Stable Isotopes, Geochemical Cycles, Water Chemistry and Environmental Geochemistry

This chapter answers the second half of the Geochemistry heading of Part B1: “Stable isotope geochemistry (H, O, C), isotope fractionation; geochemical cycles; Elements of high-temperature and low-temperature geochemical thermodynamics; Geochemical and isotopic (C-O, Sr isotopes) composition of river water, groundwater and seawater, residence time, and water-rock interaction; Principles of environmental geochemistry (pH, Salinity, TDS, BOD, COD)”. It teaches delta notation and the Rayleigh law, the isotopes of water, carbon and strontium, box models and residence time, the thermodynamic frame at high and low temperature, the chemistry of rivers, groundwater and seawater, and the environmental measures. The numericals are delta values, Rayleigh fractionation, residence time, concentration-weighted mixing and BOD.

1. Delta notation, fractionation and the Rayleigh law

Stable isotopes (¹H and ²H or D, ¹²C and ¹³C, ¹⁶O and ¹⁸O) are measured as small differences in the ratio R of heavy to light isotope relative to a standard: δ = (R(sample)/R(standard) − 1) × 1000 ‰. A δ of +10 ‰ means the sample has 1 percent more of the heavy isotope than the standard. The standards are VSMOW for H and O in waters and VPDB for C and for O in carbonates. The fractionation factor is α = R(A)/R(B) = (1000 + δ(A))/(1000 + δ(B)), and for small δ, 1000 ln α ≈ δ(A) − δ(B).

Equilibrium fractionation comes from the different vibrational energies of heavy and light molecules: the heavy isotope concentrates where bonds are stiffest (liquid water over vapour, quartz over magnetite, calcite over water), and the fractionation shrinks roughly as 1/T² as temperature rises, vanishing at very high T. This is the basis of isotope thermometry: two minerals that crystallised in equilibrium give the temperature from their δ difference. Kinetic fractionation belongs to one-way, incomplete processes (evaporation into dry air, diffusion, bacterial reduction), in which the light isotope reacts or moves faster.

When a product is removed continuously from a reservoir (rain from vapour, crystals from a melt) the remaining reservoir follows the Rayleigh law, R/R₀ = f^(α − 1), where f is the fraction of the reservoir left and α = R(product)/R(reservoir). Worked example: vapour of δ¹⁸O = −10 ‰ rains out with α(liquid/vapour) = 1.010. At f = 0.25, R/R₀ = 0.25^0.010 = 0.98623, so δ = 0.98623 × 990 − 1000 = −23.6 ‰: the vapour has become lighter by 13.6 ‰, and each later rain is lighter still.

⚠️ δ is not a percentage of an isotope
A δ value is a relative difference from a standard, in parts per thousand, not an abundance. The heavy isotope is rare (¹⁸O is about 0.2 percent of oxygen), and a difference of a few ‰ is a change in the third or fourth significant figure of the ratio. Never add δ values of different phases unless you account for the amount of the element in each.

2. Oxygen and hydrogen isotopes: the water cycle and rocks

Evaporation from the ocean (δ¹⁸O close to 0 ‰) gives vapour depleted in heavy isotopes, and condensation takes the heavy isotopes into rain first. As an air mass cools and rains out, the remaining vapour, and so each later rain, becomes lighter by the Rayleigh law. This gives the temperature, altitude, continental (distance inland) and amount effects. World meteoric waters lie on the global meteoric water line δD = 8 δ¹⁸O + 10 ‰; the slope 8 reflects the ratio of the equilibrium fractionations of H and O, and the intercept is the deuterium excess d = δD − 8 δ¹⁸O.

Evaporation into dry air adds a kinetic effect: evaporated lake and soil waters plot below the meteoric line, on an evaporation line of slope about 4 to 6 with falling d, and become heavier. Groundwater recharged at high altitude or in a cold climate keeps a light signature, so δ¹⁸O gives recharge elevation (a few tenths of a ‰ per 100 m, calibrated locally), palaeoclimate (cold-stage water is lighter) and whether a spring is meteoric, evaporated or mixed with seawater or brine. In the tropics the amount effect dominates, so δ¹⁸O of monsoon rain and of speleothems is used as a proxy for monsoon intensity.

In rocks, mantle-derived basalts have δ¹⁸O of about +5.7 ‰ (VSMOW). Granites are heavier, about +7 to +10 ‰, and higher still where sedimentary material is melted (S-type); rocks altered by hot meteoric water become light. Marine carbonates are about +20 to +30 ‰. The δ¹⁸O of marine carbonate falls by about 0.2 ‰ for each °C of warming of the water, and ice sheets lock up light ¹⁶O, so glacial-stage foraminifera are heavier.

🎯 Why slope 8 and not 1?
Deuterium differs in mass from hydrogen by a factor of two, but ¹⁸O differs from ¹⁶O by only 12 percent, so the equilibrium fractionation of D is about eight times that of ¹⁸O at ordinary temperatures. Rain therefore changes eight times as much in δD as in δ¹⁸O during Rayleigh distillation. Kinetic evaporation affects ¹⁸O relatively more, which lowers the slope below 8.

3. Carbon and strontium isotopes, and C-O-Sr in rivers

Photosynthesis discriminates against ¹³C, so organic carbon and fossil fuels are isotopically light, while carbonates stay near 0 ‰. Burning fossil fuel adds light CO₂ and lowers atmospheric δ¹³C (the Suess effect). Marine δ¹³C rises when more organic carbon is buried (¹²C is removed) and falls when light carbon from organic matter, methane or volcanoes is added, which makes carbon-isotope excursions useful in stratigraphy. In the Siwalik palaeosols of the Himalayan foreland the shift in carbonate δ¹³C records the late Miocene spread of C4 grasses.

Typical δ¹³C values (‰, VPDB)
Reservoirδ¹³C
Marine carbonateabout 0
Atmospheric CO₂ todayabout −8
Mantle carbon (MORB CO₂, diamond)about −5
C3 plantsabout −27 (−20 to −35)
C4 plantsabout −13 (−9 to −16)
Marine organic matterabout −20
Biogenic methane−60 or lighter

Carbon and oxygen in rivers. Dissolved inorganic carbon comes from weathering by carbonic acid made from soil CO₂. Carbonate weathering, CaCO₃ + CO₂ + H₂O → Ca²⁺ + 2HCO₃⁻, takes half its carbon from the carbonate (about 0 ‰) and half from soil CO₂ (about −23 ‰ under C3 vegetation), so the DIC is about −11.5 ‰; silicate weathering uses only soil CO₂, so its DIC is much lighter. Respiration of organic matter adds light DIC and photosynthesis or exchange with air makes it heavier. The δ¹⁸O of river water follows local rain and the basin’s altitude. Mixing is weighted by the amount of the element: δ(mix) = (f₁C₁δ₁ + f₂C₂δ₂)/(f₁C₁ + f₂C₂), which is f₁δ₁ + f₂δ₂ only when the concentrations are equal.

Strontium. Measured ⁸⁷Sr/⁸⁶Sr is corrected for the small mass-dependent fractionation, so the ratio of a water records the rocks it weathered: old, high-Rb/Sr granite and gneiss terrains give high ratios, young basalt and marine carbonate give low ones. Modern seawater is 0.7092 everywhere, because the residence time of Sr (a few million years) is far longer than the ocean’s mixing time (about 10³ yr). The average river is about 0.711 and mid-ocean-ridge hydrothermal fluid about 0.703, and seawater balances these inputs. The rise from about 0.708 in the Eocene to 0.7092 today has been linked to weathering of the radiogenic Himalayan-Tibetan terrains, and marine carbonates preserve the curve as a dating tool.

4. Geochemical cycles, residence time and the thermodynamic frame

A cycle is described by reservoirs of mass M joined by fluxes F, with dM/dt = F(in) − F(out). At steady state input equals output and the residence time is τ = M/F, the amount in the reservoir divided by the flux through it, which equals 1/k when the output is first-order, F(out) = kM. Water spends about ten days in the atmosphere and a few thousand years in the ocean. If τ is much longer than the ocean’s mixing time (about 10³ yr) the element is uniformly distributed and conservative, with concentration proportional to salinity; if τ is shorter, concentrations vary with depth and place.

Ocean residence times (orders of magnitude)
Elementτ (yr)Behaviour
Na, Cl10⁷ to 10⁸Conservative, tracks salinity
Ca, Srabout 10⁶Uniform; removed in carbonate
Siabout 10⁴Nutrient type, taken up by organisms
Al, Feabout 10²Scavenged by particles; variable

The global carbon cycle has a short organic loop (photosynthesis and respiration, years to centuries) and a long carbonate-silicate loop (10⁵ to 10⁶ yr): weathering of silicates by CO₂ delivers Ca²⁺ and HCO₃⁻ to the sea, where carbonate precipitates, so the net reaction is CaSiO₃ + CO₂ → CaCO₃ + SiO₂. Volcanic degassing supplies CO₂, and warming speeds weathering and draws CO₂ down, a negative feedback that stabilises climate over millions of years.

Thermodynamics at high and low temperature. Reactions run toward minimum Gibbs energy, with ΔG = ΔG° + RT ln Q and ΔG° = −RT ln K (the machinery is developed in the Thermodynamics chapters). The two regimes differ as follows. Isotope exchange is one more equilibrium, with K close to α.

High-temperature and low-temperature geochemical thermodynamics
FeatureHigh T (magmas, metamorphism)Low T (surface and diagenetic waters)
PhasesMelts, solid solutions, supercritical fluidsDilute aqueous solutions with clays, carbonates, oxides
ActivitiesMole fractions and gas fugacitiesActivity coefficients set by ionic strength
Master variablesT, P and oxygen fugacity (buffer 3 fayalite + O₂ = 2 magnetite + 3 quartz)pH, Eh and P(CO₂)
KineticsFast: equilibrium is a good modelSlow: metastable phases persist (aragonite, amorphous silica)
Isotope fractionationSmall, since it falls as 1/T²Large, so δ¹⁸O is a sensitive thermometer

5. River water, groundwater, seawater and water-rock interaction

World-average river water is dilute (TDS of about 100 mg/L) with Ca²⁺ the main cation and HCO₃⁻ the main anion, because carbonic-acid weathering dominates. The Gibbs diagram (TDS against Na/(Na + Ca) or Cl/(Cl + HCO₃), by weight) separates three controls: precipitation dominance (very low TDS, Na- and Cl-rich, sea-salt rain), rock dominance (moderate TDS, Ca-HCO₃, low Na/(Na + Ca)) and evaporation-crystallisation dominance (high TDS, Na-Cl).

Seawater has a salinity of about 35 g/kg, with Cl⁻ (about 19 g/kg) and Na⁺ (about 10.8 g/kg) making roughly 86 percent of the salts, followed by SO₄²⁻, Mg²⁺, Ca²⁺ and K⁺. The ions occur in constant proportions (Marcet’s principle), so salinity can be found from one of them, and the ionic strength is about 0.7 mol/kg. The pH of about 8.1 is buffered by the carbonate system.

Groundwater evolves along its flow path. Dilute, CO₂-rich Ca-HCO₃ recharge water becomes Na-HCO₃ through silicate weathering and cation exchange, and old, slow, deep water turns SO₄- and then Cl-dominated, the Chebotarev sequence HCO₃ → SO₄ → Cl, with TDS rising as residence time grows. The Piper diagram classifies these facies. In India, high fluoride is typical of alkaline, low-Ca Na-HCO₃ groundwater in granitic terrains, arsenic occurs in reducing shallow aquifers of the Ganga-Brahmaputra delta where iron oxyhydroxide is dissolved, and coastal and arid-zone salinity comes from seawater intrusion and evaporation.

Water-rock interaction includes: hydrolysis of silicates by carbonic acid, for example 2NaAlSi₃O₈ + 2CO₂ + 11H₂O → Al₂Si₂O₅(OH)₄ + 2Na⁺ + 2HCO₃⁻ + 4H₄SiO₄ (albite to kaolinite, an incongruent dissolution); dissolution of carbonate; dissolution of evaporites (adds Na, Cl, SO₄); oxidation of pyrite (adds SO₄ and acid); cation exchange; and microbial redox. A reliable analysis obeys electroneutrality: the sums of cations and anions in meq/L agree, with a balance error 100(Σ+ − Σ−)/(Σ+ + Σ−) within about ±5 percent.

⚠️ Carbonate dissolution is CO₂-neutral; silicate weathering is a long-term sink
Carbonate dissolution takes CO₂ from the soil but returns it when the carbonate re-precipitates in the sea, so it is neutral over geological time. Weathering of silicates followed by carbonate burial gives a net removal of atmospheric CO₂; burial of organic carbon is the other long-term sink.

6. Environmental geochemistry: pH, salinity, TDS, BOD and COD

pH = −log a(H⁺). Neutral water is pH 7 at 25 °C, but unpolluted rain is about 5.6 because it equilibrates with atmospheric CO₂; most natural and drinking waters lie between 6.5 and 8.5. pH controls metal mobility: cations such as Pb²⁺, Cu²⁺ and Zn²⁺ are mobile in acid water and sorb or precipitate as pH rises, while oxyanions such as arsenate and chromate and the fluoride ion desorb at high pH. Acid mine drainage arises from pyrite oxidation, FeS₂ + 3.5O₂ + H₂O → Fe²⁺ + 2SO₄²⁻ + 2H⁺.

Salinity is the mass of dissolved salts per kilogram of water (about 35 g/kg for seawater). TDS is the mass of dissolved solids per litre, in mg/L, found by summing the ions or by evaporating a filtered sample at 180 °C, and estimated from electrical conductivity as TDS ≈ k × EC, with k about 0.55 to 0.75 for EC in µS/cm depending on the ionic mix. Fresh water has TDS below 1,000 mg/L, the Indian drinking-water standard is 500 mg/L (up to 2,000 mg/L where no alternative exists) and seawater is about 35,000 mg/L.

Dissolved oxygen falls as temperature or salinity rises and is consumed by organic waste. BOD is the oxygen used by microbes to oxidise biodegradable organic matter, standardly over 5 days at 20 °C (BOD₅), and follows BOD(t) = L₀(1 − e^(−kt)), where L₀ is the ultimate BOD and k the rate constant with base e. COD is the oxygen equivalent of organic matter oxidised by a strong chemical oxidant (hot acid potassium dichromate) in a few hours. It includes non-biodegradable organics, so COD is normally larger than BOD₅ and the BOD₅/COD ratio indicates how biodegradable a waste is.

Worked example. A waste has L₀ = 200 mg/L and k = 0.20 d⁻¹. Then BOD₅ = 200 × (1 − e^(−0.20 × 5)) = 200 × (1 − 0.3679) = 126.4 mg/L, about 63 percent of the ultimate demand.

⚠️ BOD₅ is not the ultimate BOD, and TDS is not salinity
BOD₅ is only the oxygen used in the first five days, so it is smaller than L₀ (with k = 0.23 d⁻¹ it is about two-thirds). A very low BOD₅/COD ratio signals a large non-biodegradable fraction, not a clean water. The factor k that converts EC to TDS is not universal, and TDS (per litre of solution) differs from salinity (per kilogram) at high concentration.

Key takeaways

  • δ = (R(sample)/R(standard) − 1) × 1000 ‰, with VSMOW for waters and VPDB for carbon. Equilibrium fractionation falls as 1/T²; kinetic fractionation favours the light isotope; a Rayleigh process gives R/R₀ = f^(α − 1).
  • Meteoric waters follow δD = 8 δ¹⁸O + 10 ‰; temperature, altitude, continental and amount effects make rain lighter; evaporation lowers the slope to about 4 to 6. Basalt is about +5.7 ‰ in δ¹⁸O and granite is heavier.
  • Organic carbon is light (C3 about −27 ‰, C4 about −13 ‰) and carbonate is near 0 ‰. River DIC from carbonate weathering by soil CO₂ is about −11.5 ‰. Mixing is weighted by concentration, so Sr isotope ratios of waters mix by Sr mass, not by volume alone. Seawater ⁸⁷Sr/⁸⁶Sr is 0.7092 everywhere.
  • Residence time is τ = M/F at steady state; Na and Cl are conservative with τ of 10⁷ to 10⁸ yr, Al and Fe are scavenged with τ of about 10² yr. Silicate weathering followed by carbonate burial (and burial of organic carbon) removes atmospheric CO₂ over geological time, whereas carbonate dissolution does not.
  • Rivers are Ca-HCO₃, seawater Na-Cl with constant ion proportions and pH about 8.1, and groundwater evolves from Ca-HCO₃ to Na-HCO₃ to Na-Cl along the flow path. A good analysis balances cations and anions in meq/L within about 5 percent.
  • pH controls metal mobility; TDS is in mg/L and salinity in g/kg. BOD₅ = L₀(1 − e^(−5k)) is smaller than the ultimate BOD, and COD is normally larger than BOD₅ because it includes non-biodegradable organics.

Practice questions (18)

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. A sample has δ¹⁸O = +10 ‰ relative to VSMOW. This means that its ¹⁸O/¹⁶O ratio is:

    1. 10 percent higher than that of the standard, because ‰ is the same as a percentage
    2. 10 times that of the standard, because δ is a ratio of the two abundances
    3. 1 percent higher than that of the standard
    4. 1 percent lower than that of the standard, because the sample is enriched in the lighter isotope
    Show answer

    Answer: C — 1 percent higher than that of the standard

    δ = (R(sample)/R(standard) − 1) × 1000 ‰, so +10 ‰ means R(sample)/R(standard) = 1.010, that is 1 percent higher. The tempting error is to read ‰ as a percent, which overstates the difference tenfold. A positive δ means enrichment in the heavy isotope, not the light one.

  2. Which statement about equilibrium isotope fractionation between two coexisting minerals is correct?

    1. It falls roughly as 1/T² as temperature rises, so the δ difference of an equilibrium pair can give the crystallisation temperature
    2. It grows with rising temperature, because the vibrations of the lighter isotope become more energetic and separate it more strongly
    3. It is independent of temperature and depends only on the relative mass difference of the isotopes in question
    4. It occurs only in one-way, incomplete reactions such as evaporation into dry air, never between coexisting minerals
    Show answer

    Answer: A — It falls roughly as 1/T² as temperature rises, so the δ difference of an equilibrium pair can give the crystallisation temperature

    Equilibrium fractionation arises from vibrational energy differences and shrinks roughly as 1/T², vanishing at very high temperature; calibrated mineral-pair fractionations therefore act as thermometers. The statement that it rises with temperature reverses the trend. The one-way, incomplete case describes kinetic fractionation, not equilibrium.

  3. A maritime air mass moves inland across a continent and rains out steadily. How does the δ¹⁸O of successive rains change?

    1. It becomes less negative, because the heavy isotope is held back in the vapour as the air cools
    2. It stays constant, because condensate and vapour have the same isotopic composition at equilibrium
    3. It oscillates with the seasons only, because the distance travelled has no systematic effect on the isotopic composition of the vapour
    4. It becomes progressively more negative, because each rain removes the heavy isotope preferentially (Rayleigh distillation)
    Show answer

    Answer: D — It becomes progressively more negative, because each rain removes the heavy isotope preferentially (Rayleigh distillation)

    Condensate is enriched in the heavy isotope (α > 1), so each rain removes heavy isotope from the vapour and the vapour, and the next rain, become lighter: the continental effect. If vapour and condensate had the same composition no isotopic distillation would occur, and the idea that the heavy isotope is retained reverses the sign of the fractionation.

  4. Waters from several closed lakes in a semi-arid basin plot on a line of slope about 5 in δD against δ¹⁸O, below the global meteoric water line, with the most saline lake at the highest δ¹⁸O. Which process explains the trend?

    1. Rayleigh condensation during steady rainout, which moves the points along the meteoric water line itself with a slope near 8 and lowers both δ values
    2. Evaporation into unsaturated air, which leaves the residual water enriched in heavy isotopes along a line of slope below 8
    3. Melting of glaciers in the catchment, which supplies water heavier than the local rain
    4. Seasonal temperature change alone, which moves rain along the meteoric water line itself
    Show answer

    Answer: B — Evaporation into unsaturated air, which leaves the residual water enriched in heavy isotopes along a line of slope below 8

    Evaporation into dry air combines equilibrium and kinetic fractionation, enriches the residual lake water in the heavy isotopes and gives evaporation lines of slope about 4 to 6, below the meteoric line. Rainout and temperature effects move rain along the meteoric line itself, slope near 8, and glacier meltwater is isotopically light rather than heavy.

  5. A river drains a catchment of pure limestone. Its dissolved inorganic carbon forms by CaCO₃ + CO₂ + H₂O → Ca²⁺ + 2HCO₃⁻, with the CO₂ from soil under C3 vegetation at δ¹³C = −23 ‰ and the limestone at 0 ‰ (VPDB). Ignoring fractionation and exchange with air, the expected δ¹³C of the DIC is:

    1. −23 ‰, because all the carbon comes from the soil CO₂ that drives the reaction
    2. 0 ‰, because the dissolved bicarbonate is simply recycled limestone carbon and the soil CO₂ does not enter the product
    3. −11.5 ‰, because half of the carbon in the bicarbonate is from the limestone and half from soil CO₂
    4. −5.75 ‰, because three-quarters of the carbon is from the limestone and one-quarter from soil CO₂
    Show answer

    Answer: C — −11.5 ‰, because half of the carbon in the bicarbonate is from the limestone and half from soil CO₂

    Each Ca²⁺ is accompanied by 2 HCO₃⁻, one carbon from CaCO₃ and one from CO₂, so the DIC is a 1:1 mixture: (0 + (−23))/2 = −11.5 ‰. Taking all the carbon from soil CO₂ (−23 ‰) is the result for silicate weathering, not for carbonate weathering, which is the tempting confusion. The 3:1 ratio has no stoichiometric basis.

  6. A marine carbonate section shows a positive δ¹³C excursion of about +3 ‰ that is global in extent. Which cause is most consistent with it?

    1. Enhanced burial of organic carbon, which removes light ¹²C from the ocean-atmosphere system
    2. A large release of methane from gas hydrates, which adds very light carbon
    3. Increased oxidation of organic-rich rocks exposed on land, which returns light carbon to the ocean
    4. Cooling of seawater alone, which changes the δ¹⁸O of carbonate but not the δ¹³C of the carbon pool
    Show answer

    Answer: A — Enhanced burial of organic carbon, which removes light ¹²C from the ocean-atmosphere system

    Burying organic carbon, which is isotopically light, leaves the remaining dissolved carbon, and the carbonate precipitated from it, heavier, giving a positive excursion. Hydrate methane and oxidised old organic carbon add light carbon and would give a negative excursion; temperature changes mainly the δ¹⁸O of carbonate.

  7. Why do aluminium and iron have residence times of only about 10² yr in the ocean, while sodium and chloride have 10⁷ to 10⁸ yr?

    1. Aluminium and iron have the smallest river inputs of all the elements, so their inventory in the sea is exhausted within centuries
    2. Aluminium and iron are atmophile elements that escape from the sea surface as gases, whereas sodium and chloride stay dissolved
    3. Sodium and chloride are removed rapidly by particles but supplied even faster by the rivers, which lengthens their residence time
    4. Aluminium and iron are scavenged onto particles, so output is large relative to inventory; sodium and chloride lack an efficient sink
    Show answer

    Answer: D — Aluminium and iron are scavenged onto particles, so output is large relative to inventory; sodium and chloride lack an efficient sink

    τ = M/F is short when the output flux is large compared with the amount present. Al and Fe are rapidly removed on particles and as insoluble hydroxides, while Na and Cl are very soluble and removed only by slow processes such as evaporite formation and burial of pore water. A small river input would lengthen, not shorten, τ, and neither element is lost as gas.

  8. Which process gives a net removal of atmospheric CO₂ over millions of years?

    1. Dissolution of limestone by carbonic acid, since the reaction consumes CO₂ from the soil and the air
    2. Weathering of Ca-silicates, with Ca²⁺ and HCO₃⁻ carried to the sea and precipitated there as carbonate
    3. Oxidation of pyrite, which produces sulphuric acid that attacks the rocks and dissolves carbonate
    4. Dissolution of halite and gypsum, which supplies sodium, calcium, chloride and sulphate ions to the river
    Show answer

    Answer: B — Weathering of Ca-silicates, with Ca²⁺ and HCO₃⁻ carried to the sea and precipitated there as carbonate

    Silicate weathering takes two CO₂ per Ca but carbonate precipitation in the sea returns only one, so over geological time one CO₂ is removed per Ca and the carbon ends in carbonate rock. Limestone dissolution is the tempting answer, but the CO₂ it consumes is released again when the carbonate re-precipitates, so it is neutral. Pyrite oxidation and evaporite dissolution do not remove CO₂.

  9. An industrial effluent has BOD₅ = 40 mg/L and COD = 400 mg/L. What is the most reasonable inference?

    1. Most of its oxidisable organic matter is not readily degraded by microbes, so biological treatment alone will be inefficient
    2. The analysis must be in error, because BOD₅ measured over five days can never be lower than the COD of the same wastewater sample
    3. The effluent is almost free of organic matter, because a low BOD₅ shows that there is little left in the sample to oxidise chemically
    4. The sample carried a high dissolved oxygen content that suppressed the chemical oxidation of organic matter in the COD test
    Show answer

    Answer: A — Most of its oxidisable organic matter is not readily degraded by microbes, so biological treatment alone will be inefficient

    COD measures all matter that a strong oxidant can oxidise, including non-biodegradable organics, so COD is greater than BOD₅, and a BOD₅/COD ratio of 0.1 shows that little is quickly biodegradable. A low BOD₅ alone does not mean few organics, which is the tempting reading, and BOD₅ being below COD is normal, not an error.

  10. Which statement about pH and metal mobility in natural waters is correct?

    1. Raising pH from 5 to 9 raises the dissolved Pb²⁺ concentration, because lead hydroxide is highly soluble in alkaline water and sorption weakens
    2. Arsenate and fluoride adsorb more strongly onto iron and aluminium oxides at pH 9 than they do at pH 5
    3. Raising pH from 5 to 9 lowers dissolved Pb²⁺, Cu²⁺ and Zn²⁺ through sorption and precipitation, while arsenate and fluoride tend to desorb
    4. Acid mine drainage is poor in dissolved metals, because its low pH keeps the metals bound to the sediments and rock surfaces
    Show answer

    Answer: C — Raising pH from 5 to 9 lowers dissolved Pb²⁺, Cu²⁺ and Zn²⁺ through sorption and precipitation, while arsenate and fluoride tend to desorb

    Cationic metals sorb to mineral surfaces and precipitate as hydroxides as pH rises, so they are most mobile in acid water, which is why acid mine drainage carries high dissolved metals. Oxyanions such as arsenate and the fluoride ion are held on oxide surfaces more at low pH and are released as pH rises, the opposite behaviour, and the usual error is to assume that all contaminants respond alike.

  11. Which of the following make the δ¹⁸O of rain lower (more negative)?

    1. Rain falling at a higher altitude on a mountain slope
    2. Rain falling far inland, after the air mass has already shed much of its moisture
    3. Condensation at a lower temperature
    4. Partial evaporation of the falling raindrops in dry air, which removes the light isotope preferentially
    Show answer

    Answer: A — Rain falling at a higher altitude on a mountain slope; B — Rain falling far inland, after the air mass has already shed much of its moisture; C — Condensation at a lower temperature

    Altitude, continentality and lower condensation temperature all correspond to more advanced Rayleigh distillation or stronger fractionation, so the rain is lighter. Partial evaporation of falling drops removes the light isotope and leaves the drop heavier, raising δ¹⁸O, so it works in the opposite direction.

  12. Which of the following statements about residence time in the ocean are correct?

    1. At steady state τ = M/F, where F is the input flux, which equals the output flux
    2. Elements whose τ is much longer than the ocean mixing time (about 10³ yr) are almost uniformly distributed
    3. For a fixed ocean inventory, a larger input flux of an element means a longer residence time
    4. Calcium has a shorter residence time in the ocean than sodium
    Show answer

    Answer: A — At steady state τ = M/F, where F is the input flux, which equals the output flux; B — Elements whose τ is much longer than the ocean mixing time (about 10³ yr) are almost uniformly distributed; D — Calcium has a shorter residence time in the ocean than sodium

    τ = M/F is the inventory divided by the flux, and a long τ compared with mixing gives a well-mixed, conservative element. Ca (about 10⁶ yr) is removed in carbonate and so has a much shorter τ than Na (about 10⁷ to 10⁸ yr). At fixed M a larger F gives a shorter τ, so the statement about a longer residence time is reversed.

  13. Which of the following statements about BOD, COD and dissolved oxygen are correct?

    1. BOD₅ is the oxygen consumed by microorganisms in 5 days at 20 °C while they oxidise biodegradable organic matter
    2. COD uses a strong chemical oxidant and therefore also measures organic matter that microbes cannot degrade
    3. For any waste, BOD₅ is greater than COD, because microbes oxidise more completely than dichromate
    4. The saturation concentration of dissolved oxygen in water rises as the water warms
    Show answer

    Answer: A — BOD₅ is the oxygen consumed by microorganisms in 5 days at 20 °C while they oxidise biodegradable organic matter; B — COD uses a strong chemical oxidant and therefore also measures organic matter that microbes cannot degrade

    BOD₅ is defined as five days of microbial oxygen demand at 20 °C, and COD uses a chemical oxidant that also attacks refractory organics, so COD is normally larger than BOD₅, not smaller. Gases become less soluble as water warms, so the saturation concentration of oxygen falls, not rises.

  14. A water sample has ¹⁸O/¹⁶O = 2.0252 × 10⁻³. Taking the VSMOW ratio as 2.0052 × 10⁻³, the δ¹⁸O of the sample in ‰, to the nearest whole number, is ____.

    Numerical answer — type the value.

    Show answer

    Answer: 10

    δ = (R(sample)/R(standard) − 1) × 1000 = (2.0252/2.0052 − 1) × 1000 = (1.009974 − 1) × 1000 = 9.97 ‰, which is 10 to the nearest whole number.

  15. Water vapour with δ¹⁸O = −12 ‰ loses 80 percent of its mass as rain, so that f = 0.20 of the vapour remains. Take α(liquid/vapour) = 1.010 and the Rayleigh law R/R₀ = f^(α − 1), with R₀ corresponding to (1000 + δ₀). By how many ‰ has the δ¹⁸O of the remaining vapour fallen? Give the answer to one decimal place.

    Numerical answer — type the value.

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    Answer: 15.8

    R/R₀ = 0.20^0.010 = e^(0.010 × ln 0.20) = e^(−0.016094) = 0.984034. Then 1000 + δ = 988 × 0.984034 = 972.23, so δ = −27.77 ‰. The fall is 27.77 − 12 = 15.77, which is 15.8 ‰ to one decimal place.

  16. For a hypothetical element X, the ocean (1.4 × 10²¹ L) contains X at a mean concentration of 0.40 g/L, and the steady-state input of X from rivers is 2.8 × 10¹⁴ g/yr. The residence time of X in the ocean, in million years to one decimal place, is ____.

    Numerical answer — type the value.

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    Answer: 2.0

    Inventory M = 1.4 × 10²¹ L × 0.40 g/L = 5.6 × 10²⁰ g. At steady state τ = M/F = 5.6 × 10²⁰/2.8 × 10¹⁴ = 2.0 × 10⁶ yr, which is 2.0 million years.

  17. Two river waters are mixed in equal volumes. Water A has 100 µg/L Sr with ⁸⁷Sr/⁸⁶Sr = 0.7300; water B has 400 µg/L Sr with ⁸⁷Sr/⁸⁶Sr = 0.7080. The ⁸⁷Sr/⁸⁶Sr of the mixture, to four decimal places, is ____.

    Numerical answer — type the value.

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    Answer: 0.7124

    Mixing is weighted by the mass of Sr contributed. For 1 L of each: Sr from A = 100 µg, from B = 400 µg, total 500 µg. R = (100 × 0.7300 + 400 × 0.7080)/500 = (73.0 + 283.2)/500 = 0.7124. The unweighted mean 0.7190 is the trap, since water B supplies four times as much Sr.

  18. A wastewater has an ultimate BOD L₀ = 300 mg/L and a first-order rate constant k = 0.23 d⁻¹ (base e), with BOD(t) = L₀(1 − e^(−kt)). The BOD₅, in mg/L to the nearest whole number, is ____.

    Numerical answer — type the value.

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    Answer: 205

    kt = 0.23 × 5 = 1.15, and e^(−1.15) = 0.3166. BOD₅ = 300 × (1 − 0.3166) = 300 × 0.6834 = 205.0 mg/L, so 205. Giving L₀ itself (300) or forgetting the exponential are the usual slips.