Question

One method of removing phosphate from wastewater effluent is to precipitate itwith aluminium sulfate (alum). A possible stoichiometry (but not exact because aluminium andphosphate can form many different chemical materials) is:
2 PO43- + Al2(SO4)3  2 AlPO4 + 3 SO42-

If the concentration of phosphate (PO43-) is 30 mg/L, how many kg of alum must be purchased annually to treat 40L/s of wastewater? How many kg/yr of solid precipitate will be formed (drybasis) if all of the phosphate is precipitated as AlPO4?

Answer 1

The volume of waste water to be treated annually = 40 L / second X 3600 x 24 X 365

[As the number of days in an year = 365, the number of hours in a day = 24 and number of seconds in an hour = 3600]

The volume of waste water to be treated annyally = 1.26 X 10^9 L

Phosphate per litres = 30mg

so phosphate in the waste water to be treated annually = 1.26 X 10^9 X 30 (mg) = 3.78 X 10^7 grams

Molecular mass of phosphate = 95 g / mole

So moles of phosphate to be removed = Mass / Molecular weight = 3.78 X 10^7 / 95 = 3.98 X 10^5 moles

As per stoichiometry of equation, for each moles of phosphate we need 0.5 moles of aluminium sulphate

so for 3.98 X 10^5 moles phosphate we need = 0.5 X 3.98 X 10^5 moles of aluminium sulphate = 1.99 X 10^5 moles

Molecular weight of aluminium sulphate = 342 g / mole

So mass of aluminium sulphate required = Moles X molecular weight = 1.99 X 10^5 X 342 grams = 680.58 X 10^5 grams

Kilograms of ppt formed as sludge = Moles of aluminium sulphate X molecular weight of phosphate = 3.98 X 10^5 X 122 = 4.86 X 10^4 Kg