Question

1. (35 pts) The following balanced chemical equations represent the use of lime [Ca(OH)2] for softening of drinking water. First step: lime added and all CO2 precipitates as CaCO3 Ca(OH)2 + + CO2 – CaCO3 + H20 Second step: more lime added and all Ca²+ and HCO3- precipitate as CaCO3 Ca(OH)2 + Ca²+ + 2HCO3 – 2CaCO3 + 2H20 Third step: more lime added and Mg2+ and HCO3 are converted to MgCO3 which does not precipitate (also more Cacos precipitate is formed) Mg2+ + HCO3 + Ca(OH)2 – CaCO3 + MgCO3 + 2H20 Fourth step: more lime added reacts with the MgCO3 to form two precipitates, Mg(OH)2 and CaCO3 MgCO3 + Ca(OH)2+ CaCO3 + Mg(OH)2 Given the following concentrations of CO2 (8.5 mg/L), Ca2+ (48 mg/L), Mg2+ (39 mg/L), HCO3- (84 mg/L), and flow rate (1 MGD), determine the following: a) the required lime (Ca(OH)2], Ib/day b) the total mass of precipitate (sludge) formed, Ib/day c) the total amount of water formed, gal/day d) the residual concentrations of CO2, Cazt, and Mg2+ (mg/L).

Answer 1

Sol: Given that 1st Step :- Ca(OH)₂+ co caco, + H₂O mal step!- Ca (OH)2 + cat + 2HCO3- 2 cacozt 2H₂O gird Stepi- mget + 2H2O + Hcq- + Ca(OH), Cacoz + mgco3 gth step:- MgCO3 + Ca(OH)+ Cacoz + Mg Comby low ratei | MGO ㅛ Coz Concentration to 8.5 mg/l ca²t concentration - 48 mg/L concentration: 39 mg/L HC Concentration : – 84 mg IL Mgat IMGD = 3785411-98 LT. yees (a) Required lime [ce.com)]: moles of Ca(OH)₂ 2t ca concentration * Flow rate molar mass of CaCOH)2 T 48 mgin * 378541118 4day 74.093 de Olmol 3 48 mgIL * 3785411-8 L day 740093 * 100 mg/mol 25920267 mol (day Moles of Ca (OH)2 = In given four steps we have four lime moles, 3. The required lime - 4* moles of Ca(OH)2 * molar mass

= 4* 2592.267 * 74.093. 10 768275. 355 many day Required lime 1693.75 l day The required lime (CacoH)2) - 1693.75 lbday (6) The total mass of precipitate (sludge) formed'- Moles of caco, ca²t concentration * Flow rate molar mass of caCoz - 48 mgl * 3785411.8 l day x103 mg/mol 100. 0869 x103 moles of ca.coz 1815.42 mol l day Total mass of precipitate (sludge) formed in the given four stages is 5 moles, total mass of precipitate formed ". The Il 5* moles of cacoz * molar mass 5 * 1815.42 * 100.089 g/mol 9085170 861 8/day 2002,94 eb/day :. The total mass of precipitate (sludge) formed - 2002.94 lb day

(C) The total amount of water formed Amount moles of H20 * molecular mass of H₂O 13 5 moles * 18.01528 g/mol - 90.076 * 106 galton day 0.1986 * 106 galton day = = 237541 gal Iday 8. The total amount of water CH₂O ) formed! 231541 gal /day 2 (d) Residual Concentrations- CO2 concentration * 1000L Coz = mola mass of Co2 8.5 mgil * loool 44001 tente x103 mg/mol i. Coz Vesidual concentration mg/l 00192 mg/L at ca Ca 2+ = concentration * ICOOL molar mass of cast lcool = 48 mg/L * 40:08 x 103 mg/mol ca²t residual concentration 1. 197 mg/L

Mgat ingat concentration * 1000L molar mass of mg 27 pl 39 ng/L * 1000L 24.305 *103 mg lanol ...Mg2t residual concentration = 1•604 mg im

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