ForestShadow
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I recently did this question on the practice papers set 2 for biological diversity for OCR A2 level.

It gave the concs of acid a student was making : 0,0.1,0.2,0.4 and 0.8 M

and asked how I would make each 50cm^3 of each solution given a solution of 1M acid.

our teacher said the answer was to:

-use 100cm^3 measuring cylinder to mix 80 of acid with 20 water for 0.8M
(why 100cm^3 though? the question clearly asks for 50cm^3 so why double it? is it because 100cm^3 is standard? wouldnt the method work with just a mix of 40 acid and 10 water, the ratios are always the same?)
-take 50 of resulting solution and add 50cm^3 water for 0.4M conc
(I understand adding half of water everytime cuts the conc down by a half)
-repeat further giving 0.2


thanks for any help
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M0nkey Thunder
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(Original post by ForestShadow)
I recently did this question on the practice papers set 2 for biological diversity for OCR A2 level.

It gave the concs of acid a student was making : 0,0.1,0.2,0.4 and 0.8 M

and asked how I would make each 50cm^3 of each solution given a solution of 1M acid.

our teacher said the answer was to:

-use 100cm^3 measuring cylinder to mix 80 of acid with 20 water for 0.8M
(why 100cm^3 though? the question clearly asks for 50cm^3 so why double it? is it because 100cm^3 is standard? wouldnt the method work with just a mix of 40 acid and 10 water, the ratios are always the same?)
-take 50 of resulting solution and add 50cm^3 water for 0.4M conc
(I understand adding half of water everytime cuts the conc down by a half)
-repeat further giving 0.2


thanks for any help
If I understand what you're asking correctly, you're supposed to be adding 50cm^3 of the produced 100cm^3 0.8M solution to another test tube and 50cm^3 of distilled water to that test tube also. By the end of this, you'll have 50cm^3 of 0.8M solution and 100cm^3 of 0.4M solution. You're going to add 50cm^3 of this to the next test tube along with 50cm^3 of distilled water to get 50cm^3 of 0.4M solution and 100cm^3 of 0.2M solution and so on. You essentially need double the initial volume to then dilute and produce the other concentrations.
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Gerry-Atricks
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(Original post by ForestShadow)
I recently did this question on the practice papers set 2 for biological diversity for OCR A2 level.

It gave the concs of acid a student was making : 0,0.1,0.2,0.4 and 0.8 M

and asked how I would make each 50cm^3 of each solution given a solution of 1M acid.

our teacher said the answer was to:

-use 100cm^3 measuring cylinder to mix 80 of acid with 20 water for 0.8M
(why 100cm^3 though? the question clearly asks for 50cm^3 so why double it? is it because 100cm^3 is standard? wouldnt the method work with just a mix of 40 acid and 10 water, the ratios are always the same?)
-take 50 of resulting solution and add 50cm^3 water for 0.4M conc
(I understand adding half of water everytime cuts the conc down by a half)
-repeat further giving 0.2


thanks for any help
Yes it would work as long as it is in ratio, you can use whatever number you want, just bigger ones reduce uncertainty in concentration measurements from a practical standpoint
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ForestShadow
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(Original post by M0nkey Thunder)
If I understand what you're asking correctly, you're supposed to be adding 50cm^3 of the produced 100cm^3 0.8M solution to another test tube and 50cm^3 of distilled water to that test tube also. By the end of this, you'll have 50cm^3 of 0.8M solution and 100cm^3 of 0.4M solution. You're going to add 50cm^3 of this to the next test tube along with 50cm^3 of distilled water to get 50cm^3 of 0.4M solution and 100cm^3 of 0.2M solution and so on. You essentially need double the initial volume to then dilute and produce the other concentrations.
that makes a ton of sense but is the 100cm^3 of 0.8 still crucial to the method and the only way to do it?

couldnt you just start with 50cm^3 0.8M solution (made from 40/10 split) and then do the same method after that?

I guess its all the same in the end and the important part is using half of the previous solution and adding an equal vol of water :holmes:

(Original post by glad-he-ate-her)
Yes it would work as long as it is in ratio, you can use whatever number you want, just bigger ones reduce uncertainty in concentration measurements from a practical standpoint
ah thanks
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M0nkey Thunder
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(Original post by ForestShadow)
that makes a ton of sense but is the 100cm^3 of 0.8 still crucial to the method and the only way to do it?

couldnt you just start with 50cm^3 0.8M solution (made from 40/10 split) and then do the same method after that?

I guess its all the same in the end and the important part is using half of the previous solution and adding an equal vol of water :holmes:



ah thanks
That wouldn't work because when you pour half of the 50cm^3 0.8M solution into the next test tube, as you would during serial dilutions, you'll only have 25cm^3 of the 0.8M solution left- but you actually want 50cm^3. The only way to ensure this happens is to start with double 50cm^3, 100, so when we pour half into the next test tube we still have the desired 50cm^3 of solution.
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max smith
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(Original post by ForestShadow)
that makes a ton of sense but is the 100cm^3 of 0.8 still crucial to the method and the only way to do it?

couldnt you just start with 50cm^3 0.8M solution (made from 40/10 split) and then do the same method after that?

I guess its all the same in the end and the important part is using half of the previous solution and adding an equal vol of water :holmes:



ah thanks
Yeah the exam board wants you to understand serial dilution which is all about making a solution which you then use to make a new solution of lower concentration, which you will then use to make an even lower concentration. This way you won't have to work with tiny volumes to get very small concenctrations. I once did a serial dilution to get: 0.5M, 0.05M, 0.005M and 0.0005M. Imagine making 0.0005 from scratch.
I think that's why your teacher wants you to do it this way
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ForestShadow
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(Original post by M0nkey Thunder)
That wouldn't work because when you pour half of the 50cm^3 0.8M solution into the next test tube, as you would during serial dilutions, you'll only have 25cm^3 of the 0.8M solution left- but you actually want 50cm^3. The only way to ensure this happens is to start with double 50cm^3, 100, so when we pour half into the next test tube we still have the desired 50cm^3 of solution.
lol thats so true didnt think of that, thank you

(Original post by Jules kemps)
Yeah the exam board wants you to understand serial dilution which is all about making a solution which you then use to make a new solution of lower concentration, which you will then use to make an even lower concentration. This way you won't have to work with tiny volumes to get very small concenctrations. I once did a serial dilution to get: 0.5M, 0.05M, 0.005M and 0.0005M. Imagine making 0.0005 from scratch.
I think that's why your teacher wants you to do it this way
ah ok I guess that method does make more sense, you need to use the solution made previously each time thanks
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