10 FE practice problems: environmental chemistry, with solutions
These ten original problems practice environmental chemistry, a topic from the FE Environmental exam specification, using the Environmental Engineering, Half-Life part of the FE Reference Handbook 10.6. Each problem gives the situation and the values with units. Work it with the handbook PDF open and commit to an answer before you open the solution. Every solution shows the handbook page, the equation, the substitution with units, a size check, and the mistake behind each wrong option, so a wrong pick tells you exactly what to fix.
How to use these problems
Give each problem an honest attempt before you open the solution: write the given values with units, find the equation in the FE Reference Handbook PDF, and commit to an answer. Then compare line by line. If you picked a wrong option, read the note for that option; each one names the mistake that produces it.
The problems
Problem 1 · FE Environmental, Environmental chemistry
A pesticide in a pond degrades by first-order decay with a half-life of 7 days. The initial concentration is 159 µg/L. The concentration after 26 days is most nearly:
Handbook: Environmental Engineering, Half-Life, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (12.1 µg/L)
- Given
- C0 = 159 µg/L, thalf = 7 days, t = 26 days
- Find
- concentration after t (µg/L)
- Handbook
- Environmental Engineering, Half-Life (first-order decay), page 335
- Equation
N = N0 e^(-0.693t/τ), with τ = half-life- Substitute
k = 0.693/t½ = 0.693/7 = 0.09900 per dayC = C0 e^(-kt) = 159 e^(-0.09900 × 26) = 12.1 µg/L
- Result
- 12.1 µg/L, 3 significant figures
- Check
- 26 days is 3.714 half-lives; (1/2)^3.714 × 159 gives nearly the same value, since 0.693 ≈ ln 2.
- Why the others are wrong
- A: is the amount that decayed, not what remains
- B: counted only whole half-lives
- D: inverted t/t½ in the exponent
Problem 2 · FE Environmental, Environmental chemistry
A groundwater contains 53.1 mg/L of calcium (Ca2+) and 9.2 mg/L of magnesium (Mg2+). The total hardness is most nearly:
Handbook: Environmental Engineering, Common Radicals in Water, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (170 mg/L as CaCO3)
- Given
- Ca = 53.1 mg/L, Mg = 9.2 mg/L
- Find
- total hardness (mg/L as CaCO3)
- Handbook
- Environmental Engineering, Common Radicals in Water (equivalent weights), page 349
- Equation
mg/L as CaCO3 = mg/L of ion × (50.0/EW of the ion)- Substitute
Ca hardness = 53.1 × 50.0/20.0 = 132.8 mg/L as CaCO3Mg hardness = 9.2 × 50.0/12.2 = 37.70 mg/L as CaCO3Total hardness = 170 mg/L as CaCO3
- Result
- 170 mg/L as CaCO3, 3 significant figures
- Check
- each ion is scaled by (equivalent weight of CaCO3)/(equivalent weight of the ion), from the handbook's radicals table.
- Why the others are wrong
- A: used the calcium factor for magnesium too
- B: added the ion concentrations without converting to CaCO3
- D: used the molecular weight of CaCO3 (100.1) with equivalent weights of the ions
Problem 3 · FE Environmental, Environmental chemistry
A groundwater contains 13.0 mg/L of calcium (Ca2+) and 3.3 mg/L of magnesium (Mg2+). The total hardness is most nearly:
Handbook: Environmental Engineering, Common Radicals in Water, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (46.0 mg/L as CaCO3)
- Given
- Ca = 13.0 mg/L, Mg = 3.3 mg/L
- Find
- total hardness (mg/L as CaCO3)
- Handbook
- Environmental Engineering, Common Radicals in Water (equivalent weights), page 349
- Equation
mg/L as CaCO3 = mg/L of ion × (50.0/EW of the ion)- Substitute
Ca hardness = 13.0 × 50.0/20.0 = 32.50 mg/L as CaCO3Mg hardness = 3.3 × 50.0/12.2 = 13.52 mg/L as CaCO3Total hardness = 46.0 mg/L as CaCO3
- Result
- 46.0 mg/L as CaCO3, 3 significant figures
- Check
- each ion is scaled by (equivalent weight of CaCO3)/(equivalent weight of the ion), from the handbook's radicals table.
- Why the others are wrong
- A: added the ion concentrations without converting to CaCO3
- B: left out the magnesium hardness
- D: used the molecular weight of CaCO3 (100.1) with equivalent weights of the ions
Problem 4 · FE Environmental, Environmental chemistry
A pesticide in a pond degrades by first-order decay with a half-life of 26 days. The initial concentration is 304 µg/L. The concentration after 37 days is most nearly:
Handbook: Environmental Engineering, Half-Life, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (113 µg/L)
- Given
- C0 = 304 µg/L, thalf = 26 days, t = 37 days
- Find
- concentration after t (µg/L)
- Handbook
- Environmental Engineering, Half-Life (first-order decay), page 335
- Equation
N = N0 e^(-0.693t/τ), with τ = half-life- Substitute
k = 0.693/t½ = 0.693/26 = 0.02665 per dayC = C0 e^(-kt) = 304 e^(-0.02665 × 37) = 113 µg/L
- Result
- 113 µg/L, 3 significant figures
- Check
- 37 days is 1.423 half-lives; (1/2)^1.423 × 304 gives nearly the same value, since 0.693 ≈ ln 2.
- Why the others are wrong
- A: used k = 1/t½ instead of 0.693/t½
- B: counted only whole half-lives
- D: assumed a straight-line loss of half per half-life
Problem 5 · FE Environmental, Environmental chemistry
A pesticide in a pond degrades by first-order decay with a half-life of 33 days. The initial concentration is 68 µg/L. The concentration after 45 days is most nearly:
Handbook: Environmental Engineering, Half-Life, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (26.4 µg/L)
- Given
- C0 = 68 µg/L, thalf = 33 days, t = 45 days
- Find
- concentration after t (µg/L)
- Handbook
- Environmental Engineering, Half-Life (first-order decay), page 335
- Equation
N = N0 e^(-0.693t/τ), with τ = half-life- Substitute
k = 0.693/t½ = 0.693/33 = 0.02100 per dayC = C0 e^(-kt) = 68 e^(-0.02100 × 45) = 26.4 µg/L
- Result
- 26.4 µg/L, 3 significant figures
- Check
- 45 days is 1.364 half-lives; (1/2)^1.364 × 68 gives nearly the same value, since 0.693 ≈ ln 2.
- Why the others are wrong
- A: counted only whole half-lives
- C: inverted t/t½ in the exponent
- D: assumed a straight-line loss of half per half-life
Problem 6 · FE Environmental, Environmental chemistry
A groundwater contains 72.1 mg/L of calcium (Ca2+) and 24.1 mg/L of magnesium (Mg2+). The total hardness is most nearly:
Handbook: Environmental Engineering, Common Radicals in Water, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (279 mg/L as CaCO3)
- Given
- Ca = 72.1 mg/L, Mg = 24.1 mg/L
- Find
- total hardness (mg/L as CaCO3)
- Handbook
- Environmental Engineering, Common Radicals in Water (equivalent weights), page 349
- Equation
mg/L as CaCO3 = mg/L of ion × (50.0/EW of the ion)- Substitute
Ca hardness = 72.1 × 50.0/20.0 = 180.3 mg/L as CaCO3Mg hardness = 24.1 × 50.0/12.2 = 98.77 mg/L as CaCO3Total hardness = 279 mg/L as CaCO3
- Result
- 279 mg/L as CaCO3, 3 significant figures
- Check
- each ion is scaled by (equivalent weight of CaCO3)/(equivalent weight of the ion), from the handbook's radicals table.
- Why the others are wrong
- A: used the molecular weight of CaCO3 (100.1) with equivalent weights of the ions
- C: inverted the conversion factors
- D: left out the magnesium hardness
Problem 7 · FE Environmental, Environmental chemistry
At 25°C a solution has a hydroxide ion concentration [OH-] of 7.6 × 10⁻⁵ mol/L. The pH is most nearly:
Handbook: Chemistry and Biology, Acids, Bases, and pH, FE Reference Handbook 10.6
Show the worked solution
- Answer
- B (9.88)
- Given
- OH = 7.6 × 10⁻⁵ mol/L, T = 25°C
- Find
- pH
- Handbook
- Chemistry and Biology, Acids, Bases, and pH, page 87
- Equation
pH = log10(1/[H+]); [H+][OH-] = 10^-14 at 25°C, so pH = 14 - pOH- Substitute
pOH = -log10[OH-] = -log10(7.6 × 10⁻⁵) = 4.119pH = 14 - pOH = 14 - 4.119 = 9.88
- Result
- 9.88, 3 significant figures
- Check
- [H+] = 10^-14/[OH-] = 1.32 × 10⁻¹⁰ mol/L gives the same pH; a basic solution has pH > 7.
- Why the others are wrong
- A: is the pOH, not the pH
- C: used the natural log instead of log10
- D: added 7 instead of using pH = 14 - pOH
Problem 8 · FE Environmental, Environmental chemistry
A groundwater contains 81.3 mg/L of calcium (Ca2+) and 49.8 mg/L of magnesium (Mg2+). The total hardness is most nearly:
Handbook: Environmental Engineering, Common Radicals in Water, FE Reference Handbook 10.6
Show the worked solution
- Answer
- A (407 mg/L as CaCO3)
- Given
- Ca = 81.3 mg/L, Mg = 49.8 mg/L
- Find
- total hardness (mg/L as CaCO3)
- Handbook
- Environmental Engineering, Common Radicals in Water (equivalent weights), page 349
- Equation
mg/L as CaCO3 = mg/L of ion × (50.0/EW of the ion)- Substitute
Ca hardness = 81.3 × 50.0/20.0 = 203.3 mg/L as CaCO3Mg hardness = 49.8 × 50.0/12.2 = 204.1 mg/L as CaCO3Total hardness = 407 mg/L as CaCO3
- Result
- 407 mg/L as CaCO3, 3 significant figures
- Check
- each ion is scaled by (equivalent weight of CaCO3)/(equivalent weight of the ion), from the handbook's radicals table.
- Why the others are wrong
- B: left out the magnesium hardness
- C: inverted the conversion factors
- D: used the calcium factor for magnesium too
Problem 9 · FE Environmental, Environmental chemistry
A groundwater contains 34.7 mg/L of calcium (Ca2+) and 39.4 mg/L of magnesium (Mg2+). The total hardness is most nearly:
Handbook: Environmental Engineering, Common Radicals in Water, FE Reference Handbook 10.6
Show the worked solution
- Answer
- C (248 mg/L as CaCO3)
- Given
- Ca = 34.7 mg/L, Mg = 39.4 mg/L
- Find
- total hardness (mg/L as CaCO3)
- Handbook
- Environmental Engineering, Common Radicals in Water (equivalent weights), page 349
- Equation
mg/L as CaCO3 = mg/L of ion × (50.0/EW of the ion)- Substitute
Ca hardness = 34.7 × 50.0/20.0 = 86.75 mg/L as CaCO3Mg hardness = 39.4 × 50.0/12.2 = 161.5 mg/L as CaCO3Total hardness = 248 mg/L as CaCO3
- Result
- 248 mg/L as CaCO3, 3 significant figures
- Check
- each ion is scaled by (equivalent weight of CaCO3)/(equivalent weight of the ion), from the handbook's radicals table.
- Why the others are wrong
- A: used the calcium factor for magnesium too
- B: used the molecular weight of CaCO3 (100.1) with equivalent weights of the ions
- D: inverted the conversion factors
Problem 10 · FE Environmental, Environmental chemistry
A groundwater contains 25.8 mg/L of calcium (Ca2+) and 20.7 mg/L of magnesium (Mg2+). The total hardness is most nearly:
Handbook: Environmental Engineering, Common Radicals in Water, FE Reference Handbook 10.6
Show the worked solution
- Answer
- D (149 mg/L as CaCO3)
- Given
- Ca = 25.8 mg/L, Mg = 20.7 mg/L
- Find
- total hardness (mg/L as CaCO3)
- Handbook
- Environmental Engineering, Common Radicals in Water (equivalent weights), page 349
- Equation
mg/L as CaCO3 = mg/L of ion × (50.0/EW of the ion)- Substitute
Ca hardness = 25.8 × 50.0/20.0 = 64.50 mg/L as CaCO3Mg hardness = 20.7 × 50.0/12.2 = 84.84 mg/L as CaCO3Total hardness = 149 mg/L as CaCO3
- Result
- 149 mg/L as CaCO3, 3 significant figures
- Check
- each ion is scaled by (equivalent weight of CaCO3)/(equivalent weight of the ion), from the handbook's radicals table.
- Why the others are wrong
- A: used the calcium factor for magnesium too
- B: left out the magnesium hardness
- C: inverted the conversion factors
A new problem posts every day inside the lab, with the full worked solution the same evening.
Frequently asked questions
Where is environmental chemistry in the FE Reference Handbook?
Look in the Environmental Engineering, Half-Life part of FE Reference Handbook 10.6. Each solution gives the exact page, so you can practice finding it in the PDF the way you will on exam day.
Are these real FE exam questions?
No. They are original problems generated from the handbook formulas and checked by code. Real exam questions are confidential, and sharing them breaks the NCEES agreement every examinee accepts.
How do I check my answer before opening the solution?
Check the units of your result and whether its size makes sense for the situation. Each solution ends with the same kind of check, so you can compare your habit with ours.
Sources
- NCEES FE Environmental CBT exam specifications (PDF). Retrieved October 3, 2026.
- NCEES FE Reference Handbook 10.6 (free PDF in MyNCEES). Retrieved October 3, 2026.