Analytical Chemistry: A standard solution for Fe^2+ determination can be prepare

Question

Analytical Chemistry:

A standard solution for Fe^2+ determination can be prepared from ferrous ammonium sulfate (Fe(NH_4)_2 (SO_4)2.6H_2 O). Calculate the formula weight of ferrous ammonium sulfate. (Units = g/mol) Calculate the mass of Fe in 160 mg of ferrous ammonium sulfate. Calculate the mass of ferrous ammonium sulfate you would need to add to a 5.00 mL volumetric flask to make a solution that is 75 ppm in Fe. Stock solution A is prepared by dissolving 1.70 e + 02 mg of ferrous ammonium sulfate in water and diluting to volume in a 500-mL volumetric flask. Calculate the concentration of iron, in ppm, in stock solution A. Stock solution B is prepared by diluting 8.00 mL of stock solution A to 100.00 mL. Calculate the concentration of iron, in ppm, in stock solution B.

Explanation / Answer

a) Formula weight of ferrous ammonium sulphate = Atomic weight of Fe + 2 X atomic weight of nitrogen + 2 X atomic weight of sulphur + 20 X atomic weight of hydrogen + 14 X atomic weight of O

Formula weight = 55.84 + 2 X 14 + 2 X 32.1 + 20 x 1 + 14 X 16 = 392.14 g / mole (approx)

b) The mass of Fe in 392.14 grams of ferrous ammonium sulphate = 55.84 grams

So mass of Fe in 1 gram of ferrous ammonium sulphate = 55.84 / 392.14 grams

So mass of Fe in 0.160 grams (160mg) = 55.84 X 0.160 / 392.14 = 0.022784 grams = 22.8 mg

c) the concentration of Fe in ppm = 75 ppm = 75mg / Litre

For 55.84 grams we need 392.14 grams of ferrous ammonium sulphate

So for 1 gram we will need = 392.14 / 55.84 grams of ferrous ammonium sulphate

So for 75mg / Litre we will need = 392.14 X 75 mg / 55.84 (per litre) ferrous ammonium sulphate = 526.69 mg / Litre

For one litre = 526.69 grams

for 1 mL = 0.52669 grams Or 526.69mg

For 5mL = 5 X 0.52669 grams = 2.63 grams of ferrous ammonium sulphate

d) The mass of ferrous ammonium sulphate in stock = 1.7 X 102 mg

Volume = 0.5 Litres

concentration of iron in the stock solution :

The mass of Fe in 392.14 grams of ferrous ammonium sulphate = 55.84 grams

So mass of Fe in 1 gram of ferrous ammonium sulphate = 55.84 / 392.14 grams

So mass of Fe in 170mg = 55.84 X 170 / 392.14 mg = 24.21 mg

this mass is present in 0.5 Litres

Mass of fe in one litre = 24.21 X 2 = 48.42mg

So concentration of Fe in ppm = 48.42mg / L = 48.42 ppm

e) Initial concentration of Fe in stock solution A = M1= 48.42 ppm

Volume of stock solution A taken = 8mL = V1

Final concentraiton of Fe in stock solution B = M2 = ?

Final volume of stock solution B = V2 = 100mL

M1 V1 = M2V2

M2 = M1V1/ V2 = 48.42 X 8 / 100 = 3.87 ppm

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