Question

1. A standard solution for Fe2+ determination can be prepared from ferrous ammonium sulfate (Fe(NH4)2(SO4)2.6H2O).

a) Stock solution A is prepared by dissolving 156 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.

b) Stock solution B is prepared by diluting 5.00 mL of stock solution A to 100.00 mL. Calculate the concentration of iron, in ppm, in stock solution B.

Answer 1

Molar mass of Fe[NH4]*2[SO4]*2.6H2O is 312.8484 g/mol

comccentration = 156 x 500/ 312.8484 x 1000 =   0.2495 mole

1 ppm = 1 μg X /( mL solution)x(1 L/1000 mL)
1 ppm = 1000 μg X / L solution
1 ppm = 1 mg X/L solution

We know the molarity of the solution, which is in moles/L. We need to find mg/L. To do this, convert moles to mg.

moles/L of Fe²⁺ = 0.2493 M

moles/L of Fe²⁺ = (0.2493 mol x 312.848 g/mol)/L
moles/L of Fe²⁺ = 7.96 x 10^-4 g/L

We want mg of Fe²⁺ , so

moles/L of Fe²⁺ = 7.96 x 10^-4 g/L x 1000 mg/1 g
moles/L of Fe²⁺ = 0.79 mg/L

Since 1 ppm = 1 mg/L

moles/L of Cu²⁺ = 0.796 ppm

concentration of iron, in ppm, in stock solution A is 0.79 ppm

20 times concentration will reduce sock solution B

Hence concentration of iron, in ppm, in stock solution B = 0.796/20 = 0.0398 ppm