mineral dissolution/precipitation to determine whether or not a water is saturated with an...
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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:
• from Gibbs free energies of formation. The IAP could then be determined from a water analysis, and the saturation index calculated.
4
3
443
H
SiOHAlK
a
aaaK
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.
log aH4SiO40
-6 -5 -4 -3 -2 -1
log
(aK
+/a
H+)
0
1
2
3
4
5
6
7
KaoliniteGibbsite
Muscovite
K-feldspar
Pyrophyllite
Qua
rtz
Am
orph
ous
silic
a
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.
2 3 4 5 6 7 8 9 10 11 12-10
-8
-6
-4
-2
0
2
pH
log
a A
l++
+
Al+++
Al(OH)2
+
Al(OH)4-
AlOH++
Gibbsite
25oC
Greg Mon Nov 01 2004
Dia
gram
Al+
++,
T =
25
C,
P
= 1
.013
bar
s, a
[H
2O
] =
1
Aluminum
–10 –8 –6 –4 –2 00
2
4
6
8
10
12
14
log a Fe+++
pH
Fe+++
Fe(OH)2+
Fe(OH)3
Fe(OH)4-
Fe2(OH)2++++
FeOH++
Fe(OH)3(ppd)
25°C
Lab user Fri Oct 02 2009
Dia
gram
Fe+
++,
T
= 2
5 °C
, P
=
1.
013
bars
, a
[H2O
] =
1;
Sup
pres
sed:
FeO
(c),
He
mat
ite,
Go
ethi
te
P cycling and sediment microbes
Blue Green Algae blooms?
FeOOH
PO43-
PO43-PO4
3-Iron reducers
H2SFeS2
PO43- PO4
3-
PO43-PO4
3-
Sulfate Reducers
Fe2+
N cycling and sediment microbes
NO3- NH4
+
Ammonifyingbacteria
Linking nutrient cycling and redox chemistry in fresh water lakes
Solid State Analysis
• X-Ray Fluorescence Spectroscopy (XRF)
• X-ray Diffraction (XRD)
Extractions• Using selected acids, bases, organics, chelators
to dissolve a subset of minerals• Many possibilities for this• We will look at 2 used to investigate Fe, Mn, P:
– Aqua Regia – Combination of Nitric Acid and Hydrochloric acid, heated to just below boiling – dissolved most stuff (but not all!!)
– Ascorbic Acid extraction – Combination of ascorbic acid and bicarbonate
FeOOH/MnOOH in sediments
• Ascorbic acid extraction technique specific to ‘reactive’ Fe and Mn (Kostka and Luther, 1994)
• XRD data confirms reactive stuff is ferrihydrite
• Size is nanloparticulate, sensitive to degree of recycling!
Where is the Phosphorus?
2008 Sediment Extraction Correlation Statistics
Ascorbic Acid (reactive) Aqua Regia (total)
R2 p-value n R2 p-value n
P vs. Fe 0.8952 6.4 x 10-100 202 0.1940 3.5 x 10-10 185
P vs. Mn 0.6561 3.1 x 10-48 202 0.3026 5.0 x 10-16 185
P vs. Ca 0.6895 2.7 x 10-27 184 0.1709 5.0 x 10-9 185
P vs. Al 0.4653 1.5 x 10-26 184 0.6883 3.4 x 10-48 185
P mobility measured through hundreds of sediment digests for samples in Missisquoi Bay collected at different times, places, and depths
Only strong and consistent correlation is with reactive Fe, present predominantly as nanoparticulate FeOOH particles
0 2 4 6 8 10 12 14 160
0.2
0.4
0.6
0.8
1
1.2
f(x) = 0.0626224969662733 x + 0.119484657750947R² = 0.89520492809295
2008 Reactive Fe vs P
Fe mg/g dry sed.
mg
P/g
dry
sed
.0.050 0.100 0.150 0.200 0.250 0.300 0.350 0.400 0.450 0.500 0.550
0.000
0.100
0.200
0.300
0.400
0.500
0.600
0.700
f(x) = 0.4726378589848 x + 0.251869077652964R² = 0.568113478655601
f(x) = NaN x + NaNR² = 0 2007 Reactive Fe & Mn vs Reactive P
Fe vs PLinear (Fe vs P)P vs. Mn
0.00 10.00 20.00 30.00 40.00 50.00 60.000.00
0.50
1.00
1.50
2.00
2.50
f(x) = 0.505676950561494 x + 0.69559540238016R² = 0.632482196728678
f(x) = NaN x + NaNR² = 0 2008 Total Fe and Mn vs P
mg Mn or Fe/ g dry sed.
mg
P/g
dry
sed
.
p < 0.001
0.00 10.0020.0030.0040.0050.0060.0070.0080.0090.000.00
0.50
1.00
1.50
2.00
2.50
f(x) = 0.616797706312578 x + 0.681734924140724R² = 0.65365053479501f(x) = 0.026036614142036 x + 0.08102698241454
R² = 0.909017783071506
2007 Total Fe vs Total P
T Fe vs. P
mg Mn or Fe/g dry sed
mg
P/g
dry
sed