Chemical dilution conversion

David Lingham · Aug 25, 2021

Could anyone help me with a chemical conversion. I have a toning formula that I want to try that requires 100mls of 10% Ammonia, and I can only find Ammonia sold in a 33% solution. My question is how to calculate the amount from the stronger 33% stock that is needed to end up with 100mls @ 10%. What little maths I ever possessed has deserted me.

mohmad khatab · Aug 25, 2021

David Lingham said:
Could anyone help me with a chemical conversion. I have a toning formula that I want to try that requires 100mls of 10% Ammonia, and I can only find Ammonia sold in a 33% solution. My question is how to calculate the amount from the stronger 33% stock that is needed to end up with 100mls @ 10%. What little maths I ever possessed has deserted me.

If you need a solution of 100 ml with a concentration of 10% ammonia
You will need 3.3 ml of ammonia from that available to you (at a concentration of 33%), and you put this amount in a glass beaker and pour water on it until the amount of the solution reaches 100 ml.
- You may find it difficult to measure the amount of 3.3 ml, but I advise you to use an old syringe on which the syringe is accurate.

David Lingham · Aug 25, 2021

thank you Mohmad.

Raghu Kuvempunagar · Aug 25, 2021

I think you need to use 30.03ml of 33.3%.

100ml of 10% = 100/(33.3/10) ml of 33.3%
= 30.03 ml of 33.3%

Sirius Glass · Aug 25, 2021

Post 4 is correct, the other is just wrong.

David Lingham · Aug 25, 2021

Raghu, Sirius, thank you both.

AgX · Aug 25, 2021

David, there exists a rule for such calculations. Over here we call it "Cross-Rule":
https://de.wikipedia.org/wiki/Mischungskreuz

(Maybe someone can come up with an English description.)

runswithsizzers · Aug 25, 2021

I would set it up like this:
Word problem: "How many mL of a 33 percent solution equals 100 mL of a 10 percent solution?"
I would express the percentages as: 10/100 = 0.10, and 33/100 = 0.33, so,
X mL(0.33) = 100mL(0.10)
solving for X, divide both sides by 0.33, so
X mL = [100mL x 0.10] / 0.33
X = 30.3 mL

My answer (30.3mL) is slightly different from @Raghu Kuvempunagar's answer (30.03mL) because I interpret 33% to be 33 parts out of 100 and he used 33.3 parts out of 100. As stated, the precision of the answer is limited to two significant digits, so the two answers are, for all practical purposes, the same (30 mL) - but I think 33% is more accurately represented as 0.33 and not 0.333 - yes? no?

MattKing · Aug 25, 2021

runswithsizzers said:
My answer (30.3mL) is slightly different from @Raghu Kuvempunagar's answer (30.03mL) because I interpret 33% to be 33 parts out of 100 and he used 33.3 parts out of 100. As stated, the precision of the answer is limited to two significant digits, so the two answers are, for all practical purposes, the same (30 mL) - but I think 33% is more accurately represented as 0.33 and not 0.333 - yes? no?

There is a decent chance that the actual concentration of the 33% Ammonia is 1/3 or 33.33333....%.
The formula requires 10 units of Ammonia (10% of 100).
To get that from 33.33333...% Ammonia, you need 30ml (1/3 of 30 ml, for 10 units).

Raghu Kuvempunagar · Aug 25, 2021

MattKing said:
There is a decent chance that the actual concentration of the 33% Ammonia is 1/3 or 33.33333....%.
The formula requires 10 units of Ammonia (10% of 100).
To get that from 33.33333...% Ammonia, you need 1/3 of 30 ml, or 10 ml.

This is incorrect. OP's toner formula needs 100ml of 10% Ammonia. If you're going to use 33.3333..% Ammonia in place of 10% Ammonia, you need to use 30ml and not 10ml.

100ml of 10% = 100/(33.33333../10) ml of 33.33333..%
= 30 ml of 33.33333..%

@runswithsizzers: I misread 33% as 33.3% and hence arrived at 30.03 instead of 30.3. Actually, 33% Ammonia can be anywhere between 32-34% if you read manufacturer's description.

MattKing · Aug 25, 2021

Raghu Kuvempunagar said:
his is incorrect. OP's toner formula needs 100ml of 10% Ammonia. If you're going to use 33.3333..% Ammonia in place of 10% Ammonia, you need to use 30ml and not 10ml.

100ml of 10% = 100/(33.33333../10) ml of 33.33333..%
= 30 ml of 33.33333..%

Thanks - I've reworded my post to actually make some sense!

runswithsizzers · Aug 25, 2021

One-third of 100mL is 33.333...etc. But the solution is not described as being one-third ammonia, rather it is given as a 33% solution.

A 33% solution can be said to deliver 33 mL of solute per 100 mL of solution. But since the solution is given as 33% and not 33.3% we cannot assume any greater precision, right?

Based on @Raghu Kuvempunagar's statement "Actually, 33% Ammonia can be anywhere between 32-34% if you read manufacturer's description." ...
- if the solution contains 32% ammonia, then it would require 31mL to deliver the required 10mL of ammonia
- if the solution contains 34% ammonia, then 29 mL will deliver 10mL of ammonia

So, the correct answer is 30mL +/- 1 mL. You wont a need Class A volumetric pipette for this one.

Raghu Kuvempunagar · Aug 25, 2021

Strictly speaking it's incorrect to assume that 33% Ammonia is actually 33% w/w. Manufacturers generally give a range, say 32-34%, for the Ammonia sold as 33%. Essentially, manufacturer is saying that w/w of Ammonia can be anywhere in that range. Given that w/w of Ammonia is specified as a range by the manufacturer, the volume to be used by OP can only be specified as a range. However, from the point of view of the application at hand, this is probably hair splitting.

fgorga · Aug 25, 2021

AgX said:
David, there exists a rule for such calculations. Over here we call it "Cross-Rule":
https://de.wikipedia.org/wiki/Mischungskreuz

(Maybe someone can come up with an English description.)

Here it is in English as I used to teach it...

Where "C" stand for concentration and "V" stands for "volume" and subscript "F" stands for "final" (as in final solution) and subscript "S" stands for "stock" (as in stock solution).

The top line (shown above) is the basic relationship. This equation can be algebraically rearranged in many ways. Perhaps the most useful rearrangement, and the one relevant to the problem at hand here, is shown in the second line (above).

In the specific case mentioned the original post, the concentration of the stock solution (CS) is 33%, the desired final concentration (CF) is 10% and the desired final volume (VF) is 100 mL.

The applying the second equation show above, VS = (10/33) x 100 = 33 mL, as others have mentioned.

Thus to make this solution I would take a 100 mL graduated cylinder and add stock (33%) solution to the cylinder to the 33 mL mark. I would then add water to the 100 mL mark and mix thoroughly.

Given the volumes involved, the uncertainty in the concentration of the stock (it is nominally 33% but there are many factors that could alter this a bit) and the fact that for the large majority of photographic solutions compounds are present in excess there is no need for any greater precision.

Hope this is useful,

mohmad khatab · Aug 25, 2021

I apologize guys.
I made a mistake in writing (3.3) and I intended to write (33.3) - but it is a naive mistake on my part. I apologize.

David Lingham · Aug 26, 2021

Thank you too all who responded to my original question. The equation that forge mentioned, I did find in one of Tim Rudmans books, unfortunately his explanation completely lost me.

runswithsizzers · Aug 26, 2021

I believe the formula and discussion in post #14 describe making dilutions. That is, you could take the 33% solution and dilute it to 10%, and then use that as specified.

If you add 30ml of 33% solution, your toner will have the same amount of ammonia as if you had added 100 ml of 10% ammonia, but the final volume of your toner will be less.

If the total volume of the toner is critical then it would be preferable to dilute the 33% ammonia to 10% and add 100ml as specified. The instructions for making the toner may give quantities for the other ingredients based on the assumption that there will be an exact final volume - which is going to be about 70ml short if you use the 33% ammonia undiluted.

Don_ih · Aug 26, 2021

Just think of it as 33% is 3.3 times 10%, so you need to divide the quantity (100ml) by 3.3 - giving 30.3ml of 33% ammonia that needs to be topped up with water to 100ml.

fgorga · Aug 26, 2021

runswithsizzers said:
I believe the formula and discussion in post #14 describe making dilutions. That is, you could take the 33% solution and dilute it to 10%, and then use that as specified.

If you add 30ml of 33% solution, your toner will have the same amount of ammonia as if you had added 100 ml of 10% ammonia, but the final volume of your toner will be less.

If the total volume of the toner is critical then it would be preferable to dilute the 33% ammonia to 10% and add 100ml as specified. The instructions for making the toner may give quantities for the other ingredients based on the assumption that there will be an exact final volume - which is going to be about 70ml short if you use the 33% ammonia undiluted.

In most applications, the final volume is fairly important... if you are short by a significant volume, the concentrations of all the components will be too high.

Thus, if you do not want to "pre-dilute", the ammonia as in the original post, it is critical to make up the volume to what the recipe calls for by adding the "missing" water.

Vaughn · Aug 26, 2021

That is my approach, too. Pretty much along with Matt's thinking and taking into account the total water required.

Other than that, I suggest watching out for those significant figures. As presented, no way are hundredths of a ml significant...even tenths of a ml is questionable...

Chemical dilution conversion

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