Transcript Slide 1

Lecture 8: Ocean Carbonate Chemistry:
Carbonate Reactions
Reactions
Solutions – numerical
graphical
K versus K’
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
CO2 + rocks = HCO3- + clays
CO2
Gas Exchange
Atm
River Flux
Ocn
CO2 → H2CO3 → HCO3- → CO32-
Upwelling/
Mixing
+ H2O = CH2O + O2
+ Ca2+ = CaCO3
CO2
BorgC
BCaCO3
Biological Pump
Application:
Paleooceanography
Controls:
pH of ocean
Sediment diagenesis
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
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
Example:
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 C y-1 x 12 g mol-1 = 396 x 1012 g C y-1 = 0.4 Pg C y-1
What happens to the CO2 that dissolves in water?
CO2 is taken up by ocean biology to produce a flux of organic mater to the
deep sea (BorgC)
CO2 + H2O = CH2O + O2
Some carbon is taken up to make a particulate flux of CaCO3 (BCaCO3)
Ca2+ + 2HCO3- = CaCO3(s) + CO2 + H2O
The biologically driven flux is called the “Biological Pump”.
The sediment record of BorgC and BCaCO3 are used to unravel paleoproductivity.
The flux of BorgC to sediments drives an extensive set of
oxidation-reduction reactions that are part of sediment diagenesis.
Carbonate chemistry controls the pH of seawater which is a master
Variable for many geochemical processes.
Bronsted Concept of Acids
Acids are compounds that can donate a proton to another substance
which is a proton acceptor.
Acid1 = Base1 + proton
Proton + Base2 = Acid 2
--------------------------------------Acid1 + Base2 = Acid2 + Base1
Examples:
HCl + H2O = H3O+ + ClH2O + H2O = H3O+ + OHH2CO3 + H2O = H3O+ + HCO3-
Free H+ don’t exist in solution
To simplify we will write acids as Arrhenius Acids where acids react to
produce excess H+ in solution:
HCl = H+ + Cl-
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+
mol kg-1
pK = -logK
-logC
2.4 x 10-3 2.6
4.25 x 10-4 3.37
5.32 x 10-2 1.27
1.5 x 10-4 3.82
3.0 x 10-6 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 5.0
10 x 10-6 5.0
pK'
-13.9)
13.9 (e.g. K’ = 10
6.0
9.1
8.7
12.5
9.4
1.6
6.0
8.6
1.5
2.5
13.0
9.5
7.0
13.4
Q. Which is larger? pK = 6.0 or 9.1
CO2 reacts with H2O to make H2CO3
CO2 (g) + H2O = H2CO3
KH
H2CO3 is a weak acid
H2CO3 = H+ + HCO3-
K1
HCO3- = H+ + CO32-
K2
H2O is also a weak acid
H2O = H+ + OH-
KW
H2CO3 = carbonic acid
HCO3- = bicarbonate
CO32- = carbonate
H+ = proton or hydrogen ion
OH- = hydroxyl
Equilibrium Constants on Different Scales
Equilibrium constants can be defined on the infinite dilution activity scale
and the ionic medium scale
Equilibrium constants can be defined differently on these two scales.
There are both advantages and disadvantages of the infinite dilution and
ionic medium approaches.
Consider the following reaction:
HA = H+ + AWe define the equilibrium constants as follows.
A. On the infinite dilution scale – K is expressed in terms of activities.
Calculated from DGr
K = (H+)(A-) / (HA)
Activities = ()
B. On the ionic medium scale the equilibrium constant (K')
is defined in terms of concentrations in the ionic medium of interest.
Measured.
K' = [H+] [A-] / [HA]
Concentrations = [ ]
infinite dilution scale (K)
pros
K can be calculated from DGr
One K for all solutions
cons
Need to express concentration as activity
using free ion activity coefficients (gi) and %free (fi)
ionic medium scale (K’)
pros
Use concentrations. Don’t need to assume activity coefficients.
When K’s have been determined, they are usually very well known.
cons
K’s need to have been determined experimentally as function of T, P and S.
When pH is measured as the activity of (H+), as it is commonly done,
the mixed constant is defined in terms of (H+).
K' = (H+) [A-] / [HA]
The difference between K and K' is the ratio of the total activity
coefficients.
K' = (H+) [A-] / [HA]
K = (H+)(A-) / (HA)
K = (H+)[A-] gi fi / [HA] gi fi
K = (H+)[A-]/[HA] x gi fi /gi fi
K = K' g A / gHA
Values of K and K’
Representative values for these constants are given below.
Equations are given in Millero (1995) with which you can calculate all K's
for any salinity and T, P conditions.
The values here are for S = 35, T = 25C and P = 1 atm.
Constant
KH
K1
K2
Kw
S=0
Thermodynamic Constant (K)
10-1.47
10-6.35
10-10.33
10-14.0
S = 35
Apparent Seawater Constant (K')
10-1.53
10-6.00
10-9.10
10-13.9
Example of calculating gT from K and K’
The difference between K and K' can be illustrated by this simple example.
K = (H+)(A-) / (HA) = (H+)[A-]gT,A- / [HA] gT,HA
Rearrange to get:
K = (H+) [A-] gT,A- = K' gT,A[HA]
gT,HA
gT,HA
The difference in magnitude of K and K' is the ratio of the total activity coefficients
of the base to the acid.
If you know both K and K' , you can learn something about the activity corrections.
For:
H2CO3 = HCO3- + H+
K1 = (HCO3-)(H+) / (H2CO3)
or
K1 = [HCO3-] gT,HCO3 (H+)
[H2CO3] gT,H2CO3
or
K1 = [HCO3](H+) g T,HCO3 = 10-6.3 from tables of DGr at 25C and 1 atm
[H2CO3]
gT,H2CO3
The value of K' has been determined for the same reaction. At S = 35, 25C and 1 atm
K1' = [HCO3-] (H+) = 10-6.0
[H2CO3]
If we set K1 = K1' gT,HCO3 / gT, H2CO3
We can solve for gT,HCO3 / gT, H2CO3 = K1 / K1' = 10-6.3 / 10-6.0 = 10-0.3 = 5.0 x 10-1
K vs K’ Example, continued
We can compare this ratio with that obtained from the Garrels and Thompson
(1962) speciation model of surface seawater where:
gT,HCO3 / gT, H2CO3 = fHCO3 gHCO3
f H2CO3 gH2CO3
= (0.69)(0.68) = 4.1 x 10-1
(1.00)(1.13)
Not too bad. Not too good. These two estimates differ by about 20%.
Is that good enough?
The comparison is not so good for K2 and K2’
Apparent 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-)
( ) Activity = effective concentration
[ ] 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
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 (SCO2 or CT ) – Dissolved Inorganic carbon (DIC)
DIC = [H2CO3] + [HCO3-] + [CO32-]
defined as concentrations!
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
Calculations:
Graphical Approach
Algebraic Approach
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
The Proton Balance
The balance of species that have excess protons to species deficient in protons
relative to a stated reference level.
Reference Levels
Proton Balance
For H2CO3/H2O H+ = HCO3- + 2 CO32- + OHFor HCO3-/H2O H+ + H2CO3 = CO32- + OHFor CO32-/H2O
H+ + 2 H2CO3 + HCO3- = OHThe proton conditions define three equivalance points on the graph (see circles)
and these are used to define 6 capacity factors for the solution.
You can titrate to 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.
Open System Distributions
Assume equilibrium with a constant composition gas phase with PCO2 = 10-3.5
H2CO3 = KH x PCO2
= 10-1.5 x 10-3.5
= 10-5.0
Carbonic Acid – 6 unknowns
Carbonic acid is the classic example of a diprotic acid
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
H2CO3*
Carbonic Acid (H2CO3* = CO2 (aq) + H2CO3)
HCO3Bicarbonate
CO32Carbonate
H+
Proton
OHHydroxide
To solve for six unknowns we need six equations
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 of H+ or as pH = -log (H+)
b) Total CO2
CT = [H2CO3] + [HCO3-] + [CO32-]
c) Alkalinity
Alkalinity = [HCO3-] + 2[CO32-] + [OH-] - [H+] + [B(OH)4-] + any other bases present
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-) to the alkalinity endpoint which occurs
where (H+) = (HCO3-) (see graph)
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 (a) express the fraction each carbonate species is
of the total DIC. These a values are a function of pH
only for a given set of acidity constants. Thus:
H2CO3 = ao CT
HCO3- = a1 CT
CO32- = a2 CT
The derivations of the equations are as follows:
ao = 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 a1 and a2 can be derived in a similar
manner.
a1 = HK1 / (H2 + H K1 + K1K2)
a2 = 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:
= CT a1 + 2 CT a2
= CT (a1 + 2a2)
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 ao ( 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 ao
So
CT = KH PCO2 / ao
Now:
Alk = (KH PCO2 / ao ) ( a1 + 2a2)
Alk = KH PCO2 ( (a1 + 2 a2 ) / ao )
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-
Ks0 (calcite) = 4.26 x 10-7 = 10-6.37
Ks0 (aragonite) = 6.46 x 10-7 = 10-6.19
or
CaCO3 + CO2 + H2O = Ca2+ + 2 HCO3Ion Concentration Product = ICP = [Ca2+][CO32-]
Omega = Ω = ICP / K’s0
If water at equilibrium (saturation)
If water oversaturated
If water undersaturated
Ω=1
Ω>1
Ω <1
CaCO3 precipitates
CaCO3 dissolves
What controls the pH of seawater?
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 a1 + 2 CT a2
Alk = CT (HK1' + 2 K1' K2' ) / (H2 + H K1' + K1'K2')
Rearranging, we can calculate pH from Alk and CT.
(H+) = -K1' (Alk-CT) + [(K1')2 (Alk-CT)2 - 4 Alk K1' K2' (Alk - 2CT)] 1/2 / 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.
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
Algebraic Approach – Monoprotic Acid
Let acetic acid (CH3COOH) = HA.
The base form (CH3COO-) = AWe need to determine the concentrations of 4 species = HA, A- , H+ and OH-.
The 4 key equations are:
1. The reaction
HA =
H+ +
AAcid Hydrogen Anion (or base)
The Equilibrium Constant = K = (H+)(A-) / (HA)
2. The reaction H2O = H+ + OHKw = (H+)(OH-)
3. Mass balance on A
CA = [HA] + [A-]
4. Charge Balance
[H+] = [A-] + [OH-]
By combining equations 2 and 3 given above we can write algebraic
to solve for the main species of acetic acid (HA) and acetate (A-).
expressions
[HA] = CA H+ / K + H+
[A-] = CA K / K + H+
The master variable is pH – How do these species vary with pH?
Graphical Approach (log – log diagram)
There are three regions for these graphs ; pH = pK (system point) ; pH >> pK; pH << pK
K versus K’ - pros and cons
Oceanographers frequently use an equilibrium constant defined in terms of
concentrations. These are called apparent or operation equilibrium constants.
We use the symbol K' to distinguish them from K.
Formally they are equilibrium constants determined on the seawater activity scale.
Apparent equilibrium constants (K') are written in the same form as K
except that all species are written as concentrations.
The exception is H+ which is usually written as the activity (H+).
There are pros and cons for both the K and K' approaches.
K the pro is that we can calculate the K from DGr and one value of K
can be used for all problems in all solutions (one K fits all).
The cons are that to use K we need to obtain values for the free ion activity
coefficients (gi) and the %free (fi) for each solution.
For K‘, the con is that there needs to have been experimental determination of this
constant for enough values of S, T and P that equations can be derived to calculate
K' for the S, T and P of interest. The pro is that when this has been done the values
of K' are usually more precise than the corresponding value of K. The other pro is
that we do not need values of gi and fi when we use K'.