Question

2. A groundwater contains 3.6 mg/L Fe+2 and 0.78 mg/L Mn+2.

a. What dosage of ozone in mg/L is required to oxidize the soluble iron and manganese.
(Show balanced half reactions, balanced overall reactions, and stoichiometry derived
from overall reactions).

b. What mass concentration of solids is produced?

c. Considering only the reaction of ozone and iron: If the pH is held constant at 6, the
dissolved oxygen concentration is near saturation (assume 10 mg/L), and the residual
ozone concentration is 0.5 mg/L, how much dissolved iron remains at equilibrium?
Provide your calculation and what the answer means in practical terms.

Answer 1

a)

The overall reaction of oxidation of iron(II) by ozone is as follows:

2Fe²⁺ + O₃ + H₂O → 2Fe³⁺ + 2OH^– + O₂

Fe³⁺ so formed reacts with OH^- generated and gets precipitated as ferric hydroxide.

Oxidation half-cell:

2Fe²⁺ ---> 2Fe³⁺ + 2e^-

Reduction half cell:

O₃ + H₂O + 2e^- ---> O₂ + 2OH^-

2 moles of Fe(II) uses one mole of ozone

==> 2 x 55.845 g of Fe(II) uses 48 g of ozone

==> 3.6 mg/L of Fe(II) uses 48 x 3.6/(2 x 55.845) mg of ozone = 1.547 mg/L of ozone

The overall oxidation of Manganese is as follows:

Mn²⁺ + O₃ + H₂O → Mn⁴⁺ + 2OH^– + O₂

Mn⁴⁺ so formed precipiates as MnO₂

Oxidation half cell:

Mn²⁺ ---> Mn⁴⁺ + 2e^-

Reduction half cell:

O₃ + H₂O + 2e^- ---> 2OH^- + O₂

1 mole of Mn(II) uses 1 mole of ozone

==> 54.938 g of Mn(II) uses 48 g of ozone

==> 0.78 mg/L of Mn(II) uses 0.78*48/54.938 mg/L of Ozone = 0.6815 mg/L of ozone

Therefore total ozone demand = 1.547 + 0.6815 = 2.2286 mg/L of ozone

b)

1 mole of Fe(II) produces 1 mole of ferric hydroxide

55.845 g of Fe(II) produces 106.867 g of ferric hydroxide

3.6 mg/L of Fe(II) produces 106.867*3.6/55.845 = 6.889 mg/L of Ferric hydroxide

1 mole of Mn(II) produces 1 mole of MnO₂

54.938 g of Mn(II) produces 86.9368 g of MnO₂

Therefore 0.78 mg/L of Mn(II) produces 0.78*86.9368/54.938 = 1.234 mg/L of Manganese dioxide

c)

Amount of ozone present in water = 0.5 mg/L

48 g of ozone precipitate 2 x 55.845 g of Fe(II)

Therefore 0.5 mg/L of ozone precipitates 2*55.845*0.5/48 = 1.163 mg/L of Fe(II)

Initial concentration of Fe(II) = 3.6 mg/L

Concentration of Fe(II) precipitated = 1.163 mg/L

Therefore concentration of Fe(II) remaining = 3.6 - 1.163 = 2.437 mg/L

In practical terms, this concentration of ozone (0.5 mg/L) is not sufficient to completely precipitate Fe(II). The minimum concentration of ozone required to precipitate Fe(II) is 1.547 mg/L