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

344

3

H

SiOHAlK

aaaa

K

INCONGRUENT DISSOLUTION

• Aluminosilicate minerals usually dissolve incongruently, e.g.,

2KAlSi3O8 + 2H+ + 9H2O Al2Si2O5(OH)4 + 2K+ + 4H4SiO4

0

• 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

+ /aH

+ )

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

=

25

°C ,

P

= 1

.013

bar

s, a

[H

2O]

=

1; S

uppr

esse

d: F

eO(c

), H

emat

ite,

Goe

thite

P cycling and sediment microbes

Blue Green Algae blooms?

FeOOHPO4

3-

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 0.5500.000

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