Like duh? Substituting one cation for another question.

It has been decades since I took any chemistry course and I have forgotten almost everything from that experience. I want to attempt substituting one salt for another in a photographic formula. For example, today I acquired some sodium bromide that I'm hoping will replace either ammonium bromide or potassium bromide in several wetplate collodion formulas.

Does the substitution somehow involve the molecular weights of the compounds in grams/mole? If the compound weights are:

NaBr = 102.894 g/mol
NH4Br = 97.943 g/mol
KBr = 119.01 g/mol

is the substitution (e.g., NaBr for KBr) something simple like 119.01/102.894 = 1.157 and does this indicate I'd have to use 1.157 times the amount of the Sodium salt when substituting it for the Potassium salt? So if the formula called for 1 gram KBr I would need to use 1.157 grams NaBr, or no?

Do I need to calculate something else?

Where's Avogadro when you need him?

Help and thanks for any input.

Joe

Joe;

Substitution for these simple compounds is the ratio as you say, but you cannot just substitute ammonium salts for sodium or potassium salts due to several other factors.

Ammonium salts tend to be far more acidic than the comparable sodium or potassium salts, and they have a far different reaction in photographic processing solutions or sensitized materials.

PE

PE,

Thanks for the quick response.

I know that at least ten salts have been used in various collodion formulas. These are are the bromides and iodides of lithium, ammonium, cadmium, sodium and potassium.

The ammonium salts are very soluble and lead to a quick-ripening collodion with a fairly short shelf-life. I'm assuming Lithium salts would lead to similar results. The ammonium salts also tend to have irritating fumes.

Cadmium bromide and cadmium iodide tend to give a long-lasting, stable collodion but one that is extremely fragile and thick. Cadmium is also carcinogenic.

The other common choices in these formulas are usually KBr and KI but there is a solubility problem with KBr and it is speculated that the precipitate that forms in a collodion containing any potassium salt is KBr falling out of solution. This requires a long "clearing" process for the collodion. But, it would seem to me that this reaction negates the purpose (i.e., increased spectral sensitivity) of adding bromides to the collodion in the first place. Put a Potassium salt in and it grabs the bromine and drops out of the mix. If this is so, why use it or KI in the first place?

So, the NaBr seems a reasonable substitute that will (should?) extend the spectral sensitivity without the need to clear or ripen the collodion. And, I'm assuming the shelf life will be extended and similar to a collodion compounded with Potassium salts.

I'm not sure about the acidic factor in the collodion having any effect, but I do know acidity plays a role in the silver nitrate sensitizing bath.

I have several historical formulae on hand with which I will experiment. You've given me a starting point for compounding them with sodium salts.

Thanks,

Joe

Joe, I cannot help you with collodion, sorry.

I can say that general sensitivity is in the order Cl < Br < BrI < I with silver halide in gelatin. And, development rate is in the opposite order with AgI being very difficult to develop.

PE

smieglitz said:

Does the substitution somehow involve the molecular weights of the compounds in grams/mole? If the compound weights are:

NaBr = 102.894 g/mol
NH4Br = 97.943 g/mol
KBr = 119.01 g/mol

is the substitution (e.g., NaBr for KBr) something simple like 119.01/102.894 = 1.157 and does this indicate I'd have to use 1.157 times the amount of the Sodium salt when substituting it for the Potassium salt? So if the formula called for 1 gram KBr I would need to use 1.157 grams NaBr, or no?

Click to expand...

it's the other way 'round:

Since NaBr is lighter than KBr, there is more Br per gram of the salt. So you need less of it - 0.865 g NaBr can be substituted for 1 g KBr.

I'm not saying anything about the other effects of the substitution - only that if the amount of Br- is what you're after this is the way to do it. All else being equal, and so on...

Ole;

Thanks for that catch. I completely missed the inversion of the ratio. Again, my age is catching up with me.

PE