A standard solution for Fe2+ determination can be prepared from ferrous ammonium

Question

A standard solution for Fe2+ determination can be prepared from ferrous ammonium sulfate (Fe(NH4)2(S04)2 6H20) Calculate the formula weight of ferrous ammonium sulfate. (Units g/mol) 392.15 Calculate the mass of Fe in 158 mg of ferrous ammonium sulfate 4 22.5 4922.5 Calculate the mass of ferrous ammonium sulfate you would need to add to a 3.00 mL volumetric flask to make a solution that is 53 ppm in Fe. 4 .16 Stock solution A is prepared by dissolving 182 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 9.00 mL of stock solution A to 100.00 mL. Calculate the concentration of iron, in ppm, in stock solution B 402

Explanation / Answer

Formula is Fe(NH4)2(SO4)2. 6H2O

Atomic weights : Fe=56, N=14, H=1 O=16, S =32

Molecular weight = 56+2*(14+4)+2*(32+64) +6*(2+16)=392

392 gm of sample containd 56 gms of iron

158 mg of sample contains 158*56/392=22.57mg

3. 53 ppm= 53 mg/L (1ppm =1mg/L)

1 L contains 53 mg of Fe

3ml =3*10-3 L contains 53*3*10-3=0.159 mg Fe

Ratio of Ferrous ammonium sulfate to iron = 392/56 =7

3ml containd 0.159 mg

1mg of Fe correspond to 7 mg of Ferrous ammonium sulfate

0.159 mg of Fe correspoond to 0.159*7/3=0.371 mg

d)     182 mg of Ferrous ammonium sulfate contains 182*56/392 mg of Fe.=26 mg of Fe

26 mg of Fe is dissolved in 500ml = 26mg/0.5L= 52

ppm =mg/L, Concentration of iron =52 ppm

e) 52*9= 100* concentration of iron

Concentration of iron =52*9/100=4.68 ppm

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