Environmental Chemistry: Fundamentals, Water Chemistry, Soil Chemistry and Atmospheric Chemistry

Section 2 of the GATE Environmental Science and Engineering (ES) paper, in the order its four sub-headings give it. The fundamentals come first — bonding, chemical equations, concentration and activity, the structure of organic molecules, equilibria, and the thermodynamics and kinetics of reactions — because every later topic is one of them applied. Water chemistry follows: how quality parameters are measured, acid–base equilibria and buffers, the carbonate system that sets the pH and alkalinity of natural water, the solubility of gases, complexation, precipitation and redox, and the contaminants that result. Soil chemistry covers organic matter, nitrogen, phosphorus and potassium, cation exchange capacity, base saturation and the sodium adsorption ratio. Atmospheric chemistry closes with the composition of the atmosphere, the reactivity of its trace gases and the chemistry of ozone formation.

1. Fundamentals: bonding, concentration and activity, organic structure, equilibrium, thermodynamics and kinetics

Ionic bonds form by electron transfer between atoms of very different electronegativity (Na⁺Cl⁻) and give salts that dissociate in water; covalent bonds share electron pairs, and when the electronegativities differ the bond is polar. Water is a bent, polar molecule (H–O–H angle about 104.5°) that hydrogen-bonds to its neighbours, which is why it has an unusually high boiling point, specific heat and dielectric constant and dissolves ionic solids so readily. Non-polar organics such as hydrocarbons dissolve poorly in it and partition instead into fats, soil organic matter and air — the basis of the octanol–water partition coefficient K_ow used to predict where an organic pollutant ends up.

Concentrations are expressed as molarity (mol/L), normality (equivalents/L, where the equivalent weight is the molar mass divided by the charge or the electrons transferred), mg/L (equal to ppm by mass in dilute water), and, for hardness and alkalinity, mg/L as CaCO₃: multiply the concentration of a species by 50/(its equivalent weight), since the equivalent weight of CaCO₃ is 100/2 = 50. In solutions that are not very dilute, ions behave as though less concentrated than they are; the effective concentration is the activity {i} = γᵢ[i]. The activity coefficient depends on the ionic strength, I = ½Σcᵢzᵢ², through the Debye–Hückel limiting law log γᵢ = −0.51zᵢ²√I (water, 25 °C, valid for I below about 0.005 M, with extended forms beyond). Divalent ions are affected four times as strongly as monovalent ones, because of the z².

Organic molecules are classified by their carbon skeleton — aliphatic (alkanes, alkenes, alkynes) or aromatic (benzene rings) — and by their functional groups: alcohols (–OH), aldehydes (–CHO), ketones (C=O), carboxylic acids (–COOH), amines (–NH₂), phenols (–OH on a ring) and halides (–Cl, –Br). The functional group sets polarity and reactivity: carboxylic acids ionise and dissolve, chlorinated aromatics resist biodegradation and bioaccumulate, and polycyclic aromatic hydrocarbons (fused rings from incomplete combustion) include known carcinogens. Isomers share a formula but not a structure, and can differ sharply in toxicity.

Equilibrium, thermodynamics and kinetics — the working relations
RelationStatementWhat it tells you
Equilibrium constantaA + bB ⇌ cC + dD: K = {C}ᶜ{D}ᵈ/({A}ᵃ{B}ᵇ)compare the reaction quotient Q with K: Q < K proceeds forward, Q > K backward (Le Chatelier)
Gibbs energyΔG = ΔH − TΔS; ΔG = ΔG° + RT ln Q; ΔG° = −RT ln KΔG < 0 means spontaneous as written; a large K means a strongly negative ΔG°
van ’t Hoffln(K₂/K₁) = −(ΔH°/R)(1/T₂ − 1/T₁)how an equilibrium shifts with temperature
Rate lawszero order C = C₀ − kt; first order C = C₀e^(−kt), t½ = 0.693/k; second order 1/C = 1/C₀ + ktidentify the order from which plot is straight: C, ln C or 1/C against t
Arrheniusk = A e^(−Ea/RT); over a narrow range k_T = k₂₀θ^(T − 20)warmer water reacts faster; θ is the engineer’s compact form of Ea
⚠️ Thermodynamics says whether, kinetics says how fast
Oxidation of organic matter by dissolved oxygen has a strongly negative ΔG°, yet a bottle of river water does not consume its oxygen instantly: the reaction is thermodynamically favoured and kinetically slow until microbes catalyse it. A negative ΔG never implies a rate.

2. Water quality parameters, acid–base equilibria, buffers and the carbonate system

Water quality parameters and how they are measured
ParameterMethodReported as
pHglass electrode against a reference, calibrated with two bufferspH units
Turbiditynephelometer — light scattered at 90°NTU
Conductivity and TDSconductivity cell; TDS ≈ (0.55 to 0.7) × EC, or by evaporating a filtered sample at 180 °CμS/cm; mg/L
Alkalinitytitration with standard acid to pH 8.3 (phenolphthalein, P) and about 4.5 (total, T)mg/L as CaCO₃
HardnessEDTA titration with Eriochrome Black T at pH 10mg/L as CaCO₃
Dissolved oxygenWinkler (iodometric) titration, or membrane and optical probesmg/L
BOD, COD, TOC5-day incubation at 20 °C; dichromate reflux; combustion to CO₂mg/L
Chloride, nitrogen, phosphateMohr (argentometric) titration; Kjeldahl digestion for organic plus ammonia N; ascorbic-acid colorimetry for phosphatemg/L as Cl, N or P

Water self-ionises, H₂O ⇌ H⁺ + OH⁻, with K_w = [H⁺][OH⁻] = 10⁻¹⁴ at 25 °C, so pH + pOH = 14. A weak acid HA ⇌ H⁺ + A⁻ has K_a = [H⁺][A⁻]/[HA]; taking logarithms gives the Henderson–Hasselbalch equation, pH = pK_a + log([A⁻]/[HA]). A buffer is a weak acid with its conjugate base (or a weak base with its conjugate acid) and resists pH change because added H⁺ is taken up by A⁻ and added OH⁻ by HA. Its buffer capacity is greatest when [A⁻] = [HA], that is at pH = pK_a, and a buffer is useful within about one unit of its pK_a. At pH = pK_a the acid is exactly half dissociated; one unit above, the ratio [A⁻]/[HA] is 10.

Natural water is buffered by the carbonate system. Dissolved CO₂ forms carbonic acid, written together as H₂CO₃, which dissociates in two steps: H₂CO₃ ⇌ H⁺ + HCO₃⁻ (pK₁ ≈ 6.35 at 25 °C) and HCO₃⁻ ⇌ H⁺ + CO₃²⁻ (pK₂ ≈ 10.33). So below pH 6.35 H₂CO₃* dominates, between 6.35 and 10.33 bicarbonate dominates — which is why most natural waters, at pH 6.5 to 8.5, carry their alkalinity as HCO₃⁻ — and above 10.33 carbonate dominates. Alkalinity, the acid-neutralising capacity, is [HCO₃⁻] + 2[CO₃²⁻] + [OH⁻] − [H⁺] in equivalents. From the two titration end points the three forms are apportioned by the standard rules below, which assume hydroxide and bicarbonate cannot coexist in significant amounts.

Apportioning alkalinity from phenolphthalein (P) and total (T) alkalinity
Result of titrationOH⁻CO₃²⁻HCO₃⁻
P = 000T
P < T/202PT − 2P
P = T/202P = T0
P > T/22P − T2(T − P)0
P = TT00
🧠 Hardness and alkalinity split the hardness
Total hardness is Ca²⁺ + Mg²⁺ expressed as CaCO₃ (60 mg/L of Ca²⁺ is 60 × 50/20 = 150 mg/L as CaCO₃). Carbonate (temporary) hardness is the smaller of total hardness and alkalinity; the rest of the hardness is non-carbonate (permanent). Temporary hardness is removed by boiling or lime; permanent needs soda ash or ion exchange.

3. Gases in water, complexation, precipitation, redox, and contaminants

The solubility of a gas follows Henry’s law, C = K_H × p_gas, with K_H in mol/(L·atm): the dissolved concentration is proportional to the partial pressure of the gas above the water. For oxygen, K_H is about 1.3 × 10⁻³ mol/(L·atm) at 25 °C, so air (p_O₂ = 0.21 atm) saturates water at about 8.7 mg/L; the saturation value is higher in cold water and lower in warm and saline water — around 9.1 mg/L at 20 °C in fresh water — which is why warm, polluted rivers suffer most. Gas solubility falling with temperature is the reason thermal discharges deplete oxygen even without adding any organic load.

Complexation binds a metal ion to ligands — OH⁻, Cl⁻, CO₃²⁻, natural organic matter, or synthetic chelators such as EDTA and NTA — and changes its solubility, mobility and toxicity: free Cu²⁺ is far more toxic to fish than complexed copper, and chelators in detergents can remobilise metals from sediments. Precipitation is governed by the solubility product: for MₐXᵦ, K_sp = [M]ᵃ[X]ᵇ, and a solid forms when the ion product exceeds K_sp. For a 1:2 salt such as CaF₂, K_sp = s(2s)² = 4s³. The common-ion effect lowers solubility, which is exploited in lime softening (CaCO₃ and Mg(OH)₂ precipitation), in chemical phosphorus removal (with Al³⁺, Fe³⁺ or Ca²⁺), and in removing heavy metals as hydroxides — each metal having a pH of minimum solubility, beyond which amphoteric hydroxides redissolve.

Redox reactions transfer electrons: the oxidant gains them and is reduced, the reductant loses them and is oxidised. The tendency is measured by the electrode potential, and the Nernst equation E = E° − (0.0592/n) log Q at 25 °C corrects the standard value for concentrations. In water and sediments microbes use electron acceptors in a fixed order of decreasing energy yield — the redox ladder: O₂, then NO₃⁻ (denitrification), then Mn(IV), then Fe(III), then SO₄²⁻ (giving H₂S), and finally CO₂ (methanogenesis). The sequence explains why an anoxic hypolimnion releases dissolved iron and manganese, why sulphide odours appear only after nitrate is exhausted, and why a landfill produces methane last.

Inorganic and organic contaminants
ContaminantTypical sourceWhy it matters
Arsenicgeogenic groundwater (reducing aquifers release As adsorbed on iron oxides)skin lesions and cancers after chronic intake; As(III) more mobile and toxic than As(V)
Fluorideweathering of fluoride-bearing rocksbeneficial in small amounts, dental and skeletal fluorosis in excess
Nitratefertilisers, septic systems, manuremethaemoglobinaemia (blue-baby syndrome) in infants
Lead, mercury, cadmium, chromium(VI)batteries, chlor-alkali and mining wastes, electroplating, tanneriesneurotoxicity (Pb, Hg — methylmercury biomagnifies), kidney damage (Cd), carcinogenic Cr(VI)
Pesticides, PCBs, PAHs, VOCsagriculture, transformers, combustion, solvents and fuelspersistent, lipophilic (high K_ow), bioaccumulative; several are carcinogens
Emerging contaminantspharmaceuticals, personal-care products, microplastics, PFASincompletely removed by conventional treatment; endocrine disruption

4. Soil chemistry: organic matter, N, P, K, cation exchange capacity, base saturation and SAR

Soil organic matter is plant and microbial residue at every stage of decay; its stable, dark, colloidal end-product is humus, made of humic and fulvic acids and humin. Though usually only a few per cent of soil mass, it holds water, binds aggregates, supplies nutrients as it mineralises, and carries a large, pH-dependent negative charge. Its C:N ratio decides whether decomposition releases nitrogen or locks it up: residues wider than about 25–30:1 (straw, sawdust) cause microbes to immobilise soil nitrogen, while narrow residues (legumes, manure) release it.

Nitrogen is mostly organic in soil; mineralisation releases NH₄⁺, which nitrifying bacteria oxidise to NO₃⁻. Ammonium is held on exchange sites, but nitrate is an anion and leaches readily to groundwater, or is lost as N₂ and N₂O by denitrification in waterlogged soil, or as NH₃ by volatilisation from alkaline soil. Phosphorus is the opposite: it moves very little, because phosphate is fixed — by Fe and Al oxides in acid soils and by calcium as calcium phosphates in alkaline soils — so plant-available P is greatest near pH 6–7, and P reaches water mainly attached to eroded soil. Potassium exists in mineral lattices (unavailable), as fixed K between clay layers (slowly available), as exchangeable K⁺, and in solution.

Clays and humus carry negative charge that holds exchangeable cations. The cation exchange capacity (CEC), in cmol(+)/kg (numerically equal to meq/100 g), is the total of those sites: low for kaolinite, far higher for the 2:1 expanding clay montmorillonite, and highest per unit mass for humus. Base saturation is the share of the CEC occupied by the base cations: BS = (Ca²⁺ + Mg²⁺ + K⁺ + Na⁺)/CEC × 100; the remainder is held by H⁺ and Al³⁺, so low base saturation marks an acid soil. For irrigation water and sodic soils the sodium adsorption ratio, SAR = Na⁺/√[(Ca²⁺ + Mg²⁺)/2] with all concentrations in meq/L, predicts how far sodium will displace calcium and magnesium on the exchange sites; high exchangeable sodium disperses clay, seals the surface and destroys permeability. The soil-side measure is the exchangeable sodium percentage, ESP = exchangeable Na/CEC × 100.

⚠️ SAR needs meq/L, not mg/L
Convert first: Na⁺ 230 mg/L ÷ 23 = 10 meq/L; Ca²⁺ 40 mg/L ÷ 20 = 2 meq/L; Mg²⁺ 24 mg/L ÷ 12 = 2 meq/L. SAR = 10/√[(2 + 2)/2] = 10/√2 = 7.07. Putting mg/L straight into the formula gives 230/√32 = 40.7, a sixfold error.

5. Atmospheric chemistry: composition, trace-gas reactivity and ozone formation

Dry air is about 78.08% N₂, 20.95% O₂ and 0.93% Ar by volume; everything else is a trace gas — CO₂ at roughly 420 ppm in the 2020s, CH₄ near 2 ppm, N₂O about a third of a ppm, ozone, CO, SO₂, NOₓ and volatile organics at ppb levels or below — plus variable water vapour. Trace gases matter out of all proportion to their abundance because they absorb infrared radiation, react, and form particles. How long one persists is its residence time, τ = atmospheric burden/removal rate: short-lived species (hours to days, such as SO₂ and NOₓ) are regional problems concentrated near their sources, while long-lived ones (years to centuries, such as CO₂, N₂O and CFCs) mix globally.

The troposphere cleans itself chiefly through the hydroxyl radical, OH, formed when ozone is photolysed by ultraviolet light to an excited oxygen atom that reacts with water vapour. OH oxidises CO to CO₂, CH₄ ultimately to CO₂, SO₂ towards sulphuric acid and NO₂ to nitric acid, so it sets the lifetime of most reactive gases — and gases that OH cannot attack, such as CFCs and N₂O, survive long enough to reach the stratosphere.

  • Photolysis: NO₂ + hν (λ < about 420 nm) → NO + O.
  • Ozone formation: O + O₂ + M → O₃ + M, where M is any third molecule that carries away the energy.
  • Titration: NO + O₃ → NO₂ + O₂. These three reactions alone reach a photostationary state, [O₃] = J[NO₂]/(k[NO]), with no net ozone gain.
  • The VOC short-circuit: hydrocarbons oxidised by OH form peroxy radicals (RO₂, HO₂) that convert NO to NO₂ without consuming ozone — RO₂ + NO → RO + NO₂ — so the NO₂/NO ratio rises and ozone accumulates. Peroxyacetyl nitrate (PAN) is a by-product and an eye irritant.
🎯 Why ozone is highest downwind, and why cutting NOₓ can raise it
Fresh NO from traffic destroys ozone locally (the titration step), so city centres at rush hour have less ozone than suburbs downwind a few hours later, where the VOC chemistry has had time to work. Where VOCs are scarce relative to NOₓ (VOC-limited, typical of city cores) cutting NOₓ removes the titrant and can increase ozone; where NOₓ is scarce (NOₓ-limited, typical of rural areas) cutting NOₓ lowers it. In the stratosphere ozone is made by photolysis of O₂ itself (the Chapman cycle) and destroyed catalytically by chlorine: Cl + O₃ → ClO + O₂, then ClO + O → Cl + O₂, freeing the chlorine to repeat.

Key takeaways

  • Activity is γ × concentration; ionic strength I = ½Σcz², and log γ = −0.51z²√I shows divalent ions are affected four times as much as monovalent.
  • ΔG° = −RT ln K decides direction, not rate; identify reaction order by which of C, ln C or 1/C is linear in time, and correct rates with θ^(T − 20).
  • pH = pK_a + log([A⁻]/[HA]); buffers work best at pH = pK_a. In the carbonate system (pK₁ 6.35, pK₂ 10.33) bicarbonate dominates natural waters; split alkalinity with the P and T rules.
  • Henry’s law C = K_H p; solubility products decide precipitation (4s³ for a 1:2 salt); microbes climb down the redox ladder O₂, NO₃⁻, Mn(IV), Fe(III), SO₄²⁻, CO₂.
  • Base saturation = base cations/CEC; SAR = Na/√[(Ca + Mg)/2] in meq/L. Tropospheric ozone is secondary: NO₂ photolysis makes it, NO titrates it, and VOC peroxy radicals tip the balance.

Practice questions (17)

Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.

  1. What is the ionic strength of a 0.01 M solution of CaCl₂, assuming complete dissociation, in mol/L?

    Numerical answer — type the value.

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    Answer: 0.03

    CaCl₂ gives 0.01 M Ca²⁺ and 0.02 M Cl⁻. I = ½(0.01 × 2² + 0.02 × 1²) = ½(0.04 + 0.02) = 0.03 M. Forgetting to square the charge of Ca²⁺ gives 0.02; forgetting that there are two chlorides gives 0.025.
  2. Using the Debye–Hückel limiting law, log γ = −0.51z²√I, what is the activity coefficient of a divalent ion at an ionic strength of 0.01 M? Give the answer to two decimal places.

    Numerical answer — type the value.

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    Answer: 0.63

    log γ = −0.51 × 2² × √0.01 = −0.51 × 4 × 0.1 = −0.204, so γ = 10^(−0.204) = 0.625, i.e. 0.63. A monovalent ion at the same I has log γ = −0.051 and γ = 0.89: the z² makes the divalent ion far less "active".
  3. A reaction has an equilibrium constant of 1000 at 298 K. What is its standard Gibbs energy change, in kJ/mol, to one decimal place? (R = 8.314 J/(mol·K))

    Numerical answer — type the value.

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    Answer: -17.1

    ΔG° = −RT ln K = −8.314 × 298 × ln 1000 = −8.314 × 298 × 6.908 = −17115 J/mol, which is -17.1 kJ/mol. The negative sign follows from K > 1; using log₁₀ instead of ln gives −7.4, a factor of 2.303 too small.
  4. A pesticide degrades in soil by first-order kinetics with k = 0.1 per day. What is its half-life, in days, to two decimal places?

    Numerical answer — type the value.

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    Answer: 6.93

    t½ = ln 2/k = 0.693/0.1 = 6.93 days, independent of the starting concentration, which is the signature of first order. Using 1/k = 10 days gives the mean lifetime, not the half-life.
  5. A buffer contains 0.1 M acetic acid and 0.2 M sodium acetate. If the pKa of acetic acid is 4.76, what is the pH? Give the answer to two decimal places.

    Numerical answer — type the value.

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    Answer: 5.06

    pH = pKa + log([A⁻]/[HA]) = 4.76 + log(0.2/0.1) = 4.76 + 0.301 = 5.06. Inverting the ratio gives 4.46, which would describe a buffer richer in acid than in base.
  6. A water contains 60 mg/L of calcium ion and no magnesium. What is its hardness in mg/L as CaCO₃?

    Numerical answer — type the value.

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    Answer: 150

    Equivalent weight of Ca²⁺ = 40/2 = 20 and of CaCO₃ = 50, so hardness = 60 × 50/20 = 150 mg/L as CaCO₃. Using the molar masses, 60 × 100/40, gives the same 150 because both are divalent; mixing them, 60 × 100/20, gives 300.
  7. Titration of a water sample gives a phenolphthalein alkalinity of 20 mg/L and a total alkalinity of 100 mg/L, both as CaCO₃. What is the bicarbonate alkalinity, in mg/L as CaCO₃?

    Numerical answer — type the value.

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    Answer: 60

    P = 20 is less than T/2 = 50, so hydroxide is absent, carbonate = 2P = 40 and bicarbonate = T − 2P = 100 − 40 = 60 mg/L as CaCO₃. Taking bicarbonate as T − P = 80 forgets that the phenolphthalein end point converts carbonate only halfway, to bicarbonate.
  8. Taking the first dissociation constant of carbonic acid as pK₁ = 6.35, what is the ratio [HCO₃⁻]/[H₂CO₃*] in a water at pH 7.35?

    Numerical answer — type the value.

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    Answer: 10

    From K₁ = [H⁺][HCO₃⁻]/[H₂CO₃], the ratio is 10^(pH − pK₁) = 10^(7.35 − 6.35) = 10¹ = 10: one pH unit above pK₁, bicarbonate outweighs carbonic acid tenfold. Only about 9% of the inorganic carbon is then H₂CO₃.
  9. Henry’s constant for oxygen is 1.3 × 10⁻³ mol/(L·atm). What is the saturation concentration of dissolved oxygen in water exposed to air with an oxygen partial pressure of 0.21 atm, in mg/L, to two decimal places?

    Numerical answer — type the value.

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    Answer: 8.74

    C = K_H × p = 1.3 × 10⁻³ × 0.21 = 2.73 × 10⁻⁴ mol/L; × 32 000 mg/mol = 8.74 mg/L. Using 1 atm instead of the partial pressure gives 41.6 mg/L, the saturation against pure oxygen.
  10. The solubility product of CaF₂ is 3.9 × 10⁻¹¹. What is the fluoride concentration of water in equilibrium with solid CaF₂ in pure water, in mg/L, to two decimal places? (F = 19)

    Numerical answer — type the value.

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    Answer: 8.12

    CaF₂ ⇌ Ca²⁺ + 2F⁻: K_sp = s(2s)² = 4s³, so s = (3.9 × 10⁻¹¹/4)^(1/3) = 2.136 × 10⁻⁴ M. [F⁻] = 2s = 4.27 × 10⁻⁴ M × 19 000 mg/mol = 8.12 mg/L. Taking K_sp = s² (as for a 1:1 salt) gives s = 6.2 × 10⁻⁶ M and a fluoride far too low.
  11. In the sediment of a eutrophic lake, after dissolved oxygen and nitrate are exhausted, which electron acceptors are used next, in order?

    1. sulphate, then iron(III), then manganese(IV)
    2. manganese(IV), then iron(III), then sulphate, then CO₂
    3. CO₂, then sulphate, then iron(III)
    4. iron(III), then CO₂, then manganese(IV)
    Show answer

    Answer: B — manganese(IV), then iron(III), then sulphate, then CO₂

    Microbes use acceptors in order of decreasing energy yield: O₂, NO₃⁻, Mn(IV), Fe(III), SO₄²⁻ and finally CO₂ (methanogenesis). That is why an anoxic lake bottom releases Mn and Fe before it smells of H₂S, and methane appears last. Option A reverses the middle of the ladder.
  12. Which of the following statements about water-quality measurement are correct?

    1. For most wastewaters COD exceeds BOD₅
    2. The Winkler method determines dissolved oxygen by iodometric titration
    3. Hardness is determined by EDTA titration with Eriochrome Black T
    4. Turbidity is reported in mg/L as CaCO₃
    Show answer

    Answer: A — For most wastewaters COD exceeds BOD₅; B — The Winkler method determines dissolved oxygen by iodometric titration; C — Hardness is determined by EDTA titration with Eriochrome Black T

    (a) COD oxidises nearly all organic matter chemically, including what microbes cannot degrade in five days, so COD > BOD₅. (b) and (c) are the standard methods. (d) is false: turbidity is an optical property reported in NTU from a nephelometer; mg/L as CaCO₃ is the unit of hardness and alkalinity.
  13. An irrigation water contains 230 mg/L of Na⁺, 40 mg/L of Ca²⁺ and 24 mg/L of Mg²⁺. What is its sodium adsorption ratio, to two decimal places? (Na = 23, Ca = 40, Mg = 24)

    Numerical answer — type the value.

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    Answer: 7.07

    Convert to meq/L: Na⁺ = 230/23 = 10, Ca²⁺ = 40/20 = 2, Mg²⁺ = 24/12 = 2. SAR = 10/√[(2 + 2)/2] = 10/√2 = 7.07. Using mmol/L for Ca and Mg (1 and 1) gives 10/1 = 10, and using mg/L gives 40.7.
  14. A soil has a cation exchange capacity of 20 cmol(+)/kg and exchangeable Ca²⁺, Mg²⁺, K⁺ and Na⁺ of 10, 3, 1 and 1 cmol(+)/kg. What is its base saturation, in per cent?

    Numerical answer — type the value.

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    Answer: 75

    BS = (10 + 3 + 1 + 1)/20 × 100 = 15/20 × 100 = 75%. The remaining 5 cmol(+)/kg is held by H⁺ and Al³⁺, the acidic cations. Leaving out Na⁺ gives 70%, but sodium is a base cation.
  15. Per unit mass, which soil constituent generally has the highest cation exchange capacity?

    1. kaolinite
    2. quartz sand
    3. montmorillonite
    4. humus
    Show answer

    Answer: D — humus

    Humus carries abundant carboxylic and phenolic groups whose negative charge gives it a CEC per kilogram higher than any clay. Montmorillonite, an expanding 2:1 clay, is next and far above kaolinite, a non-expanding 1:1 clay; quartz sand has almost none.
  16. In a sunlit urban air mass the photolysis rate of NO₂ is 0.5 min⁻¹, the rate constant of NO + O₃ → NO₂ + O₂ is 25 ppm⁻¹ min⁻¹, and [NO₂]/[NO] = 2. Assuming the photostationary state, what is the ozone concentration, in ppm?

    Numerical answer — type the value.

    Show answer

    Answer: 0.04

    At steady state ozone formation equals its titration: J[NO₂] = k[NO][O₃], so [O₃] = (J/k)([NO₂]/[NO]) = (0.5/25) × 2 = 0.04 ppm. Doubling the NO₂/NO ratio — which is what VOC peroxy radicals do — doubles the ozone.
  17. Which of the following statements about atmospheric chemistry are correct?

    1. The hydroxyl radical is the principal daytime oxidant of the troposphere
    2. Stratospheric ozone destruction by chlorine is catalytic
    3. Ground-level ozone is a primary pollutant emitted mainly by vehicles
    4. Nitrous oxide is long-lived in the troposphere and is destroyed mainly in the stratosphere
    Show answer

    Answer: A — The hydroxyl radical is the principal daytime oxidant of the troposphere; B — Stratospheric ozone destruction by chlorine is catalytic; D — Nitrous oxide is long-lived in the troposphere and is destroyed mainly in the stratosphere

    (a) OH oxidises CO, CH₄, SO₂ and NO₂ and sets most tropospheric lifetimes. (b) Cl + O₃ → ClO + O₂ and ClO + O → Cl + O₂ regenerate the chlorine, so one atom destroys many ozone molecules. (d) N₂O does not react with OH and is removed by photolysis in the stratosphere. (c) is false: ozone is not emitted but formed from NOₓ and VOCs in sunlight, so it is secondary.