Citric acid standard for lab proficiency testing

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#1 tessalouwho

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Posted 28 March 2018 - 04:10 PM

For lab proficiency testing, I would like to make some citric acid standards since that is what we are checking for in our fruit juices. Not exactly sure how to make for example a 0.83% standard. Would i measure 0.83 grams into a 100 ml volumetric flask then fill to 100 ml with water. I am not sure if that is correct. Is it that straight forward? Or is it 0.83 g in a 100 ml of water to get 0.83% solution.

If anyone can help me out that would be great.

#2 Scampi

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Posted 28 March 2018 - 04:49 PM

Perhaps this will help?

https://www.rpc-rabr...ts/Dilution.pdf

Or you can purchase from reputable lab supply company

Because we always have is never an appropriate response!

#3 tessalouwho

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Posted 28 March 2018 - 04:59 PM

Thank you, I am still looking for the actual math. I need to make a lower solution and the chart it clearly isn't to make a 10% solution just "double" the recipe. 5% is 25.5 grams and for 10% solution its 52 grams. So I would like to see the math.

#4 FurFarmandFork

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Posted 28 March 2018 - 05:46 PM

You will have to correct for the purity of your sourced citric acid. But to create any solution, determine your desired concentration first.

I want 50% CA solution. Let's say 50g CA/100mL solution so that it's 50% CA by weight/weight. (I don't know if you can make this strong of a solution using water or not, this is just an example).

The table linked has pre-made conversions from g to volume to correct for the volume change from adding CA, since you can't just add 50g CA to 100mL DI water without changing the final volume.

Add DI water in small amounts, swirling to suspend/dissolve the CA as you go (e.g. add 30mL initially then 5 mL at a time)

Continue to add DI water until your final volume is 100mL. You now have a solution that contains 50g CA/100mL of the solution you made. You can convert that to molarity, concentration, whatever.

Once you have this concentrated solution, you can make further dilutions using this equation:

C1V1=C2V2

C1=Concentration of solution 1

V1=Initial volume of solution 1

C2=Concentration of new solution

V2=Volume of new solution

Solve for whatever variable you want.

Austin Bouck
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Consulting for companies needing effective, lean food safety systems and solutions.

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#5 jdpaul

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Posted 11 May 2018 - 10:01 PM

For 83% it would actually 83 grams of citric acid to 100 grams of water (density of water depends on temperature but it is very close to 1g/mL). This would be %w/w

#6 FurFarmandFork

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Posted 14 May 2018 - 07:26 PM

For 83% it would actually 83 grams of citric acid to 100 grams of water (density of water depends on temperature but it is very close to 1g/mL). This would be %w/w

This is incorrect.

83g citric acid/ (183g of material total) would be 45% citric acid by weight. You need to divide by the final material content.

Austin Bouck
Owner/Consultant at Fur, Farm, and Fork.
Consulting for companies needing effective, lean food safety systems and solutions.

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#7 jdpaul

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Posted 14 May 2018 - 07:29 PM

Excuse me, yes

83 grams in 17 grams water would be 83% w/w%

#8 jdpaul

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Posted 14 May 2018 - 07:32 PM

8.3 grams in 2.7 grams water would be 0.83% w/w%

#9 jdpaul

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Posted 14 May 2018 - 07:32 PM

8.3 grams in 2.7 grams water would be 0.83% w/w%

1.7 grams*

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