D-72 and sodium metabisulfite

[Dmosher]

I was going to mix up some D72 but all I have is Sodium Metabisulfite. I dug through a fair number of threads on sodium sulfite vs sodium metabisulfite and I'm still note sure if I can substitute the metabisulfite for sulfite..

Can I? At what ratio?

keep it simple please, I was a political science major (no hard science involved)…

Thanks!

 

[Anon Ymous]

No, you can't, although you can use a metabisulfite + sodium hydroxide mix that would practically give you sulfite in solution.

 

[Dmosher]

I have sodium hydroxide. Any thoughts on going that route with d72?

 

[Anon Ymous]

I'm not at home at the moment, I'll get back to you later.

 

[Ian Grant]

The Sulphite in Rodinal starts life as a concentrated Metabisulphite solution, original RO9 Rodinal just neutralise the Metabisulphite, modern Rodinal has a higher pH and uses excess Hydroxide.

I have the conversion for Potassium Metabisulphet and Potassium Hydroxide to form Potassium Sulphite in a spreadsheet but not the Sodium version. It's easy to calculate from Molecular Weights.

Ian

 

[Rudeofus]

Molar weight of Sodium Sulfite anhydrous is about 126, that of Sodium Metabisulfite is 190.1, that of lye is 40. You need half a mol of Sodium Metabisulfite and one mol of lye to make one mol of Sodium Sulfite, which means 95.05g Sodium Metabisulfite and 40g lye put into solution are equivalent to 126g of Sodium Sulfite anhydrous. The 45g of Sodium Sulfite in D-72 equal 0.357 mols, and you would require 33.95g Sodium Metabisulfite and 14.29g Sodium Hydroxide.

 

[Anon Ymous]

Molar weight of Sodium Sulfite anhydrous is about 126, that of Sodium Metabisulfite is 190.1, that of lye is 40. You need half a mol of Sodium Metabisulfite and one mol of lye to make one mol of Sodium Sulfite, which means 95.05g Sodium Metabisulfite and 40g lye put into solution are equivalent to 126g of Sodium Sulfite anhydrous. The 45g of Sodium Sulfite in D-72 equal 0.357 mols, and you would require 33.95g Sodium Metabisulfite and 14.29g Sodium Hydroxide.

This.

Alternatively, you can use sodium carbonate instead of sodium hydroxide, but keep in mind that it will fizz a lot and you need a larger mixing vessel. 1 mol of sodium metabisulfite (190,1g) needs 1 mol of sodium carbonate (106g) and gives 2 mols of sodium sulfite (252g). So, in order to get 45g of sodium sulfite, you will need 33,95g of metabisulfite and 18,93g of anhydrous sodium carbonate (22,14 of the monohydrate). Since you are going to use carbonate to mix D72, you may as well try it.

 

[Dmosher]

I'll be mixing shortly!! Thanks a lot!

 

[Raghu Kuvempunagar]

Alternatively, you can use sodium carbonate instead of sodium hydroxide.

Wouldn't using carbonate and metabisulphite together as you suggested produce bicarbonate as a biproduct and hence might have some effect on the pH of the developer?

 

[Anon Ymous]

Wouldn't using carbonate and metabisulphite together as you suggested produce bicarbonate as a biproduct and hence might have some effect on the pH of the developer?

This reaction between metabisulfite and carbonate gives sulfite and carbonic acid. This acid decomposes to water and carbon dioxide. If you mixed two solutions (of carbonate and metabisulfite), you would get less fizzing than when adding a powder to a solution. In the latter case, powder probably acts as a nucleation site where bubbles are formed, so less carbon dioxide is dissolved. When considering the warm water used for mixing the developer solution, carbon dioxide solubility is low. The reaction of carbon dioxide with water to form carbonic acid is an equilibrium reaction, meaning than not all of the carbon dioxide is converted to acid. Even then, there is a large excess of sodium carbonate, so any carbonate - bicarbonate buffer created would be very close to carbonate pH. In the end, it may have a measurable difference, which is probably indifferent in a non critical application like a print developer.