Alkalinity Requirement

 

 

Alkalinity Requirement for Aluminum and Iron
Coagulants.

For
aluminum sulfate:

 if one assumes the majority of the alkalinity
is available as calcium bicarbonate, then:

Al2(SO4)3 · 14 H2O + 3Ca(HCO3)2       ⟶        2Al(OH)3 
+ 3CaSO4 + 6CO2 + 14H2O

 

One Mole of hydrated aluminum sulfate
(alum) reacts with 3 moles of calcium hydrogen carbonate (calcium bicarbonate).

 

One mole of alum = 621 mg; one mole
of calcium bicarbonate = 342 mg; and, one mole of calcium carbonate (CaCO3) =
100 mg.

 

Since one typically expresses alkalinity as CaCO3:

 

One mole Ca(HCO3)2 as CaCO3 = 342 mg
Ca(HCO3)2 x (100 mg CaCO3 /342 mg Ca(HCO3)2)

 

One mole Ca(HCO3)2 as CaCO3 = 100 mg

 

So, 621 mg (1 mole) of alum reacts
with 300 mg of Ca(HCO3) as CaCO3. Thus the alkalinity required is:

 

moles of alkalinity / 1 mole of
alum=3 moles of Ca(HCO3)2 as CaCO3 / Al2(SO4)3 · 14 H2O

= 300 mg/621 mg=0.49  moles of alkalinity/1 mole of alum

 

Similarly: Ferric Sulfate (1 mole = 272 mg):

Fe2SO4  · 9 H2O 
+ 3Ca(HCO3)2       ⟶         2Fe(OH)3  + 3CaSO4 + 6CO2 + 9H2O

moles of alkalinity/ 1 mole of Ferric
Sulfate = 3 moles of Ca(HCO3)2 as CaCO3/ Fe2(SO4)3

=300 mg/562 mg = 0.53 moles of
alkalinity / 1 mole of Ferric Sulfate

 

And for ferric chloride (1 mole = 325 mg):

2FeCl3 + 3Ca(HCO3)2         ⟶       2Fe(OH)3  + 3CaCl2 + 6CO2

moles of alkalinity / mole of Ferric
Sulfate = 3 moles of Ca(HCO3)2 as CaCO3 / 2 moles FeCl3

300 mg / 325 mg = 0.92 moles of
alkalinity /  1 mole of Ferric chloride

 

References:

o  
Hudson, Herbert E., Water
Clarification Processes, Practical Design and Evaluation, Van Nostrand
Reinhold, 1981.

o  
Kerri, Kenneth, Water
Treatment Plant Operation, Vol. 1, California State University, 1990.