6. An electroplating facility produces a hazardous waste stream containing 1.2 g

Question

6. An electroplating facility produces a hazardous waste stream containing 1.2 g/L copper (Cu2+). The stream enters a continuous-flow precipitation reactor operated at a pH of 9.5. Will the effluent stream (the treated water, after sedimentation of the precipitate) meet the drinking water standard for copper of 1.3 mg/L? Support your answer with calculations Cu(OH)2(s) ? Cu2+ + 2 OH. Ks 2x10-19 The reactor described in Problem 6 receives 40 Umin of the 1.2 g/L Cu2+ stream. The sedimentation tank following the reactor produces a sludge stream that is 5% Cu(OH)2(s) by mass. By what percentage did you reduce the volume of the hazardous waste stream? You may assume the density of the sludge stream is 1.04 g/cm3 7.

Explanation / Answer

Problem 6

pH of precipitator = 9.5 thus pOH = 14-9.5 = 4.5

Thus [OH] = 1.1*10^-2 M

Ks= 2*10^-19

Thus [Cu] *[OH]^2 = 2*10^-19 thus [Cu] = 1.653* 10^-15 M

Thus concentration of copper in precipitation unit = 1.653*10^-15 M = 64* 1.653*10^-15 g/l = 105.792*10^-15 g/l = 1.0579*10^-10 mg/l which will be the copper ion concentration in the effluent stream from the precipitator and thus it will meet the drinking water limit as it is below that limit of 1.3 mg/l.

Problem 7

Density of sludge stream = 1.04g/cm3

Discharge of stream = 40L/ min

Copper ion concentration in discharge= 1.2g/l

Cu (OH)2 = 5*1.04/100 g /ml = 0.052 g/ml= 52 mg/l = 52/98 = 5.31* 10^-4 M

Thus copper ion removed = 5.31*10^-4 * 64 g/l = 339.84 *10^-4 g/l

Thus percentage removal = (0.034)/1.2 = 2.83%

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