Question

You wish to adapt the AA method to measure the amount of iron in leaf tissues. The minimum amount of iron in the tissues is expeted to be about 0.0025% by mass. The minimum concentration for AA measurements is 0.30 ppm. Your plan is to weigh out 4.0g leaf tissue samples, digest them in acid, filter and dilute them to 50mL. This solution is your "sample stock solution". You will then pipet a portion of this solution into a 25-mL volumetric flask and dilute to volume. This solution is your "diluted sample solution" and you will make your AA measurements on this solution. The question is, how much of the sample stock solution should you use if the dilute sample solution needs to have a concentration of 0.20 ppm?

a) How many milligrams of Fe are in 4.0g of a leaf tissue that is 0.0025% Fe by mass? *Remember, 0.0025% by mass = 0.0025g Fe in 100g of sample

b) If all of the iron from the 4.0g leaf sample in the previous question is diluted in a 50 mL flask, what is the concentration of the resulting stock solution (in ppm)?

c) What volume of the stock solution made in the previous question is needed to prepare 25.0 mL of a dilute sample solution with a concentration of 0.30 ppm Fe?

Answer 1

a) Mass of Fe in leaf tissue = Mass % of Fe * Mass of leaf tissue

=> 0.0025% of 4.0g

=> 0.0025/100 * 4

=> 0.0001g = 0.1 mg

b) 1 ppm = 1 mg/L

Here we are dissolving 0.1 mg in 50 mL of flask, hence the concentration will be 0.1 mg/(0.050)L = 2 ppm

Hence the concentration of the stock solution (in ppm) will be 2 ppm

c) We need to create diluted sample solution of 0.30 ppm Fe

M1 = Molarity of stock solution = 2 ppm

V1 = Volume of stock solution = V1

M2 = Molarity of stock solution = 0.30 ppm

V2 = Volume of stock solution = 25.0 mL

2 * V1 = 0.30 * 25

V1 = 3.75 mL