what waS your last photography related purchase?

[fgorga]

It's possible I somehow forgot to account for this; I really don't know. Kind of hard to imagine since it's a pretty silly thing to overlook, but we all make mistakes.

Not accounting for this difference would certainly account for a weak developer.

The pH of the final solution made with KOH instead of NaOH and not taking into account the difference in molar mass would be significantly lower than expected. Thus, any phenolic developing agent would be much less effective.

I did the math just to satisfy my curiosity...

Parodinal calls for 20g of sodium hyroxide which has a molar weight of 40g/mol

So that works out to be 0.5mol of NaOH.

Potassium hydroxide as a molar weight of 56.1g/mol.

0.5mol of KOH is thus 28.05g.

This assumes they work identically in the developer and all you have to do is achieve an equivalent reactive amount.

I have now exhausted my 50+ year old chemistry knowledge and am more than open to being refuted or corrected

This is correct! You did well!!!!

 

[chuckroast]

Not accounting for this difference would certainly account for a weak developer.

The pH of the final solution made with KOH instead of NaOH and not taking into account the difference in molar mass would be significantly lower than expected. Thus, any phenolic developing agent would be much less effective.



This is correct! You did well!!!!

Thanks! Even the elderly can still do the maffs

Now then, riddle me this Reaction Man. If we only added 20g of KOH instead of the prescribed 28.05g, is there a direct way to compute what effect it would have on the resulting developer pH assuming the correct amount would have yielded a pH of 10?

IOW, is there a way to determine whether this alone would have accounted for @koraks outcomes?

 

[thinkbrown]

Not accounting for this difference would certainly account for a weak developer.

The pH of the final solution made with KOH instead of NaOH and not taking into account the difference in molar mass would be significantly lower than expected. Thus, any phenolic developing agent would be much less effective.



This is correct! You did well!!!!

Okay, not a chemist but I did spend a while reading up on parodinal. The hydroxide isn't just responsible for the the pH. It's partially consumed converting the paracetamol to 4-aminophenol. The typical recipe that calls for 15g of paracetamol is approximately 0.1 mole of paracetamol. IIRC the stoichiometry of the hydrolysis of paracetamol consumes an equal molar amount of the hydroxide, and then the acetic acid resulting from that hydrolysis consumes an additional equivalent.

In the recipe as written you end up with a solution of 250ml of ~1.2m hydroxide which the internet tells me has a pH of around 14. If you used 20g of potassium hydroxide instead you end up with a solution that's around 0.6m hydroxide that's allegedly around 13.8. I'm not sure how big the difference ends up being post dilution

 

[fgorga]

Thanks! Even the elderly can still do the maffs

Now then, riddle me this Reaction Man. If we only added 20g of KOH instead of the prescribed 28.05g, is there a direct way to compute what effect it would have on the resulting developer pH assuming the correct amount would have yielded a pH of 10?

Yes, but it is not straightforward.

One would need to know all of the other components and the pKa of each titratable functional group. You could then derive a set of simultaneous equations that would need to be solved.

IOW, is there a way to determine whether this alone would have accounted for @koraks outcomes?

Again, this would not be straightforward and the result of any calculation would be an approximation with an unknown uncertainty. Doing the experiment would yield a better answer.

 

[chuckroast]

Yes, but it is not straightforward.

One would need to know all of the other components and the pKa of each titratable functional group. You could then derive a set of simultaneous equations that would need to be solved.



Again, this would not be straightforward and the result of any calculation would be an approximation with an unknown uncertainty. Doing the experiment would yield a better answer.

That's kind of what I figured, but I do love solving simultaneous eqns... unless they're differentials.

 

[fgorga]

Okay, not a chemist but I did spend a while reading up on parodinal. The hydroxide isn't just responsible for the the pH. It's partially consumed converting the paracetamol to 4-aminophenol. The typical recipe that calls for 15g of paracetamol is approximately 0.1 mole of paracetamol. IIRC the stoichiometry of the hydrolysis of paracetamol consumes an equal molar amount of the hydroxide, and then the acetic acid resulting from that hydrolysis consumes an additional equivalent.

In the recipe as written you end up with a solution of 250ml of ~1.2m hydroxide which the internet tells me has a pH of around 14. If you used 20g of potassium hydroxide instead you end up with a solution that's around 0.6m hydroxide that's allegedly around 13.8. I'm not sure how big the difference ends up being post dilution

Your pH calculations are certainly not accurate. They ignore all of the other components of the solution that contain titratable groups.

We can agree that the solution made with KOH will have a lower pH than that made with NaOH. Exactly how much is difficult to say... see my other post above.

Furthermore, titration curves are very steep when the pH of the solution is near (roughly +/-1 pH unit) the pKa of the important functional group. Thus, small changes in pH can have large effects on reactivity.

 

[eli griggs]

Good to see this discussion, I simply considered a 1:1 ratio for both.

I'll speak up if I have any problems with my mixes, but I don't believe that'll be the case.

Eli

 

[thinkbrown]

Your pH calculations are certainly not accurate. They ignore all of the other components of the solution that contain titratable groups.

Almost certainly. Given that standard rodinal has a pH of 14 (according to the MSDS) I figured they're probably fairly close though.