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Chemical feed and solutions: California T2 water treatment operator math

Solution strength, specific gravity, dilution, blending, feeders, jar tests, chemical cost.

Below are 3 worked examples. Each answer is computed with the formulas and constants on the CA Exam Conversion Sheet (revised 4/30/2026), and the solution names the line it uses.

Formulas from the sheet for this topic

Example 1. A plant treats 13.2 MGD and feeds sodium hydroxide solution at 35.5 mg/L. The product is 50% pure and weighs 12.7 lbs/gal. How many gallons per day of product are needed?

  1. A) 308 GPD
  2. B) 461 GPD
  3. C) 615 GPD
  4. D) 205 GPD
Show the answer and worked solution

Answer: C) 615 GPD

1. GPD = MGD × mg/L × 8.34 ÷ ((% purity/100) × Chemical Wt.) [Sheet p.1, Solutions: "GPD = MGD x mg/L x 8.34 / ((% purity/100) x Chemical Wt. (lbs/gal))"]. 2. Numerator: 13.2 × 35.5 × 8.34 = 3,908 lbs/day. 3. Denominator: 0.5 × 12.7 = 6.35 lbs active per gal. 4. GPD = 3,908 ÷ 6.35 = 615.4 GPD. 5. Answer: 615 GPD (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "GPD = MGD x mg/L x 8.34 / ((% purity/100) x Chemical Wt. (lbs/gal))"

Example 2. An operator dissolves 90 lbs of alum in 85 gallons of water. What is the percent strength of the solution by weight?

  1. A) 12.7%
  2. B) 11.3%
  3. C) 51.4%
  4. D) 22.6%
Show the answer and worked solution

Answer: B) 11.3%

1. Water weight = 85 gal × 8.34 lbs/gal = 708.9 lbs [Sheet p.1, Units and Conversion Factors, row 16: "1 Gal (H2O) = 8.34 lbs"]. 2. Solution total weight = 90 + 708.9 = 798.9 lbs. 3. Chemical in Solution (%) = Dry Chemical × 100 ÷ Solution Total Weight [Sheet p.1, Solutions: "Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)"] = 90 × 100 ÷ 798.9 = 11.27%. 4. Answer: 11.3% (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Units and Conversion Factors, row 16: "1 Gal (H2O) = 8.34 lbs"; p.1, Solutions: "Chemical in Solution (%) = Dry Chemical (Lbs) x 100 / Solution Total Weight (lbs)"

Example 3. How many gallons of 15% sodium hypochlorite are needed to make 200 gallons of a 2.5% solution?

  1. A) 25.0 gal
  2. B) 33.3 gal
  3. C) 11.1 gal
  4. D) 16.7 gal
Show the answer and worked solution

Answer: B) 33.3 gal

1. C1 × V1 = C2 × V2 [Sheet p.1, Dilution/Blending: "C1 x V1 = C2 x V2"], so V1 = C2 × V2 ÷ C1. 2. V1 = 2.5 × 200 ÷ 15 = 33.33 gal. 3. Answer: 33.3 gal (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Dilution/Blending: "C1 x V1 = C2 x V2"

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