Question

2. A dose of 41 mg/L of alum is used in coagulating a turbid water at a water treatment plant treating a flow rate of 32 million gallons per day (mgd). The natural alkalinity of the source water is 10 mg/L (as CaCO3). a. Will the alkalinity be completely consumed in the reaction? If not, then how much alkalinity will be left? If so, then what concentration of caustic soda (NaOH) must be added to avoid a change in the pH? b. How many kg of aluminum hydroxide (Al(OH)3) is produced per day?

Answer 1

1 0 Alum Alloy)3 18 H20 Weight of alumi = 666 gm Imol Alkalinity Celor Weight = 800gm/mol 666 г. в Арт Форсаrе е 3xtoo yvnot alkal.ruht 41.gan mglar of alum reppure = 3x100 xul mguur of atkelinoty edenog - 666 v Alkelenty required = 18.488 mglaer ascecay Natural alkelinity = 10 mgler as Calos Alkaenity required a natural alkalinity Completely consumed ite ilakul albaunety. → b) Weight of Al@H)2 -- 27+ 3(16+2) = 78. 666 gan of clum produce = 2x78 gm of ALOIT)3 41 mgler of cellon prodecce = 2x78x4l on 1 englap of Allots 666 y = 9.603 mglet of 46C014), Totul Al6H), produceed = 4xQ G= 32x106 gallyday Igallen = 3.785 let @ - (3283.785) xcom eer/day Total Alcht)g = 9.603 * @2x3-785)xco* mo. (seychand CS Scanned with a 1163.18 kg/day . . CamScanner