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How can you prepare 1dm3 solution of 0.025m EDTA solution from its dipotassium salt?
Accepted Answer
Step 1: Calculate the required mass of EDTA dipotassium salt
To prepare 1 dm3 of 0.025 M EDTA solution, the number of moles of EDTA required is:
Moles of EDTA = Volume of solution (in L) × Molarity
= 1 dm3 × 0.025 mol/dm3 = 0.025 mol
The molar mass of EDTA dipotassium salt (C10H14O8K2·2H2O) is 372.24 g/mol. Therefore, the mass of EDTA dipotassium salt required is:
Mass of EDTA dipotassium salt = Moles of EDTA × Molar mass of EDTA dipotassium salt
= 0.025 mol × 372.24 g/mol = 9.306 g
Step 2: Dissolve the EDTA dipotassium salt in water
Weigh out 9.306 g of EDTA dipotassium salt and dissolve it in a small volume of deionized water.
Step 3: Transfer the solution to a volumetric flask
Transfer the dissolved EDTA dipotassium salt solution to a 1 dm3 volumetric flask.
Step 4: Add water to the mark
Add deionized water to the volumetric flask until the solution reaches the 1 dm3 mark.
Step 5: Mix thoroughly
Stopper the volumetric flask and mix the solution thoroughly.
The prepared solution is a 1 dm3 solution of 0.025 M EDTA.
To prepare 1 dm3 of 0.025 M EDTA solution, the number of moles of EDTA required is:
Moles of EDTA = Volume of solution (in L) × Molarity
= 1 dm3 × 0.025 mol/dm3 = 0.025 mol
The molar mass of EDTA dipotassium salt (C10H14O8K2·2H2O) is 372.24 g/mol. Therefore, the mass of EDTA dipotassium salt required is:
Mass of EDTA dipotassium salt = Moles of EDTA × Molar mass of EDTA dipotassium salt
= 0.025 mol × 372.24 g/mol = 9.306 g
Step 2: Dissolve the EDTA dipotassium salt in water
Weigh out 9.306 g of EDTA dipotassium salt and dissolve it in a small volume of deionized water.
Step 3: Transfer the solution to a volumetric flask
Transfer the dissolved EDTA dipotassium salt solution to a 1 dm3 volumetric flask.
Step 4: Add water to the mark
Add deionized water to the volumetric flask until the solution reaches the 1 dm3 mark.
Step 5: Mix thoroughly
Stopper the volumetric flask and mix the solution thoroughly.
The prepared solution is a 1 dm3 solution of 0.025 M EDTA.
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