Question

I am making a working reagent that requires a 50:1 ratio (50 of reagent A and 1 of reagent B) and a total volume of 600uL. Therefore, I need 588uL of reagent A (water) and 12uL of reagent B (lead acetate).

I need to make a solution of lead acetate (reagentB) with: 3mM,7.7mM, and 46mM. Upon calculating the mass needed to make 3mM, 7.7mM, and 46mM I got: 0.9756 mg, 2.504 mg, and 14.9592 mg respectively and to each mass I will add 1000mL of H2O to dissolve. Now from here, how would I get a 50:1 ratio for a total volume of 600uL? Will adding more water change millimolar?

I want to learn how to go about this and any formulas/calculation will be greatly appreciated. Thank you

Answer 1

Here,

The calculations for preparing: 3 mM, 7.7mM, and 46 mM by using anhydrous Lead(II) acetate of molar mass 325.29 g/mol is correct. [Double check if the lead acetate is anhydrous or not. If its hydrate, then the mass required will be different]

Also, for preparing a 50:1 ratio, need 588 uL: 12 uL ratio of the reagents are correct.

However, instead of preparing all the concentrations separately, the highest concentration can be prepared and other concentrations can be made from it by dilution (Here, C1V1 =C2V2 formula can be used) as follows,

For preparing 7.7 mM stock solution, add 167.39 mL of the 46 mM lead acetate solution to 832.61 mL water

Similarly, for preparing a 3 mM stock solution, add 65.21 mL of the 46 mM lead acetate solution to 934.79 mL water

Here, since reagent A is water, the purpose of preparing a 50:1 ratio of the reagents is simply dilution (need not bother about the concentration after dilution). Also, adding more water will definitely change the concentration

[Use comment box if it’s not clear or if further clarifications are required]