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Chemical feed and solutions: California T3 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 tank holds 2,400 gallons of a 12.5% solution with a specific gravity of 1.2. How many pounds of active chemical does it contain?

  1. A) 4,500 lbs
  2. B) 24,000 lbs
  3. C) 6,000 lbs
  4. D) 3,000 lbs
Show the answer and worked solution

Answer: D) 3,000 lbs

1. (lbs) Chemical = SG × 8.34 × Solution (Gal) × Solution (%/100) [Sheet p.1, Solutions: "(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)"]. 2. = 1.2 × 8.34 × 2,400 × 0.125 = 3,002.4 lbs. 3. Answer: 3,000 lbs (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Solutions: "(lbs) Chemical = SG x 8.34 x Solution (Gal) x Solution (%/100)"

Example 2. Well 1 pumps 250 GPM with 17 mg/L nitrate (as N). Well 2 pumps 1,900 GPM with 2 mg/L nitrate (as N). The two flows are blended. What is the nitrate concentration of the blend?

  1. A) 0.209 mg/L
  2. B) 2.81 mg/L
  3. C) 1.98 mg/L
  4. D) 3.74 mg/L
Show the answer and worked solution

Answer: D) 3.74 mg/L

1. (C1 × V1) + (C2 × V2) = C3 × V3 [Sheet p.1, Dilution/Blending: "(C1 x V1) + (C2 x V2) = C3 x V3"], with flows in place of volumes. 2. C1 × V1 = 17 × 250 = 4,250; C2 × V2 = 2 × 1,900 = 3,800. 3. V3 = 250 + 1,900 = 2,150 GPM; C3 = 8,050 ÷ 2,150 = 3.744 mg/L. 4. That is below the 10 mg/L nitrate (as N) MCL, which is the point of blending. 5. Answer: 3.74 mg/L (rounded to 3 significant figures).

Source: CA Exam Conversion Sheet (Revised 4/30/2026), p.1, Dilution/Blending: "(C1 x V1) + (C2 x V2) = C3 x V3"

Example 3. A plant treats 2.9 MGD and feeds sodium hydroxide solution at 12.5 mg/L. The product is 50% pure and weighs 12.7 lbs/gal. How many gallons per day of product are needed?

  1. A) 47.6 GPD
  2. B) 23.8 GPD
  3. C) 15.9 GPD
  4. D) 35.7 GPD
Show the answer and worked solution

Answer: A) 47.6 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: 2.9 × 12.5 × 8.34 = 302.325 lbs/day. 3. Denominator: 0.5 × 12.7 = 6.35 lbs active per gal. 4. GPD = 302.325 ÷ 6.35 = 47.61 GPD. 5. Answer: 47.6 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))"

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