Ocean Carbonate Chemistry:
Carbonate Reactions
Reactions and equilibrium constants (K)
Solutions – numerical
graphical
What can you measure?
Theme 1 (continuation) – Interior Ocean Carbon Cycle
Theme 2 – Ocean Acidification (man’s alteration of the ocean)
Sarmiento and Gruber (2002) Sinks for Anthropogenic Carbon
Physics Today August 2002 30-36
1Pg = 1015g
Weathering and River Flux
Atmospheric CO2 is converted to HCO3- in rivers and transported to the ocean
Examples:
Weathering of CaCO3
CaCO3(s) + CO2(g) + H2O = Ca2+ + 2 HCO31
+ 1
2
Weathering of alumino-silicate minerals to clay minerals.
silicate minerals + CO2(g) + H2O == clay minerals + HCO3- + 2 H4SiO4° + cation
1
1
2
A specific example of orthoclase to kaolinite
KAlSi3O8(s) + CO2(g) + 11/2H2O
= 1/2 Al2Si2O5(OH)4(s) + K+ + HCO3- + 2H4SiO4°
Calculate Global Flux::
Global River Flux = River Flow x global average HCO3 concentration
Global River Flux = 3.7 x 1016 l y-1 x 0.9 mM = 33.3 x 1012 mol y-1 x 12 g/mol = 0.4 x 1015 g y-1 = 0.4 Pg y-1
S&G (2002) give 0.8 Pg y-1 with 0.4 Pg y-1 from weathering
CO2 + rocks = HCO3- + clays
CO2
Gas Exchange
Atm
River Flux
Ocn
CO2 → H2CO3 → HCO3- → CO32Upwelling/
Mixing
+ H2O = CH2O + O2
+ Ca2+ = CaCO3
CO2
BorgC
BCaCO3
Biological Pump
Controls:
pH of ocean
Sediment diagenesis
CO2 reacts with H2O to make H2CO3
CO2 (g) + H2O = H2CO3
K’H
H2CO3 is a weak acid
H2CO3 = H+ + HCO3-
K’1
HCO3- = H+ + CO32-
K’2
H2O is also a weak acid
H2O = H+ + OH-
K’W
Species n = 6
CO2(g)
H2CO3 = carbonic acid
HCO3- = bicarbonate
CO32- = carbonate
H+ = proton or hydrogen ion
OH- = hydroxyl
Equilibrium Constants:
4 equilibrium constants in seawater = K’ = f (S,T,P)
These are expressed as K'.
1. CO2(g) + H2O = H2CO3* (Henry's Law)
K’H = [H2CO3*] / PCO2
(note that gas concentrations are given as partial pressure;
e.g. atmospheric PCO2 = 10-3.5)
2. H2CO3* = H+ + HCO3K’1 = [HCO3-][H+] / [H2CO3*]
3. HCO3- = H+ + CO32K’2 = [H+][CO32-] / [HCO3-]
4. H2O = H+ + OHK’w = [H+][OH-]
[ ] Concentration
Values of K’
The values here are for S = 35, T = 25°C and P = 1 atm.
Constant
K’H
K’1
K’2
K’w
Apparent Seawater Constant (K')
10-1.53
10-6.00
10-9.10
10-13.9
K’ vary with T, S and P
Freshwater Constants (K)
10-1.5
10-6.3
10-10.3
10-14.0
pH
H+ from pH = -log H+
at pH = 6; [H+] = 10-6
OH- from OH- = KW / H+
at pH = 6; [OH-] = 10-8
Total CO2 (ΣCO2 or CT ) – Dissolved Inorganic carbon (DIC)
DIC = [H2CO3] + [HCO3-] + [CO32-]
Example: If you add reactions what is the K for the new reaction?
H2CO3 = H+ + HCO3- K1 = 10-6.0
plus
HCO3- = H+ + CO32- K2 = 10-9.1
-----------------------------------------------H2CO3 = 2H+ + CO32- K12 = 10-15.1
Example: Say we want the K for the reaction
CO32- + H2CO3 = 2 HCO3Then we have to reverse one of the reactions. Its K will change sign as well!!
So:
H2CO3 = H+ + HCO3K = 10-6.0
H+ + CO32- = HCO3K = 10+9.1
--------------------------------------------------------------------( HCO − )2
3
H2CO3 + CO32- = 2HCO3- K = 103.1 ⇒ ( H CO )(CO 2− )
2
3
3
Construct a Distribution Diagram for H2CO3 – Closed System
a. First specify the total CO2 (e.g. CT = 2.0 x 10-3 = 10-2.7 M)
b. Locate CT on the graph and draw a horizontal line for that value.
c. Locate the two system points on that line where pH = pK1 and pH = pK2.
d. Make the crossover point, which is 0.3 log units less than CT
e. Sketch the lines for the species
(not open to the
atmosphere)
Table of acids in seawater
Element
H2O
C
B
Mg
Si
P
S(VI)
F
Ca
Reaction
H2O = H+ + OHH2CO3 = HCO3- + H+
HCO3- = CO32- + H+
B(OH)3 + H2O = B(OH)4- + H+
Mg2+ + H2O = MgOH+ + H+
H4SiO4 = SiO(OH)3- + H+
H3PO4 = H2PO4- + H+
H2PO4- = HPO 2- + H+
HPO42- = PO43- + H+
HSO4- = SO42- + H+
HF = F- + H+
Ca2+ + H2O = CaOH+ + H+
And in anoxic systems
N
NH4+ = NH3 + H+
S(-II)
H2S = HS- + H+
HS- = S2- + H+
pK = -logK
mol kg-1
-logC
2.4 x 10-3
2.6
4.25 x 10-4
5.32 x 10-2
1.5 x 10-4
3.0 x 10-6
3.37
1.27
3.82
5.52
2.82 x 10-2 1.55
5.2 x 10-5 4.28
1.03 x 10-2 1.99
10 x 10-6
10 x 10-6
5.0
5.0
pK'
13.9
6.0
9.1
8.7
12.5
9.4
1.6
6.0
8.6
1.5
2.5
13.0
(e.g. K’ = 10-13.9)
9.5
7.0
13.4
Q. Which is larger? pK = 6.0 or 9.1
Q. If K is larger, what does that mean?
Carbonic Acid – 6 unknowns
Carbonic acid is the classic example of a diprotic acid (it has two H+)
and it can have a gaseous form.
It also can be expressed as open or closed to the atmosphere
(or a gas phase)
There are 6 species we need to solve for:
CO2(g)
Carbon Dioxide Gas
Carbonic Acid (H2CO3* = CO2 (aq) + H2CO3)
H2CO3*
HCO3Bicarbonate
CO32Carbonate
H+
Proton
OHHydroxide
To solve for six unknowns we need six equations
Four of the equations are equilibrium constants!
What can you measure?
We can not measure these species directly. What we can measure are:
a) pH
pH is defined in terms of the activity or concentration of H+. Depends on calibration.
Written as pH = -log (H+)
b) Total CO2 or DIC
DIC = CT = [H2CO3] + [HCO3-] + [CO32-]
c) Alkalinity (defined by the proton balance for a pure solution of CO2)
Alkalinity = [HCO3-] + 2[CO32-] + [OH-] - [H+] + [B(OH)4-] + other bases present (e.g. DOC)
Use Carbonate Alkalinity for calculations (AC).
The alkalinity is defined as the amount of acid necessary to titrate all the weak bases in
seawater (e.g. HCO3-, CO32-, B(OH)4-, NH3) to the alkalinity endpoint which occurs
where (H+) = (HCO3-) (see graph). This is at about pH = 4.3.
Filter samples. Cells contribute alkalinity.
d) PCO2
The PCO2 in a sample is the PCO2 that a water would have if it were in
equilibrium with a gas phase.
Carbonate System Calculations
pH and CT
Alkalinity and PCO2
A useful shorthand is the alpha notation, where the
alpha (α) express the fraction each carbonate species is
of the total DIC. These α values are a function of pH
only for a given set of acidity constants. Thus:
H2CO3 = αo CT
HCO3- = α1 CT
CO32- = α2 CT
The derivations of the equations are as follows:
αo = H2CO3 / CT = H2CO3 / (H2CO3 + HCO3 + CO3)
= 1 / ( 1+ HCO3 / H2CO3 + CO3/H2CO3)
= 1 / ( 1 + K1/H + K1K2/H2)
= H2 / ( H2 + HK1 + K1K2)
The values for α1 and α2 can be derived in a similar
manner.
α1 = HK1 / (H2 + H K1 + K1K2)
α2 = K1K2 / ( H2 + H K1 + K1K2)
For example:
Assume pH = 8, CT = 10-3, pK1' = 6.0 and pK2' = 9.0
[H2CO3*] = 10-5 mol kg-1
(note the answer is in concentration because we used K')
[HCO3-] = 10-3 mol kg-1
[CO32-] = 10-4 mol kg-1
Alk = HCO3 + 2 CO3 + OH - H
For this problem neglect H and OH (a good assumption ),
then:
Alk = CT α1 + 2 CT α2
Alk = CT (α1 + 2α2)
We can use this equation if we have a closed system and
we know 2 of the 3 variables (Alk, CT or pH).
For an open system we can express CT in terms of PCO2 as
follows:
We know that H2CO3* = CT αo ( you can also use this
equation if you know pH and PCO2)
But H2CO3 can be expressed in terms of the Henry's Law:
KH PCO2 = CT αo
So
CT = KH PCO2 / αo
Now:
Alk = (KH PCO2 / αo ) ( α1 + 2α2)
Alk = KH PCO2 ( (α1 + 2 α2 ) / αo ))
Alk = KH PCO2 ( (HK1 + 2 K1K2 )/ H2 )
Assume:
Alk = 10-3
PCO2 = 10-3.5
pK1' = 6.0
pK2' = 9.0
Then: pH = 8.3
CaCO3 solubility calculations
CaCO3 = Ca2+ + CO32-
K’s0 (calcite) = 4.26 x 10-7 = 10-6.37
K’s0 (aragonite) = 6.46 x 10-7 = 10-6.19
Ion Concentration Product = ICP = [Ca2+][CO32-]
or
CaCO3 + CO2(g) + H2O = Ca2+ + 2 HCO3Ion Concentration Product = ICP = [Ca2+][HCO3-]2 / CO2(g)
Degree of Saturation (Ω):
Omega = Ω = ICP / K’s0
Ω=1
Ω > 1 CaCO3 precipitates
There are more products than there should be
Ω <1
CaCO3 dissolves
There are not enough products than there should be
If water at equilibrium (saturation)
If water oversaturated
If water undersaturated
What controls the pH of seawater?
pH and PCO2 are not conservative. DIC and Alk are capacity factors.
pH in seawater is controlled by alkalinity and DIC and can be calculated from these
two parameters as shown below.
Alk ≈ HCO3- + 2 CO32Alk ≈ CT α1 + 2 CT α2 = CT (α1 + 2α2 )
Alk = CT (HK1' + 2 K1' K2' ) / (H2 + H K1' + K1'K2')
Rearranging, we can calculate pH from Alk and CT (use the quadratic equation)
(H+) = {-K1' (Alk-CT) + [(K1')2 (Alk-CT)2 - 4 Alk K1' K2' (Alk - CT)] } / 2 Alk
So the question boils down to what controls alkalinity and total CO2.
Internal variations of pH in the ocean and controlled by internal variations
in DIC and alkalinity which are controlled by photosynthesis, respiration
and CaCO3 dissolution and precipitation.
The long term controls on alkalinity and DIC are the balance between the sources and sinks
and these are the weathering (sources) and burial (sinks) of silicate and carbonate rocks
and organic matter.
The Proton Balance
The balance of species that have excess protons to species deficient in protons
relative to a stated reference species.
Reference species
Proton Balance
For H2CO3:
H+ = HCO3- + 2 CO32- + OHFor HCO3-:
H+ + H2CO3 = CO32- + OHFor CO32-:
H+ + 2 H2CO3 + HCO3- = OHThe proton conditions define three equivalence points on the graph and
these are used to define 6 capacity factors for the solution.
You can approach each equivalence point from either the acid or base direction.
If you add strong acid (e,g, HCl ) it is represented as CA
Strong base (e.g. NaOH) is represented as CB.
For Example:
For a pure solution of H2CO3:
CB + H+ = HCO3- + 2 CO32- + OH- + CA
Then:
CB – CA = HCO3- + 2CO32- + OH- - H+ = Alkalinity
CA – CB = H+ - HCO3- + 2CO32- + OH- = H+-Acidity
Open System - Gas Solubility – Henry’s Law
The exchange or chemical equilibrium of a gas between gaseous and liquid phases
can be written as:
A (g) ===== A (aq)
At equilibrium we can define
K = [A(aq)] / [A(g)]
Henry's Law:
We can express the gas concentration in terms of partial pressure using the
ideal gas law:
PV = nRT
so that [A(g)] is equal to the number of moles n divided by the volume
n/V = [A(g)] = PA / RT where PA is the partial pressure of A
Then
or
K = [A(aq)] / PA / RT
[A(aq)] = (K/RT) PA
[A(aq)] = KH PA
units for K are mol kg-1 atm-1;
for PA are atm in mol kg-1
Henry's Law states that the solubility of a gas is proportional its overlying partial
pressure.
Example: Gas concentrations in equilibrium with the atmosphere
Atmosphere Composition
Gas
N2
O2
Ar
CO2
Gas
N2
O2
Ar
CO2
Mole Fraction in Dry Air (fG) (where fG = moles gas i/total moles)
0.78080
0.20952
9.34 x 10-3
3.3 x 10-4
Pi
0.7808
0.2095
0.0093
0.00033
KH (0°C , S = 35)
0.80 x 10-3
1.69 x 10-3
1.83 x 10-3
63 x 10-3
Ci (0°C, S = 35; P = 1 Atm
62.4 x 10-3 mol kg-1
35.4 x 10-3
0.017 x 10-3
0.021 x 10-3
Open System Distributions
Assume equilibrium with a constant composition gas phase with PCO2 = 10-3.5
