How do you prepare 0.02N EDTA solution?

Answer 1

Molecular weight of EDTA disodium salt dihydrate: 372.24 grams/mole

This molecular weight is used for preparation is because most hydrate salts are easier to dissolve in solution rather than anhydrous salts.

0.1N EDTA

250mL solution

Normality 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.

Answer 2

You mast to first accurate wight 3.7224 ,then transfere to flask mayer add 400 ml water dianoes until dissolve then volume with water to marl 1 liter.

Answer 3

Molecular weight of EDTA disodium salt dihydrate: 372.24 grams/mole

This molecular weight is used for preparation is because most hydrate salts are easier to dissolve in solution rather than anhydrous salts.

0.1N EDTA

250mL solution

Normality 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.

Answer 4

3.8 grams of EDTA salt in 1 liter of DI water made up using a volumetric flask will give you 0.02n or 0.01m of EDTA solution.

normality*eq.wt*volume rqrd

weight= 1000

then will get weight of the compound required for that normality

Answer 5

1. Weigh 5,8448 g of EDTA.

2. Quantitatively transfer EDTA in a clean and dry volumetric flask of 1L.

3. Add 900 mLof demineralized water.

4. Place the the volumetric flask in a thermostat at 20 0C.

5. Wait half hour.

6. Add demineralized water to the mark.

7. Put a glass stopper and stir the content.

8. Add a label with: data, name of the solution, concentration of the solution, operator.

Answer 6

The molar mass of EDTA is 292.24 g/mol. To make 1 liter of 0.01 N EDTA proceed as follows, considering that 0.01 N means 0.01 equivalents/liter and there are 4 equivalent wts/mole.
0.01 equiv/L x 1 L x 1 mol/4 equiv x 292.24 g/mole = 0.73 grams EDTA in a final volume of 1 liter.

Answer 7

You mast to first accurate wight 3.7224 ,then transfere to flask mayer add 400 ml water dianoes until dissolve then volume with water to marl 1 liter.

Answer 8

Dissolve 5,8448 g EDTA in 0,9 L demineralized water in a clean volumetric flask. At 20 0C add water to the mark and stir.

Answer 9

Dissolve 58,448 g dried EDTA in distilled water using a volumetric flask.