Question

A municipal water treatment plant is designed to produce 20 million gallons per day (MGD) of potable water. The plant treats surface water using coagulation followed by sedimentation, filtration, and chlorination. Although the water is hard, it is not presently softened. The water is coagulated using a 30 mg/L dosage of alum (aluminum sulfate, Al2(SO4)3•14.3H2O, MW = 600). This chemical reacts in water to form Al(OH)3 as follows:

Al2(SO4)3 + 6HCO3- → 2Al(OH)3(s) + 3SO4-2 + 6CO2

How many mg/L of bicarbonate (alkalinity) will be "consumed" (converted to CO2) when the

alumisadded? (AWs: Al=27,S=32,C=12,0=16,H=1,Ca=40)

Estimate the pH of the water following coagulant addition. Assume an initial bicarbonate concentration of 244 mg/L and water initially in equilibrium with the atmosphere (i.e., H2CO3* = 10-5 M). (Hints: assume negligible loss of CO2 to the atmosphere from the deep basins used for treatment; adjust the concentrations of HCO3- and H2CO3*, based on your answer to Problem 1; then use the expression for K1 to solve for [H+] following coagulant addition. Assume [OH-] and [CO3-2] are negligible.)

How many mg/L of hydrated lime, Ca(OH)2, would be needed to keep the pH constant when the alum is added? Hint: the neutralization reaction can be represented as follows:

Al2(SO4)3 + 3Ca(OH)2 → 2Al(OH)3 + 3Ca+2 + 3SO4-2

Answer 1

From the first reaction,

1 mole of alum reacts with 6 moles of HCO3-

molarity of alum = 30 mg/L x 1/1000 x 1/600 = 5 x 10^-5 moles/L

Volume = 20 MGD = 20 x 694.4 gpm = 694.4 x 3.785 L/min

moles of alum = molarity x volume = 5 x 10^-5 x 20 x 694.4 x 3.785 L = 2.63 moles/min

So, moles of HCO3- consumed would be = 6 x 2.63 = 15.78 moles/min

[HCO3-] = 15.78/20 x 694.4 x 3.785 = 0.0003 M

K1 = [H+][HCO3-]/[H2CO3]

[H2CO3] = 244 mg/L = 244 x 1/1000 x 1/62.03 = 0.004 M

So, at equlibrium,

[H2CO3] = 0.004-0.0003 = 0.0037 M

let [H+] = x M

K1 = 2 x 10^-4 [from literature]

Feed values,

K1 = 2 x 10^-4 = (x)(0.0003)/0.0037

x = [H+] = 1.233 x 10^-3 M

pH = -log[H+]

     = -log(1.233 x 10^-3)

     = 2.91

From the second equation given above,

1 mole of alum requires 3 moles of Ca(OH)2 for neutralization reaction

moles of alum = 2.63 moles

so, moles of Ca(OH)2 = 3 x 2.63 = 7.89 mole

mg/L of Ca(OH)2 needed = 7.89/(74.093 x 20 x 694.4 x 3.785 = 0.002 mg/L