# Learn to calculate the solubility of carbonate minerals such as calcite.

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```THE GEOCHEMISTRY OF
NATURAL WATERS
THE CARBONATE SYSTEM
CHAPTER 3 - Kehew (2001)
Alkalinity
1
LEARNING OBJECTIVES
z Understand sources of CO2 in natural
waters.
z Define and understand alkalinity.
z Learn to calculate the solubility of carbonate
minerals such as calcite.
z Understand the common-ion effect.
z Become familiar with the concept of incongruent
dissolution.
z Apply these concepts to some case studies.
2
BRIEF REVIEW
z We saw in Lecture 3 that pH, pCO2 and
bicarbonate ion concentrations are all
interrelated.
z Rearrangement of the equations we have
worked with previously yields:
pCO2 =
a H + aHCO 3
K1KCO2
z Thus, if we measure pH and bicarbonate ion
concentration, we can calculate pCO2.
3
SOURCES OF CO2 IN NATURAL
WATERS
z When the equation in the previous slide is applied to
natural waters, particularly ground waters and soil
solutions, pCO2 values greater than atmospheric are
commonly obtained. Why?
z A system closed to atmospheric CO2 is implied.
z Respiration by plant roots and microbes consumes
organic matter and produces CO2:
CH2O + O2  CO2 + H2O
z Amount of CO2 production depends on temperature, soil
moisture content, and the amount of organic matter.
4
ALKALINITY - I
z In aqueous solutions, positive and negative charges
must balance.
z In a pure CO2-H2O system, the charge-balance condition
is:
M H + = M HCO - + 2 M CO2- + M OH 3
3
z This equation shows that, as H2CO3* dissociates to form
HCO3-, the concentration of H+ also increases to
maintain charge balance.
z Often, because CO32- and OH- are negligible, the chargebalance expression can be approximated as:
M H + &raquo; M HCO 3
5
ALKALINITY - II
z Reactions with minerals can affect this relationship. For
example, dissolution of calcite would result in:
2 M Ca2 + + M H + = M HCO - + 2 M CO2 - + M OH 3
3
z If this solution is removed from contact with calcite, and
strong acid is added, the concentration of H+ and all the
carbonate species would change, but the concentration
of Ca2+ would not change.
z Thus, Ca2+ is a conservative ion, and HCO3-, CO32-, H+
and OH- are non-conservative. Grouping these ions
accordingly we get:
2 M Ca2 + = M HCO - + 2 M CO 2- + M OH - - M H +
3
3
6
ALKALINITY - III
z The quantity:
is
M HCO - + 2 M CO2- + M OH - - M H +
3
called the total alkalinity.
3
z Another definition of total alkalinity: the equivalent sum
of bases titratable with a strong acid.
z Total alkalinity is the neutralizing capacity of a solution;
the greater the total alkalinity, the more acid the
solution can neutralize.
z For a general natural water, the charge-balance can be
written:
2 M Ca 2 + + 2 M Mg 2 + + M Na + + M K + - M Cl - - 2 M SO 24
= M HCO - + 2 M CO2 - + M OH - - M H + = [alk ]
3
3
7
ALKALINITY - IV
z Because all the terms on the left-hand side of the
previous charge-balance expression are
conservative, then the alkalinity must also be
conservative.
z The only way to change alkalinity is to either add
strong acid or base, or for solids to dissolve or
precipitate. This is why it is important to measure
alkalinity in the field before precipitation can occur.
z Alkalinity is measured by titration with strong acid.
A known volume of sample is titrated (usually with
H2SO4) until an endpoint.
8
Titration curve for a 510-3 m Na2CO3 solution, together with a Bjerrum plot for
the same solution. A is the beginning of the titration, B is the carbonate endpoint,
C is the region of strong carbonate buffering, and D is the bicarbonate endpoint.
HCO3
-
CO3
2-
D
OH-
HCO3
C
2-
CO3
HCO
2
3
B
HCO3-
-
V (mL of acid)
log ai (Bjerrum plot)
HCO
2
3
H+
alkalinity
p-alkalinity
pH
A
9
ALKALINITY - V
z Alkalinity is often expressed as the equivalent
weight of calcium carbonate (mg L-1 CaCO3).
z Calculation of alkalinity from a titration is according
to:
mLacid &acute; N acid &acute; eq. wt. CaCO3 &acute; (1000 mg g -1 )
Total alk =
mLsample
z The equivalent weight of CaCO3 is 50 g eq-1.
z Example: A 100 mL sample is titrated to the methyl
orange end point with 2 mL of 0.5 N H2SO4. What is
the total alkalinity in mg L-1 as CaCO3 and what is
the concentration of HCO3- in mg L-1?
10
ALKALINITY - VI
z The total alkalinity in mg L-1 as CaCO3 is given by:
-1
2 mL &acute; 0.5 eq L &acute; 50 g eq &acute; (1000 mg g )
Total alk =
100 mL
= 500 mg L-1 as CaCO3
-1
-1
z In most natural waters, bicarbonate is the dominant
contributor to the total alkalinity, so the
concentration of HCO3- is given as:
-1
-1
-1
2
mL
&acute;
0
.
5
eq
L
&acute;
61
g
eq
&acute;
(
1000
mg
g
)
-1
mg L HCO3 =
100 mL
= 610 mg L-1
11
LEARNING OBJECTIVES
z Understand sources of CO2 in natural waters.
z Define and understand alkalinity.
z Learn to calculate the solubility of
carbonate minerals such as calcite.
z Understand the common-ion effect.
z Become familiar with the concept of incongruent
dissolution.
z Apply these concepts to some case studies.
12
CARBONATE MINERAL
EQUILIBRIA
z The solubility of calcite at 25&deg;C is governed by:
Kcal = aCa2+ aCO2 - = 10
z For aragonite we have:
-8.40
3
Karag = aCa2+ aCO2- = 10
-8.16
3
z Aragonite is more soluble (less stable) than
calcite.
z The solubility of dolomite at 25&deg;C is governed
by:
2
-17.0
Kdol = aCa2+ aMg 2+ aCO2- = 10
3
13
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - I
We have six dissolved species: H+, OH-, H2CO3*, HCO3-,
CO32- and Ca2+ whose concentrations are unknown.
We need six independent equations to solve for these
concentrations.
Mass Action Expressions:
K w = aH + aOH K1 =
aHCO - aH +
3
aH 2CO3*
KCO2 =
K2 =
aH 2CO3*
pCO2
aCO 2 - aH +
3
aHCO 3
Kcal = aCa 2 + aCO 23
14
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - II
The sixth constraint is the charge-balance equation:
2 M Ca2 + + M H + = M HCO - + 2 M CO2 - + M OH 3
This can be simplified to:
3
2 M Ca 2+ &raquo; M HCO 3
At a constant value of pCO2, the logarithms of the concentrations of each of the
species can be expressed as a straight-line function of pH. For example:
log aH 2CO3* = log KCO2 + log pCO2 = -1.46 + ( -3.5) = -4.96
15
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - III
Bicarbonate can be calculated from:
K1 =
aHCO - aH +
3
aH 2CO3*
aHCO - =
K1aH 2CO3*
aH +
3
log aHCO - = log K1 + pH + log aH 2CO3*
3
= -6.35 + pH - 4.96 = -11.31 + pH
And carbonate from:
K2 =
aCO 2 - aH +
3
aHCO 3
aCO 2 - =
3
K 2aHCO 3
aH +
16
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - IV
log aCO 2- = log K2 + pH + log aHCO 3
3
= -10.33 + pH - 11.31 + pH = -21.64 + 2 pH
Calcium ion concentration is obtained from:
Kcal = aCa 2 + aCO 2-
aCa 2+
3
K cal
=
aCO 2 3
log aCa 2+ = log Kcal - log aCO 2 3
= -8.37 - ( -21.64 + 2 pH ) = 13.27 - 2 pH
17
Log-log plot of concentrations of species in solution in equilibrium
with calcite vs. pH at constant pCO2 = 10-3.5 atm.
0
2
4
6
8
10
12
14
0
0
H+
) M
-2
-2
-4
-4
H2CO3*
-6
log Concentratio
HCO3
-8
-6
-
OH
Ca
-
-8
2+
2-
CO3
-10
-10
0
2
4
6
8
10
12
14
pH
18
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - V
In addition to the graphical solution, we have the numerical solution based on:
2 M Ca 2+ &raquo; M HCO 3
&aelig;
Kcal aH2 +
2&ccedil;
&ccedil; K1K2 KCO pCO
2
2
&egrave;
a
3
H+
(
KK
=
1
&aelig; K &ouml;
2&ccedil; cal &divide; = M HCO 3
&ccedil; M CO 2- &divide;
3
&egrave;
&oslash;
&ouml; K1KCO2 pCO2
&divide;=
&divide;
aH +
&oslash;
CO2 pCO2 ) K 2
2
2 Kcal
19
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - VI
Substituting the appropriate values for the K’s we get:
aH3 +
(
10
=
-6.35
10
10 ) 10-10.33
- 24.85
=
10
2(10-8.4 )
-1.46
-3.5 2
aH + = 10-8.28
pH = 8.28
Now based on the equation: log M 2+
Ca
we obtain
= 13.27 - 2 pH
log M Ca2+ = 13.27 - 2(8.28) = -3.30
-4
M Ca 2+ = 5.04 &acute; 10 mol L
-1
20
SOLUBILITY OF CALCITE IN AN
OPEN SYSTEM - VII
If we take into account activity coefficients, then the following expressions can
be derived (see pp. 55-57 in Kehew, 2001 for details):
a
3
H+
(
KK
=
1
pCO2 ) K2g Ca 2 +
2
CO2
2 K calg HCO 3
These equations may be used to derive the plots on the next two slides
(assuming activity coefficients are unity).
M
3
Ca 2 +
=
K1K cal KCO2 pCO2
2
4 K 2g Ca 2+ g HCO
3
21
The pH of pure water in equilibrium with calcite at 25&deg;C as a
function of the partial pressure of CO2. Note that pH decreases linearly
with increasing CO2 partial pressure.
8.5
8.0
7.5
pH
7.0
6.5
-4.0
-3.5
-3.0
-2.5
log pCO
-2.0
2
-1.5
-1.0
22
Plot of calcium concentration vs. partial pressure of CO2 for a water
in equilibrium with calcite at 25&deg;C. Mixing of two saturated waters
A and B can lead to undersaturation and calcite dissolution.
3.5
Supersaturated
3.0
A
) -1
2.5
2.0
1.5
2+ Ca L
(mmol
1.0
B
Undersaturated
0.5
0.0
0.00
0.02
0.04
0.06
pCO
0.08
0.10
2
23
LEARNING OBJECTIVES
z Understand sources of CO2 in natural waters.
z Define and understand alkalinity.
z Learn to calculate the solubility of carbonate
minerals such as calcite.
z Understand the common-ion effect.
z Become familiar with the concept of
incongruent dissolution.
z Apply these concepts to some case studies.
24
THE COMMON-ION EFFECT - I
Calcite solubility is governed by the reaction:
CaCO3(s)  Ca2+ + CO32(1)
Suppose we added a second compound containing
carbonate, and this compound is more soluble
than calcite, e.g., Na2CO3. This compound will
dissolve according to:
Na2CO3(s)  2Na+ + CO32- (2)
To the extent that reaction (2) proceeds to the
right, by Le Chatlier’s principle, this will force
reaction (1) to the left, precipitating calcite.
25
THE COMMON-ION EFFECT - II
The effect of adding sodium carbonate to the
solution can be demonstrated by adjusting the
charge-balance expression to be:
2 M Ca 2+ + M Na + &raquo; M HCO 3
By repeating the derivation of the equations on a
previous slide using this charge-balance
expression we obtain:
K1K cal KCO2 pCO2
2
M Ca 2+ (M Na + + 2 M Ca 2+ ) =
2
K2g Ca 2 + g HCO
3
Increasing Na+ concentration leads to decreased
Ca2+ concentration.
26
Ca Concentration (mmol L )
-1
2+
=
+
Na
0
-3
0
1
=
+
Na
+
m
-3
Na
10
x
=5
-2
+
Na
pCO2
(atm)
0
1
=
m
m
Figure 3-14 from Kehew
(2001). Curves showing
Ca concentration in
equilibrium with calcite
as increasing amounts of
NaHCO3 are added to
solution. Addition of the
common ion (HCO3-) in
the form of sodium
bicarbonate causes
precipitation of calcite
and a consequent
decrease in the
concentration of
dissolved Ca.
27
ANOTHER EXAMPLE OF THE
COMMON-ION EFFECT
Consider a groundwater just saturated with
respect to calcite. This water encounters a rock
formation containing gypsum.
Gypsum is more soluble than calcite; it dissolves
according to:
CaSO4&middot;2H2O  Ca2+ + SO42- + 2H2O(l)
To the extent that this reaction goes to the right,
it pushes the following reaction to the left:
CaCO3(s)  Ca2+ + CO32causing calcite to precipitate.
28
INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - I
z Incongruent dissolution - when one mineral
dissolves simultaneously with the precipitation
of another.
z Example: when calcite and dolomite are both
encountered along a ground water flow path.
z How do we determine what will happen when
both dolomite and calcite are present?
z Start by rearranging the KSP for dolomite:
(
)(
Kdol = aCa 2+ aCO2- aMg 2+ aCO23
3
)
29
INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - II
z If a solution were in equilibrium with dolomite
alone, then the activities of Ca2+ and Mg2+
would be equal so that:
(
Kdol = aCa 2+ aCO 23
)
2
(Kdol )
1
2
= aCa 2 + aCO 2 -
z At 10&deg;C we have Kdol&frac12; = 10-8.355, which is
exactly equal to Kcal = 10-8.355 for this
temperature. If dolomite had first reached
equilibrium, then calcite would not be able to
dissolve because IAP = Kcal!
3
30
INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - III
z However, at other temperatures, in general IAP
would not be equal to Kcal.
z For example, at 30&deg;C we have: Kdol&frac12; = 10-8.950,
and Kcal = 10-8.510.
z In this case calcite would dissolve, because the
ion activity product would be less than the
solubility product for calcite.
z Dissolution of calcite would then cause dolomite
to precipitate via the common-ion effect.
31
INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - IV
z The latter process would be termed incongruent
dissolution of calcite.
z At 0&deg;C we have: Kdol&frac12; = 10-8.28, Kcal = 10-8.34.
z In this case, calcite would precipitate and
dolomite would dissolve incongruently.
z We might also get incongruent dissolution
because calcite dissolves more rapidly than
dolomite. In this case, Ca2+ and CO32concentrations increase more rapidly than Mg2+,
so calcite may reach supersaturation while
dolomite is still undersaturated.
32
SOLUBILITY PRODUCTS FOR CALCITE AND
DOLOMITE IN PURE WATER AT 1 BAR
pKcal
pKdol
pKdol&frac12;
0
8.340
16.56
8.280
5
8.345
16.63
8.315
10
8.355
16.71
8.355
15
8.370
16.79
8.395
20
8.385
16.89
8.445
25
8.400
17.00
8.500
30
8.510
17.90
8.950
Temp (&deg;C)
Source: Freeze and Cherry (1979)
33
LEARNING OBJECTIVES
z Understand sources of CO2 in natural waters.
z Define and understand alkalinity.
z Learn to calculate the solubility of carbonate
minerals such as calcite.
z Understand the common-ion effect.
z Become familiar with the concept of incongruent
dissolution.
z Apply these concepts to some case
studies.
34
THE MADISON AQUIFER - I
z The Madison aquifer is located east of the Rocky
Mountains.
z In this aquifer, we can see the effects of both
the common-ion effect, and incongruent
dissolution.
z The aquifer comprises Mississippian carbonates
in which the primary minerals are calcite,
dolomite and anhydrite.
35
Predevelopment potentiometric surface of the Madison aquifer. Contours in meters
above sea level (from Plummer et al., 1990, Water Resources Research, v. 26, pp.
1981-2014).
36
THE MADISON AQUIFER - II
z Along the flow path to the northeast, dissolution
of anhydrite induces calcite precipitation via the
common-ion effect.
z Precipitation of calcite results in decreased pH
and Ca2+ concentration and increased pCO2
according to:
Ca2+ + 2HCO3-  CaCO3(s) + CO2(s) + H2O
z This results in dolomite becoming
undersaturated, so it dissolves.
37
THE MADISON AQUIFER - III
z The dissolution of dolomite is a type of
incongruent dissolution called dedolomitization.
z Sulfate concentration increases along the flow
path until saturation with respect to anhydrite,
thus sulfate serves as a measure of distance
from the recharge area.
z Calcite is close to saturation throughout the
aquifer, so very little anhydrite dissolution is
required to precipitate calcite.
38
Saturation indices of gypsum as a function of SO42- concentration for the Madison
aquifer. The relationship between these variables shows that dissolution of gypsum
/anhydrite is the source of sulfate in these waters. (Data from Plummer et al., 1990,
Water Resources Research, v. 26, pp. 1981-2014)
1.0
0.5
Equilibrium
0.0
-0.5
-1.0
-1.5
-2.0
-2.5
Saturation Index
-3.0
Gypsum
-3.5
-4.0
0
5
10
15
20
25
Sulfate (mmol Kg -1 H2O)
39
Saturation indices of calcite as a function of SO42- concentration for the Madison
aquifer. The majority of the waters are saturated to slightly oversaturated. (Data
from Plummer et al., 1990, Water Resources Research, v. 26, pp. 1981-2014)
1.0
Calcite
0.5
0.0
Equilibrium
Saturation Index
-0.5
-1.0
0
5
10
15
20
25
Sulfate (mmol Kg -1 H2O)
40
Saturation indices of dolomite as a function of SO42- concentration for the Madison
aquifer. The SI for dolomite is more variable, with undersaturation present across
the aquifer. (Data from Plummer et al., 1990, Water Resources Research, v. 26, pp.
1981-2014)
1.0
0.5
0.0
Equilibrium
-0.5
-1.0
Saturation Index
-1.5
Dolomite
-2.0
0
5
10
15
20
25
Sulfate (mmol Kg -1 H2O)
41
Precipitation of calcite by the common-ion effect as a function of anhydrite
dissolution in the Madison aquifer. The dashed line shows the trend due to
dedolomitization alone, and the solid arrow shows dedolomitization plus cation
exchange reactions.
24
2
O) H-1
Flow path 1
Flow path 2
Flow path 3
Flow path 4
Flow path 5
Flow path 6
Flow path 7
Flow path 8
20
16
Flow path 2
12
8
4
Precipitated C
0
0
2
4
6
8
10
12
14
16
18
20
22
-1
Dissolved Anhydrite (mmmol kg H2O)
42
Dissolution of dolomite by the common-ion effect as a function of anhydrite
dissolution in the Madison aquifer. The dashed line shows the trend due to
dedolomitization alone, and the solid arrow shows dedolomitization plus cation
exchange reactions.
12
2
O) H-1
Flow path 1
Flow path 2
Flow path 3
Flow path 4
Flow path 5
Flow path 6
Flow path 7
Flow path 8
10
8
Flow path 2
6
4
2
Dissolved Dolo
0
0
2
4
6
8
10
12
14
16
18
20
22
-1
Dissolved Anhydrite (mmmol kg H2O)
43
Plot of calcite precipitated vs. dolomite dissolved in the Madison aquifer. The
remarkably linear relationship demonstrates the nature of the incongruent
dissolution.
24
2
O) H-1
Y = 0.729 + 2.090X
r2 = 0.9662
20
16
12
8
4
Precipitated C
0
0
2
4
6
8
10
12
Dissolved Dolomite (mmmol kg -1 H2O)
44
IONIC STRENGTH EFFECT ON
SOLUBILITY - I
z If NaCl is added to a solution saturated with
calcite, what will happen?
z NaCl contains no ions in common with calcite,
so we would not expect solubility to decrease
from a direct common-ion effect.
z On the other hand, NaCl will increase the ionic
strength of the solution. What will this do?
z Consider the equation:
(
Kcal = aCa 2+ aCO2- = (g Ca 2+ M Ca 2+ ) g CO2- M CO 23
3
3
)
45
IONIC STRENGTH EFFECT ON
SOLUBILITY - II
(
Kcal = aCa 2+ aCO2- = (g Ca 2+ M Ca 2+ ) g CO2- M CO 23
3
3
)
z Addition of NaCl will increase ionic strength, which
in general decreases the activity coefficients.
z To keep the solubility product constant, if the
activity coefficients decrease, the concentration
terms must increase. Thus, addition of NaCl will
generally increase calcite solubility.
z Minerals tend to be more soluble in concentrated
solutions than in dilute ones (providing there is no
common-ion effect).
46
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