What was the resulting chlorine dose in mg/L when 22.6 gallons of 9.5% sodium hypochlorite solution were used to treat 725,000 gallons of water?

• A. 2.96 mg/L
• B. 3.12 mg/L
• C. 3.3 mg/L
• D. 3.5 mg/L

Answer: (A)

The main idea is to convert the dosing solution to a mass of disinfectant and spread it over the treated water to find mg/L. You add up the total mass of NaOCl delivered and divide by the total volume of water being treated.

First convert volumes: 22.6 gallons is about 85.56 liters. The water to be treated, 725,000 gallons, is about 2,744,422 liters.

Next, determine how much NaOCl is in the dosing solution. 9.5% by weight means 0.095 g NaOCl per g of solution. With density close to 1 g/mL (roughly 1,000 g per liter), that corresponds to about 95 g of NaOCl per liter of solution. So the total NaOCl added is 85.56 L × 95 g/L ≈ 8,128 g (≈ 8,128,000 mg).

Now divide by the treated water volume: 8,128,000 mg / 2,744,422 L ≈ 2.96 mg/L.

The resulting chlorine dose is about 2.96 mg/L.