Interpreting profiles of pore water solutes

First, solute transport (simple)

1. Diffusive Transport:

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Flux =−φDseddCdx

2. Sediment BurialGenerally: Assume a constant mass flux (…not always true)

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massvolume

=ρsolid 1 −φ( ) ingsolid

cmsed3

Mass accumulation rate:

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MAR =wρsolid 1 −φ( ) ingsolid

cmsed2 ⋅y

Solute burial, cont.

Below the “compacting layer”

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MAR =w∞ρsolid 1 −φ( )

Which must be the same as MAR at shallower depths…

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wρsolid 1 −φ( ) =w∞ρsolid 1 −φ∞( )

w x( ) =w∞

1 −φ∞( )

1 −φ x( )( )

That’s for solids…

For solutes…

The burial rate for solutes is slightly smaller than for solids…

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v x( ) =φ∞φ x( )

w∞ AND

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F =φvC

Interpreting pore water profiles…

In general: we consider rates of change of concentration over space and time

We derive descriptive expressions by considering the mass balance in a layer of sediment

x1

x2

R

F1

F2

In the box:

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x2 −x1( )∂φC∂t

=F1 −F2 + φR x2 −x1( )

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x1 −x2( )

Interpreting, cont.

(x2 - x1) --> small…

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∂φC∂t

=−∂F∂x

+ φR

Since…

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F =−φDsed∂C∂x

+ φvC

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−∂F∂x

=∂∂x

φDsed∂C∂x

⎧ ⎨ ⎩

⎫ ⎬ ⎭−

∂∂x

φvC{ }

Interpreting, cont…

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∂∂t

φC{ } =∂∂x

φDsed∂C∂x

⎧ ⎨ ⎩

⎫ ⎬ ⎭−

∂∂x

φvC{ } + φR

Then…

Simplify: steady state ; constant ; constant D ; burial << transport:

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0 =Dsedd2C

dx2+ R

==> A simple interpretation of pw profiles

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d2C

dx2=−

RDsed

60

50

40

30

20

10

0

403020100

NO3- μ /mol l

- CVP Site M 3 NO data : & .Fit A S stoich

Interpretation of profile shapes : quantitative

Steady-state mass balance in a sediment layer: Rate of reaction within the layer = net flux out of the layer

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R = Fout − Fin

Diffusive flux :

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F = −φDseddC

dxoxic

denitrification

12

Flux at pt. 1 (x=0) : gives total, net NO3 Production in sediment column

Flux at pt. 2 : gives rate of NO3 consump.By denitrification

Sum of absolute values of Flux at 1 + Flux at 2:Gives rate of NO3 production by oxicDecomposition of organic matter

But we can get more information…

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0 =Dsedd2C

dx2+ R

What else do we need to solve this equation?

But we can get more information…

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0 =Dsedd2C

dx2+ R

Boundary conditions!

At sediment-water interface (x=0)

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C x =0( ) =Cbw

At depth in the sediments:

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dCdx

⎛

⎝ ⎜

⎞

⎠ ⎟x=x max

=0 or Cx = xmax( ) =0

What about R ?

Example : organic matter oxidation by O2

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CH 2O( )106NH 3( )16

H 3PO4( ) +138O2 ⇒ 106HCO3

− +16NO3

− + H 2PO4

− +124H + +16H 2O

What solutes could you measure to define reaction rate?

For O2, it has been convenient to use

R = P(x)i.e., with no dependence on [O2], even though it’s a reactant

Can that be justified?

Devol (1978) DSR 25, 137-146

Cultured marine bacteria from low-O2 waters…Found O2 consumption followedMichaelis-Menten kinetics:

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dO2

dt= Vm

O2[ ]Ks + O2[ ]

And found a “Critical O2 Concentration” belowWhich rate depended on [O2]Of ~ 2.4 µmol/l

What about R ?

Pore water profiles :O2 all done by in situ microelectrode profiling

2

1

0

Depth (cm)

200150100500

O2 (µmol/l)

BBay August 2003 dep_22 dep_12 dep_13

4

3

2

1

0

-1

Depth (cm)

2001000O2 (µmol/l)

616m JSL II Dive 2949-a JSL II Diev 2949 -c JSL II Dive 2949 -d Fit Range of x*

7

6

5

4

3

2

1

0

Depth (cm)

250200150100500

O2 (µmol/l)

Ceara Rise3200 m

(Hales et al., 1996)

TotalCorg ox.Rate(µmol/cm2/y)

14 45 350

Continental margin sediments:* large organic matter flux

* electron acceptors other than O2

Let’s consider a sediment dominated by sulfate reduction:

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CH 2O( )106NH 3( )16

H 3PO4( ) + 53SO4

2− →106HCO3

− +16NH 4

+ + HPO4

2− + 53HS− + 39H +

Defining P as the production rate of a solute,

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PNH 4

PTCO 2

=16

106

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PHPO 4

PTCO 2

=1

106

What would we predict pore water profiles of these 3 solutes to look like?

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0 = Dd 2Cdx 2

+ P0e− p1x

C x = 0( ) = Cbw

dCdx

x → ∞( ) = 0

Solution:

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C = Cbw +P0

p1

2D1− e− p1x( )

Solve the equation for each solute:

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PTCO 2 = P0e− p1x

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PNH 4 =16106

PTCO 2

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PHPO 4 =1

106PTCO 2

Assume porosity = 0.8 and Dsed = Dsw x (porosity)2 … then

DHCO3- = 323 cm2/y , DNH4+ = 543 cm2/y, DHPO42- = 208 cm2/y

Assume P0 = 100,p1 = 0.2

25

20

15

10

5

0

800050002000TCO2 (µmol/l)

5000NH4 (µmol/l)

1000PO4 (µmol/l)

Plotting the concentration of one solute vs. another…

600

400

200

0

8000600040002000TCO2 (µmol/l)

s = 0.08979

100

80

60

40

20

0

8000600040002000TCO2 (µmol/l)

s = 0.01465

Interpreting the slopes:At any depth,

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FTC = −DTC

dTCdx

FN = −DN

dNdx

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FTC

FN

=DTC

dTCdx

DNdN

dx~

FTC

FN

=DTC

DN

×ΔCΔN

Therefore, the slopes imply

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PN

PTC

=16106

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PP

PTC

=1

106

A mineral,undersaturated in seawater

apparently simple dissolution kinetics…

What do we expect [Si(OH)4] in pore water to look like?

Depth

ConcentrationCbw Csat

Diagenesis of a solid Undersaturated in bottom water Asymptotic approach to saturation in pore water

CBW CSATConcentration

Diagenesis of a solid, undersaturated in bw Asymptotic approach to

Saturation in pore water

Csat = 100-120

Csat = 600

Csat = 550-830Csat = 500-750

N. Atlantic (Bermuda)

PeruMargin

Southern Ocean

Observations: Si(OH)4

In pore waters