Lecture 21 - Earth and Atmospheric Sciences

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Carbonate System
Alkalinity
Lecture 21
TOTH
• TOTH is the total amount of component H+, rather
than the total of the species H+.
•
o Every species containing H+ contributes positively to TOTH while every
species formed by subtracting H+ contributes negatively to TOTH.
For pure water: TOTH = [H+] - [OH–]
o Of course in pure water [H+] = [OH–] so TOTH = 0.
• Now we dissolve CaCO3 to our solution and chose
Ca2+ and CO32- as components.
•
o In near neutral pH, almost all the CO32- will react to form HCO3–:
CO3+ + H2O = HCO3- + OH–
o some Ca2+ (though generally not much) will form Ca(OH)+, so our mole
balance equation will be
TOTH = [H+] - [OH–] + [HCO3–] - [Ca(OH)+]
Since we have not added [H+], TOTH remains 0.
TOTH
• Now we dissolve CO2 in our solution:
H2O + CO2 ⇋ H2CO3
o In near neutral pH, almost all the H2CO3 will react to form HCO3–:
H2CO3 = HCO3- + H+
o If we chose CO2 as our component,
HCO3– = CO2 + H2O - H+
TOTH = [H+] - [OH–] - [HCO3–]
• This time HCO3- contributes negatively.
•
•
Every species containing H+ contributes positively to TOTH while every
species formed by subtracting H+ contributes negatively to TOTH.
How we write the TOTH equation depends on how we defined
components.
• Since we have not added [H+], TOTH remains 0.
Conservation Equations
• Remember our problem set problem on the dissolution of
galena:
PbS + 2H+ ⇋ Pb2+ + H2S
• We used mass balance to solve this.
• A number of other reactions are possible:
H2S ⇋ H+ + HS–
HS– ⇋ H+ + S2Pb2+ + S2– ⇋ PbSaq
• However, we may still write a conservation equation
• ΣS = PbSaq + H2S + HS– + S2–
• And if galena is the only source of Pb and S, then
ΣPb = ΣS
• Mass balance, or conservation, equations can be useful
constraints on our system. However, they often come with
caveats, like the above (galena the only source of Pb and S).
Charge Balance
• Aqueous solutions are always electrically neutral
(period, no caveats).
• Thus the following constraint always holds:
åm z
i i
=0
• Often, however, charge and mass conservation
equations end up being the same (since we can
only add electrically neutral substances to our
solution).
The Carbonate System
•
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•
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We now turn our attention to carbonate.
Water at the surface of the Earth inevitably contains dissolved CO2, either
as a result of equilibration with the atmosphere or because of respiration
by organisms. CO2 reacts with water to form carbonic acid:
CO2 + H2O ⇄ H2CO3
Some of the carbonic acid dissociates to form bicarbonate and
hydrogen ions:
H2CO3 ⇄ H+ + HCO3–
Some of the bicarbonate will dissociate to an additional hydrogen ion
and a carbonate ion:
HCO3– ⇄ H+ + CO32–
We can write three equilibrium constant expressions for these reactions:
aH CO
K sp = 2 3
ƒCO2
K1 =
K2 =
aH + aHCO3
aH 2CO3
aH + aCO23
aHCO 3
Carbonate Minerals
• Another important reaction in the carbonate
system is precipitation of carbonate minerals
(mainly calcite) in veins in rocks, in soils (caliche), as
shells, on your faucet, etc., and dissolution of
carbonate, as in limestone caverns and sinkholes:
CaCO3 ⇄ Ca2+ + CO32-
Carbonate Speciation
• Suppose now that we
have a known fixed
total carbonate
activity, e.g.,
o ΣCO2 = H2CO3 + HCO3– + CO32-
• Combining this with our
equilibrium constant
expressions, we can
solve for the species
activities as a function
of pH, e.g.:
aHCO- =
3
aH +
SCO2
/ K1 +1+ K 2 / aH +
pH at fixed carbonate
concentration
• If we have a solution with fixed ΣCO2 that is closed and
contains no other dissolved species, the pH is also fixed.
• We can calculate pH by simultaneously solving charge and
mass balance equations together with equilibrium constant
expressions:
[H + ]4 + K1[H + ]3 + {K1K2 - KW - K1SCO2 }[H + ]2 - {KW + 2K2 SCO2 }[H + ] - K1K2 KW = 0
o
(note typo in book -missing exponent 2)
• We can guess our solution will be acidic, in which case we
can ignore CO32– and OH–, we means we can drop terms
containing K2 and Kw. Therefore:
[H + ]2
+ [H + ] SCO2
K1
• This illustrates a key part of solving such equations – knowing
when and how to simplify them by neglecting terms.
Equivalence Points
•
•
•
Particularly simple relationships occur
when the activities of two species are
equal.
These are determined by equilibrium
constant expressions:
CO2 E.P.:
[H+] = [HCO3–]
a 2H + = K1aH 2CO3
•
•
Bicarbonate E.P.
[CO32–] = [H2CO3]
aH2 + = K1K 2
•
•
Carbonate E.P.
[OH–] = [HCO3–]
aH2 + - aH + K1K 2 { SCO2KW -1} + K1 = 0
•
Two others [H2CO3] = [HCO3–],
[HCO3–] = [CO32-]
Conservative and NonConservative Ions
• Conservative ions are those whose concentrations are
not affected by changes in pH, temperature, and
pressure, assuming no precipitation or dissolution.
In natural waters, the principal conservative ions are Na+, K+, Ca2+, Mg2+, Cl–,
and .
o These ions are conservative because they are fully dissociated from their
conjugate acids and bases over the normal range of pH of natural waters.
o
• Non-conservative ions are those that will undergo
association and dissociation reactions in this pH range.
These include the proton, hydroxyl, and carbonate
species as well as B(OH)4–, H3SiO4–, HS–, NH4OH,
phosphate species, and many organic anions.
• Virtually all the non-conservative species are anions, the
two principle exceptions being H+ and NH4OH.
Alkalinity
• Alkalinity is an important and fairly readily measured
property of natural waters.
• Alkalinity is a measure of acid-neutralizing capacity
of a solution and is defined as the sum of the
concentration (in equivalents) of bases that are
titratable with strong acid.
• Mathematically, we define alkalinity as the
negative of TOTH when the components are the
principal species of the solution at the CO2
equivalence point (which carbonate species
there?) The acidity can be defined as the negative
of alkalinity, and hence equal to TOTH.
Carbonate Alkalinity
• Consider a solution containing a fixed total dissolved
concentration of CaCO3. At the CO2 equivalence point,
H2CO3 is the principal carbonate species, so we choose our
components as H+, H2O, CO2, and Ca2+.
H2CO3 = H2O + CO2
HCO3- = H2O + CO2 - H+
CO32- = H2O + CO2 - 2H+
• The proton mole balance equation is then:
TOTH = [H+] - [HCO3- ] - 2[CO32- ] - [OH–]
• The alkalinity is then:
Alk = -TOTH = -[H+] + [HCO3-] + 2[CO32- ] + [OH–]
• The above is also the carbonate alkalinity: the alkalinity due to
the presence of carbonate ions.
o
In many solutions, carbonate alkalinity is very nearly equal total alkalinity. To be
clear, we sometimes call the alkalinity as defined previously Total Alkalinity.
Alkalinity & NonConservative Ions
Charge balance requires:
Σcations - Σ anions = 0
Σcons. cations-Σcons. anions+Σnon-cons. cations-Σnon-cons. anions = 0
Σcons. cations-Σcons. anions= -Σnon-cons. cations+Σnon-cons. anions
The right hand side is equal to the alkalinity:
Alk = Σcons. cations-Σcons. anions
= -Σnon-cons. cations+Σnon-cons. anions
This equation emphasizes an important point. The difference of
the sum of conservative anions and cations is a conservative
property (they cannot be changed except by the addition or
removal of components).
• Since alkalinity is equal to this difference, alkalinity is also a
conservative quantity (i.e., independent of pH, pressure and
temperature). Thus total alkalinity is conservative, even though
concentrations of individual species are not.
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The Tableau Method of Morel
& Hering
• In determining alkalinity, we need to correctly write the TOTH
equation.
• To do so, we need to decide on a proper system of
components (as usual, we want the minimum number of
components) and then decide which species are formed by
adding H+ and which are formed by subtracting H+.
• The Morel & Hering method is to produce a matrix-like table (a
tableau) with species across the top and components listed
vertically.
• Entries in the Tableau are the stoichiometric coefficient of
each component needed to form each species.
• Consider a solution containing H3SiO4–, H4SiO4, B(OH)3, B(OH)4–,
H2S, HS–, HPO42-, H2CO3, HCO3–, CO32–, and, H+, and OH–. We
also recall that we always chose H2O as a component.
H+
H+
1
OH–
-1
H2CO3
H2O
CO2
1
1
HCO3–
-1
1
1
CO32–
-2
1
1
HPO42-
-1
1
H2PO4–
H3SiO4–
1
-1
1
H4SiO4
B(OH)4–
1
-1
B(OH)3
1
1
1
H2S
HS–
H2PO4– H4SiO4 B(OH)3 H2S
1
-1
1
Alk = -TOTH = -{[H+] - [OH-] - [HCO3–] - 2[CO32-] - [HPO42-] - [H3SiO4-] - [HS-]}
Titration Determination of
Alkalinity
• We have learned how to calculate alkalinity, but it
can also be measured by titration.
• Titration is the process of progressively adding a
strong acid or base to a solution until a specified
pH, known as an end-point, is reached.
• In the case of the determination of alkalinity, this
end-point is the CO2 equivalence point, as the
definition suggests.
• The analytical definition of alkalinity is its acid
neutralizing capacity when the end-point of the
titration is the CO2 equivalence point.
Titration Example
• Consider a solution containing sodium carbonate
(Na2CO3). Because the carbonate ion can act as a
proton acceptor, Na2CO3 is a base.
o We assume ideal behavior, complete dissociation, and a volume of
solution large enough that the titration results in trivial dilution.
• The charge balance equation during the titration is:
[Na+] + [H+] = [Cl–] + [ HCO3–] + 2[CO32- ] + [OH–]
o Since the Cl– concentration is conservative, it will be equal to the total
amount of HCl added.
• Using equilibrium constant expressions to eliminate
some species, e.g.,
SCO
K [H CO ]
[HCO3- ] =
1
2
+
[H ]
3
[H 2CO3 ] =
+
2
1+ K1 /[H ] + K1K 2 / [H + ]2
• and noting mass balance requires 2[Na]=ΣCO2, we
derive the following:
2[Na + ]
K1K 2 ü K w
ì
[Cl ] = [Na ] + [H ] K
- +
í
1
+
+ 2
+ 2 ý
1+ K1 /[H ] + K1K 2 /[H ] î
[H ] þ [H ]
-
+
+
Titrating Alkalinity
• This plot shows the how
carbonate species in a 5
mM Na2CO3 solution
change with pH as HCl is
added during titration.
• Notice how the HCl curve
flattens at the
equivalence points - pH
changes rapidly with
small additions of HCl.
o This means we don’t have to hit
the pH of the E.P. with great
precision to obtain the alkalinity.
o In practice, a solution indicator or
pH meter would be used.
• The end-point is reached
here at 10 mM HCl.
Calculated and Titrated
Alkalinity
• In this example, titration yielded an alkalinity of 10
mM.
• Can we obtain the same result from calculation?
TOTH = [H+] - [OH-] - [HCO3–] - 2[CO32-]
• Charge balance is (before addition of HCl):
[Na+] + [H+] = [HCO3–] + 2[CO32-] + [OH–]
• Solving the two equations, we find:
Alk = -TOTH = [Na+] = 10 mM

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