Transcript Bjerrum plot showing the activities of inorganic carbon species as... function of pH for a value of total inorganic carbon...

Bjerrum plot showing the activities of inorganic carbon species as a
function of pH for a value of total inorganic carbon of 10-3 mol L-1.
-2
Common pH
range in nature
6.35
H2CO3*
-3
-
HCO3
10.33
2-
CO3
log ai
-4
OH-
-5
+
H
-6
-7
-8
0
2
4
6
8
10
12
14
pH
In most natural waters, bicarbonate is the dominant carbonate species!
SPECIATION IN OPEN CO2-H2O
SYSTEMS - I
• In an open system, the system is in contact with its
surroundings and components such as CO2 can migrate
in and out of the system. Therefore, the total carbonate
concentration will not be constant.
• Let us consider a natural water open to the atmosphere,
for which pCO2 = 10-3.5 atm. We can calculate the
concentration of H2CO3* directly from KCO2:
KCO2 
M H 2CO3*
M H 2CO3*  pCO2 KCO2
pCO2
log M H 2CO3*  log pCO2  log KCO2
Note that M H2CO3* is independent of pH!
SPECIATION IN OPEN CO2H2O SYSTEMS - II
• The concentration of HCO3- as a function of pH is next
calculated from K1:
K1 
M HCO  aH 
M HCO  
3
M H 2CO3*
K1M H 2CO3*
aH 
3
but we have already calculated M H2CO3*:
so
M H 2CO3*  pCO2 KCO2
M HCO  
aH 
 log K1KCO2 pCO2  pH
3
log M HCO 
3
K1KCO2 pCO2


SPECIATION IN OPEN CO2H2O SYSTEMS - III
• The concentration of CO32- as a function of pH is next
calculated from K2:
K2 
M CO 2  aH 
M CO 2 
3
M HCO 
3
3
K 2 M HCO 
3
aH 
but we have already calculated M HCO3- so:
M HCO  
3
K1KCO2 pCO2
aH 
and
M CO 2 
3
K 2 K1KCO2 pCO2
aH2 
log M CO 2  log K2 K1KCO2 pCO2   2 pH
3
SPECIATION IN OPEN CO2H2O SYSTEMS - IV
• The total concentration of carbonate CT is
obtained by summing:
CT  M H 2CO3*  M HCO   M CO 2
3
CT  pCO2 KCO2 
K1 pCO2 KCO2
aH 
3

K1K 2 pCO2 KCO2
a
2
H




K
K
K
log CT  log  pCO2 KCO2 1  1  12 2 
 a 

a


H
H


Plot of log concentrations of inorganic carbon species H+ and OH-,
for open-system conditions with a fixed pCO2 = 10-3.5 atm.
pK1
pK2
log concentration (molar)
0
CO32-
-2
CT
H+
OH-
-4
H2CO3*
-6
HCO3-
-8
2
3
4
5
6
7
pH
8
9
10
11
12
Plot of log concentrations of inorganic carbon species H+ and OH-,
for open-system conditions with a fixed pCO2 = 10-2.0 atm.
pK1
pK2
log concentration (molar)
0
H+
-2
-4
CO32CT
H2CO3*
OH-
-6
HCO3-
-8
2
3
4
5
6
7
pH
8
9
10
11
12
Methods of solving equations that
are ‘linked’
• Sequential (stepwise) or simultaneous methods
• Sequential – assume rxns reach equilibrium in
sequence:
• 0.1 moles H3PO4 in water:
–
–
–
–
H3PO4 = H+ + H2PO42pK=2.1
[H3PO4]=0.1-x , [H+]=[HPO42-]=x
Apply mass action: K=10-2.1=[H+][HPO42-] / [H3PO4]
Substitute x  x2 / (0.1 – x) = 0.0079  x2+0.0079x-0.00079
= 0, solve via quadratic equation
– x=0.024  pH would be 1.61
• Next solve for H2PO42-=H+ + HPO4-…
Calcite Solubility
• CaCO3 -> Ca2+ + CO32- log K=8.48
• We consider minerals to dissolve so that 1 Ca2+
dissolves with 1 CO32• If dissolving into dilute water (effectively no Ca2+ or
CO32- present): x2=10-8.48, x= aCa2+ = aCO32• If controlled by atmospheric CO2, substitute CO32for expression
log M CO 2  log K2 K1KCO2 pCO2   2 pH
3
• What happens in real natural waters??
Charge Balance
• Principle of electroneutrality  For any solution, the
total charge of positively charged ions will equal the
total charge of negatively charged ions.
– Net charge for any solution must = 0
• Charge Balance Error (CBE)
m z  m

CBE 
m z  m
c c
a
za
c c
a
za
– Tells you how far off the analyses are (greater than 5% is
not good, greater than 10% is terrible…)
• Models adjust concentration of an anion or cation to
make the charges balance before each iteration!
Using Keq to define equilibrium
concentrations
DG0R = -RT ln Keq
K eq 
n
[
products
]

i
n
[
reactants
]

i
• Keq sets the amount of ions present relative
to one another for any equilibrium condition
AT Equilibrium
K eq  Q 
n
[
products
]

i
n
[
reactants
]

i
Speciation
• Any element exists in a solution, solid, or
gas as 1 to n ions, molecules, or solids
• Example: Ca2+ can exist in solution as:
Ca++
Ca(H3SiO4)2
Ca(O-phth)
CaB(OH)4+
CaCH3COO+
CaCO30
CaCl+
CaF+
CaH2SiO4
CaH3SiO4+
CaHCO3+
CaNO3+
CaOH+
CaPO4CaSO4
CaHPO40
• Plus more species  gases and
minerals!!
How do we know about all those
species??
• Based on complexation  how any ion
interacts with another ion to form a
molecule, or complex (many of these are
still in solution)
• Yet we do not measure how much
CaNO3+, CaF+, or CaPO4- there is in a
particular water sample
• We measure Ca2+  But is that Ca2+ really
how the Ca exists in a water??
Aqueous Complexes
•
Why do we care??
1. Complexation of an ion also occuring in a
mineral increases solubility
2. Some elements occur as complexes more
commonly than as free ions
3. Adsorption of elements greatly determined
by the complex it resides in
4. Toxicity/ bioavailability of elements depends
on the complexation
Defining Complexes
• Use equilibrium expressions:
0
0
0
0
D
G

n
G
(
products
)

n
G
i i
 i i (reactants )
DG R = -RT ln Keq
R
i
• cC + lHL  CL + lH+
c
 n
[CL] [ H ]
i 
c
l
[C ] [ HL ]
• Where B is just like Keq!
i
Mass Action & Mass Balance
c
 n
[CL] [ H ]
i 
c
l
[C ] [ HL ]
mCa   mCa L
2
2 n
x
• mCa2+=mCa2++MCaCl+ + mCaCl20 + CaCL3- +
CaHCO3+ + CaCO30 + CaF+ + CaSO40 +
CaHSO4+ + CaOH+ +…
• Final equation to solve the problem sees the
mass action for each complex substituted into
the mass balance equation
Mineral dissolution/precipitation
• To determine whether or not a water is saturated with
an aluminosilicate such as K-feldspar, we could write a
dissolution reaction such as:
• KAlSi3O8 + 4H+ + 4H2O  K+ + Al3+ + 3H4SiO40
• We could then determine the equilibrium constant:
aK  a Al3 aH3 4 SiO4
K
4
aH 
• from Gibbs free energies of formation. The IAP could
then be determined from a water analysis, and the
saturation index calculated.
INCONGRUENT
DISSOLUTION
• Aluminosilicate minerals usually dissolve
incongruently, e.g.,
2KAlSi3O8 + 2H+ + 9H2O
 Al2Si2O5(OH)4 + 2K+ + 4H4SiO40
• As a result of these factors, relations among
solutions and aluminosilicate minerals are often
depicted graphically on a type of mineral
stability diagram called an activity diagram.
ACTIVITY DIAGRAMS: THE
K2O-Al2O3-SiO2-H2O SYSTEM
We will now calculate an activity diagram for the
following phases: gibbsite {Al(OH)3}, kaolinite
{Al2Si2O5(OH)4}, pyrophyllite
{Al2Si4O10(OH)2}, muscovite
{KAl3Si3O10(OH)2}, and K-feldspar {KAlSi3O8}.
The axes will be a K+/a H+ vs. a H4SiO40.
The diagram is divided up into fields where only
one of the above phases is stable, separated by
straight line boundaries.
6
Quartz
7
Amorphous
silica
Activity diagram showing the stability relationships among some
minerals in the system K2O-Al2O3-SiO2-H2O at 25°C. The dashed
lines represent saturation with respect to quartz and amorphous silica.
Muscovite
log (aK+/aH+)
5
K-feldspar
4
3
Gibbsite
Kaolinite
2
1
Pyrophyllite
0
-6
-5
-4
-3
log aH SiO 0
4
4
-2
-1
Seeing this, what are the reactions these lines represent?
2
0
Gibbsite
Al
+++
, T = 25 C , P = 1. 013 bars , a [ H 2O] = 1
-4
-
-6
Al(OH)4
-8
++
+++
+
AlOH
Al(OH)2
Diagram A l
log a Al
+++
-2
25oC
-10
2
3
4
5
6
7
pH
8
9
10
11
12
Greg Mon Nov 01 2004