Why the ocean contains 98% of the carbon stored in the combined reservoir ocean-atmosphere?
 For most other gases like oxygen and nitrogen this distribution pattern is reversed.
 The key to explain this fact is to understand the complex chemistry of CO2 in seawater.
Solution chemistry of CO2
 when CO2 dissolves in seawater, it first gets hydrated o form the aqueous species (CO2(aq))
 CO2(aq) reacts with water to produce carbonic acid H2CO3
 It is difficult to distinguish analytically between the two species CO2(aq) and H2CO3
 It is usual to combine the two species and express them as a hypothetical species H2CO3*
-
 This hypothetical acid dissociates in two steps to form HCO3 and CO3
2-
 These reactions are very fast, so we can assume that thermodynamic equilibrium between
the species is established
 CO2(aq) + H2O
k0
 H2CO3*
k1
HCO3 + H+
-
(2)
k2
CO32- + H+
(3)
 HCO3

-
H2CO3*
(1)
The equilibrium relationships between these species are given by the equations:
 K0 = [H2CO3*] / pCO2
(4)
(pCO2 is the partial pressure of CO2 in the air)
-
 K1 = [H+] [HCO3 ] / [H2CO3*]
-
2-
 K2 = [H+] [CO3 ] / [HCO3 ]
-1
K0 in mol Kg atm
(5)
(6)
-1
K1 & K2 in mol Kg-1
•
The carbon dioxide system consists so far of 5 unknowns:
-
pCO2, [H2CO3*], [HCO3 ], [CO32-] and [H+]
And 3 equations, 4, 5 & 6
In order to determine the system, we have to specify any two of the five unknowns
Atmospheric pCO2 and pH are frequently measured but could not be used because neither is
conservative with respect to change in state (temperature, salinity and pressure).
The two parameters used are dissolved inorganic carbon (DIC) and total alkalinity (ALK):
-
DIC = [H2CO3*] + [HCO3 ] + [CO32-]
(7)
-
ALK = [HCO3 ] + 2 [CO32-] + [OH-] – [H+] + [B(OH)4-] + minor bases
(8)
The contribution of minor bases (equation 8) like phosphate, silicate and sulphate to Alk is less than
1% and could be neglected
Both DIC and Alk are conservative with respect to changes in state
Total alkalinity (Alk) is a measure of excess bases (proton acceptors) over acids (proton donors) and is
operationally defined by the titration with H of all weak bases present in solution.
Alternatively Alk can be viewed as the charge balance of all strong acids and bases un affected by this
titration
ALK = [Na+] + [K+] + 2[Mg2+] + 2[Ca2+] + minor cations – [Cl-]- 2[SO42-] – [Br-] – [NO3-] – minor anions (9)
 The definition of DIC and Alk introduces four new unknowns (DIC, Alk, OH- & B(OH)4-)
(total of 9)
 And only 2 equations (7 & 8) for a total of 5
 The hydroxide ion originates from the self-dissociation of water
H2O
H+ + OH-
Kw = [H+] [OH-]
(9)
(10)
 Borate is formed by the dissociation of boric acid
H3BO3 + H2O
H+ + B(OH)4-
KB = [H+] [B(OH)4-] / [H3BO3]
(11)
(12)
The total boron concentration is very nearly conservative, its concentrations is assumed to
be proportional to salinity
TB = [B(OH)4-] + [H3BO3] = c . S = 1.185 . S
(13)
Equations 10, 12 and 13 add three new equations and one new variable [H3BO3]
We end up with 8 equations and 10 unknowns, the specification of two unknowns completely
determines the inorganic carbon system
Unknowns
-
CO2(gas), H2CO3*, HCO3 CO32- H+, OH- , B(OH)4- , H3BO3 , DIC, Alk
 Reactions:
CO2(gas) + H2O
H2CO3*
k1
HCO3
H2O
-
k2
(1)
(2)
CO32- + H+
(3)
(5)
(6)
H+ + OH-
kw
H3BO3 + H2O
Equilibrium constants:
K0 = [H2CO3*] / pCO2
K1 = [H+] [HCO3 ] / [H2CO3*]
K2 = [H+] [CO32-] / [HCO3 ]
Kw = [H+] [OH-]
k0
H2CO3*
HCO3 + H+
kB
H+ + B(OH)4-
(7)
(8)
(9)
(10)
KB = [H+] [B(OH)4-] / [H3BO3] (11)
Concentration definitions:
DIC = [H2CO3*] + [HCO3-] + [CO32-]
ALK = [HCO3-] + 2 [CO32-] + [OH-] – [H+] + [B(OH)4-]
(12)
(13)
TB = [B(OH)4-] + [H3BO3] = c . S = 1.185 . S
(14)
Solution: DIC and ALK
 The goal is to rewrite the equation for Alk (13) in terms of the known parameters DIC
and TB and then solve the resulting equation for unknown [H+]
 From equation (8):
-
[H2CO3*] = ([H+]/ K1 ) . [HCO3 ]
(15)
From equation (9):
-
[CO32-] = (K2 / [H+]) . [HCO3 ]
(16)
Introduce 15 & 16 in equation 12, we get:
DIC = ([H+]/ K1 ) . [HCO3 ] + [HCO3 ] + (K2 / [H+]) . [HCO3 ]
(17)
DIC = [HCO3 ] (1 + [H+]/ K1 + K2 / [H+])
(18)
[HCO3 ] = DIC / (1 + [H+]/ K1 + K2 / [H+])
(19)
Multiplying 8&9 gives:
K1K2 = ([H+]2 [CO32-]) / [H2CO3*]
(20)
[H2CO3*] = ([H+]2 [CO32-]) / K1K2
(21)
From equation 9:
[HCO3 ] = [H+] [CO32-] / K2 (22)
inserting19&20 in 12
DIC = ([H+]2 [CO32-]) / K1K2 + [H+] [CO32-] / K2 + [CO32-] (23)
DIC = [CO32-] (1 + ([H+]2 / K1K2) + ([H+] / K2)
(24)
[CO32-] = DIC / (1 + ([H+]2 / K1K2) + ([H+] / K2)
(25)
Solution: continued
 From equation 10:
[OH-] = Kw / [H+]
(26)
Using equations 11 & 14:
TB = [B(OH)4-] + [H3BO3]
(27)
TB = [B(OH)4-] + ([H+] [B(OH)4-])/ KB
TB = [B(OH)4-] (1+ [H+]/KB)
(28)
(29)
[B(OH)4-] = TB / (1+ [H+]/KB)
(30)
TB = 1.185 . S
in µmol kg-1
[B(OH)4-] = 1.185 . S / (1+ [H+]/KB)
(31)
 Inserting 19, 25, 26 and 31 in equation 13 results in the final equation which is forth
order in the unknown H+
 This equation can be solved using an iterative approach
 Once [H+] has been calculated , pCO2 can be calculated from the equation:
[DIC]
[H+]2
pCO2 =
(32)
K0
[H+]2 + K1 [H+] + K1 K2
 ALK and pCO2
 Rewrite the equations for the three carbon species 7, 8 &9 in terms of [H+] and pCO2
[H2CO3*] = K0 pCO2
(33)
-
•
[HCO3 ] = K0K1 pCO2 / [H+]
(34)
[CO32-] = K0K1K2 pCO2 / [H+]2
(35)
Substitute these terms into the equation of ALK (13) and also replace the [OH-] and
[B(OH)4-] in terms of [H+] and total concentration to calculate [H+]
•
Once [H+] is found, DIC is calculated using equations 11, 33, 34 and 35.
•
Solving the equations for global mean surface seawater properties yields:
•
DIC = [H2CO3*] + [HCO3-] + [CO32-]
0.5%
88.6%
10.9%
ALK = [HCO3-] + 2 [CO32-] + [OH-] – [H+] + [B(OH)4-]
76.8%
18.8%
0.2%
4.2%
continued
 This shows that:


Only a very small fraction of the DIC exists as dissolved CO2 , and that the majority exists as
bicarbonate ion
We can approximate DIC as the sum of carbonate and bicarbonate
DIC ≈ [HCO3-] + [CO32-]
(36)
 Computations reveal also:

Contribution of the dissociation of water to ALK is negligible

Contribution of borate to alkalinity is very small

ALK can reasonably be approximated by the carbonate alkalinity
ALK ≈ [HCO3-] + 2 [CO32-]
(37)
 Combining equations 36 and 37 allows to express the concentration of bicarbonate and
carbonate in terms of DIC and ALK:
[HCO3-] ≈ 2.DIC – ALK
[CO32-] ≈ ALK – DIC
(38)
(39)
Distribution of DIC species as a function of pH
 At pH below pK1, [H2CO3*] dominates
-
 At pH = pK1, [H2CO3*] is equal to [HCO3 ]
-
 When pK1 < pH < pK2 , [HCO3 ] is the dominant species of the DIC
-
 When pH = pK2, [HCO3 ] = [CO32-]
 At pH > pK2, [CO32-] dominates
-
 Ocean has a mean surface pH slightly above 8, in between pK1 & pK2 which explains why HCO3
is the dominant DIC species

pH is also closer to pK2 than to pK1, this is why CO32- is the second most important DIC species
DIC and ALK are normalized to a constant salinity:
•
Fresh water contains very little DIC and ALK
•
Excess evaporation over precipitation leads to a net removal of fresh water
•
Net fresh water exchange at the surface leads to variations in DIC and ALK that can mask the
changes driven by ocean biology, chemistry and mixing/transport