(20 pts) The following problem comes from Environmental Engineering: Principles
ID: 884115 • Letter: #
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(20 pts) The following problem comes from Environmental Engineering: Principles and Practice, by RO. Mines, Jr. (whose nephew, by the way, graduated from USF with a degree in Civil Engineering - he was a student of mine in ENV 4001!). A solids analysis is performed on a wastewater sample. The abbreviated procedure is outlined as follows: a. A Gooch crucible and filter pad are dried at 105 C to a constant mass of 25.439 g. b. 200 mL of a well-mixed sample of the wastewater is passed through the filter pad. e. The crucible, filter pad, and solids collected on the pad are dried at 105 oc to a constant mass of 25.645 g. d. 100 mL of the filtrate that passes through the filter pad in step (b) is placed in an evaporation dish that has been pre-weighed at 275.410 g. e. The sample in step (d) is evaporated to dryness at 105 oc and the dish and the residue are weighed at 276.227 g. f. Both the crucible from step (e) and the evaporation dish from step (e) are placed in a muffle furnace at 550 degree C for an hour. After cooling in a dessicator, the mass of the crucible is 25.50 1 g and the mass of the dish is 275.944 g. Determine the following: suspended solids (mg/L), dissolved solids (mg/L), total solids (mg/L), organic or volatile fraction of the suspended solids (mg/E), and the organic or volatile fraction of the dissolved solids (mg/L).Explanation / Answer
a. weight of crucible + filter pad = 25.439 g
b. Volume of waste water passed throught filter pad = 200 ml
c. Weight of crucible + filter pad + residue on filter pad = 25.645 g
So, weight of suspended solids = 25.645 - 25.439
= 0.206 g/200 ml = 1.03 mg/ml
d. weight of evaporating dish = 275.410 g
e. evaporating dish + 100 ml filtrate = after dryness weighs 276.227 g
So weight of dissolved solids = 276.227 - 275.410 = 0.817 g/100 ml = 8.17 mg/ml
Total solids = 1.03 + 8.17 = 9.20 mg/ml
f. Heating c. and e. at 550 oC for 1 h.
final weight of crucible with residue = 25.501 g
So, weight of organic or volatile fraction of the suspended solids = (25.645 - 25.439) - (25.501 - 25.439) = 0.144 g/200 ml = 7.2 mg/ml
final weight of evaporating dish with residue = 275.944 g
So, weight of organic or volatile fraction of the dissolved solids = (276.227 - 275.410) - (275.994 - 275.410) = 0.233 g/100 ml = 2.33 mg/ml
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