Two inversly related Q's re. gram / ml concentration conversions (;

Hi,

I am ordering chemicals for reversal processing of Plus-x 16mm reversal film but in two instances I am unsure of the amounts I should be using in the formula and therefore cant say how much I need to order in the first place...


The first instance calls for 'Sodium Thiocyanate 9 mL' to be added to a liter of standard D-19 to make a '9mL/liter' concentration ...

*I can only order Sodium Thiocyanate in a solid form measured in grams*

>does the 9ml refer to Sodium Thiocyanate in a super-saturated mixture ?

If this is the case how much Sodium Thiocyanate in gm's does it take to get 9ml of the liquid form that makes sense in the above instructions ?

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the second problem lies in that I need to add '9.5 g Potassium Dichromate' to another 1 liter of water - which would be simple its just that to get around hazmat issues I have to order my Potassium Dichromate in solution

I want to order some sort of 2^n (2x,4x etc...) solution of the original percentage solution so I could simply just add water (double the volume, quadruple the volume etc...) to get the original 9.5 g/liter solution -

> question is: what exactly is the % solution is 9.5g Pot dichromate per liter ?


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original formula instructions here:

http://lavender.fortunecity.com/lavender/569/negtoposdeveloper.html


I remember seeing a page that helped with such conversions for many photographic chemicals - wish I bookmarked it !


any help appreciated

The aim level of NaSCN in D-94 is 6 g/L as the pure salt.

Kodak have H-24 on their website if you would rather refer to that than someone's dodgy re-type of the formulae. I guess that's it's not quite as cool though. I mean Kodak! The scoundrels. What do they know about film anyway?

Best,
Helen

Helen B said:

Kodak have H-24 on their website if you would rather refer to that than someone's dodgy re-type of the formulae. I guess that's it's not quite as cool though. I mean Kodak! The scoundrels. What do they know about film anyway?

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Yeah really, what do they know about film anyway?

You are doing it the hard way! just figure out how much of your starting solution has 9.5 grams of dichromate in it, pour it into a 1l graduate and fill it to the 1l mark.

nick mulder said:

> question is: what exactly is the % solution is 9.5g Pot dichromate per liter ?

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Both sodium and potassium thiocyanate solids are extremely deliquescent and will rapidly absord water and partially dissolve. For this reason they are often supplied as a 50 - 55% solution. Since your formula specifies 9 ml of solution, halve that amount to get 4.5 grams.

avandesande said:

You are doing it the hard way! just figure out how much of your starting solution has 9.5 grams of dichromate in it, pour it into a 1l graduate and fill it to the 1l mark.

Click to expand...

I believe you have misunderstood me - either way I'm still stuck with the same question - how do I calculate % solutions ?
(ie, figuring out 'how much of your starting solution has 9.5 grams of dichromate in it')

if I add X gram amounts of XYZ chemical to water what % solution do I get ?

Are % solutions in terms of %volume, %mass or %'active molecules' ?

I would have thought % 'active molecules' would be the more helpful answer in terms of chemistry (but not in terms of mixing it in the first place) - do we then need to know the relative size of the molecule ? or at least its density in a solid state ? I think I need to know the mole/gram ratio thingy - A mole is a discrete quantity of chemical measured in individual molecules as opposed to mass or volume if I remember back to school (could be mistaken here tho) ? ?

What i am saying is that you shouldn't calculate the percent solution, because doing it on weight basis accuratly is difficult. The formula asks for a certain weight in a <i>volume</i> of water, which is easy. You are asking the wrong question. You don't need to calculate the percent solution.

The other poster's suggestion just to use the 50% stock solution and reducing the amount is a good one.

There are two ways of making percentage solutions. You can have weight per volume or weight per weight.

If you dissolve 10 grams of potassium dichromate in enough water to make 1 liter you have a 1% w/v solution. If you dissolve 100 grams you have a 10% w/v solution.

With weight per weight solutions you have to weight out an amount of the solution rather than mesure it by volume/

Gerald Koch said:

Both sodium and potassium thiocyanate solids are extremely deliquescent and will rapidly absord water and partially dissolve. For this reason they are often supplied as a 50 - 55% solution. Since your formula specifies 9 ml of solution, halve that amount to get 4.5 grams.

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Gerald,

I'll repeat myself:
"The aim level of NaSCN in D-94 is 6 g/L as the pure salt."
That's from the horse's mouth, and the trusty horse is called Kodak. There is no need to guess what concentration the starting solution might be. Kodak did supply NaSCN in solution for D-94, but the procedures for checking the concentration in new and replenished solutions are based on 6 g/L as the aim level. This I know from direct experience.

Best,
Helen

Gerald Koch said:

There are two ways of making percentage solutions. You can have weight per volume or weight per weight.

Click to expand...

Actually, there are 4 types of percent solutions:

w/w = weight per weight
w/v = weight per volume
v/w = volume per weight
v/v = volume per volume

I once suggested that people try to include the appropriate designation when discussing percent solutions, but I was accused of being too nit-picky. Perhaps, but it would remove some of the ambiguity.

there are 4 types of percent solutions:

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Yes, but only two were pertinent to this discussion and I did not want to complicate things further.

Helen B said:

"The aim level of NaSCN in D-94 is 6 g/L as the pure salt."

Click to expand...

Sorry, someone mentioned 9 ml for D-19 and I went from there with my example.

avandesande said:

What i am saying is that you shouldn't calculate the percent solution, because doing it on weight basis accuratly is difficult. The formula asks for a certain weight in a <i>volume</i> of water, which is easy. You are asking the wrong question. You don't need to calculate the percent solution.

The other poster's suggestion just to use the 50% stock solution and reducing the amount is a good one.

Click to expand...

ah hah - ok got ya

the reason I stuck with the % solution question is that the people sending it chems over just directly asked the question "what percentage solution would you like ?" (Bostick and Sullivan)

I'll explain what I have learned here and get it done how you suggest

"the reason I stuck with the % solution question is that the people sending it chems over just directly asked the question "what percentage solution would you like ?" (Bostick and Sullivan)"

Hi Nick,

If you had mentioned that you were going to get the stuff from them in the first place... Odd how they say that you can have 30 g of pot. dichromate without Hazmat charges, but the minimum size they offer is 50 g.

You asked a question about numbers of molecules and all that. Avogadro's got your number, mate. In the Bostick and Sullivan catalog you'll notice that in the description there is 'F.W. 294.18' which is the 'formula weight'. Solutions made using the formula weight are not generally known as percentage solutions, but as x.x molar solutions, eg a 0.1 M solution contains a tenth of a mol (29.4 g in this case) per litre. If you are making up developers etc, it is more useful to have percentage solutions because the calculations are easy.

Best,
Helen