影響碳酸鹽沉澱的因素
離子濃度與活度
pH值與CO2分壓的影響
溫度效應
生物、蒸發作用
同離子效應
離子強度
非共溶性
ACTIVITY AND ACTIVITY
COEFFICIENTS
In thermodynamic expressions, the activity takes
the place of concentration. (an ‘effective’
concentration)
 Activity and concentration are related:
ai = i·Mi
where ai is the activity, Mi is the concentration and
i is the activity coefficient.
 In dilute solutions, i  1, so ai  Mi. However, in
concentrated solutions activity and concentration
may be far from equal.

pH and Carbonic acid
H2O = H+ + OH-, Kw = aH+ *aOHpH = -logKw
 CO2 + H2O = H2CO3 , KH = aH2CO3/fCO2
 水中的電離作用:
 H2CO3 = H+ + HCO3- ,
Ka1 = aH+*aHCO3-/ aH2CO3
 HCO3- = H+ + CO32- ,
Ka2 = aH+*aCO3-/ aHCO3
Range of pH values in the natural
environment



Most natural waters have pH between 4-9.
The acids are usually weak, including carbonic acid and organic acids (e.g.
fulvic and humic)
pH values > 8.5 are rare, occurring only in evaporitic lakes, lakes clogged with
photosynthetic plants, and springs discharging from serpentine or ultramafic
rocks.
pH值的影響

Most reactions in gas/water/rock systems involve or are
controlled by pH:





Aqueous acid-base equilibria, including hydrolysis and
polymerization.
Adsorption, because protons compete with cations and hydroxyl
ions compete with anions for adsorption sites. Also, the surface
charge of most minerals is pH dependent.
The formation of metal ligand complexes, because protons
compete with metal ions to bond with weak-acid ions, and OHcompetes with other ligands that would form complexes.
Oxidation-reduction reactions, whether abiological or biologically
mediated. Oxidation usually produced protons, whereas
reduction consumes them.
The solubility rate of dissolution of most minerals is strongly pHdependent. Weathering of carbonate, silicate, and aluminosilicate minerals consumes protons and releases metal cations.
Acidity酸度


Capacity of water to give or donate protons
Contributions are:
 Immediate acidity – species present in solution:
 Strong and weak acids
 Salts of strong acids and weak bases
 Hydrolysis of Fe3+ and Al3+
 Oxidation and hydrolysis of Fe2+ and Mn2+
 Long-term acidity
 Due to reactions of water with solids in system


Acidity gives water a greater capacity to attack
geological material and is usually accompanied
by high total dissolved solids (TDS), including
hardness (the sum of the concentrations of the
multivalent cations, such as Ca2+ and Mg2+).
Acidity increases the solubility of hazardous
substances such as heavy metals and it is
corrosive and toxic to fish and other aquatic life.
Alkalinity碱度





Alkalinity is the capacity of water to accept protons.
Due primarily to bicarbonate ion (HCO3-) and to a
minor extent, the carbonate ion (CO32-).
 Carbonate alkalinity = HCO3- + 2 CO32The contribution of OH- is important above pH=10.
Other bases that contribute to the total alkalinity are:
 Ligands of fulvic acid
 Organic anions such as formate, acetate, and
propionate
 Bisulfides, orthophosphates, ammonia, and
silicates.
Usually reported as mg/L CaCO3 or meq/L CaCO3
Conversions转换常数
• Equivalent wt. = Formula wt. / charge of ion
• Na+ = 22.99g/1 = 22.99g
• Ca2+ = 40.078g/2 = 20.039g
• meq/l = mg/l / (equivalent wt. in g)
• 10 mg/l Na+ / 22.99 = 0.435 meq
• 10 mg/l Ca2+ / 20.039 = 0.499 meq
• Molality (m) = (mg/l * 10-3) / (formula wt. in g)
• 10 mg/l Na+ * 10-3 / 22.99 = 4.35 x 10-4 m
• 10 mg/l Ca2+ * 10-3 / 40.078 = 9.98 x 10-4 m
• Molality (m) = (meq/l * 10-3) / (valence of ion)
CO2分壓的影響 Solubility of CO2
K w  a H  aOH 
Gas
K a1 
CO2
CO2
K a2 
a HCO  a H 
3
a H 2CO3
aCO  a H 
3
a HCO 
3
Water
KH 
a H 2CO3
f CO2
mH   mOH   mHCO   2  mCO 
3
3
Constant PCO2
2
1
固定CO2分壓
0
-1
logC
-2
H
OH
HCO3
H2CO3
CO3
TotCO3
-3
-4
-5
-6
-7
-8
2
3
4
5
6
7
pH
8
9
10
11
12
Constant Total carbonate
2
1
固定含碳總離子濃度
0
-1
水體中含碳離子隨pH的變化
logC
-2
H
OH
HCO3
H2CO3
CO3
TotCO3
-3
-4
-5
-6
-7
-8
2
3
4
5
6
7
pH
8
9
10
11
12
Solubility of Calcite in Water
CaCO3 = Ca2+ + CO32Water
Cc
Closed system:
No CO2 exchange
Ksp
K w  a H  aOH 
K a1 
a HCO  a H 
3
a H 2CO3
K a2 
aCO  a H 
3
a HCO 
3
K sp  aCO  aCa 2 
3
2  mCa 2   mH   mOH   mHCO   2  mCO 
3
3
mCa 2   CT  mH 2CO3  mHCO   mCO 
3
3
碳酸鹽溶解度受pH與CO2分壓的影響
Calcite in Water
1
各种碳酸盐共生
0
-1
-2
log C
-3
H
OH
Ca
H2CO3
HCO3
CO3
-4
-5
-6
Soly
-7
-8
2
3
4
5
6
7
pH
8
9
10
11
12
CT=mH2CO3 + mHCO3- + mCO3=
= Constant
Calcite Sol'y Constant CT
1
0
-1
-2
H
OH
Ca
H2CO3
HCO3
CO3
CT
log C
-3
-4
-5
-6
-7
-8
2
3
4
5
6
7
pH
8
9
10
11
12
Solubility of Calcite Open to CO2
Gas
CO2
CO2
Cc
Water
K w  a H  aOH 
K a1 
a HCO  a H 
3
a H 2CO3
K a2 
aCO  a H 
3
a HCO 
3
K sp  aCO  aCa 2 
3
2  mCa 2   mH   mOH   mHCO   2  mCO 
3
KH 
a H 2CO3
f CO2
3
Open System碳酸鹽溶解
Open System Dissolution of Calcite
1
0
-1
-2
H
OH
Ca
H2CO3
HCO3
CO3
CT
log C
-3
-4
-5
-6
-7
-8
2
3
4
5
6
7
pH
8
9
10
11
12
Comparison of Open and Closed
System Equilibrium
System
pH
pCa
pH2CO3
pHCO3
pCO3
pCO3
Closed
9.9
3.9
7.65
4.05
4.4
3.9
Open
8.26
3.34
4.97
3.06
5.13
3.05
溫度效應
VARIATION OF LOG K WITH
TEMPERATURE
 The
solubility product is determined at
T = 25°C.
K is a function of temperature and cannot be
used at temperatures < or > 25°C
 How do we then find solubility relationships at
other temperatures?

The following is a generally valid
relationship:
rG° = rH° - TrS°
If we assume that rH° and rS° are
approximately constant (true over a limited
temperature range), then because:
o
  rG
log K 
2.303RT
we can write
  r H o  T r S o
log K 
2.303RT
THE VAN’T HOFF EQUATION
As an alternative, we can use the Van’t Hoff
equation:
If we assume again that rH° is approximately
constant, we can write the expression:
r H  1 1 
  
log KT2  log KT1 
2.303R  T1 T2 
o
We can calculate rH° and rS° according to:
o

H
 f 
r H o 
products
r S o 
o
S
 
products
o

H
 f
reactants
o
S

reactants
Example

Determine the solubility product of calcite at 40°C







CaCO3  Ca2+ + CO32From tables: Species ΔHf°(kJ/mol)
ΔSf° (kJ/mol)
Ca2+
-543.0
-0.0562
CO32CaCO3
-675.2
-1207.4
-0.0500
0.09197
ΔHr° = ΔHf°products - ΔHf°reactants
ΔHr° = -543.0 -675.2 –(-1207.4) = -10.8
ΔSr° = ΔSf°products - ΔSf°reactants
ΔSr° = -0.0562 - 0.0500 –(0.09197) = -0.19817
Using ln Keq = ΔHr° - T ΔSr° / -RT

Log K = -10.8 – (313.15)(-0.19817) / 2.303(-8.3143  10-3)(313.15) = -8.56
Log K = -8.56
K = 10-8.56

So what is the point?
For calcite K40 = 10-8.56 and K25 = 10-8.37
 Calcite is less soluble at 40°C than at 25°C
 Implications?

碳酸鹽溶解度受溫度的影響
logK
Temperature Variation of Ksp
-8
-8.2
-8.4
-8.6
-8.8
-9
-9.2
-9.4
-9.6
-9.8
-10
0
20
40
60
t°C
calcite
aragonite
dolomite
80
100
Temperature Variation
25°C
10°C
log P
-3.5
-2.5
-1.5
-3.5
-2.5
-1.5
Ca2+
20
44
100
25
55
127
HCO3
58
131
298
74
166
380
pH
8.29 7.62 6.97 8.31 7.65 7.00
綜合效應
Relationship of pH and PCO2 at calcite saturation
9.00
8.50
pH
7.50
溶解态
7.00
6.50
6.00
5.50
5.00
-5.00
-4.50
-4.00
固态
正常大气二氧化碳分压
8.00
-3.50
-3.00
-2.50
log(PCO2)
-2.00
-1.50
-1.00
-0.50
0.00
Calcite solubility as a function of PCO2
3.50
3.00
Calcite Precipitates
[Ca] (mM)
2.50
2.00
B
S
1.50
C
Calcite Dissolves
1.00
A
0.50
0.00
0.000
0.010
0.020
0.030
0.040
0.050
PCO2
0.060
0.070
0.080
0.090
0.100
Exsolution/Dissolution
CO2逸出使得碳酸鈣沉澱
Exsolution of CO2
Log[Ca2+]
B
A
E´
C
D
E
Dissolution of CO2
CO2溶解使得碳酸鈣溶解
Log PCO2
生物Photosynthesis/Respiration
光合作用使得碳酸鈣沉澱
Log[Ca2+]
Photosynthesis
B
A
Respiration Decay
D
E
呼吸作用使得碳酸鈣溶解
Log PCO2
Log[Ca2+]
Evaporation蒸发
B
A
Log PCO2
Common Ion Effect同離子效應
Log[Ca2+]
C
B
A
Log PCO2
IONIC STRENGTH - I
Recall that activity and concentration are
related through the activity coefficient
according to:
ai = i·Mi
 Activity coefficients different from unity
arise because of the interaction of ions as
concentration rises.
 The degree of ion interaction depends on
ionic charge as well as concentration.

IONIC STRENGTH - II

Ionic strength (I ) is a quantity that is
required to estimate activity coefficients. It
takes into account both concentration and
charge:
2
1
I

2
M z
i i
The calculation of ionic strength must take
into account all major ions:
I  12 [ M Na   4 M Ca 2  4 M Mg 2  M HCO  M Cl   4 M SO2 ]
3
4
Ionic Strength III

Calculation example:



A river water has the following composition:
Calculate the Ionic Strength:
ion
mg/l
234
First, convert mg/l to molality Ca2+
I
1
2
M z
2
i i
m/l
0.0058
4
Mg2+
39
0.0016
HCO3-
290
0.0047
5
SO42-
498
0.0051
8
H4SiO4
48
0.0066
7
I = ½(0.00584 * 22 + 0.0016 * 22 + 0.00475 * 12 + 0.00518 * 22)
I = 0.0276
DEBYE-HÜCKEL EQUATION
 Az I
log  i 
1  Ba o I
2
i
Used to calculate activity coefficients for
ions at ionic strengths < 0.1 mol L-1.
 A, B are functions of temperature and
pressure and are given in Geochemistry
tables.
 ao is the distance of closest approach and
it is a property of the specific ion.
 ionic charge, zi.

DEBYE-HÜCKEL
PARAMETERS
T(C)
A
B(108)
0
0.4883
0.3241
5
0.4921
0.3249
10
0.4960
0.3258
15
0.5000
0.3262
20
0.5042
0.3273
25
0.5085
0.3281
30
0.5130
0.3290
40
0.5221
0.3305
50
0.5319
0.3321
60
0.5425
0.3338
DISTANCES OF CLOSEST
APPROACH FOR SELECTED
IONS
Ion
a0 (10-8)
Ion
a0 (10-8)
Ca2+
5.0
HCO3-, CO32-
5.4
Mg2+
5.5
NH4+
2.5
Na+
4.0
Sr2+, Ba2+
5.0
K+, Cl-
3.5
Fe2+, Mn2+, Li+
6.0
SO42-
5.0
H+, Al3+, Fe3+
9.0
Example Calculation



Given the analysis we used
earlier to calculate the Ionic
strength:
I = 0.0276
What is the activity of Ca2+?
 Azi2 I
log  i 
1  Ba o I
ion
mg/l
m/l
Ca2+
234
0.00584
Mg2+
39
0.0016
HCO3-
290
0.00475
SO42-
498
0.00518
H4SiO4
48
0.00667
log i = (- 0.5085 * (2)2 * (0.0276)1/2) / (1+ (0.3281 x 108)*(5 x 10-8)* (0.0276)1/2)
log i = -0.3379 / 1.2725
log i = - 0.2655
i = 0.543
THE DAVIES EQUATION


I
log  i   Az 
 0.2 I 
1 I

2
i
Used to calculate activity coefficients for
ions at ionic strengths < 0.5 mol L-1.
 The value of A is the same as the one
employed in the Debye-Hückel equation.
 The advantage over the D-H equation is
that the only ion-dependent parameter is
the ionic charge, zi.

ACTIVITY COEFFICIENTS
1.2
H+
-
HCO3
Activity coefficient, 
1.0
+
Na
+
K , Cl
0.8
0.6
0.4
Mg2+, CO322+
2-
Ca , SO4
0.2
0.0
0.2
0.4
0.6
Ionic strength (I)
0.8
1.0
1.2
EFFECT OF ACTIVITY COEFFICIENTS
ON GYPSUM SOLUBILITY
Question: what is the solubility of gypsum in pure water at
25°C and 1 bar? For the equilibrium:
CaSO4·2H2O(s)  Ca2+ + SO42- + 2H2O(l)
we can calculate log KSP = -4.41. We can also see that, if
gypsum dissolves in pure water, then the stoichiometry
of the reaction is such that the molarity of calcium
should equal the molarity of sulfate.
Thus, solubility of gypsum = MCa2+.
So what’s the problem? Catch 22! We don’t know the
concentrations, so we can’t calculate the ionic strength,
so we can’t calculate the activity coefficients, so we can’t
calculate the concentrations!
WHAT TO DO?
We start by making an initial assumption that the activity
coefficients are equal to 1 and solve the problem by
iteration.
We write:


K  aCa 2 aSO2   Ca 2 M Ca 2   SO2 M SO2  104.41
4
4
4
but because we assume activity coefficients are equal to 1,
we write:
2
4.41
K  M Ca 2 M SO2  M Ca

10
2
4
This is what the solubility would be if we ignored activity
coefficients altogether.
M Ca 2  102.205  6.24  103
THE NEXT STEP
Now, having the concentration of Ca2+ and SO42-, we can
calculate the ionic strength according to:
I
1
2
46.24 10   46.24 10   2.5 10
3
3
2
mol L
Applying the Debye-Hückel formula we get:
 Ca   SO  0.548
2
2
4
which are about half the originally assumed values. We
calculate a new estimate for the molality of Ca2+:
4.41
M Ca 2
1
10
3

 11.40  10
0.548  0.548
This is used to calculate a new ionic strength and the
whole process is repeated until convergence.
RESULTS OF ITERATIONS
Iteration
1
2
3
4
5
6
7

1
0.548
0.476
0.460
0.454
0.452
0.453
MCa2+
6.2410-3
11.4010-3
13.1110-3
13.5610-3
13.7510-3
13.7810-3
13.7810-3
I
2.5010-2
4.5610-2
5.2410-2
5.4210-2
5.5010-2
5.5110-2
5.5110-2
FINAL ANSWER
The final calculated molarity of Ca2+ is
13.7810-3.
 This is 2.21 times the calculated molality of
Ca2+ assuming activity coefficients are unity.
 We see that activity coefficient corrections
are very important for this solution.
 It is customary to express solubility in g L-1 of
gypsum:
(13.7810-3 mol L-1)(172.1 g mol-1) = 2.37 g L-1

Charge Balance

More than 90% of all the dissolved solids
in waters can be attributed to eight ions:

Ca2+, Mg2+, Na+, K+, Cl-, SO42-, HCO3-, and CO2Direct analyses can be done for the first six.
HCO3-, and CO2- are determined by titrating with an acid to an
endpoint of ~4.4





This is reported a total alkalinity
The proportion of carbonate and bicarbonate are calculated using the
sampling temperature and pH.
To check on the analyses a cation-anion balanced in performed.
Charge Balance

To perform a charge balance
Convert all ionic concentrations to equivalents
per liter
 Sum charges on cations zmc
 Sum charges on anions zma
 Determine charge balance error:

 CBE%
= zmc - zma / zmc + zma
Common Ion Effect
A groundwater saturated with calcite
encounters a rock formation containing
gypsum
 Gypsum is more soluble than calcite. It
dissolves according to;

 CaSO4•2H2O

↔ Ca2+ + SO42- + 2H2O
To the extent this reaction goes to the right, it
pushes the following reaction to the left:
↔ Ca2+ + CO32 Resulting in the precipitation of calcite
 CaCO3
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.
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   M HCO 
3
By repeating the derivation of the equations on a
previous slide using this charge-balance
expression we obtain:
K1Kcal KCO2 pCO2
2
M Ca 2 M Na   2 M Ca 2  
2
K2 Ca 2  HCO

3
Increasing Na+ concentration leads to decreased
Ca2+ concentration.
-1
Ca Concentration (mmol L )
+
Na
=
0
-3
0
1
=
+
Na
m
-3
+
Na
=
1
x
5
0
-2
+
2+
Na
=1
pCO2 (atm)
0
m
m
Figure 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.
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:
CaSO4·2H2O  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.
INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - I




Incongruent dissolution - when one mineral
dissolves simultaneously with the precipitation of
another.
Example: when calcite and dolomite are both
encountered along a ground water flow path.
How do we determine what will happen when
both dolomite and calcite are present?
Start by rearranging the KSP for dolomite:


Kdol  aCa 2 aCO 2 aMg 2 aCO 2
3
3

INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - II

If a solution were in equilibrium with
dolomite alone, then the activities of Ca2+
and Mg2+ would be equal so that:

K dol  aCa 2 aCO 2

3

2
Kdol 
1
2
 aCa 2 aCO 2
3
At 10°C we have Kdol½ = 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!
INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - III
However, at other temperatures, in general
IAP would not be equal to Kcal.
 For example, at 30°C we have: Kdol½ = 108.950, and K
-8.510.
=
10
cal
 In this case calcite would dissolve,
because the ion activity product would be
less than the solubility product for calcite.
 Dissolution of calcite would then cause
dolomite to precipitate via the common-ion
effect.

INCONGRUENT DISSOLUTION OF
CALCITE AND DOLOMITE - IV




The latter process would be termed incongruent
dissolution of calcite.
At 0°C we have: Kdol½ = 10-8.28, Kcal = 10-8.34.
In this case, calcite would precipitate and
dolomite would dissolve incongruently.
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.
SOLUBILITY PRODUCTS FOR CALCITE
AND DOLOMITE IN PURE WATER AT 1 BAR
Temp (C)
0
5
10
15
20
25
30
pKcal
8.340
8.345
8.355
8.370
8.385
8.400
8.510
Source: Freeze and Cherry (1979)
pKdol
16.56
16.63
16.71
16.79
16.89
17.00
17.90
½
dol
pK
8.280
8.315
8.355
8.395
8.445
8.500
8.950
Dolomite
Dolomite dissolution
CaMg CO3 2  2CO2  2 H 2O  Ca 2  Mg 2  4 HCO3
Calcite - dolomite equilibrium
2CaCO3  Mg 2  CaMg CO3 2  Ca 2
K
aCa 2 
aMg 2 

K sp2 cc
K spdol

mCa 2 
mMg 2 
Mg-Calcite
Cation replacement reaction
CaCO3calcite  Mg 2  MgCO3calcite  Ca2
 aMgCO3

 aCaCO
3


 aMg 2

 K Mg Ca 
 a 2

 Ca
calcite




Stoichiometric saturation
Mg xCa1 x CO3  xMg 2  1  x Ca2  CO32
 a Mg 2 
K sp  
 a 2
 Ca
x

 a 2 a 2
 Ca CO3

Ca2+ + 2HCO3- → CaCO3 + H2O + CO2
牡蠣的效用何在
餐食的用途
剩餘的用途 : 外殼因富含氮素
之故製造蠔殼粉，當有機肥中
氮素的主要原料,也可以賣給
飼料廠當飼料的原料，也有池
塘養殖人家買回去，飄撒在池
塘內以改善水質