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What is Na2EDTA?

Updated: 4/28/2022
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Na2EDTA is disodium ethylenediamine tetraacetate.

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How do you prepare 0.02N EDTA solution?

Molecular weight of EDTA disodium salt dihydrate: 372.24 grams/moleThis molecular weight is used for preparation is because most hydrate salts are easier to dissolve in solution rather than anhydrous salts.0.1N EDTA250mL solutionNormality is described by how many equivalents of protons are available to participate in your reaction. For example, strong acids like a 1M H2SO4 automatically deprotonate in water to release two protons per one molecule of H2SO4, which means a 1M H2SO4 = 2N H2SO4.EDTA has the ability to deprotonate 4 times, but it does NOT all deprotonate the moment it hits the solution. It's pkas are approximately ~2, 3, 6, 10. If your solution is to be at pH = 8, then you will have 3 equivalents of protons released into the solution (pkas of 2,3,6 are more than 1 unit lower than pH = 8, so protons are released). If it is at pH = 5, then you will release 2 equivalents. Most molecular Biology work working with enzymes near neutral pH will buffer EDTA at pH = 8. However, since this question is asking for a preparation of EDTA without a buffering effect, I will assume all four protons will be used.(0.1N EDTA) x (1M equivalent Na2EDTA.2H2O/ 4 N equivalent protons ) x (1L/1000ml) x (250ml solutionl) x (372.24 grams/ mole) = 2.33 grams Na2EDTA.2H2O required.Dissolve 2.33 grams Na2EDTA.2H2O into 150ml ddH2O, increase pH to 8 with NaOH pellets and maintain this pH until all EDTA is dissolved. Add more NaOH pellets to pH 12, then bring volume up to 250ml withddH2O.


Related questions

How do you pronounce Na2EDTA?

u say it like Na2EDTA pretty simple


What is molecular weight of edta?

To accurately calculate the molarity of an EDTA solution you need to determine the number of moles present and liters of solution. Take these two totals and divide them to determine molarity.


How do you prepare 0.02N EDTA solution?

Molecular weight of EDTA disodium salt dihydrate: 372.24 grams/moleThis molecular weight is used for preparation is because most hydrate salts are easier to dissolve in solution rather than anhydrous salts.0.1N EDTA250mL solutionNormality is described by how many equivalents of protons are available to participate in your reaction. For example, strong acids like a 1M H2SO4 automatically deprotonate in water to release two protons per one molecule of H2SO4, which means a 1M H2SO4 = 2N H2SO4.EDTA has the ability to deprotonate 4 times, but it does NOT all deprotonate the moment it hits the solution. It's pkas are approximately ~2, 3, 6, 10. If your solution is to be at pH = 8, then you will have 3 equivalents of protons released into the solution (pkas of 2,3,6 are more than 1 unit lower than pH = 8, so protons are released). If it is at pH = 5, then you will release 2 equivalents. Most molecular Biology work working with enzymes near neutral pH will buffer EDTA at pH = 8. However, since this question is asking for a preparation of EDTA without a buffering effect, I will assume all four protons will be used.(0.1N EDTA) x (1M equivalent Na2EDTA.2H2O/ 4 N equivalent protons ) x (1L/1000ml) x (250ml solutionl) x (372.24 grams/ mole) = 2.33 grams Na2EDTA.2H2O required.Dissolve 2.33 grams Na2EDTA.2H2O into 150ml ddH2O, increase pH to 8 with NaOH pellets and maintain this pH until all EDTA is dissolved. Add more NaOH pellets to pH 12, then bring volume up to 250ml withddH2O.