Well-Water Chlorination and Iron Removal
A village vocational health group tests a well used by a 5000 L storage tank. Results: pH 6.3; iron 0.5 mg/L; nitrate 12 mg/L as NO3-; E. coli 30 CFU/100 mL; free chlorine after a trial dose is below detection. Local limits: pH 6.5-8.5; iron <=0.3 mg/L; nitrate <=45 mg/L; E. coli 0 per 100 mL; free chlorine residual >=0.2 mg/L after 30 min. The group has bleaching powder with 30% available chlorine. A jar test shows a chlorine demand of 1.2 mg/L, but that test used raw water before iron removal. The group plans to dose to 2.0 mg/L free chlorine. Iron oxidation by chlorine follows 2Fe2+ + Cl2 -> 2Fe3+ + 2Cl-. The group must decide the dose, the order of treatment, and whether the water can be declared safe after a single chlorination step.
1. Calculate the bleaching powder needed for 5000 L to achieve a 2.0 mg/L free chlorine residual, ignoring iron. Compare the demand and the applied dose.
Read the worked answer
The applied chlorine dose must cover demand plus residual: dose = 1.2 + 2.0 = 3.2 mg/L as Cl2.
For 5000 L: mass of Cl2 = 3.2 mg/L x 5000 L = 16000 mg = 16.0 g Cl2.
Bleaching powder is 30% available chlorine: mass of powder = 16.0 / 0.30 = 53.3 g.
So about 53 g of bleaching powder is needed if the raw-water demand of 1.2 mg/L is the only demand. The 3.2 mg/L dose is larger than the 1.2 mg/L demand because the extra 2.0 mg/L is the intended residual.
2. Recalculate the dose and bleaching powder mass after accounting for chlorine consumed by 0.5 mg/L iron(II). State any assumption.
Read the worked answer
From 2Fe2+ + Cl2 -> 2Fe3+ + 2Cl-, 2 mol Fe (2 x 55.8 = 111.6 g) consumes 1 mol Cl2 (70.9 g). So chlorine required per mg Fe2+ is about 70.9 / 111.6 = 0.635 mg Cl2 per mg Fe.
For 0.5 mg/L Fe2+: chlorine consumed = 0.5 x 0.635 = 0.32 mg/L.
Revised demand = 1.2 + 0.32 = 1.52 mg/L. Revised dose for 2.0 mg/L residual = 1.52 + 2.0 = 3.52 mg/L.
For 5000 L: mass Cl2 = 3.52 x 5000 = 17600 mg = 17.6 g. Bleaching powder = 17.6 / 0.30 = 58.7 g.
This assumes all iron is dissolved Fe2+ and that the 1.2 mg/L jar-test demand did not already include iron oxidation. If some iron was already oxidized, the extra demand would be lower.
3. Judge whether the water can be declared safe after one chlorination step. Recommend the treatment order and the post-treatment checks.
Read the worked answer
The water cannot be declared safe from the dose calculation alone. E. coli is 30 CFU/100 mL, above the limit of 0 per 100 mL, so disinfection is required. Iron is 0.5 mg/L, above the 0.3 mg/L limit, and pH is 6.3, below the 6.5-8.5 range. Nitrate at 12 mg/L is below the 45 mg/L limit, so it is not the immediate problem.
A single chlorination of raw water is a poor order because chlorine will oxidise Fe2+ to Fe3+, creating turbidity and consuming part of the dose. The precipitate can shield microbes and reduce disinfection efficiency. Better order: aerate/oxidise and filter the iron first, correct pH into range, then chlorinate with enough contact time, and finally check free chlorine residual.
After treatment, verify E. coli = 0 per 100 mL, free chlorine >=0.2 mg/L after 30 min, iron <=0.3 mg/L, and pH 6.5-8.5. If these are not met, repeat or adjust treatment. Chlorination alone at pH 6.3 may disinfect, but it does not fix the iron or pH violations.