Solution % Help - Reversal Processing Bleach - Sulfuric Acid Content

[holmburgers]

After some bad math last night, I came to the conclusion that I needed to add 380mL of sulfuric acid to 570mL of water for my bleach in reversal processing. Luckily this seemed absurd to me and I am happy to report that I do not look like the Joker from Batman....

But seriously, I need help with solution percentages. I posted this in the thread entitled "BW reversal options", but figured it would get more traffic if I reposted and I obviously need help.

In short, I'm using a K-dichromate bleach.

I'm using deep tanks which have a capacity of 1900mL.

I have 48% sulfuric acid from Photoformulary.

My bleach is going to be a 2 part bleach; A is 13g (≈7g/L) of K-dichromate and 950mL of water, B is going to be ____mL sulfuric acid & ____mL water, for a total of 950mL. Add equal parts to get 1900mL of bleach.

In short, what percentage of acid do I want in my bleach? Ilford says "10% concentrated sulfuric acid". As I understood that, it means for my 48% acid I'll need ≈20% in the final solution to get 10% concentrated (≈98%). However, that brings me to an amount of about 380mL of acid, and every account I've seen doesn't add more than 20mL per liter, whether it's battery acid or whatever.

What am I missing? It seems like I actually need a 1% solution of concentrated sulfuric acid, assuming concentrated sulfuric acid is 95-98% pure.

And as a side note; it would be convenient if my part B solution can also double as a stock solution for sensitizing carbon tissues. That requires a concentration of 2-5% K-dichromate. If someone could help me with the math required to do that, I'd be appreciative; understanding that adding equal parts A & B for a reversal bleach might not be possible.

But primarily, help me with my bleach!

Yours truly, a failed student of mathematics, Dog Bless our U.S. of A's Math Education.... vive le avoirdupois! Down with metric!

 

[Kirk Keyes]

I'm not sure where you got your formula, but Ilford has one here:
http://www.ilfordphoto.com/Webfiles/20061291034093.pdf
that says:

Bleach
Two solutions, A and B are mixed as follows:

for solution A add 2g of potassium permanganate to 500ml water

for solution B add 10ml of concentrated sulphuric acid to 490ml water.

If concentrated sulphuric acid cannot be obtained use dilute acid. This can be a 10% solution. If 10% sulphuric acid is used add 100ml to 400ml of water to make the part B bleach solution.

These stock solutions will keep for a long period of time before they are mixed together. For use, mix equal parts of A and B, making a fresh working solution for each film, and discard after use.

So if you have 48% conc sulfuric (you could call it 50% when talking about conc. sulfuric, if you want to), that's a 2x dilution of the conc. sulfuric.

That means you need 20 mls of your 48% sulfuric added to 480 mls water. (10 mls of conc. sulfuric from the instructions times the 2x dilution that your sulfuric already has and that gives you a 20 ml volume you need to use to get the same final conc. in the example.

If those are the instructions you are following, I think you are confusing the 10% acid in Ilfords example as the conc of the stock solution. It's not, they are saying as an example that IF you are starting with 10% sulfuric then to use the volumes they are suggesting.

So to make 1900 mls, Ilford's example is diluting 10 mls of conc sulfuric into 500 mls final volume. That's a 500ml/10ml dilution, which equals a 50X dilution. To make 1900 mls using conc. sulfuric, you would make a 50X dilution with conc sulfuric (38 mls conc into 1900 mls final volume).

With your 48% sulfuric, you have an extra 2x dilution from conc that you need to consider, so you take the 50X in Ilfords formula and divide it by your already diluted by 2X 48% sulfuric. That means you need (50X/2X) = 25X. (Note you need half as much dilution with your already 2X diluted acid.) So you need to use 76 mls of the 48% sulfuric into 1900 mls final volume.

Here's the math two ways:
1900mls/(50X/2X) = 76 mls

(1900 mls*2X)/50X = 76 mls

I find it easiest to work these things out by figuring out dilution factors and then going from there with final volumes.

I hope that covered the bases there for you. (Err, I mean the acids there. )
;^)

 

[holmburgers]

Thank you Kirk, that was extremely helpful. I was indeed misreading and mistaking 10% dilute acid for the final concentration %. It always helps to have someone more experienced break it down for you! I appreciate it, thanks again,

Chris