Question

Primary clarifier effluent was analyzed for solids content by an analytical laboratory. A 250-mL sample of the effluent was filtered through a laboratory (1 micrometer) filter. A 241-mL sample of the filtrate contained 89 mg of salts as dissolved solids (Residue 1). Solid residue from the filtration was dried overnight at 105 deg C and weighed 37 mg (Residue 2). What were the concentrations of total dissolved solids, total suspended solids, and total solids in the primary clarifier effluent?

Water Sample Volume (mL)	250
Filtrate Volume (mL)	241
Filtrate Residue (Residue 1, mg)	89
Filter Residue (Residue 2, mg)	37

Answer 1

-> we know that, Total solids = Total suspended solids + Total dissolved solids.

-> Suspended solids can be find using filtration. Total suspended solids equal to the weight of solid residue from filtration was dried overnight at 105°C.

-> Dissolved solids can be find using electric conductivity through filtrate obtained in filtration.

-> so it is clear that, after filtration we have residue and filtrate. From residue we calculate suspended solids and from filtrate we calculate dissolved solids.

-> From given data, residue 1 gives total dissolved solids = 89 grams for 250 ml of sample.

So, concentration of total dissolved solids = weight of total dissolved solids/ sample quantity.

= 89/250 = 0.356 mg/ ml= 356 mg/ litre.

-> The Residue 2 is equal to the total suspended solids= 37 mg for 250 ml of sample.

So, concentration= 37/ 250 = 0.148 mg/ ml= 148 mg/ litre.

-> Finally total solids= 89+37= 126 mg for 250 ml sample.

And it's concentration= 126/250= 0.504 mg/ ml= 504 mg/litre

( Please give rating. Thanks)