A question in chemistry.

[mohmad khatab]

Hello respected teachers and experts.
I want to prepare the compound, Fe(III)EDTA manually as it is not available in Egypt.
And my Russian friend sent me that attached document.
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Dissolve 0.4 g (0.01 mol) of NaOH in 10 cm 3 of water, and then add 3.8 g (0.01mol) of Na 2 H 2 EDTA⋅2H 2 O
Gently heat the solution until the solid dissolves to give a clear solution.
Dissolve 2.5 g (0.009 mol) of iron(III) chloride hexahydrate in 5 cm 3 of water,
which is then added to the EDTA solution with swirling.
Gently boil off the water until most of the yellow powder precipitates out.
Cool down the solution and collect the precipitate by suction filtration.
Wash the product thoroughly with ice water until it is free of iron(III) ions.
Wash the product with ethanol twice and dry it with filter paper.
Weigh your product, calculate its theoretical and percentage yields.
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The recipe is based on (Na2) - this element is not available to me at the moment, while I have (Na4) only
the question is ..
How are quantities calculated in that case?
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Actually, I'm basically an accountant, and I'm trying to learn the basics of chemistry in order to understand these dilemmas and learn how to calculate weights, but I think I'm learning very slowly
So I hope you will kindly find a solution to this chemical dilemma. .
God bless you and thank you in advance.

 

[Rudeofus]

The original recipe starts with 0.01 mol NaOH and 0.01 mol Na2-EDTA. This amounts to 0.03 mol Na+ and 0.01 mol EDTA. If you start with 0.01 mol Na4-EDTA, you already have 0.04 mol Na+ and 0.01 mol EDTA, i.e. 0.01 mol Na+ too much. The only way to work around this is by adding 0.01 mol of some acid, i.e. you start with 0.6 g glacial Acetic Acid and 3.8g Na4-EDTA. After that, you continue as described in the instructions.

 

[mohmad khatab]

The original recipe starts with 0.01 mol NaOH and 0.01 mol Na2-EDTA. This amounts to 0.03 mol Na+ and 0.01 mol EDTA. If you start with 0.01 mol Na4-EDTA, you already have 0.04 mol Na+ and 0.01 mol EDTA, i.e. 0.01 mol Na+ too much. The only way to work around this is by adding 0.01 mol of some acid, i.e. you start with 0.6 g glacial Acetic Acid and 3.8g Na4-EDTA. After that, you continue as described in the instructions.

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I find it difficult to understand.
Why should we add acetic acid to the formula?
Can the chemical equation be balanced in both cases? The first case and the case to be implemented as an alternative solution?Thank you for the response, regards.

 

[Rudeofus]

The Na4-EDTA has too much Na+ for the final product. Therefore you have to add Acetic Acid to get NaFe-EDTA and Acetic Acid Sodium Salt. If you do not add Acetic Acid, the product will be too alkaline.

 

[mohmad khatab]

The Na4-EDTA has too much Na+ for the final product. Therefore you have to add Acetic Acid to get NaFe-EDTA and Acetic Acid Sodium Salt. If you do not add Acetic Acid, the product will be too alkaline.

Okay ,,
I want to produce 1 gram Fe(III)EDTA
What is the final recipe or final steps for the production of Fe(III)EDTA if only (Na4) is available, while the element (Na2) is not available.