, you will need to increase it by 80/51=1.569 or 156.9 ml. To make things slightly simpler. The answer to your question is put 157 ml of 51% solution into a graduate and fill it up to 1000ml. This will give you an 8% solution.

Thanks so much, guys! It's like my mind hits a brick wall when it comes to these things. If I get the solution to work, I'll post pics.

No, because you can't divide different units of measure (grams = mass, ml = volume) and get a meaningful answer. According to Kodak's formula for bleach replenisher, a 51% sodium thiocyanate solution contains 19.6 grams per 14.9 mL, or about 1.33 grams/ml. To get 100 grams, you would need 100/1.33 or 75ml. http://www.kodak.com/global/en/professional/support/techPubs/cis111/cis111.jhtml This was the only resource I could find to determine the # of grams per ml in solution. If anyone has a better one, or could check my math, I'd appreciate it.

Actually, it is common practice to prepare percentage solutions by dividing the mass of the chemical by the volume of the solvent it is dissolved in (you need to keep the units in a common system). Thus a 1% solution of Gold Chloride can be prepared by dissolving 1 gram of gold chloride in 100ml of water. The ratio of gold chloride to water is: 1.0/100 = .01 Note that this (.01) is also the number of grams of gold chloride in 1ml To convert to percent, multiply .01 x 100 = 1% 1 ml of the 1% solution thus contains .01 gram of gold chloride. See Percentage Solutions on page 8, Conversion Tables on page 186 and Formula #92 on page 156 of: The Darkroom Cookbook, Stephen G. Anchell.

Tom, I'll defer to your better knowledge of chemistry conventions. Am I misreading Kodak's statement about the grams vs ml's in the 51% solution?

Moose, I think Kodak has pitched you a curve ball. Kodak apparently want you to measure out 14.9ml of the 51% solution, 14.9 ml of which (i.e. the solution) apparently weighs 19.6 grams. The amount of sodium thiocyanate in the 14.9 ml of 51% solution should be about 7.6 grams (.51 x 14.9 = 7.599). Kodak inconsistently intermixes weight and volume specifications for liquid formula comonents in the referenced bulletin - very confusing. There is a very similar Kodak recipe in my Compact Photo Lab Index wherin all liquid measurements are given by volume and all dry chemistry is given by weight.

I made a handy formula that works, hope I can explain. Lets say you are you using 157ml of the 51% solution and want to add water to that. Then it goes like this ---> 157x(51/8-1) = 843.875mL of water. Which adds up to 1000.875mL of 8% solution. I use the same numbers as Dimitri so you can check it against his method

Has it already been said? I don't know but since we are mixing with water, then we know that 1 ml of water weighs 1gm. So 1250ml or water is 1250gm. Add 100gm to that and you have 1350gm total. 1% of that is 13.50gm. Then divide 100/13.5 and you get 7.074% solution by weight(not by volume which is trickier).

Well the original poster said disolve 100gm into 1250ml of water but if the formula says disolve 100gm into water and bring the volume upto 1250ml then it would be nearer to 8%.

As I commented on a thread some months ago, a quick way to convert % solutions to weight of contained solid is to take the percentage (in this case 51%) and multiply the percentage by ten and call it milligrams/millilitre. So, you have 510mg/ml (or .51g/ml). You want 80g in the liter you want to make up, therefore you take 80/.51 ml and make up to 1 litre. The answer is 156.8ml to be exact, but 160ml would be close enough, surely?. Murray