Question

If you were given a 1000 mg/L stock solution of your metal and a 100 mM solution of Na2EDTA, how much of each solution would you need to make a 25 mL 1mM mixed solution of these two chemicals, where both the metal and EDTA have the same concentration? Describe how you will prepare this solution in addition to your calculation.

Answer 1

EDTA:

For stock solution: M1 = 100 mM

Let the volume of EDTA stock dolution added = V1

We need to prepare 25 mL(0.025L) of the solution mixture with 1 mM each of EDTA and the metal.

Hence M2 = 1 mM

V2 = 25 mL = 0.025 L

Since the number of moles remains same we can apply the relation

M1V1 = M1V2

=> V1 = M2V2 / M1 = (1 mM x 0.025L) / 100 mM = 0.00025 L = 0.25 mL

Hence we need to take 0.25 mL of the EDTA stock solution.

Metal:

For stock solution of metal, concentration, C1 = 1000 mg/L = 1000 mg / M mmol / L = (1000/M) mM

where M = molar mass of the metal.

Let the volume of metal stock dolution added = V1 L

We need to prepare 25 mL(0.025L) of the solution mixture with 1 mM each of EDTA and the metal.

Hence C2 = 1 mM

V2 = 25 mL = 0.025 L

Since the number of moles remains same we can apply the relation

M1V1 = M1V2

=> V1 = M2V2 / M1 = (1 mM x 0.025L) / (1000/M) mM = 0.000025M L = 0.025M mL

Hence we need to add 0.025M mL of the metal stock solution.

Note: Here M = molar mass in gram. We only need to multiply 0.025 by M(in gram) to get the volume of the stock solution in mL.

Procedure to prepare:

1. Take 0.025M mL of the metal stock slution and put it to a 25 mL flask. Rinse it completely so that all of the metal stock solution are poured into the 25 mL flask.

2. Also take 0.25 mL of the EDTA stock slution and put it to a 25 mL flask. Rinse it completely so that all of the EDTA stock solution are poured into the 25 mL flask.

3. Add distilled water to dilute the solution upto 25 mL.