Environmental chemistry carries 7–11 questions. It is the chemistry of water, soil and air rather than of a process plant: equilibria in natural water, how fast a compound breaks down and where it ends up.
Updated Oct 29, 2026 · Not affiliated with NCEES.
ExamFE Environmental
questions on the exam7–11
problems on this page15
areas in the specification15
Checked
What the area covers
Four parts, as the specification lists them:
Stoichiometry and reactions: equilibrium, acid-base, oxidation-reduction, precipitation, pC-pH diagrams.
Kinetics: chemical conversion, growth and decay.
Organic chemistry: nomenclature and functional-group reactions.
Multimedia equilibrium partitioning: Henry's law, the octanol-water partition coefficient.
15 equilibrium and fate problems
0 / 15 answered · 0 correct
Question 1 of 15 · Environmental Chemistry · 7–11 items
In water quality testing for trace contaminants in drinking water, which analytical method is most suitable for accurately measuring heavy metals at very low concentrations?
Solution
Trace metals at µg/L levels and below call for an elemental method with very low detection limits; ICP-MS is the standard choice.
GC-ECD is built for volatile halogenated organics such as pesticides and solvents, not for dissolved metals.
ICP-MS measures many metals at once down to parts per trillion, which makes it the standard choice for trace metals in drinking water.
Turbidity measures scattered light from suspended particles; it cannot identify or quantify a dissolved metal.
Multiple-tube fermentation is a microbiological test for coliform bacteria and says nothing about metals.
Question 2 of 15 · Environmental Chemistry · 7–11 items
Experimental data for the degradation of a contaminant are: t = 0 min, C = 200 mg/L; t = 20 min, C = 100 mg/L; t = 60 min, C = 50 mg/L. Which rate constant k (its units show the reaction order) fits these data?
Given
C = 200 mg/L at t = 0 min
C = 100 mg/L at t = 20 min
C = 50 mg/L at t = 60 min
Find
Rate constant k with units
Solution
Check which plot is linear: the half-life doubles as C halves, which is the second-order signature (Environmental chemistry: reaction kinetics).
1/C − 1/C0 = kt
k = (1/100 − 1/200) L/mg ÷ 20 min = 2.5 × 10−4 L/(mg·min)
The number is right, but mg/(L·min) are zero-order units; a second-order constant has units of L/(mg·min).
ln 2/20 min fits only the first half-life; a first-order reaction would keep a constant half-life, but here the second half-life doubles to 40 min.
100 mg/L lost in 20 min is a zero-order slope, but the next 50 mg/L take 40 min, so the rate is not constant.
1/C rises by 0.005 L/mg in the first 20 min and by 0.015 L/mg over 60 min, a constant slope, so the reaction is second order with k = 2.5 × 10−4 L/(mg·min).
This divides 1/C at 20 min by the time and forgets to subtract 1/C0; the integrated law is 1/C − 1/C0 = kt.
Question 3 of 15 · Environmental Chemistry · 7–11 items
A pesticide in a surface water reservoir undergoes simultaneous hydrolysis and photolysis. The first-order hydrolysis rate constant is 0.04 d−1 and the first-order photolysis rate constant is 0.01 hr−1. Calculate the percentage of the initial concentration remaining after 48 hours.
Given
Hydrolysis rate constant kh = 0.04 d−1
Photolysis rate constant kp = 0.01 hr−1
Time t = 48 h
Find
Percent of initial concentration remaining (%)
Solution
Parallel first-order losses add their rate constants once the units match (Environmental chemistry: reaction kinetics).
ktotal = 0.04 + 0.01 × 24 = 0.28 d−1, and e−0.28 × 2 = 0.571, so 57.1% remains after 48 h.
61.9% is e−0.24 × 2, photolysis alone; parallel first-order processes add their rate constants.
75.6% averages the two rate constants (0.14 d−1) instead of adding them; parallel losses add.
90.5% adds 0.04 and 0.01 directly, mixing d−1 with hr−1; the photolysis constant must first become 0.24 d−1.
92.3% is e−0.04 × 2, hydrolysis alone, leaving out photolysis.
Question 4 of 15 · Environmental Chemistry · 7–11 items
A nonpolar organic compound has a log organic carbon partition coefficient (log Koc) of 4.0 (Koc in L/kg). In a sediment-water system where the sediment contains 5% organic carbon (foc) and the aqueous concentration is 0.02 mg/L, calculate the equilibrium sorbed concentration in the sediment.
Given
log Koc = 4.0
Organic carbon fraction foc = 5%
Aqueous concentration Cw = 0.02 mg/L
Find
Equilibrium sorbed concentration (mg/kg)
Solution
Sorbed concentration from the organic-carbon partition coefficient (Environmental chemistry: sorption).
200 mg/kg is Koc × Cw, which skips the organic carbon fraction; Kd = Koc × foc.
500 is Kd in L/kg; it still has to be multiplied by the 0.02 mg/L aqueous concentration.
1,000 mg/kg uses foc = 5 instead of 0.05, so the percent was never converted to a fraction.
25,000 mg/kg divides Kd (500 L/kg) by the 0.02 mg/L aqueous concentration instead of multiplying.
Question 5 of 15 · Environmental Chemistry · 7–11 items
A chemical spill occurs in a shallow, turbulent river. The chemical has a dimensionless Henry's law constant of 0.5 and a log Koc of 2.5. Which mechanism is expected to be the primary removal pathway from the water?
Solution
A high Henry's constant (0.5 dimensionless) plus a turbulent, shallow channel favors gas transfer; log Koc 2.5 means only moderate sorption.
Log Koc = 2.5 means only moderate sorption, and a shallow river carries little organic-rich sediment per volume of water, so sorption removes a minor share.
Nothing in the data points to strong light absorption, and turbulence speeds volatilization, not photolysis.
No hydrolysis rate is given, and turbulence does not speed hydrolysis; the high Henry's constant points to the air.
Uptake by organisms is slow and small next to volatilization from a turbulent stream.
A dimensionless Henry's constant of 0.5 is high, and a shallow, turbulent river renews the air–water interface quickly, so loss to air dominates.
Question 6 of 15 · Environmental Chemistry · 7–11 items
A batch sorption experiment is conducted by adding 5.0 mg of a contaminant to a system containing 1.0 L of water and 0.05 kg of soil. After equilibrium, the aqueous concentration is measured as 2.0 mg/L. The soil-water partition coefficient (Kd) is determined to be 20 L/kg. Calculate the mass of contaminant unaccounted for in the mass balance.
Given
Contaminant added = 5.0 mg
Water volume = 1.0 L
Soil mass = 0.05 kg
Equilibrium aqueous concentration = 2.0 mg/L
Partition coefficient Kd = 20 L/kg
Find
Contaminant mass unaccounted for (mg)
Solution
Close the mass balance over water and soil (Environmental chemistry: sorption).
Mw = CwV = 2.0 mg/L × 1.0 L = 2.0 mg
Ms = KdCwms = 20 L/kg × 2.0 mg/L × 0.05 kg = 2.0 mg
Mmissing = 5.0 − 2.0 − 2.0 = 1.0 mg
Water holds 2.0 mg/L × 1.0 L = 2.0 mg, the soil 20 L/kg × 2.0 mg/L × 0.05 kg = 2.0 mg, so 5.0 − 4.0 = 1.0 mg is missing.
2.0 mg is the sorbed mass (or the dissolved mass) alone, not what remains after subtracting both from 5.0 mg.
2.5 mg computes the sorbed concentration as Kd/Cw = 10 mg/kg instead of Kd × Cw = 40 mg/kg, putting only 0.5 mg on the soil.
3.0 mg subtracts only the dissolved 2.0 mg and forgets the 2.0 mg sorbed on the soil.
4.0 mg is the mass accounted for (2.0 mg dissolved + 2.0 mg sorbed), not the mass missing from the 5.0 mg added.
Question 7 of 15 · Environmental Chemistry · 7–11 items
Calculate the cell potential at 25°C for the half-reaction MnO4− + 8H+ + 5e− → Mn2+ + 4H2O (E° = 1.51 V) if the pH is 3.0 and all other species are at standard unit concentrations.
Given
Half-reaction: MnO4− + 8H+ + 5e− → Mn2+ + 4H2O
E° = 1.51 V
pH = 3.0
Temperature = 25 °C
Find
Cell potential E (V)
Solution
Nernst equation at 25 °C with only H+ away from unit activity (Chemistry: electrochemistry).
E = E° − (0.05916/n)·log Q, Q = [Mn2+]/([MnO4−][H+]8) = 1/(10−3)8 = 1024
E = 1.51 − (0.05916/5) × 24 = 1.51 − 0.284 = 1.23 V
1.79 V adds the pH term instead of subtracting it; removing H+, a reactant, makes the reduction less favorable.
1.51 V is E°, valid only at [H+] = 1 M (pH 0); at pH 3 the potential drops.
1.47 V uses [H+] to the first power; the eight protons in the half-reaction put [H+]8 in Q.
1.42 V takes log Q as 8 (the H+ exponent) instead of 8 × pH = 24.
Q = 1/[H+]8 = 1024, so E = 1.51 − (0.05916/5) × 24 = 1.51 − 0.284 = 1.23 V.
Question 8 of 15 · Environmental Chemistry · 7–11 items
Equal volumes (50 mL each) of 2.0 × 10−3 M Pb(NO3)2 and 2.0 × 10−3 M Na2SO4 are mixed. Given Ksp for PbSO4 is 6.3 × 10−7, determine the reaction quotient Q immediately upon mixing and the resulting state.
Given
Volume of each solution = 50 mL
Pb(NO3)2 = 2.0 × 10−3 M
Na2SO4 = 2.0 × 10−3 M
Ksp of PbSO4 = 6.3 × 10−7
Find
Reaction quotient Q and whether precipitate forms
Solution
Compare the ion product after mixing with Ksp (Chemistry: solubility product).
[Pb2+] = [SO42−] = 2.0 × 10−3 M × 50/100 = 1.0 × 10−3 M
This halves the concentrations twice (to 5.0 × 10−4 M); one dilution by the equal volume is all that happens.
The conclusion holds, but Q uses the unmixed 2.0 × 10−3 M concentrations; mixing equal volumes halves each one.
Mixing equal volumes halves each concentration to 1.0 × 10−3 M, so Q = 1.0 × 10−6, which exceeds Ksp = 6.3 × 10−7.
Q is right, but Q > Ksp means the solution is supersaturated, so solid PbSO4 forms.
1.0 × 10−3 is a single ion concentration; Q is the product [Pb2+][SO42−].
Question 9 of 15 · Environmental Chemistry · 7–11 items
Calculate the equilibrium constant K at 25° C for a redox reaction involving the transfer of n=2 electrons with a standard cell potential E° = 1.10 V.
Given
Electrons transferred n = 2
Standard cell potential E° = 1.10 V
Temperature = 25 °C
Find
Equilibrium constant K
Solution
At 25 °C the standard potential fixes the equilibrium constant (Chemistry: electrochemistry).
log K = nE°/0.05916 = (2 × 1.10 V)/0.05916 V = 37.19
K = 1037.19 = 1.5 × 1037
2.4 × 1074 is K squared, as if four electrons were transferred instead of two.
6.5 × 10−38 is 1/K: a positive E° means a spontaneous reaction with K ≫ 1, not K ≪ 1.
log K = nE°/0.05916 = 2 × 1.10/0.05916 = 37.19, so K = 1037.19 = 1.5 × 1037.
1.4 × 1016 is e37.19: the base-10 exponent nE°/0.05916 was used with the natural exponential.
3.9 × 1018 uses n = 1; two electrons are transferred, which doubles log K.
Question 10 of 15 · Environmental Chemistry · 7–11 items
A non-aqueous phase liquid (NAPL) spill contains trichloroethylene (TCE) with a mole fraction of 0.02. The pure component vapor pressure of TCE is 0.08 atm and its Henry's law constant is 8.0 × 10−3 atm · m3 mol−1. Given the molecular weight of TCE is 131.4 g/mol, determine the equilibrium aqueous concentration of TCE.
Given
TCE mole fraction in NAPL x = 0.02
Pure vapor pressure Psat = 0.08 atm
Henry's constant H = 8.0 × 10−3 atm·m3/mol
Molecular weight = 131.4 g/mol
Find
Equilibrium aqueous TCE concentration (mg/L)
Solution
Raoult's law sets the partial pressure over the NAPL; Henry's law converts it to the dissolved concentration (Environmental chemistry).
0.200 is the molar concentration in mol/m3; it still has to be multiplied by 131.4 g/mol.
Raoult's law gives P = 0.02 × 0.08 atm = 0.0016 atm; Henry's law gives C = 0.0016/(8.0 × 10−3) = 0.200 mol/m3 = 26.3 g/m3 = 26.3 mg/L.
657 mg/L divides H by the partial pressure, which inverts Henry's law; Cw = P/H.
1,310 mg/L uses the pure vapor pressure (0.08 atm) and forgets the 0.02 mole fraction from Raoult's law.
Question 11 of 15 · Environmental Chemistry · 7–11 items
Calculate the log octanol-air partition coefficient (log Koa) for a compound with a Henry's law constant of 5.0 × 10−4 atm · m3 mol−1 at 25°C (298 K) and a log Kow of 5.5. Assume the universal gas constant R = 8.206 × 10−5 atm · m3/(mol · K).
Given
Henry's law constant H = 5.0 × 10−4 atm·m3/mol
Temperature T = 298 K
log Kow = 5.5
Gas constant R = 8.206 × 10−5 atm·m3/(mol·K)
Find
log Koa
Solution
Koa = Kow/Kaw with the dimensionless air–water coefficient Kaw = H/RT.
6.11 uses T = 25 instead of 298 K, which makes RT about twelve times too small.
3.81 subtracts log(RT/H) = 1.69; since Koa = Kow·RT/H, the term is added.
5.50 is log Kow alone; it ignores the air–water partitioning step.
8.80 adds log(1/H) = 3.30 without RT, so the Henry's constant was never made dimensionless.
Question 12 of 15 · Environmental Chemistry · 7–11 items
A wastewater stream contains 25 mg/L of dissolved calcium (Ca2+). Sodium fluoride (NaF) is added to precipitate calcium fluoride (CaF2). Given Ksp = 3.9 × 10−11 for CaF2 and assuming the calcium concentration remains effectively constant due to the common-ion effect, what is the theoretical maximum concentration of dissolved fluoride (F−) in mg L−1?
Given
Dissolved Ca2+ = 25 mg/L
Ksp of CaF2 = 3.9 × 10−11
Find
Maximum dissolved F− (mg/L)
Solution
Treat [Ca2+] as fixed and solve the solubility product for fluoride (Chemistry: solubility product).
2.38 mg/L comes from writing (2[F−])2 in Ksp; fluoride dosed as NaF is simply the free ion concentration, so no factor of 2 belongs there.
9.50 mg/L doubles the answer to count two fluorides per calcium, but the stoichiometric 2 is already carried by the square in Ksp.
0.0237 mg/L comes from plugging 25 mg/L into Ksp as if it were mol/L; Ca2+ must first be converted to 6.24 × 10−4 mol/L.
19.5 mg/L converts 2.50 × 10−4 mol/L with the molar mass of CaF2 (78.1 g/mol) instead of fluoride (19.0 g/mol).
[Ca2+] = 6.24 × 10−4 mol/L, so [F−] = √(Ksp/[Ca2+]) = 2.50 × 10−4 mol/L, which is 4.75 mg/L at 19.0 g/mol.
Question 13 of 15 · Environmental Chemistry · 7–11 items
A sealed system contains 2 L of water, 20 L of air, and 1 kg of soil with 300 mg of a contaminant. Calculate the equilibrium aqueous concentration given a dimensionless Henry's constant (H) of 0.4 (air/water), Koc = 500 L/kg, and a soil organic carbon fraction of 0.04.
Given
Water volume = 2 L
Air volume = 20 L
Soil mass = 1 kg
Total contaminant mass = 300 mg
Dimensionless Henry's constant H = 0.4
Koc = 500 L/kg
Organic carbon fraction foc = 0.04
Find
Equilibrium aqueous concentration (mg/L)
Solution
Equilibrium mass balance over three phases (Environmental chemistry: partitioning).
0.588 mg/L uses Koc = 500 L/kg in place of Kd = 20 L/kg, making the soil capacity 25 times too large.
4.17 mg/L divides Vair by H (20/0.4 = 50) instead of multiplying; the air term is H·Vair = 8 L.
300 mg = Cw(2 L + 0.4 × 20 L + 20 L/kg × 1 kg) = Cw × 30 L, so Cw = 10.0 mg/L.
13.6 mg/L is 300/(2 + 20), which leaves the air phase out of the mass balance.
30.0 mg/L is 300/(2 + 8), which leaves out the soil, the phase holding two thirds of the mass.
Question 14 of 15 · Environmental Chemistry · 7–11 items
Parameter
Value
Original groundwater volume taken
25.0 mL
Deionized water added for dilution
75.0 mL
Volume of diluted mixture titrated
50.0 mL
Volume of 0.020 M H2SO4 required
14.2 mL
An environmental engineer is testing a contaminated groundwater sample for alkalinity. The original groundwater sample is diluted prior to analysis to conserve standard reagents. A specified volume of this diluted sample is then titrated with a standard 0.020 M sulfuric acid (H2SO4) solution to an endpoint pH of 4.5. Using the procedural data in the table, determine the total alkalinity of the original undiluted groundwater sample, expressed as mg/L of calcium carbonate (CaCO3, MW = 100 g/mol).
Given
Original groundwater volume = 25.0 mL
Deionized water added = 75.0 mL
Diluted volume titrated = 50.0 mL
Titrant H2SO4 concentration = 0.020 M
Titrant volume = 14.2 mL
Molar mass of CaCO3 = 100 g/mol
Find
Total alkalinity of the original sample (mg/L as CaCO3)
Solution
Alkalinity as CaCO3 = acid equivalents per liter of sample × 50,000 mg/eq, then undo the dilution (Environmental chemistry: alkalinity).
N = 2 eq/mol × 0.020 mol/L = 0.040 eq/L
Alkdiluted = (14.2 mL × 0.040 eq/L × 50,000 mg/eq)/50.0 mL = 568 mg/L as CaCO3
284 mg/L makes two errors: it takes 0.020 M as the normality (H2SO4 gives 2 eq/mol) and skips the fourfold dilution.
568 mg/L is the alkalinity of the diluted mixture; it still has to be multiplied by the dilution factor (25.0 + 75.0)/25.0 = 4.
1,136 mg/L applies the dilution but takes the acid as 0.020 N instead of 0.040 N, which halves the result.
1,704 mg/L multiplies by 75.0/25.0 = 3, the water added over the sample; the dilution factor is total volume over sample volume, 100/25 = 4.
With N = 2 × 0.020 = 0.040 eq/L, the titrated mixture holds 568 mg/L as CaCO3, and the 4× dilution gives 2,272 mg/L in the original sample.
Question 15 of 15 · Environmental Chemistry · 7–11 items
A closed water sample has a total carbonate concentration CT = 2.0 × 10−3 M at pH 7.35. Using pK1 = 6.35 and pK2 = 10.33 for the carbonate system, what is the bicarbonate concentration [HCO3−]?
Given
Total carbonate CT = 2.0 × 10−3 M
pH = 7.35
pK1 = 6.35
pK2 = 10.33
Find
[HCO3−] (M)
Solution
Ionization fraction of bicarbonate (Environmental chemistry: carbonate system).
2.0 × 10−3 M assigns all carbonate to bicarbonate, but about 9% is still H2CO3*.
At pH 7.35, [HCO3−]/[H2CO3*] = 107.35 − 6.35 = 10 and CO32− is negligible, so α1 = 10/11 = 0.91 and [HCO3−] = 1.8 × 10−3 M.
1.0 × 10−3 M splits CT evenly, which is true only at pH = pK1 = 6.35.
1.8 × 10−4 M is H2CO3*, the share left one unit above pK1.
1.9 × 10−6 M is the carbonate ion, CO32−, which is tiny three units below pK2.
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Common traps
pH + pOH = 14 only at 25 °C. The ion product of water changes with temperature, so at other temperatures neutral is not pH 7.
Henry's constant comes in several forms. It may be dimensionless (gas over liquid concentration), in atm·m3/mol, or inverted. Check the units before you use it, and convert between the first two with
Hdimensionless = H / (RT)
Two alkalinities. Phenolphthalein alkalinity is titrated to about pH 8.3, total alkalinity to about pH 4.5; below pH 8.3 a sample has no hydroxide or carbonate alkalinity.
Koc from Kow is an empirical fit. Published log Koc versus log Kow correlations differ by compound class; use the one the problem gives rather than one from memory.
Reporting hardness and alkalinity as CaCO3 is explained on water and wastewater.