How would a student make 500.0 mL of a 0.08500 M solution of iron(III) using a stock solution that is 0.3180 M in iron(III)? Explain how this would be done.

Question

DrBob222 · Accepted Answer

I know not how a student would do it but I would use c1v1 = c2v2
0.08500M*500 mL = 0.3180M*v2
v2 = 133.6 mL
The way you do this is to pipet or measure accurately 133.6 mL of the stock solution and transfer quantitatively to a 500 mL volumetric flask. Then add distilled water to the mark of the volumetric flask.