how to go from a 40% solution to 4%

Peter Schrager · Jun 28, 2020

I want to take a 40% solution down to 4% for glyoxal
Formulary sells this as 40% in 100ml; but I need it to be a 4% solution
thanks in Advance!!!
Best,Peter

138S · Jun 28, 2020

This is straight, throw to the 100ml 40% solution in a graduated recipient and fill to 1L.

If you have 40ml of glyoxal in 100ml final solution (Volume to Volume)... then if you throw the 100ml and fill to 1L you have 40ml of active produt in 1000ml final solution, so 4% Vol/Vol

This is is the usual way for chemists to kown what dose they are using, it simplifies calculations. Think that many times final volume of a mix is not the addition of the volumes of the ingredients, using content in the final mix, without considering the (in this case) the water volume is what makes calculations easy.

Tom Taylor · Jun 28, 2020

(100mL / 40 gm) X 4 gm = 10mL (i.e., there are 4 gms in the 10 mL.) No add 90mL of water to the 10mL and you have 4 gms in 100MmL which is 4%.

Dimensional Analysis is one of the most valuable courses that is taught.

Thomas

138S · Jun 28, 2020

Tom Taylor said:
Now add 90mL of water to the 10mL and you have 4 gms in 100MmL which is 4%.

This is the inaccurate way, don't add 90ml, fill to 100ml. You don't know if you have to add 86ml or 94ml to get 100ml final mix amount, because volume of the mixture may not be the addition of the substances mixed.

See this: http://www.departments.bucknell.edu/chem_eng/eLEAPS/cheg200/DVofMixing/DVmix.pdf

First principle is: whole ≠ sum of its parts

Tom Taylor · Jun 28, 2020

Well topping off to 100mL is the correct method and the one that I would follow. But the difference for photographic applications is essentially insignificant. In practice I find myself usually going to 100mL rather than the other way - it offers more choices down the line as you would need to use advanced chemistry and lab equipment to increase the dilution.

Thomas

pentaxuser · Jun 28, 2020

I have a feeling that someone in the past provided a simple sheet that helpfully does this for you. Useful for the darkroom board.

pentaxuser

138S · Jun 28, 2020

Tom Taylor said:
Well topping off to 100mL is the correct method and the one that I would follow. But the difference for photographic applications is essentially insignificant.

This time you are right, because in this case the final solution is 4% and probably for photography difference won't be significant, but if final solution was (say) 20% from (say) 70% then a large error could be there in many non ideal liquid mixtures !!

It's interesting learning to do things in a conceptually correct way, if not...

Peter Schrager · Jun 29, 2020

Tom and 1385 thank you mucho!!!
Have a great day!!

Tom Taylor · Jun 29, 2020

(100mL/70gms) X 20 Gms = 28.6 mL. So 28.6mL of the 70% solution contains 20 grams of what's needed. If you top off to 100 mL the resultant would be a 20% mixture. If you instead add 71.4 mL of distilled water the resultant would still be close enough to 20% that the difference would be negligible and in many cases not even discernable using the ordinary graduates commonly used in photography. Topping off is the preferred method but many amateur photographers add the difference.

Have a great day.

Thomas

Doremus Scudder · Jun 29, 2020

Seems like I post this every so often...

A change in dilution strength can be described by a simple proportion:

Q1/Q2 :: C2/C1

Where Q1 = Quantity 1, Q2 = Quantity 2, C1 = Concentration 1 and C2 = Concentration 2

To find an unknown, fill in the known values and a symbol (X is always good) for the unknown. Then cross multiply and solve the equation.

In your case, say you have 100ml of a 40% solution and you want to know the amount of a 4% solution it will produce:
(Q1 = 100, Q2 = X, C1 = 40 and C2 = 4). So:

100/X :: 4/40.

Cross multiplying results in:

100 x 40 = 4X, or 4000 = 4X

Divide both sides by 4 to solve for X:

1000 = X.

So, Q2 = 1000ml

This method works for any change in concentration or volume and one unknown in any position.

Say you need 350 ml of a 4% solution and you have a 40% stock solution. The question is, how much of the stock solution do I need in the 350mll total volume? So, Q1 = X, Q2 = 350, C1 = 40 and C2 = 4. Our proportion is now:

X/350 :: 4/40

We cross multiply, resulting in:

40X = 350 x 4, or 40X = 1,400

Dividing both sides by 40 to solve for X gets us:

X = 35ml

So, you need 35ml of the 40% stock (and 315ml of water) to make 350ml of a 4% solution.

EZPZ

Best,

Doremus

Tom Taylor · Jun 29, 2020

4 into 40 goes 10 so 10 X the initial volume is the end volume. But in most cases you only want a specific volume - say 100mL of 4% from a volume of 40% - so why convert all of the 40% into a 4%? You can always dilute but you can't easily go back the other way which always leads to a diminishing return (e.g., returning a 4% back to a 40% results in a volume that is 10 times less (100mL) even if you could do it. Dimensional analysis is the way to go - it's far easier and once proficient you can string long chains together to solve very complex problems.

Have a great day,

Thomas

138S · Jun 29, 2020

Doremus Scudder said:
Seems like I post this every so often...

A change in dilution strength can be described by a simple proportion:

Q1/Q2 :: C2/C1

Doremus, dilutions usually do not follow total porportionality, errors that can be there may vary, this is the excess volume (deficit in this case) graph for water + ethanol

If you mix

1L of Ethanol 90%
+
1L of Ethanol 10%

then you won't get 2L , and it may not be Ethanol 50%, so not 50ml of pure ethanol in 100ml solution.

For this reason when an active substance is diluted in an "excipient" we use topping off, in that way we overcome the variable excess/deficit volume of mixtures that may be present, so we get accurate doses and exact calculations. You know, most darkroom mixtures use topping off.

Still, as Tom pointed, many times we may get well tolerable deviations for what's photography... But sometimes a pitfall can be there.

David Lyga · Jun 29, 2020

You divide the percentage that you HAVE by the percentage that you WANT.

Thus, 40 / 4 = 10

The quotient is the factor that you must multiply the original VOLUME by, If you had, for example, 100 mL of the 40%, you would now have to have 10X more: Thus, 10 X 100 = 1000.mL In other words, you will add water to make the 1000 mL (i.e., 10x more than you had originally).

- David Lyga

Doremus Scudder · Jun 30, 2020

138S said:
Doremus, dilutions usually do not follow total porportionality, errors that can be there may vary, this is the excess volume (deficit in this case) graph for water + ethanol

View attachment 249351

If you mix

1L of Ethanol 90%
+
1L of Ethanol 10%

then you won't get 2L , and it may not be Ethanol 50%, so not 50ml of pure ethanol in 100ml solution.

For this reason when an active substance is diluted in an "excipient" we use topping off, in that way we overcome the variable excess/deficit volume of mixtures that may be present, so we get accurate doses and exact calculations. You know, most darkroom mixtures use topping off.

Still, as Tom pointed, many times we may get well tolerable deviations for what's photography... But sometimes a pitfall can be there.

Well...

Sure it works, and reliably, for simple w/v (weight-to-total-volume) solutions, which is what the OP was asking about. Of course, starting with the more concentrated solution and then topping off is assumed. I wasn't addressing mixing two solutions, just dilution one more concentrated solution to end up with a weaker one. The total volume of the compound in solution is all in one solution. And, all that matters here is the weight of the solute in the total volume of the solvent.

Note, I'm not addressing mixing a w/v solution from dry chemicals, but diluting an already-mixed solution. When starting from dry chemicals, one always mixes the chemicals in a smaller volume of water and then tops up to the final desired volume. That's another topic and goes without saying for most of us here.

For v/v (volume-to-volume) solutions where the solute (in your example, ethanol) can slip between the water molecules and take up varying amounts of room depending on the dilution ratio, things get more complicated. However, note that the w/v proportion works here...

In most photographic applications, we use w/v solutions of dry chemicals. For that purpose, the proportional method works splendidly.

Doremus

138S · Jun 30, 2020

Doremus Scudder said:
For v/v (volume-to-volume) solutions where the solute (in your example, ethanol) can slip between the water molecules and take up varying amounts of room depending on the dilution ratio, things get more complicated.

Doremus, you know, if topping off with the "excipient" we always keep track of what exact amount of active substance we have, both if active substance was solid or liquid.

For example "40% glyoxal solution" we know we have 40ml of glyoxal in 100ml solution but we don't know how much water we take. A bit we loss the accurate water amount but we know how much glyoxal we are dosing.

A Lab trick: If water content is important then we can recover the water content with an scale.

Supose we have those 100ml of 40% glyoxal solution.... we know for sure that we have 40ml of glyoxal but... How much water do it contains ?

We know 40ml of glyoxal weights 40 * 1.27 g/cm³ = 50.08g Then we weight those 100ml of 40% solution, water content in grams will be the total weight minus 50.08g the glyoxal weights.

Doremus Scudder · Jun 30, 2020

138S said:
Doremus, you know, if topping off with the "excipient" we always keep track of what exact amount of active substance we have, both if active substance was solid or liquid.

For example "40% glyoxal solution" we know we have 40ml of glyoxal in 100ml solution but we don't know how much water we take. A bit we loss the accurate water amount but we know how much glyoxal we are dosing.

A Lab trick: If water content is important then we can recover the water content with an scale.

Supose we have those 100ml of 40% glyoxal solution.... we know for sure that we have 40ml of glyoxal but... How much water do it contains ?

We know 40ml of glyoxal weights 40 * 1.27 g/cm³ = 50.08g Then we weight those 100ml of 40% solution, water content in grams will be the total weight minus 50.08g the glyoxal weights.

138S,

I was assuming (correctly, I think) that the 40% glyoxal solution was a w/v solution, i.e., 40 grams (not ml) of crystalline glyoxal in 100ml of water. AFAIK, glyoxal is a dry chemical and usually mixed w/v, not v/v. Maybe I'm wrong here??

However, if I'm not wrong that the original solution is w/v, then everything should be just fine using the proportional method of diluting that I posted.

Best,

Doremus

Lachlan Young · Jun 30, 2020

@Doremus Scudder - 40% w/v is pretty much the standard way glyoxal is supplied - in and of itself, it is a crystalline solid, but is usually (near universally?) supplied in aqueous solution.

Doremus Scudder · Jun 30, 2020

Lachlan,

I imagine that glyoxal is supplied in solution due to its rather low melting point (15°C). Still I was under the impression, which you confirm, that the solutions available were w/v, not v/v. If this is still in question, we need to clear this up to help the OP.

Doremus

138S · Jun 30, 2020

Doremus Scudder said:
glyoxal is a dry chemical and usually mixed w/v, not v/v. Maybe I'm wrong here??

Glyoxal is a liquid, at 20ºC. It melts at 15ºC.

It is a pitfall that the PF label does not state if 40% was Weight or Vol, but they say (MSDS) that source is Sigma-Aldrich/Merk and that people sell it in w/vol %

______

Yes... your proportional method is totally correct... if volumes are stated in topping off conditions both in the "before" and "after" situations, and only if what varies is the "excipient".

______

But now imagine we mix 1L of glyoxal 20% w/vol + 1L glyoxal 40% w/vol, in this case proportionality cannot be applied, we won't have 2L and mixture won't 30% w/vol.

We would have for sure 200 + 400 = 600gr of pure glyoxal, but we should measure what is the final mixture volume is, and dividing 600 by the real final volume in deciltres to have the new w/vol %

Lachlan Young · Jun 30, 2020

@Doremus Scudder Pretty much everyone sells it as 40% w/v - in many cases they are likely acting as re-sellers for BASF who make the stuff in Ludwigshafen.

Doremus Scudder · Jun 30, 2020

138S said:
... But now imagine we mix 1L of glyoxal 20% w/vol + 1L glyoxal 40% w/vol, in this case proportionality cannot be applied, we won't have 2L and mixture won't 30% w/vol.

We would have for sure 200 + 400 = 600gr of pure glyoxal, but we should measure what is the final mixture volume is, and dividing 600 by the real final volume in deciltres to have the new w/vol %

Yes, but...

I was discussing just diluting something further, not mixing two different dilutions of anything. However, if we want to do that, we can use the proportional method as well to find the total final volume:

Using your example: We mix 1 liter of glyoxal 20% w/vol and 1 liter glyoxal 40% w/vol. Whatever final volume we end up with, we know, with certainty, what the amount of glyoxal is in the solution. It's 600 grams. We just then use the proportional method I posted above to find what that final volume should be for a 30% w/v solution since it works that way too. we just have to set up the problem a little differently. (I'm sure you know how to do this; I'm showing my work for the benefit of those that might need it.)

The formula is now: W1/Q1 :: W2/Q2, where W1 and W2 are weights 1 and 2 respectively and Q1 and Q2 are quantities 1 and 2 respectively.

Let's use 30 g / 100ml as our stand-in 30% w/v solution and as W1 and Q1. W2 is our known 600 grams. Q2 is our X:

Inserting the values we get: 30/100 :: 600/X. Cross-multiplying to get our equation gets us: 30X = 60,000. Dividing both sides by 30 gets us our solution (pun intended): 2000ml.

So, whatever volume we end up with when we mix 1 liter of glyoxal 20% w/vol and 1 liter glyoxal 40% w/vol, it's certainly not going to be more than 2 liters; we just top it up to 2 liters with our excipient (in this case, water) and, voilà, we've got 2 liters of a 30% solution.

Best,

Doremus

138S · Jun 30, 2020

Doremus Scudder said:
w/vol, it's certainly not going to be more than 2 liters;

Doremus, it depends on the substances we mix, there are mixtures of substances that have a positive deviation of the Raoult's law, a mixture may have an excess volume, instead a deficit.

Doremus Scudder · Jul 1, 2020

138S said:
Doremus, it depends on the substances we mix, there are mixtures of substances that have a positive deviation of the Raoult's law, a mixture may have an excess volume, instead a deficit.

I've never encountered that, probably because I don't do much complex mixing of chemistry, just compounding my own developers and stock solutions of single reagents like BTA, sodium carbonate, potassium bromide, potassium ferricyanide, etc. I never mix two different dilutions of the same chemical together to get a third... I don't really see much need to ever do so in a typical darkroom either, so I likely never will.

I find all of this quite interesting, nevertheless. I was aware that mixing different dilutions or two different solutions could end up with a total volume that is not equal to the sum of their parts, but have never seen a case where that sum is actually greater than the individual volumes added together. Could you give me an example of something that has a positive deviation of Rauolt's law?

Best,

Doremus

138S · Jul 1, 2020

Doremus Scudder said:
example of something that has a positive deviation of Rauolt's law?

(https://www.quora.com/How-will-we-k...h-compound-show-ve-deviation-from-Raoults-law)

Quoting:

Cause of Positive Deviation:

Difference in extent of association in two liquids (H2O & CH3OH)
Association in one of the liquids through H-bonding (C2H5OH & C6H6)
Greater difference in length of hydrocarbon part of members of same homologous series (n-butane & n-heptane)
Difference in polarity of liquids: General Examples are when one is polar & Other is non-polar (CCl4 & CHCl3)
Greater Difference in molar mass of non-polar molecules (CCl4 & C6H6)

Tom Taylor · Jul 1, 2020

Doremus Scudder said:
Seems like I post this every so often...

Say you need 350 ml of a 4% solution and you have a 40% stock solution. The question is, how much of the stock solution do I need in the 350mll total volume? So, Q1 = X, Q2 = 350, C1 = 40 and C2 = 4. Our proportion is now:

X/350 :: 4/40

We cross multiply, resulting in:

40X = 350 x 4, or 40X = 1,400

Dividing both sides by 40 to solve for X gets us:

X = 35ml

So, you need 35ml of the 40% stock (and 315ml of water) to make 350ml of a 4% solution.

EZPZ
/QUOTE]

(1000mL/400gms)*(4gms/100mL)*350mL= 35mL of 40% solution. (The numerators and denominators cancel leaving only the initial numerator which is what you are looking for: How many mL of a 40%... is the question.)

Much easier and cleaner!

Have a nice day.

how to go from a 40% solution to 4%

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