Problem .
To aid sedimentation in the primary settling tank, 25 mg/L of ferrous sulfate
heptahydrate is added to the wastewater. Determine the minimum alkalinity
required to react initially with the ferrous sulfate. How many grams of lime
should be added as CaO to react with Fe(HCO3)2 and the dissolved oxygen in
the wastewater to form insoluble Fe(OH)3 ?
Solution :
FeSO4.7H2O + Ca(HCO3)2 -------- Fe(HCO3)2 + 7H2O
25 mg/L
Ca(HCO3)2 = 25 x (1/1) x (MW Ca(HCO3)2 / MW FeSO4.7H2O)
= 25 x (162/278) = 14.57 mg/L
Alkalinity = 14.57 x (100/162) = 8.99 mg/L
Alkalinity : mg/L CaCO3
HCO3- - pH less than 8.3 but more than 4.5
CO3- - pH greater than 8.3 - P Alkalinity
OH- - pH greater than 8.3 - P Alkalinity
P = ½ CO3-2 + OHM = HCO3- + CO3-2 + OHAlkalinity is 0 at pH lower than 4.5
FeSO4 .7H2O + Ca(HCO3)2 → Fe(HCO3)2 + 7H2O
Fe(HCO3)2 = 25 x (MW Fe(HCO3)2/MW FeSO4 .7H2O)
= 25 x (178/278) = 16.01 mg/L
Fe(HCO3)2 + 2Ca(OH)2 → Fe(OH)2 + 2CaCO3 + 2H2O
16.01
Ca(OH)2 = 16.01 x (2/1) x (MW Ca(OH)2/MW Fe(HCO3)2)
= 16.01 x 2 x (74/178) = 13.31 mg/L
CaO = 13.31 x (56/74) = 10.07 mg/L
4Fe(OH)2 + O2 + H2O → 4Fe(OH)3
Fe(OH)2 = 16.01 x (90/178) = 8.09 mg/L
O2 = 8.09 x (1/4) x (32/90) = 0.72 mg/L