The Busytown drinking water treatment plant supplies 4 ML d^-1 of clean water to
ID: 477095 • Letter: T
Question
The Busytown drinking water treatment plant supplies 4 ML d^-1 of clean water to town residents. The plant uses ferric chloride (FeCl_3) coagulant to produce ferric hydroxide (Fe(OH)_3) floc according to this reaction: 2FeCl_3 + 3Ca^2+ + 6HCO_3^- rightarrow 2Fe(OH)_3 (s) downarrow, + 3Ca^2+ + 6Cl^- + 6CO_2 Source water to the plant contains total hardness of 150 mg L^-1 as CaCO_3 and alkalinity of 75 mg L^-1 as CaCO_3 and 15 mg L^-1 of total suspended solids. The plant operators dose 50 mg L^-1 of ferric chloride into the water to remove 60% of the suspended solids. Evaluate whether there is sufficient alkalinity in the water for the coagulation to proceed until all of the ferric chloride is consumed. Calculate the daily mass (kg) of treatment residual/sludge/solids produced by the ferric chloride coagulation process in the Busytown treatment plant.Explanation / Answer
(a)
Total hardness=150mg/L as CaCO3
Alkalinity=150mg/L as CaCO3
Total suspended solids=150mg/L
2FeCl3+3Ca2++6HCO3 2Fe(OH)3+3Ca2++6Cl_+6CO2
[(2x55.845)+(3x35.5)g] that is 218.2g of FeCl3 is needed to remove 390g of HCO3- as213.7 g of Fe(OH)3
Or
218.2mg of FeCl3 is needed to remove 390mg of HCO3
Therefore 50mg can remove=390x50/218.2=89.368mg of HCO3-
The alkalinity of hard water is due to the presence of HCO3
Since all the HCO3- is removed, there is not enough alkalinity in water for the coagulation to proceed.
(b)
2FeCl3+3Ca2++6HCO3 2Fe(OH)3+3Ca2++6Cl_+6CO2
From the above equation it is clear that only HCO3- is removed as OH. Hence all the sludge produced by the addition of ferric chloride is due to the formation of Fe(OH)3
To 1 L of water 50mg of ferric chloride is added
Therefore to 4,000,000L (4 ML )= 200,000,000mg of ferric chloride is added
Since 1,000,000mg= 1Kg
200,000,000mg=200Kg of ferric chloride is added
218.2g of FeCl3 of ferric chloride removes 213.7 g of Fe(OH)3
Therefore 200g of FeCl3 of ferric chloride removes 195.875 g of Fe(OH)3
Therefore 200Kg of FeCl3 of ferric chloride removes 195.875 Kg of Fe(OH)3
195.875 Kg of Fe(OH)3 is daily produced as sludge
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