Alkalinity Use Question

[caneng2002]

Greetings

Based on the following equation taken from Water Treatment - Principles and Design (2nd Edition)

4Fe(HCO3)2 + O2 + H2O >>> 4Fe(OH)3 + 8CO2

I'm trying to figure out how Alkalinity used (mg/ mg Fe) of 1.8 was calculated.

I can see how oxidant needed of 0.14 mg/mg Fe was calculated (32/223.2) and that sludge production is 1.9 kg/kg Fe (233 +204)/233.2.

The closest I can get is bicarbonate on lefts side = 488 subtract hydroxide on right side of 51 divided by Fe. which is more a guess.

Any advice appreciated.

 

[bimr]

Oxidation of ferrous bicarbonate with air:

4Fe(HCO3)2 + O2 + H2O >>> 4Fe(OH)3 + 8CO2

Oxidant needed of 0.14 mg/mg Fe was calculated (32/223.2) and sludge production of 1.9 kg/kg Fe was calculated (223.2 +204)/223.2.

Calculate alkalinity used (mg/ mg Fe):

Alkalinity used = (4 *61*2 ) * (0.82) /223.2 = 1.8

Alkalinity is reported in terms of equivalent CaCO3. The 0.82 factor converts the substance to CaCO3 equivalent.