Question

The Busy Town drinking water treatment plant recently changed coagulant aid from alum
to ferrous sulfate (FeSO4·7H2O) for TSS removal. The raw water contains 7 mg/L of TSS.
Ferrous sulfate coagulation proceeds as follows:
FeSO4·7H2O + 2 Ca(HCO3-) + 1⁄2O2 -> 2Fe(OH)3 (s) ↓ + 2CaSO4 + 13H2O

If an optimum dose of 30 mg/L of ferrous sulfate is required to remove 90% of the TSS, calculate the mass of residuals (mg/L) generated in this process.

Answer 1

Molecular weight of FeSO4·7H2O = 278

Molecular weight of 2Fe(OH)₃ = 213.74

As per given equation

1 molecule of ferrous sulphate produce 2 molecule of ferric hydroxide (ppt)

---> 278 mg ferrous sulphate produce 213.74 mg of ferric hydroxide (ppt)

30 mg/l ferrous sulphate will produce ferric hydroxide = (213.74 /278)×30

= 23 .06 mg /l

90% of TSS is removed

i.e. 0.9 x 7 = 6.3 mg/l

Mass of residual generated

= 23.06 + 6.3

= 29.36 mg/l