Question

1.1 Accurately (using an analytical balance ONLY) weigh 1.4 g of Fe(NH4)2(SO4) 2-6H20 1.2 Dissolve in 50 mL of 4% sylturisacid in a beaker. 1.3 Transfer into a 1 L volumetric flask; adjust the volume and mix well. Note the purity of the Fe(NH4)2(SO4)2.6H20 given on the label of the container. 1.4 Transfer this solution into a 1 L glass bottle. Preparation of working iron standard solution: 1.5 Pipet 10.00 mL of the iron standard stock solution into a 500.0 mL volumetric flask and dilute to the mark with water 1.6 Calculate the molarity of Fe in the solution from the weight of Fe(NH4)2(S04)2.6H20 used and its purity (consult the label on the container). Transfer this solution into a 1 L glass bottle. This solution is called the working iron standard solution. Label on the container as follows ACS specification Fe(NH4)2(SO4)2.6H2O...0.8% Copper.. Ferric iron (Fe+3).. Insoluble matter.... ..0.0003% .0. .005% ......0.002%

Answer 1

1.6 (asked question )

Weight of Fe(NH₄)₂(SO₄)₂·6 H₂O = 1.4 g

Fe(NH₄)₂(SO₄)₂·6 H₂O molar mass = 392.13 g/mol

Moles of Fe(NH₄)₂(SO₄)₂·6 H₂O = 1.4 g / (392.13 g/mol ) =3.5702*10^-3 mole

Initially prepared Stock solution = 1.0 L

Working iron standard solution (from 10.00 ml of Stock solution to 500.0 ml ) = 50 time dilution

Fe Molarity in iron standard solution :

Assay of Fe(NH₄)₂(SO₄)₂·6 H₂O = 100.08 %

So moles in stock solution = 100.08 % *3.570*10^-3 mole = 3.573*10^-3 mole /L

And in standard solution = 7.146*10^-5 mole /L = Fe (II) moles

( 1 mol of  Fe(NH₄)₂(SO₄)₂·6 H₂O = 1 mol of Fe (II) )

It contains Fe(III) : 0.005 % , thus its amount in standard solution = 1.4 g *0.005 % / 50

= 1.4*10^-6 g = 2.507*10^-8 mol /L

(Fe M.W. = 55.85 g / mol )

Total moles of Iron in standard solution = 7.149 *10^-5 mole /L

So Fe molarity in standard solution : 7.149 *10^-5M