Problem Set 7 12.742 Answer Key 1.a) U is conservative in ...

Problem Set 7 12.742 Answer Key
1.a)
234 U is conservative in seawater we can calculate the expected flux of 230 Th at any
depth based solely on our knowledge of the decay rate of 234 U and the depth of the water.
At steady state the production rate of 230 Th must equal its loss rate. The production rate
is equal to the decay of 234 U (or its activity), and the loss is equal to decay of 230 Th and
the scavenging rate. The decay rate of 230 Th is much slower than the scavenging rate,
therefore for 230 Th production equals scavenging, and the flux is simply equal to the
depth times the production rate:
Production = 234 A U
=1.14 ∗ 238 A U
( )
Loss : 230 Th ∗ λ 230
+ 230 Th∗ k scav
= 230 Th λ 230
+ k scav
λ 230
>decay,
Production=Scavenging). This is again true for any model.

Problem Set 7 12.742 Answer Key
Production Rate ∗ z
[ particles]= ω
ω = 1.14 ∗238 A U
∗ λ 230
∗z
particles [ ]
⎛
ω⎜
⎝
m ⎞
⎟ = 1.14 ∗2400dpm ∗ 9.24 ×10 −6 y −1 ∗ z
m 3
y⎠
⎛
[ particles] ⎜ dpm ⎞
⎟
⎝ m 3 ⎠
⎛
ω⎜
⎝
m ⎞ ⎛
⎟ = 0.0253
y
dpm ⎞
⎜
⎠ ⎝ m 3 ⎟ ∗
⋅ y⎠
z
particles
⎛
⎜
[ ] ⎝
m4 dpm
To make sure that we get an accurate sinking rate (ω) let’s calculate the sinking rate a
few different locations. One thing I’ll point out here is when converting the particle
concentrations from dpm/kg to dpm/m 3 we are talking about the density of 230 Th particles
in the water column. Not the density of what is caught in the trap. We can pull this
information of the total small particle concentration graph, then convert to dpm/m 3 using
the trap information. This makes a huge difference! I’ve done the calculations in the
excel spreadsheet. You can see that the sinking rate ranges from 42m/y at 300m to
500m/y at 3700m. Then it drops lower in the deepest samples. Why is this? Well, if we
think about how 230 Th works it is really a deep water scavenging isotope – so it isn’t good
for the upper water column. So, let’s not use the samples shallower than 1000 m. Then
we have the deepest samples. All the Th that has been scavenged out of the ocean
eventually ends up in the sediments, and can get stirred back up into the bottom waters.
So let’s not bother with the samples from the lowest depths (4500 and deeper) that are
clearly influenced by the bottom. So we are left with the middle samples, and they have
an average sinking rate of about 325 m/y.
1.b) To estimate the forward and backward scavenging rates (k 1 and k -1 , respectively)
we must use two isotopes. 230 Th and 228 Th are both affected by the same scavenging
rates, because they are the same species, chemically. They are different only in terms of
their decay constants. Therefore, we can set up scavenging equations for both isotopes
that contain k 1 and k -1 as the unknowns, and by combining the two equations solve for
each of the constants. To start, set up the equations and solve each for k 1 :
⎞
⎟
⎠

Problem Set 7 12.742 Answer Key
Now, solve for k -1 :
A 230 ( k
Th(d) 1
+ λ 230 )= P + k 230
Th −1A 230
Th(p)
k 1
>>> λ 230
A 230
Th(d) k 1
= P 230
Th + k −1A 230
Th(p)
k 1
= P 230 Th + k −1A 230
Th(p)
A 230
Th(d)
A 228 ( k
Th(d) 1
+ λ 228 )= P + k 228
Th −1A 228
Th(p)
k 1
= P 228 Th + k −1A 228
Th(p) − λ 228A 228
Th(d)
A 228
Th(d)
P 230
Th
A 230
Th(d)
+ k −1A 230
Th(p)
A 230
Th(d)
= P228 Th
+ k −1A 228
Th(p)
− λ 228
A 228 A
Th(d)
228
Th(d)
k −1
A 230
Th(p)
A 230
Th(d)
− k −1A 228
Th(p)
A 228
Th(d)
= P228 Th
− λ 228
− P230 Th
A 228 A
Th(d)
230
Th(d)
P 228
Th
A 228
Th(d)
k −1
=
− λ 228
− P 230 Th
A 230
Th(d)
A 230
Th(p)
A 230
Th(d)
− A 228 Th(p)
A 228
Th(d)
Now, all we need to know are the activities of the defined species (from the table) and
calculate the production rate of 228 Th (just like for 230 Th, but we using 228 Ra (d) as its
parent). This is all calculated in the excel spreadsheet. To calculate k 1 we just plug and
chug our value of k -1 into the above equation(s). Again, this is done in the excel
spreadsheet. The answers are dependant upon which depths you use (as you can see from
the spreadsheet), and I haven’t graded based on your numerical answer, rather on your
methodology.
1.c) Now we are dealing with a system where the Th is scavenged onto small particles,
but then the small particles aggregate into large particles, and only the large particles
sink. Our equations are set just as they were previously, with production equaling loss:

Problem Set 7 12.742 Answer Key
Production = Loss
dA D
dt
dA S
dt
= λ D
A P
− λA D
− k 1
A D
+ k −1
A S
+ β −1
A S
= k 1
A D
+ β −2
A L
− k −1
A S
− β −1
A S
− β 2
A S
− λA S
dA L
dt
= β 2
A S
− β −2
A L
− λA L
−ω ∂A L
∂z
steady state : dA
dt = 0
λ D
A P
+ ( k −1
+ β −1 )A S
= ( λ + k 1 )A D
k 1
A D
+ β −2
A L
= ( k −1
+ β −1
+ β 2
+ λ)A S
β 2
A S
= ( β −2
+ λ)A L
+ ω ∂A L
∂z
1.d) If β 2 and β- 2 are 3 y -1 and 150 y -1 , respectively, then particles must disaggregate
much faster than they aggregate. This indicates that at any moment most of the particles
are in the small form. By knowing these two values we are able to calculate the particle
sinking rate, and thus the removal of organic matter from the ocean:
β 2
A S
= ( β −2
+ λ)A L
+ ω ∂A L
∂z
ω = β A − β 2 S ( + λ −2 )A L
∂A L
∂z
To understand how this affects OM cycling let’s simplify and say that new production is
primarily fueled by the replenishing of nutrients through upwelling of nutrient-rich deep
water. Deep water nutrient concentrations are dependant upon the remineralization of
sinking organic matter, and this is dependant on the aggregation/disaggregation rates of
particles. Thus, on some long time scales, new production is dependant on the
aggregation/disaggregation rates of particles.
2.a) This is another Th problem! Why so many Th problems? Because it an
extremely useful element for measuring ocean processes, not only are its isotopes in U/Th
decay series, but they are particle reactive, and the different isotopes have a large range
of decay rates, allowing us to measure lots of different time scales. In this case we are
comparing the relatively long-lived isotopes of 230 Th (t 1/2 = 75000y) to another particle
reactive isotope, 231 Pa. Th is more susceptible to scavenging (more particle reactive) than
Pa, therefore Th has a shorter residence time in the water column. Consequently, Pa is
more likely to be affected by advection then Th, and more of it can then be removed from
the central gyre regions to areas of high scavenging rates (like the continental margins).
By comparing these two isotopes in sediment traps we can get a handle on how much of
each is vertically scavenged, laterally advected, and how well the sediment traps do in
capturing falling OM.

Problem Set 7 12.742 Answer Key
First, we need to calculate the amount of scaveneged Th and Pa the traps
collected. To do this we need to take the amount of each in the trap and subtract off the
amount of Th and Pa present in secular equilibrium with their parent isotopes ( 234 U and
235 U, respectively). We do this because we are only interested in the ‘excess’ Th and Pa
– the equilibrium Th and Pa trapped in the mineral matrix came from the decay of U
trapped in the matrix, and was never dissolved in the ocean, and therefore was never
scavenged. For Th, this is easy, 238 U is give to us, so:
Scavenged 230 Th = 3.00 dpm g − 0.20dpm g = 2.80dpm g
In this case we can assume that 234 U and 238 U are in secular equilibrium. If you make the
calculations (as below for 235 U) using the crustal abundance you will find that 234 U =
1.01* 238 U = 1.01*0.20 = 0.20 (the same within the significant figures we have available
to us).
To calculate the 235 U in the trap we use the known 235 U/ 238 U ratio and the decay
constants. The ratios given in the HBCP are atom ratios – so you can’t just multiply by
the activities! (Note: 238 U is essentially 1, which is why I only used 0.0072. It is more
rigorous is to use 0.0072/0.99275, but the answer is the same.)
235 U 238 = 0.0072 (by atoms)
U 235 A
238 A = ⎛ λ ⎞ ⎛
235
⎜ ⎟ N ⎞ ⎛
235
9.85 ×10−10 ⎞
⎜ ⎟ = ⎜
⎟ 0.0072 = 0.046
⎝ ⎠ ⎝ ⎠ ⎝ 1.551×10 −10
⎠
λ 238
N 238
235 A = 0.046 ∗ 235 A = 0.046 ∗ 0.20 dpm g = 9.15 ×10−3 dpm g
And now we can calculate the scavenged 231 Pa:
Scavenged 231 Pa = 0.125 dpm g − 9.15 ×10−3 dpm g = 0.1158dpm g
Since the trap was deployed for one year, the fluxes of 230 Th and 231 Pa:
Scavenged Flux = Amount Scavenged dpm 10.2
⎛
⎜
⎞
g ⎟
y
⎝ g
×
⎠ 0.5 m 2
Scavenged Flux ( 230 Th) = 2.80 dpm g × 10.2 g y
0.5 m 2 ≅ 57.1 dpm m 2 ⋅ y
Scavenged Flux ( 231 Pa) = 0.1158 dpm g × 10.2 g y
0.5 m 2 ≅ 2.36 dpm m 2 ⋅ y
2.b)
There are 4 equations that we can use to solve this problem:

Problem Set 7 12.742 Answer Key
P Th
=V Th
+H Th
P Pa
=V Pa
+H Pa
R V
= V Th
V Pa
R H
= H Th
H Pa
These equations are combined to solve for the vertical fluxes as follows:
(
V Th
= P Th
-R H
⋅ P Pa )R V
R V
− R H
( )
V Pa
= P Th
-R H
⋅ P Pa
R V
− R H
R H
= 2
R V
= V Th
V Pa
= 57
2.36 ≈ 24
P Th
= λ 230
∗ A 234
U = 9.22 ×10−6 y −1 ∗2.75 dpm L 103 L
m
3
( ) = 2.52 ×10−2 dpm m 3 y
P Pa
= λ 231
∗ A = 2.13 235
U ×10−5 y −1 ∗0.108 dpm L 103 L
( m
3) = 2.30 dpm ×10−3 m 3 y
⎛
2.52 ×10 −2 dpm m 3 y − 2⋅ 2.30 dpm ⎞
⎜
×10−3 m 3 ⎟ ∗24
⎝
y⎠
V Th
=
= 2.25 ×10 −2 dpm 24 − 2
m 3 y
⎛
2.52 ×10 −2 dpm m 3 y − 2 ⋅ 2.30 dpm ⎞
⎜
×10−3 m 3 ⎟
⎝
y⎠
V Pa
=
= 9.36 ×10 −4 dpm 24 − 2
m 3 y
The production (P) of thorium and protactinium is not a flux – it is a concentration per
unit time. You then need to use the depth of the trap to determine the V values, but the P
values are technically not fluxes. It works out ok here, because things cancel, but in
many other cases you do not want to multiply by depth unless you are calculating a flux.
The fraction of each isotope removed by vertical flux and the fraction removed by lateral
transport are as follows:

Problem Set 7 12.742 Answer Key
f 230
= V 2.25 ×10 −2 dpm
Th
m 3 y
=
P Th 2.52 ×10 −2 dpm = 0.89
m 3 y
f 231
= V 9.36 ×10 −4 dpm
Pa
m 3 y
=
P Pa 2.33×10 −3 dpm = 0.40
m 3 y
The Th removal by vertical flux is 89% and by lateral is 11%. The Pa removal by
vertical flux is 40% and by lateral is 60%.
2.c)
The trapping efficiency of the sediment trap is calculated as:
E= F V
Flux from sediment trap
F=
depth of sediment trap
E Th
= F 57.1
Th
= 3000 = 0.84
−2
V Th
2.24 ×10
E Pa
= F 2.36
Pa
= 3000 = 0.84
−4
V Pa
9.33×10
Since the trap is collects 84% of the vertical flux, then we know that the actual amount of
material falling at 3000 m is:
Flux collected in trap = Acutal Flux∗ Trapping Efficiency
10.2 g y
Actual Flux = ÷ 0.84 ≅ 24 g 0.5 m 2 y⋅ m 2
3.a) DIC increases as a water mass ages, both from the remineralization of OM and
the dissolution of calcium carbonate. These two process differ in that remineralization
typically requires the consumption of O 2 (except in anoxic water parcels) and dissolution
of CaCO 3 does not. Also, CaCO 3 dissolution increases the alkalinity of the water parcel,
and OM remineralization does not. These differences allows us calculate the DIC from
remineralization of OM and from the dissolution of CaCO 3 .
The Apparent Oxygen Utilization (AOU) is used with the Normal Atmospheric
Equilibrium Concentration (NAEC) to stoichiometrically calculate the OC remineralized:

Problem Set 7 12.742 Answer Key
AOU = NAEC - [ O 2 ] in situ
AOU = 331.8 μmol kg − 250μmol kg = 81.8μmol kg
⎛
OC remin. = AOU × C ⎞
⎜ ⎟
⎝ O 2 ⎠
Redfield
OC remin.= 81.8 μmol kg × ⎛ 117 ⎞
⎜ ⎟ ≈ 56 μmol ⎝ 170⎠
kg
To calculate the number of moles of CaCO 3 remineralized we can use the
relationship between CaCO 3 dissolution and alkalinity: for 1 mole of CO 3 2- added to the
water from CaCO 3 , 2 moles of alkalinity are added.
CaCO 3
dissolved = 1 2 ( CA meas
-CA pref )
CA = [HCO 3 - ] +2[CO 3 2- ]
TA = [HCO 3 - ] +2[CO 3 2- ] +[OH - ] +[B(OH) 4 - ] -[H + ] +minor species
( )
CA pref
= TA pref
− [OH - ] +[B(OH) 4 - ] -[H + ]
pH = 8.2, S = 35‰ :
TA pref
= 547.05 + ( 50.56 × S)= 2316.7 μmol kg
[H + ]=10 -8.2 mol kg
[OH - ]=10 -5.8 mol kg
[B(OH) 4 - ] = K B
∗11.9 ×10 −6 ∗ S
K B
+ [H + ]
= 10−8.91 ⋅10 −8.91 ⋅ 35
10 −8.91 +10 −8.2 = 66.1 μmol kg
CA pref
= 2316.7 μmol kg −10-5.8 mol kg
− 66.1 μmol kg +10-8.2 mol kg
= 2249 μmol kg
CaCO 3
dissolved = 1 ⎛
⎜ 2
2300 μmol ⎝ kg − 2249μmol ⎞
⎟
kg
≈ 25 μmol ⎠ kg
3.b) To calculate the preformed nutrient we use the AOU and Redfield ratios to
determine the nutrients produced through OM remineralization, and then subtract those
numbers from the measured (in situ) concentrations. For DIC we do basically the same
thing, but we also have to account for the DIC added during CaCO 3 dissolution.

Problem Set 7 12.742 Answer Key
⎛
NO - 3
recycled = AOU × N ⎞
⎜ ⎟
⎝ O 2 ⎠
Redfield
NO - 3
recycled = 81.8 μmol kg × ⎛ 16 ⎞
⎜ ⎟ = 7.7 μmolN ⎝ 170⎠
kg
NO - 3
preformed = 21.5 μmol kg − 7.7μmol kg =13.8μmol kg
⎛
PO 3- 4
recycled = AOU × P ⎞
⎜ ⎟
⎝ O 2 ⎠
Redfield
PO 3- 4
recycled = 81.8 μmol kg × ⎛ 1 ⎞
⎜ ⎟ = 0.48 μmolP ⎝ 170⎠
kg
PO 3- 4
preformed =1.45 μmol kg − 0.48μmol kg = 0.97μmol kg
DIC preformed = DIC meas
- DIC OM
− DIC CaCO3
DIC preformed = 2180 μmol kg − 56μmol kg − 25μmol kg ≈ 2100μmol kg
3.c) The Ca 2+ added from CaCO 3 dissolution is only a small addition to the total Ca 2+
present in seawater. Ca 2+ is conservative in seawater at ~10mmol/kg. Therefore, the
increase of 25µmol/kg only represents a change of ~0.25%.
3.d) In order to use up the remaining O 2 , the amount of OC we will need to
remineralize will be (assuming perfect Redfield stoichiometry):
⎛ ⎞
⎟
⎝ 170⎠
[ C org ]= ⎜
117
Redfield
∗ 250 μmol O 2
kg
=172 μmol C org
kg
3.e) To deplete the remaining nitrate we can do the same thing. But before we start
multiplying by ratios – what is the remaining NO - 3 ? It is very important to remember that
the oxic remineralization of OM in part d also generated NO - 3 . How much?
⎛ ⎞
⎟
⎝ 170⎠
−
[ NO 3 ] = 16
added
⎜
Redfield
∗ 250 μmol O 2
kg
= 23.5 μmol C org
kg
−
−
−
[ NO 3 ] = NO 3 remaining
[ ] + NO 3 measured
[ ] = 21.5 + 23.5 = 45 μmol
added kg
Now to determine the appropriate ratio for denitrification. We are only considering
carbohydrates, so that certainly makes life easier. The equation we need to use will be:
10CH 2 O + 8HNO 3 ⇔ 10CO 2 + 4N 2 + 14H 2 O
⎛ 10 mol C⎞
OC oxidized by denitrification = ⎜ ⎟ × 45 μmolN ⎝ 8 mol N ⎠ kg = 56.3μmolC kg

Problem Set 7 12.742 Answer Key
However, we know that organic matter is, on average, more reduced than a simple
carbohydrate. If the molecular formula is changed to CH 2.72 O to reflect this we have the
equation:
10CH 2.72 O + 9.44HNO 3 ⇔ 10CO 2 + 4.72N 2 + 18.32H 2 O
⎛ 10 mol C ⎞
OC oxidized by denitrification = ⎜ ⎟ × 45 μmolN ⎝ 9.44 mol N⎠
kg = 47.7μmolC kg
The amount of OC required to deplete all the NO 3 - thus decreases. This shouldn’t be a
surprise, since the OC is more reduced each molecule requires more of the oxidant (in
this case NO 3 - ) for full oxidation.
COMMON ERRORS ON Q3:
CA preformed = TA preformed
This is a bad assumption in this case! If we make this assumption then CA pref >
CA meas , indicating CaCO 3 precipitation. Remember – this water parcel is deep water –
there is no CaCO 3 formation going on since it has left the surface, and probably only
dissolution (pH drops).
DIC pref = DIC meas + DIC Corg + DIC CaCO3
This statement is entirely false. To account for the remineralization of OC and
the dissolution of CaCO 3 these terms must be subtracted from the measured
concentration. (However, if you made the CA=TA error I did accept a +DIC CaCO3
because it incorrectly calculates to an subtraction of DIC from the preformed
concentration.)
The last question I read incorrectly while grading, so there are some scribbles at the
bottom of some of your pages. Sorry about that!