Question

Iron sulfate reacts with strong bases in aqueous solutuon to form a whitish precipitate that acquires a green color on exposure to air. This compound is a component of the "green rust" sometimes observed on iron and steel surfaces.

a) Write balanced molecular and net ionic equations for the reactions of iron sulfate with aqueous NaOH. What is the identity of the "green rust" precipitate?

b) What mass of the "green rust" precipitate forms when 175 mL of 0.277 M NaOH is added to 230 mL of a solution that contains 14.5 g of iron sulfate per liter?

Answer 1

a) The balanced molecular equation is

2 NaOH + FeSO4 -------------> Fe (OH)2 + Na2SO4.

The net ionic equation is

Iron(II) sulfate when dissolved produces Fe²⁺ ions and SO₄²⁻ ions.

Sodium hydroxide produces Na⁺ ions and OH⁻ ions.

You do not write solids as ions.

The ionic equation is

Fe²⁺(aq)+SO₄²⁻(aq)+2Na⁺(aq)+2OH⁻(aq)→Fe(OH)2(s)+2Na+(aq)+SO₄²⁻(aq)

Cancel items that appear on each side of the equation (SO₄²⁻ and 2Na+) to get the net ionic equation.

The net ionic equation is

Fe²⁺(aq)+2OH⁻(aq)→Fe(OH)2(s)

=> Iron(II) hydroxide is a white solid, but even traces of oxygen impart a greenish tinge. The air-oxidized solid is sometimes known as "green rust".

=> If the solution is not deoxygenated and the iron reduced, the precipitate can vary in color starting from green to reddish brown depending on the iron(III) content. Iron(II) ions are easily substituted by iron(III) ions produced by its progressive oxidation.

b) Given 175 mL of 0.277 M NaOH, and 230 mLof 14.5 g of Iron sufate

moles of NaOH=Molarityxvolume=0.277 mol/Lx0.175 L=0.04847 moles NaOH.

Weight of Iron sulfate=14.5 g/Lx0.230L=3.335 g, molar mass =151.908 g/mol

moles of iron sulfate=3.335 g/151.908 g/mol=0.0219 moles.

The mole ratio between iron sulfate to NaOh is 1: 2, if we consider moles of iron sulfate, moles of NaOH is=2x0.0219 moles=0.0439 moles.

But here moles of NaOH is 0.0484 moles. So limiting agent is iron sulfate.

From the balanced equation 1mole of iron sulfate forms 1 mole of green rust.

So weght of green rust=0.0219 molx89.86 g/mol=1.967 g.