Chemical Properties of Seawater, Major and Minor Elements, the Carbonate System and Ocean Acidification

This chapter opens the chemical and biological oceanography of the Engineering Sciences (XE) paper. It answers the first part of the second half of Section XE7.2 of XE7 Atmospheric and Oceanic Sciences, one of the nine optional sections of which a candidate chooses any two, which reads: “Chemical Properties of Seawater, Major and Minor Elements, Ocean Acidification”. It teaches salinity and the principle of constant proportions, minor elements and residence time, the carbonate system with DIC, alkalinity and pH, ocean acidification and the saturation horizon, and dissolved gases. The numericals are salinity from chlorinity, residence time, pH from pCO₂, the saturation state and the carbonate-to-bicarbonate ratio; every constant is given in the stem.

1. Salinity, major ions and the principle of constant proportions

Seawater is a dilute solution of salts in water: typical open-ocean salinity is S ≈ 35, meaning about 35 g of dissolved salt per kg of seawater, with a range of about 33 to 37 in the open ocean (lower near rivers and ice melt, up to about 40 in the Red Sea). Practical salinity is defined from the electrical conductivity of the sample relative to a standard potassium chloride solution, so it is a ratio and carries no unit. An older measure, from titrating chlorinity (Cl, in g per kg), uses S = 1.80655 × Cl. Six ions make up over 99 % of the dissolved salts, by mass: Cl⁻ about 55 %, Na⁺ 30.6 %, SO₄²⁻ 7.7 %, Mg²⁺ 3.7 %, Ca²⁺ 1.2 % and K⁺ 1.1 %. These are the major elements; for S = 35 chloride is about 19.3 g kg⁻¹ and sodium about 10.8 g kg⁻¹.

The principle of constant proportions (Marcet’s principle, confirmed by Dittmar’s analyses) states that the ratios of the major ions to one another are the same in all open-ocean samples, whatever the total salinity. Salinity changes by adding or removing fresh water (evaporation, precipitation, river runoff, ice formation and melting), which changes all the ions in the same proportion, so a single measurement such as conductivity fixes the whole composition. The reason is that the ocean mixes in about a thousand years, much faster than the residence time of the major ions. The principle fails in estuaries, near hydrothermal vents and in the pore waters of sediments, and for ions affected by biology or precipitation, notably calcium and bicarbonate, to a small degree.

⚠️ Trap: salinity has no unit
Practical salinity is a conductivity ratio, so S = 35 is not “35 ‰” in the strict sense, although it is numerically close to 35 g kg⁻¹. Constant proportions applies to ratios of major ions, not to nutrients, oxygen or carbon dioxide, which vary with biology.

2. Minor elements, conservative behaviour and residence time

Elements present below about 1 mg per kg are minor or trace elements (iron, zinc, copper, cadmium, lead, aluminium and many others), measured in nmol or pmol per kg. A property is conservative if mixing and changes of fresh-water content are the only way to change its ratio to salinity (the major ions, and temperature and salinity themselves away from the surface), and non-conservative if chemical, biological or radioactive processes also change it (nutrients, oxygen, carbon dioxide, trace metals). Trace metals fall in three vertical profile types: nutrient-type (taken up at the surface and regenerated at depth, like cadmium with phosphate and zinc with silicate), scavenged-type (adsorbed on sinking particles, highest at the surface or near sources and lowest in old deep water, like aluminium and lead) and conservative-type.

The residence time of an element is the average time a particle spends in the ocean: τ = A/F, with A the amount (inventory) in the ocean and F the rate of input (or equal rate of removal) at steady state. Worked example. An element at 5.0 × 10⁻⁶ mol kg⁻¹ in an ocean of 1.4 × 10²¹ kg has an inventory 7.0 × 10¹⁵ mol; with a river input of 2.0 × 10⁹ mol yr⁻¹, τ = 7.0 × 10¹⁵/2.0 × 10⁹ = 3.5 × 10⁶ yr. Sodium and chloride have residence times of order 10⁸ years, calcium about 10⁶, and the scavenged metals only tens to hundreds of years. Residence times much longer than the mixing time (about 10³ yr) give a uniform concentration; short ones give strong spatial variation. Water itself has a residence time in the ocean of order 4 × 10⁴ years against river input alone (a few thousand years against evaporation or rainfall), because the ocean holds so much water.

3. The carbonate system

Carbon dioxide dissolves in seawater and reacts: CO₂(g) ⇌ CO₂(aq) with [CO₂*] = K₀ pCO₂ (K₀ the solubility constant, which falls with temperature), then CO₂* + H₂O ⇌ H⁺ + HCO₃⁻ with K₁ = [H⁺][HCO₃⁻]/[CO₂*], then HCO₃⁻ ⇌ H⁺ + CO₃²⁻ with K₂ = [H⁺][CO₃²⁻]/[HCO₃⁻]. Here [CO₂*] is dissolved CO₂ plus carbonic acid. Dissolved inorganic carbon: DIC = [CO₂*] + [HCO₃⁻] + [CO₃²⁻]. At the pH of surface seawater, about 8.1, roughly 90 % of DIC is bicarbonate, roughly 10 % carbonate, and well under 1 % dissolved CO₂. The ratio [CO₃²⁻]/[HCO₃⁻] = K₂/[H⁺] falls as pH falls.

Total alkalinity is the excess of proton acceptors over donors; to a good approximation TA ≈ [HCO₃⁻] + 2[CO₃²⁻] + [B(OH)₄⁻] + [OH⁻] − [H⁺], about 2.3 mmol kg⁻¹ in the open ocean. TA is unchanged by CO₂ gas exchange and by photosynthesis and respiration of carbon (the small nitrate effect aside), but formation of 1 mol of CaCO₃ lowers TA by 2 mol and lowers DIC by 1 mol, and dissolution raises both. Worked example (simplified: bicarbonate fixed at 1.8 × 10⁻³ mol kg⁻¹). With K₀ = 0.030 mol kg⁻¹ atm⁻¹, K₁ = 1.4 × 10⁻⁶ mol kg⁻¹ and pCO₂ = 400 µatm: [CO₂] = 0.030 × 4.0 × 10⁻⁴ = 1.2 × 10⁻⁵; [H⁺] = K₁[CO₂]/[HCO₃⁻] = 1.4 × 10⁻⁶ × 1.2 × 10⁻⁵/1.8 × 10⁻³ = 9.3 × 10⁻⁹, so pH = 8.03. In this idealisation [H⁺] ∝ pCO₂, so doubling pCO₂ lowers pH by log 2 = 0.30. The real ocean is better buffered, because dissolving CO₂ also consumes carbonate (CO₂ + CO₃²⁻ + H₂O → 2HCO₃⁻), and its Revelle factor, about 10, is the ratio of the fractional change in pCO₂ to that in DIC.

4. Ocean acidification and the saturation horizon

The ocean has taken up roughly a quarter to a third of the CO₂ emitted by humans. The extra CO₂ forms carbonic acid, adds H⁺, and lowers pH and carbonate ion: this is ocean acidification. Surface pH has fallen by about 0.1 unit since the pre-industrial era (from about 8.2 to 8.1), which, because pH is logarithmic, is a rise of 10^0.1 = 1.26, or about 26 %, in [H⁺]. The ocean is still alkaline, so “acidification” means a move towards lower pH, not a pH below 7. The fall in [CO₃²⁻] matters most for organisms that make calcium carbonate shells and skeletons.

The saturation state is Ω = [Ca²⁺][CO₃²⁻]/K_sp. Ω > 1 favours precipitation and preservation, Ω < 1 favours dissolution. Worked example. With [Ca²⁺] = 1.03 × 10⁻² mol kg⁻¹, [CO₃²⁻] = 1.5 × 10⁻⁴ mol kg⁻¹ and K_sp = 6.5 × 10⁻⁷ mol² kg⁻² (aragonite, for this stem): Ω = 1.03 × 10⁻² × 1.5 × 10⁻⁴/6.5 × 10⁻⁷ = 2.4. Ω is above 1 in warm surface water and falls with depth, because colder water and higher pressure raise the solubility of CaCO₃ and respiration adds CO₂ at depth. The depth at which Ω = 1 is the saturation horizon. It is shallower for aragonite than for calcite (aragonite is more soluble) and shallower in the old, CO₂-rich North Pacific than in the North Atlantic. Below the carbonate compensation depth (a few km) the sea floor carries almost no CaCO₃. Ocean acidification raises the saturation horizon, so more of the water becomes corrosive to carbonate.

⚠️ Trap: pH direction and CO₂
More CO₂ means more bicarbonate and H⁺ but less carbonate ion, so Ω falls even though DIC rises. Warming lowers CO₂ solubility, so warm water holds less dissolved CO₂ at a given pCO₂, and cold polar water takes up CO₂ most readily.

5. Dissolved gases

Gases dissolve according to Henry’s law, and their solubility decreases as temperature and salinity rise. Cold polar water therefore carries more oxygen at saturation (about 350 µmol kg⁻¹ at 0 °C) than warm tropical water (about 200 µmol kg⁻¹ at 25 °C). Conservative gases (N₂, argon) are changed only by physical processes; non-conservative gases (O₂, CO₂, N₂O) are also changed by biology. Gas exchange across the surface is proportional to the departure from saturation and increases strongly with wind speed.

The oxygen profile has three features: near-saturation or slight supersaturation in the mixed layer (photosynthesis and gas exchange), an oxygen minimum at about 500 to 1000 m where sinking organic matter is respired faster than the slow circulation resupplies oxygen, and higher oxygen in the cold deep water that last touched the surface at high latitude. The apparent oxygen utilisation is AOU = O₂(saturation at the water’s T and S) − O₂(observed), a measure of the oxygen consumed since the water left the surface. It is large in old water and small in newly formed water.

Air-sea CO₂ flux. The flux into the ocean is F = k K₀ (pCO₂,air − pCO₂,sea), where the gas transfer velocity k grows rapidly with wind speed (roughly as U² to U³) and a positive F means uptake. Where pCO₂,sea is below the air value the ocean is a sink: cold, high-latitude water and water in which photosynthesis has drawn down DIC. Where it is above the air value the ocean is a source: warm upwelling regions, such as the eastern equatorial Pacific, where CO₂-rich deep water reaches the surface. Because the sea-surface pCO₂ rises by roughly 4 % for each degree of warming, a warming ocean at fixed DIC is less able to take up CO₂.

Key takeaways

  • S ≈ 35; six major ions make over 99 % of the salts, and constant proportions follows from fast mixing against long residence times.
  • Residence time τ = inventory/flux; short τ means patchy, nutrient-type or scavenged profiles.
  • DIC = [CO₂*] + [HCO₃⁻] + [CO₃²⁻]; at pH 8.1 bicarbonate is about 90 %, and K₂/[H⁺] = [CO₃²⁻]/[HCO₃⁻].
  • Gas exchange changes DIC but not alkalinity; CaCO₃ formation removes 1 DIC and 2 alkalinity per mole.
  • Acidification lowers pH and [CO₃²⁻], so Ω falls and the saturation horizon shoals; aragonite dissolves before calcite.
  • Gas solubility falls with warming; the oxygen minimum arises from respiration, and AOU measures it.

Practice questions (16)

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. The principle of constant proportions states that, in open-ocean seawater,

    1. the concentration of every nutrient and gas is the same at every location
    2. the ratios of the major ions to one another are constant at any salinity
    3. the total salinity is fixed at the value 35 throughout the open ocean
    4. the residence time of every dissolved element is the same, near 1000 years
    Show answer

    Answer: B — the ratios of the major ions to one another are constant at any salinity

    Constant proportions concerns ratios of the major ions, which is why chlorinity or conductivity fixes salinity. Nutrients and gases vary with biology, and salinity itself ranges from about 33 to 37.

  2. An element has a residence time of 100 years in the ocean, short compared with the mixing time of about 1000 years. Its concentration in the deep ocean is therefore expected to be

    1. uniform everywhere, as for the long-lived sodium and chloride
    2. zero, since it is removed too fast for it to be detected at all
    3. spatially variable, with highest values near its sources
    4. equal to its surface concentration at every depth and location
    Show answer

    Answer: C — spatially variable, with highest values near its sources

    When removal is faster than mixing, the element is lost before it is spread evenly, so its concentration differs between regions and near sources; the long-lived major ions are uniform. It is not zero, as the trace metals aluminium and lead show.

  3. At the pH of surface seawater (about 8.1) the dominant form of dissolved inorganic carbon is

    1. dissolved CO₂ gas, CO₂(aq)
    2. the carbonate ion, CO₃²⁻
    3. undissociated carbonic acid
    4. the bicarbonate ion, HCO₃⁻
    Show answer

    Answer: D — the bicarbonate ion, HCO₃⁻

    About 90 % of DIC is HCO₃⁻, roughly 10 % is CO₃²⁻ and well under 1 % is dissolved CO₂ plus H₂CO₃.

  4. Seawater takes up CO₂ from the atmosphere with no other change. Which set of changes is correct?

    1. DIC rises, pH falls, [CO₃²⁻] falls, total alkalinity unchanged
    2. DIC rises, pH rises, [CO₃²⁻] rises, total alkalinity rises
    3. DIC unchanged, pH falls, [CO₃²⁻] falls, total alkalinity falls
    4. DIC rises, pH falls, [CO₃²⁻] rises, total alkalinity unchanged
    Show answer

    Answer: A — DIC rises, pH falls, [CO₃²⁻] falls, total alkalinity unchanged

    CO₂ + CO₃²⁻ + H₂O → 2HCO₃⁻ raises DIC, adds H⁺ so pH falls, and consumes carbonate. Gas exchange does not change total alkalinity.

  5. Precipitation of 1 mol of CaCO₃ by calcifying plankton changes DIC and total alkalinity in seawater by

    1. −2 mol and −1 mol
    2. −1 mol and −2 mol
    3. −1 mol and −1 mol
    4. +1 mol and −2 mol
    Show answer

    Answer: B — −1 mol and −2 mol

    Ca²⁺ + CO₃²⁻ → CaCO₃ removes one carbon from DIC and, since a carbonate counts twice in alkalinity, lowers TA by 2. Swapping the two numbers is the usual slip.

  6. Why is the aragonite saturation horizon in the North Pacific much shallower than in the North Atlantic?

    1. Pacific water has a higher salinity, and that greatly raises carbonate solubility there
    2. The Pacific is colder at depth, which lowers CaCO₃ solubility there
    3. Pacific deep water is older with more respired CO₂, so [CO₃²⁻] is lower
    4. Aragonite is more soluble only in the Pacific
    Show answer

    Answer: C — Pacific deep water is older with more respired CO₂, so [CO₃²⁻] is lower

    Water that has aged longer since leaving the surface has accumulated respired CO₂, so [CO₃²⁻] and Ω are lower at a given depth. Colder water would raise solubility, not lower it.

  7. Which pair of statements about oxygen in the ocean is correct?

    1. Cold water holds less oxygen at saturation, and the oxygen minimum lies at the very sea floor everywhere
    2. Cold water holds more oxygen at saturation; a mid-depth minimum arises where respiration outpaces supply
    3. Oxygen solubility rises with salinity, and the mid-depth minimum arises from photosynthesis there
    4. Oxygen is a conservative gas, so AOU is always zero
    Show answer

    Answer: B — Cold water holds more oxygen at saturation; a mid-depth minimum arises where respiration outpaces supply

    Solubility falls with temperature and salinity, and respiration of sinking matter beneath the euphotic zone creates the mid-depth minimum. Oxygen is not conservative, so AOU is non-zero.

  8. Two ions in an open-ocean sample have a mass ratio Cl⁻:Na⁺ of about 1.8. A water mass with S = 30 formed by diluting seawater with river water of negligible ion content would have Cl⁻:Na⁺

    1. about 1.8, unchanged, since dilution scales all ions alike
    2. about 1.5, since chloride is lost much more readily on dilution
    3. about 2.1, since sodium is lost much more readily on dilution
    4. about 1.2, since the river water adds extra sodium to the mixture
    Show answer

    Answer: A — about 1.8, unchanged, since dilution scales all ions alike

    Adding water that carries almost no ions lowers all concentrations by the same factor, so the ion ratios stay at their constant-proportions values.

  9. Which of the following statements about the carbonate system are correct?

    1. DIC = [CO₂*] + [HCO₃⁻] + [CO₃²⁻]
    2. Ω < 1 favours dissolution of CaCO₃
    3. Air-sea CO₂ exchange changes DIC but not total alkalinity
    4. Ocean acidification lowers pH by driving it below 7
    Show answer

    Answer: A — DIC = [CO₂*] + [HCO₃⁻] + [CO₃²⁻]; B — Ω < 1 favours dissolution of CaCO₃; C — Air-sea CO₂ exchange changes DIC but not total alkalinity

    The first three are correct. Surface seawater remains alkaline (pH about 8.1), so acidification is a fall in pH, not a pH below 7.

  10. Which of the following are conservative properties in the open ocean below the surface?

    1. Salinity
    2. The ratio of sodium to chloride
    3. Dissolved oxygen concentration
    4. Nitrate concentration
    Show answer

    Answer: A — Salinity; B — The ratio of sodium to chloride

    Salinity and major-ion ratios change only through mixing and fresh-water exchange. Oxygen and nitrate are altered by respiration and uptake, so they are non-conservative.

  11. Which of the following statements about gases in seawater are correct?

    1. The solubility of O₂ decreases when temperature rises
    2. AOU = O₂(saturation) − O₂(observed)
    3. N₂ and argon are changed only by physical processes
    4. Warming a water sample raises its O₂ solubility
    Show answer

    Answer: A — The solubility of O₂ decreases when temperature rises; B — AOU = O₂(saturation) − O₂(observed); C — N₂ and argon are changed only by physical processes

    The first three are correct. Warming lowers gas solubility.

  12. Using the old relation S = 1.80655 × Cl, with Cl in g per kg, a sample has a chlorinity of 19.20. Its salinity is ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 34.69

    S = 1.80655 × 19.20 = 34.686, i.e. 34.69.

  13. An element has a mean ocean concentration of 5.0 × 10⁻⁶ mol kg⁻¹. Take the ocean mass as 1.4 × 10²¹ kg and a steady total input of 2.0 × 10⁹ mol yr⁻¹. Its residence time is ____ × 10⁶ years (one decimal place).

    Numerical answer — type the value.

    Show answer

    Answer: 3.5

    A = 5.0 × 10⁻⁶ × 1.4 × 10²¹ = 7.0 × 10¹⁵ mol. τ = A/F = 7.0 × 10¹⁵/2.0 × 10⁹ = 3.5 × 10⁶ yr.

  14. In a simplified model of seawater, [HCO₃⁻] is held at 1.8 × 10⁻³ mol kg⁻¹. Take K₀ = 0.030 mol kg⁻¹ atm⁻¹ and K₁ = 1.4 × 10⁻⁶ mol kg⁻¹, with [CO₂] = K₀ pCO₂ and K₁ = [H⁺][HCO₃⁻]/[CO₂]. The pH at pCO₂ = 400 µatm is ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 8.03

    pCO₂ = 4.0 × 10⁻⁴ atm, so [CO₂*] = 0.030 × 4.0 × 10⁻⁴ = 1.2 × 10⁻⁵. [H⁺] = 1.4 × 10⁻⁶ × 1.2 × 10⁻⁵/1.8 × 10⁻³ = 9.33 × 10⁻⁹. pH = −log₁₀(9.33 × 10⁻⁹) = 8.03.

  15. Take [Ca²⁺] = 1.03 × 10⁻² mol kg⁻¹, [CO₃²⁻] = 1.5 × 10⁻⁴ mol kg⁻¹ and K_sp = 6.5 × 10⁻⁷ mol² kg⁻² for aragonite. The saturation state Ω = [Ca²⁺][CO₃²⁻]/K_sp is ____ (two decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 2.38

    Ω = 1.03 × 10⁻² × 1.5 × 10⁻⁴/6.5 × 10⁻⁷ = 1.545 × 10⁻⁶/6.5 × 10⁻⁷ = 2.38.

  16. With K₂ = [H⁺][CO₃²⁻]/[HCO₃⁻] = 1.0 × 10⁻⁹ mol kg⁻¹, the ratio [CO₃²⁻]/[HCO₃⁻] at pH 8.10 is ____ (three decimal places).

    Numerical answer — type the value.

    Show answer

    Answer: 0.126

    [H⁺] = 10⁻⁸·¹⁰ = 7.943 × 10⁻⁹. Ratio = K₂/[H⁺] = 1.0 × 10⁻⁹/7.943 × 10⁻⁹ = 0.126.