Sharp et al. (2004) fractionated the dissolved organic material in a natural wat

Question

Sharp et al. (2004) fractionated the dissolved organic material in a natural water into four components and determined the charge density of each fraction by titrating solutions with poly-DADMAC, a cationic polymer; the (diluted) polymer solution had a charge density of 6.2 meq/g of solution. The results of the fractionation and titrations are shown in the following table. The titrations were done using concentrated solutions of the different fractions, in each of which 250 mg of dissolved organic carbon (DOC) was added to 100 ml of water. Assuming that the density of the polymer solution is the same as water (1g/mL), what was the dose of polymer (mg polymer solution per l of sample) required to neutralize the charge associated in the concentrated solution containing the humic acid fraction? Assuming the polymer fed using the same diluted polymer solution described above, what dose of polymer (mg polymer solution per L of water) would be required to neutralize the charge associated with each fraction the DOC in the natural water? What would the doses be In terms of volume of polymer solution per volume of sample? The source water also has a solids concentration of 8 mg/L. For simplicity, assume that all particles are spherical with a diameter of 1.5 pm and a density of 1.5 g/cm. If the (negative) charge density on the particle surfaces is 0.2 C/m^2, what would be the additional required polymer dose. In mg/L, to neutralize these particles, if the same diluted polymer stock solution was used? If polymer were added to neutralize the charge on both the organics and the particles, what fraction of the polymer dose would be allocated to the dissolved organics? For the experiments reported, the poly-DADMAC was diluted to 0.1 % w/w solution, but the product is sold in a 40 % w/w solution. If the flow to the water treatment plant using this water as its source is 2 m^2/s, what is the daily requirement (in units of kg/d) for the product as sold?

Explanation / Answer

a)

For table 7.9 mg DOC has 3.5 g Humic acid

250 mg DOC gives 110.75 g Humic acid

Concentration of humic acid in mg/ L is 110.75 mg /110 L = 1 mg/ L

For titration concentration should be same, concentration of polymer is 1 mg/ L

b) Humic acid

3.5 * 6.8 = x* 6.2

x = 3.84 mg/ L

Fulvic acids

2.4 * 4.2 = x * 6.2

10.08 = x * 6.2

x = 1.625 mg/ L

Hydrophilic acids

2* 0.06 = 6.2 * x

x = 0.019 mg/L

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