Monthly Archives: May 2018

Preparation of 0.5 M EDTA stock solution from anhydrous EDTA free acid

Preparation of 0.5 M EDTA stock solution from anhydrous EDTA

  • EDTA (EthyleneDiamineTetraAcetic acid), a polyamino carboxylic acid, is extensively used in molecular biology experiments as a chelating agent. It sequesters metal ions such as Ca2+ and Fe3+. Metal ions are necessary for the action of many enzymes including DNases.
  • EDTA is commercially available as _ _ _ _ _
    • Anhydrous EDTA (CAS Number 60-00-4, Molecular Weight 292.24)
    • Disodium EDTA dihydrate (EDTA.Na2.2H2O, CAS Number 6381-92-6, Molecular Weight 372.24) and
    • Tetrasodium EDTA tetrahydrate (EDTA.Na4.4H2O, CAS Number 13235-36-4, Molecular Weight 452.23).
  • Anhydrous free acid EDTA is least soluble among all EDTA forms in water. To dissolve anhydrous free acid EDTA in water, a lot of NaOH (3.1 ratio) is added to bring the pH to 8.0.
  • Disodium EDTA dihydrate have better solubility than anhydrous free acid EDTA and is most commonly used for the preparation of 0.5 M EDTA solution.
  • Tetrasodium EDTA tetrahydrate is soluble in water. The resulting solution has pH above 10.0, therefore is not suitable for cell and molecular biology experiments.
  • Here we show preparation of 0.5M EDTA solution from Anhydrous free acid EDTA.

Related posts

Requirements

  • Reagents
    • EDTA
    • NaOH pallet / 10N NaOH solution (for pH adjustment)
    • Deionized / Milli-Q water
  • Equipment and disposables
    • Measuring cylinder
    • Conical flask / Beaker
    • Magnetic stirrer

Composition:

0.5 M EDTA, pH 8.0 at 25°C

Objective:  

Preparation of 1000 ml of 0.5 M EDTA solution, pH 8.0 in water from Anhydrous free acid EDTA (Molecular Weight 292.24)

Preparation

Step 1: Weigh out 146.12 grams EDTA.Na2.2H2O (Molecular Weight 292.24). Transfer to 2 L beaker / conical flask. Add 700 ml deionized / Milli-Q water.

Precaution: Do not dissolve in 1000 ml of deionized / Milli-Q water. In most cases, solution volume increases when a large amount of solute dissolves in the solvent.

Step 2: While stirring vigorously on a magnetic stirrer, add NaOH pellet to adjust the solution pH 8.0.

Tips:
  • ~40 g NaOH pellet is required to adjust the pH 8.0.
  • It is not easy to dissolve EDTA. To dissolve the EDTA completely, solution pH 8.0 is required.

Step 3:  Adjust the volume to 1000 ml with deionized / Milli-Q water. Mix it again.

Step 4:  Transfer the solution to autoclavable bottle. Sterilize the solution by autoclaving (20 minutes at 15 lb/sq.in. (psi) from 121-124°C on liquid cycle)

Tip:
  • Depending on the consumption, one can make small aliquots of solution.
Note:
  • One can sterilize the solution by passing through 0.22 μ filter unit. Filter sterilization removes all suspended particles with size more than 0.22 μ which includes most bacteria their spores but not mycoplasma. However, it does not inactivate enzyme activities (e.g., DNases). Autoclaving inactivates most enzymes except some (e.g., RNases) and kills most microorganisms including mycoplasma.

Storage:

Solution can be stored at 15 – 25 °C (room temperature) for several months.

Applications:

0.5 M EDTA solution is used for the preparation of many solutions including TAE, TBE, DNA loading dye, resuspension buffer (isolation of plasmid), Tris-EDTA, Trypsin-EDTA, etc.

Follow the table to prepare EDTA solution of specific concentration and volume from anhydrous EDTA Acid free (Molecular Weight 292.24).
Conc. / Volume 100 ml 250 ml 500 ml 1000 ml
10 mM 0.29 g 0.73 g 1.46 g 2.92 g
100 mM 2.92 g 7.306 g 14.612 g 29.224 g
0.25 M 7.306 g 18.265 g 36.53 g 73.06 g
0.5 M 14.612 g 36.53 g 73.06 g 146.12 g

 

Calculating amount of Disodium ethylenediaminetetraacetate dihydrate (EDTA.Na2.2H2O) for preparation of 1 litre of 0.5 M EDTA solution

Calculating amount of Disodium ethylenediaminetetraacetate dihydrate (EDTA.Na2.2H2O) for preparation of 1 litre of 0.5 M EDTA solution

Molecular weight of EDTA.Na2.2H2O = 372.24 g/mol
Solution volume = 1 L
Desired molarity of solution = 0.5 M
Amount of substance (in grams) = To be calculated

Formula :

Place all values in the formula
Amount of substance (in grams) = 0.5 x 372.24 x 1
                                                          = 186.12 grams

Conclusion: To prepare 1 litre of 0.5 M solution, you need 186.12 grams of Disodium ethylenediaminetetraacetate dihydrate (EDTA.Na2.2H2O).

Molarity

Molarity

Molarity is the most widely used unit to express solute concentration in a solution. Molarity refers to moles of solutes per litre of solution.

The molarity of a solution can be calculated as follows:

A solution which contains 1 mole of solute per litre of solution is called 1 molar.

A mole of a substance contains 6.02 x 1023 (Avogadro constant) molecules of that substance. 

One mole is equal to the molecular weight of the substance in grams.

                                                     1 mole = molecular weight of substance

For example, 40 gram of NaOH (Molecular weight = 40) is equal to 1 mole of NaOH.

Mole is calculated by dividing the total amount of substance (weight in gram) by molecular weight of that substance.
So if a solution contains 40 gram (1 mole) of NaOH per litre it is said to be 1 molar NaOH solution.

Preparation of 1 mM EDTA solution from stock solution of EDTA (0.5M, pH 8.0)

Overview:

  • EDTA (Ethylenediaminetetraacetic acid) is a chelating agent. Due to its ability to sequester metal ions such as Mn2+, Ca2+, Mg2+ and Fe3+, EDTA is widely used in number of cell and molecular biology experiments.
  • Metal ions are necessary for the activity of many enzymes (e.g., DNases and DNA modifying enzymes) as well as interactions of biomolecules (e.g., receptor-ligand interaction). Sequestration of metal ions disrupts metal ion-dependent interactions (e.g., cell-cell and cell substratum interaction) as well as inhibits the action of metal ion dependent enzymes.
  • Here we describe steps to prepare 1 mM EDTA solution from stock EDTA solution.

Requirement

  • Reagents
    • 0.5 M EDTA solution, pH 8.0 at 25°C
    • Deionized / Milli-Q water
  • Equipment and disposables
    • Measuring cylinder

Composition:

1 mM  EDTA, pH 8.0 at 25°C

Objective:

Preparation of 100 ml of 1 mM EDTA solution (pH 8.0) from 0.5 M EDTA solution (pH 8.0)

Preparation:

Step 1: Calculate amount of 0.5M EDTA required for the preparation of 100 ml of 1 mM EDTA.

You need to add 0.2 ml of 0.5 M EDTA in 99.8 ml water to achieve 100 ml of 1 mm EDTA solution. (See calculation)

Step 2: Take ≈50 ml water in a measuring cylinder and add 0.2 ml of 0.5M EDTA in it.

Step 3: Adjust the final volume 100 ml with deionized/milli Q water. Mix it.

Tips: To mix solution, you can cover the measuring cylinder with parafilm and mix the solution by inverting the measuring cylinder 3 – 4 times. Take care no solution leaks out during this process. Alternatively, you can transfer solution to storage bottle and mix solution by swirling the bottle or using magnetic stirrer.

Storage:

Transfer solution to a storage bottle. You can store the solution at room temperature. Solution is stable at room temperature for long time.

 

Follow the table to prepare EDTA solution of specific concentration and volume from 0.5 M EDTA solution
Conc. / Volume 100 ml 250 ml 500 ml 1000 ml
1 mM 0.2 ml 0.5 ml 1.0 ml 2.0 ml
5 mM 1.0 ml 2.5 ml 5.0 ml 10.0 ml
10 mM 2.0 ml 5.0 ml 10.0 ml 20.0 ml
25 mM 5.0 ml 12.5 ml 25.0 ml 50.0 ml

 

Calculating volume of 0.5M EDTA required for preparation of 100 ml of 1 mM EDTA

Calculating volume of 0.5M EDTA required for preparation of 100 ml of 1 mM EDTA

Concentrated 0.5 M EDTA stock solution can be diluted to prepare 1 mM EDTA solution. Following procedure can be used to calculate amount of 0.5M EDTA solution required to prepare 100 ml of 1 mM EDTA solution

Procedure:

Step 1: Convert all concentration unit to either molar or millimolar.

The SI prefix “milli” represents a factor of 10-3. Here we convert stock solution concentration (0.5M EDTA) to millimolar (mM). 

                                    1 mM   =     10-3 M = 0.001 M
 
OR                               1 M      =      1000 mM
                                    0.5 M   =      0.5 x 1000 mM 

                                    0.5 M   =      500 mM

Stock solution concentration is 500 mM.

Step 2: Calculate required amount of stock solution (500 mM EDTA) to prepare 1000 ml of 1 mM EDTA

Formula:

Cf: Concentration of final solution i.e., 1 mm EDTA
Vf: Volume of the final solution i.e., 100 ml
CS: Conc. of stock solution i.e., 500 mM

VS: Volume of stock solution – To be calculated

Place all values in the formula

                                                                        1 x 100 = 500 x VS                                                                            VS = 0.2 ml

Step 2: Calculate the amount of water

You want to prepare 100 ml solution of which 0.2 ml is 0.5M EDTA stock solution. Rest of the amount will be water which is calculated by subtracting 0.2 ml from 100 ml.

Vf     : Volume of the final solution i.e., 100 ml
Vs    : Volume of the stock solution i.e., 0.2 ml

Amount of diluent i.e., water: 100 – 0.2 = 99.8 ml

Conclusion: To prepare 100 ml of 1 mM solution of EDTA from 0.5M EDTA stock solution, you need to mix 0.2 ml of 0.5M EDTA stock solution and 99.8 ml water.

Preparing solution of specific concentration using simple dilution

Preparing solution of specific concentration using simple dilution

If you have a concentrated stock solution of known concentration, you can prepare working solution of specific concentration and volume by diluting stock solution.

Use following formula to calculate the volume of stock solution required to prepare diluted solution

                                    Cf x Vf = CS x VS

Cf: Concentration of final solution
Vf: Volume of the final solution
CS: Conc. of stock solution

VS: Volume of stock solution

Mole fraction

Mole fraction

Mole fraction refers to the fraction of a component in a solution when the amounts (component and solution) are expressed in moles. In other words, Mole fraction of a component is the ratio of moles of that particular component to the total moles of the solution.

Mole fraction (𝒳i) can be calculated by
ni = Mole of ith component of the solution

ntot = Mole of the solution

Suppose a solution contains 3 components (two solutes, A and B, and a solvent C)

Mole fraction of component A (𝒳A) =
Mole fraction of component B (𝒳B) =
Mole fraction of component C (𝒳C) =
Where
nA = Mole of component A
nB = Mole of component B

nC = Mole of component C

Since the solution contains only three components, two solutes and a solvent, the total moles (ntot) of the solution will be equal to

ntot =  nA + nB + nC

Therefore above equation can also be written

Mole fraction of component A (𝒳A)
Mole fraction of component B (𝒳B)
Mole fraction of component C (𝒳C)
Mole fraction is dimensionless:

Since value of mole fraction is obtained by dividing mass of a component by total mass of solution, it dimensionless. However, sometimes it is written as mole/mole.

The sum of all Mole fraction is equal to 1.
In above example

𝒳A + 𝒳B + 𝒳C = 1

Independent of temperature

In contrast to density, Mole fraction is independent of temperature i.e., if temperature changes, the value of Mole fraction does not change.

Mass fraction

Mass fraction

Mass fraction refers to the fraction of a component in a solution when the amounts (component and solution) are expressed in mass (weight). In other words, mass fraction is the ratio of the mass (or weight) of a component to the total mass (or weight) of the solution.

Mass fraction (ωi) can be calculated by
Where
mi = Mass of ith component of the solution

mtot = Mass of the solution

Suppose a solution is prepared by dissolving two solutes A and B in a solvent C.

Mass fraction of component A (ωA)
Mass fraction of component B (ωB)
Mass fraction of component C (ωC)
Where
mA = Mass of component A
mB = Mass of component B

mC = Mass of component C

Since the solution contains only three components, two solutes and a solvent, the total mass of the solution (mtot) will be equal to

                                                                     mtot =  mA + mB + mC

Therefore above equation can also be written
Mass fraction of component A (ωA)
Mass fraction of component B (ωB)
Mass fraction of component C (ωC)
Mass fraction is dimensionless:

Since value of mass fraction is obtained by dividing mass of a component by total mass of the solution, it dimensionless. However, sometimes it is written as gram/gram or Kg/Kg.

The sum of all mass fraction is equal to 1.
In above example

ωA + ωB + ωC = 1

Independent of temperature

In contrast to density, mass fraction is independent of temperature i.e., if temperature changes, the value of mass fraction does not change.