Question

Jar testing was performed using iron (ferric chloride, FeCl3⋅7H2O) on a raw drinking-water source that contained an initial turbidity of 20 NTU and an alkalinity of 30 mg/L as CaCO3. The optimum coagulant dosage was determined as 36 mg/L with a final turbidity of 0.2 NTU. Determine the quantity of alkalinity consumed as CaCO3. (MW of iron = 288 g/mole) FeCl3⋅7H2O + HCO3 - ↔Fe(OH)3(s)+ CO2+ Cl- + H2O (note that you must balance this equation)

Answer 1

The reaction is

FeCl3⋅7H2O + HCO3- ↔Fe(OH)3(s)+ CO2+ Cl- + H2O

balanced equation

FeCl₃·7H₂O + 3HCO₃ = Fe(OH)₃ + 3CO₂ + 3Cl- + 7H₂O

As per equation one mole of Ferric chloride will react with three moles of HCO3-

Dose of coagulant used = 36mg / L

mol wt of FeCl₃·7H₂O = 288g / moles

Molecular weight of HCO3- = 61 g / moles

So 288 grams will react with 61 X 3 grams of HCO3-

1 gram will react with = 183 / 288 grams of HCO3-

so 36mg/L will react with = 183 x 36 / 288 mg / L = 22.875mg/L

61 grams of HCO3- is equivalent to 50 grams of CaCO3 [equivalents]

1 gram is equivalent to 50/61 CaCO3 grams

22.875mg/L will be equivalent to 22.875 X 50 / 61 mg /L CaCO3 = 18.75m / L