Lecture No. 19. Determining Geochemical Background II

Let's take another look at the mixing equation for two conservative species, X and Y, from two sources, A and B.

This is again the equation of a linear plot of Y_M against X_M, with slope [(Y_A – Y_B)/(X_A – X_B)] and intercept [(X_AY_B – X_BY_A)/(X_A – X_B)].

Now let’s say that A is a suspected source of contamination and B is the background. The slope term can be thought of as representing (mainly) the ratio of Y to X in the contamination source, and the intercept represents mainly the background concentrations. As we go down the river and encounter more tributaries, f_A keeps getting smaller and smaller. X_M and Y_M should approach closer and closer to X_B and Y_B, their background values, as the water gets more and more diluted. If we plot Y_M vs. X_M for different points M₁, M₂, M₃, etc. along the river and get a straight line, then X and Y are conservative.

The overhead shows a plot of copper vs. lead for two different contamination sources. As sediment from source A is diluted with sediment of background concentration B, its concentrations will follow the dilution line from A to B. Similarly, as sediment source C is diluted with sediment B, its concentrations will follow the B-C dilution line. A sample with concentrations D comes from source A.

Example: This calculation method is valid for water as well as sediment, as long as the elements analyzed are conservative. Consider calcium and strontium levels in the Great Lakes, comparing Lake Superior water to Lake Huron water. Lake Superior is well-mixed. Lake Huron is well-mixed at some distance from the input of Lake Superior water at Sault Sainte Marie. In the region near Sault Sainte Marie, Superior water is mixing with Huron water. Samples were taken at various points out in Huron, farther and farther away from the inflow. Let Lake Huron = A and Lake Superior = B.

The calcium and strontium data were plotted.

The calcium concentration was a linear function of the strontium concentration. All data yielded the equation

[Ca²⁺] = 0.178 [Sr²⁺] + 9.81 R² = 0.989

Lake Superior had about 14 ppm Ca²⁺, while Lake Huron had 26 ppm Ca²⁺. Now consider a point not too far from Sault Sainte Marie where [Ca²⁺] = 22 ppm.

or

At the point where [Ca²⁺] is 22 ppm, the water is 2/3 Lake Huron water. At this point, the equation tells us [Sr²⁺] is 68.5 ppb.

Limitations to This Approach

All of these calculations are based on several assumptions. If any of the assumptions is not true (or approximately true), these calculations do not work. The assumptions:

1. Everything must act conservatively.
2. Sources are constant. The level of each element entering the water from the source point per volume of water must remain constant.
3. The system must be well-mixed.
4. The differences have to be big enough to measure. About the best resolution we can expect to get from our samples is ± 10%, so the source concentrations have to be at least 20-30% higher than background concentrations, preferably higher than that, for these calculations to be useful.

Fingerprinting the Sources

We can convert the Y_M-X_M equation into logarithmic form if the intercept term (which represents background concentrations) is small compared to the (slope « X_M) term and can therefore be neglected.

A plot of log Y_M vs. log X_M will have a slope of one and an intercept equal to log [(Y_A–Y_B)/(X_A–X_B)]. This is extremely useful, since (1) the intercept is the log of the ratio of Y enrichment over X enrichment, and (2) it lets us make use of sediment data even far downstream at high dilutions.

Consider the following plot:

There are three sets of sediment analyses. A and B are from two upstream source areas. C is a set of analyses of downstream sediment samples. All 3 sets are elevated in copper and lead. Where did the copper and lead in set C come from?

For any given source, the equation tells us that log [Cu] and log [Pb] fall on a line as they are diluted with (more or less) pure water. This would mean that B is the source of the copper and lead in C.

This use of the log-log diagram is valid as long as all concentrations are more than about 10« background. The box in the lower left corner delineates background samples, because its limits are the background copper and lead concentrations. The dotted lines mark off the area (to upper right) where Pb and Cu are both more than 10« background and the log-log approximation applies. In that region, we can tell at a glance whether a given contaminant assemblage came from a given source.

The next overhead is a log-log plot of lead and copper in the Clark Fork system. Notice how the Clark Fork and Silver Bow Creek analyses follow an essentially linear trend on this plot. Flint creek (FC) falls on a line, but at a higher Pb-Cu ratio. The Little Blackfoot also seems to contain some contamination at a Pb-Cu ratio between those of Flint Creek and the Clark Fork. Gold Creek is near background in both metals, and Rock Creek is near or maybe somewhat lower than the overall basin background.

Returning to the linear formula for mixing ratios, suppose we apply it to the copper and lead in the Clark Fork River at Milltown. If sediment is sampled from the Clark Fork River above and below Flint Creek and from Flint Creek above its confluence with the Clark Fork, the lead concentrations are about 256 ppm in Flint Creek sediments, 82 ppm in the Clark Fork above Flint Creek, and 135 ppm in the Clark Fork below Flint Creek. We can plug these values into the mixing equation:

f_A = 0.305

We can then calculate how much of the lead downstream from the confluence comes from Flint Creek.

Pb (Flint Creek) = 0.305« 256 = 78.0

And the lead from Flint Creek is 100% « (78/135) = 58% of the total downstream lead.

In principle, then, we can calculate how much of a contaminant came from each of two sources. But there is a problem with this approach. Analyses of sediment samples, even of samples taken from the same place, can vary significantly. One reason is that the metal-laden materials can be diluted with greater or lesser amounts of siliciclastic sediments. Averaging of a large number of individual samples can help, but there is a better way to go about it. The ratio of the concentrations of two metals varies much less than the concentrations themselves do.

The next overhead is a plot of the log of the Cu-Pb ratio against river miles (increasing upstream.) The copper and lead analyses themselves vary all over the map from one sediment sample to another. To see a difference between triplicate copper or lead numbers between two different locations, the concentrations would have to be 20-30% different. However, the ratio of the two metals shows a pronounced step change at about 415 miles — the confluence of Flint Creek and the river.

Let’s derive a formula for calculating element partitioning based on lead-copper ratios. We start with the linear formula for lead in sediment. Let A = Flint Creek, B = the Clark Fork above Flint Creek, and M = the mixed sediment in the Clark Fork below Flint Creek.

Now multiply both sides by 1/[Cu]_M:

Next, multiply the first term on the right by [Cu]_A/[Cu]_A and the second term by [Cu]_B/[Cu]_B. (This is equivalent to multiplying each term by 1.)

Finally, rearrange the variables:

We now have an equation for the lead:copper ratio of the mixed sediments in terms of the lead:copper ratios of the two component sediments and the ratios of the upstream (unmixed) and downstream (mixed) copper concentrations. Since the lead:copper ratios vary much less from sample to sample than the individual metal concentrations do, we will generally find that the random error in our calculations is smaller. The ratio of upstream copper to mixed copper is obtained from the averages of several sediment copper concentrations taken at each location.

Example: Flint Creek and the Clark Fork River

Now we calculate lead:copper ratios separately for each sediment sample from each stream:

Next we calculate mean copper concentrations for the three sediments.

The upstream copper: mixed copper ratios are then 0.10 for Flint Creek and 0.85 for the Clark Fork.

Next, plug the values for these ratios into the equation and solve for f_A:

0.199 = f_A« 3.80« 0.10 + (1-f_A)« 0.153« 0.85

0.199 = 0.38« f_A + 0.13 – 0.13« f_A

0.069 = 0.25« f_A

f_A = 0.28

So 28% of the sediment comes from Flint Creek and 72% comes from the upper Clark Fork. Now we can calculate how much of the downstream lead and copper come from each source. Mean sediment lead concentrations are 256 ppm in Flint Creek and 82 ppm in the upper Clark Fork.

Lead from Flint Creek = 0.28« 256 ppm = 71.7 ppm

Lead from the upstream Clark Fork = 0.72« 82 ppm = 59 ppm

Thus Flint Creek’s contribution to the lead downstream is 71.7/(71.7+59)« 100% = 55%.

Copper from Flint Creek = 0.28« 67 ppm = 19 ppm

Copper from the upstream Clark Fork = 0.72« 574 ppm = 413 ppm

Flint Creek’s contribution to the copper downstream is 19/(19+413) = 4%.

The result tells us that little Flint Creek is supplying only 4% of the downstream sediment copper but —perhaps surprisingly — supplies over half of the downstream lead in sediment.

Note that the metals must be acting conservatively for this method to work. For lead, copper, and many other metals, this means the samples must be taken far enough downstream for the pH to have settled in at steady, usually near-neutral, values. Manganese is likely to be non-conservative, because it can exist in lots of different forms with different solubilities. Sulfur is guaranteed to be non-conservative. Arsenic is questionable.

Assignment: Problem set passed out in class. Due in one week.

Here are some total production figures from Butte (as of January 1994) and estimates of remaining reserves (also as of 1994):

Quite a bit of arsenic was also produced, but it is hard to find hard figures for that element.

The 1994 estimates of reserves were based on 547« 10⁶ T of ore containing 0.33% Cu, 0.32% Mo, and 0.72 oz/T Ag. Note that most of the copper is already mined, but that most of the molybdenum is still in the ground, and that about 1/3 of the silver is still in the ground. The deposit now being mined is in the Continental Pit, east of the Continental Fault. Porphyry deposits come in two common varieties: copper-gold and copper-molybdenum. The underground mines and the Berkeley Pit, both west of the fault, worked a copper-gold deposit poor in molybdenum. The ore east of the fault is a copper-molybdenum ore very poor in gold. It does look richer in silver, however.

Birnessite and Arsenic

The oxidation of As (III) to As (V) by manganese oxide minerals is important to the process of altering arsenic toxicity in aquatic environments, since the toxicity of As (V) is about 60 times less than that of As (III). As (III) is a protoplasmic poison that disrupts sulfur linkages on proteins. The mechanism of As (V) toxicity apparently stems from disruption of phosphorus compounds (DNA, RNA, phospholipids, etc.). (This is somewhat complicated by the fact that arsenic can be reduced or oxidized by metabolic processes, but in general As (III) is much more toxic than As (V).)

Birnessite (named for Birness, Scotland) is a frequently encountered manganese oxide in modern sediments and soils, and plays an important role in the Mn-promoted As (III) oxidation.

Birnessite is not a clay mineral, but it is similar in its platy, layered structure. No birnessite grains exist that are large enough for single-crystal x-ray diffraction or other crystallographic study.

Jones and Milne first identified birnessite as a distinct mineral in 1956. They found it in manganese-rich pans at Birness. They equated it to d -MnO₂. The found its formula to be (Na_0.7Ca_0.3)Mn₇^3.8+O₁₄ á 2.8 H₂O.

In 1970 Giovanoli proposed a sheet structure for birnessite consisting of MnO₆ octahedra joined along their edges, and with 1 out of every 6 manganese positions empty. Each manganese atom would therefore have 2 oxygen atoms to itself, plus another 4 oxygens, each of which was shared with 3 other manganese atoms, for an overall oxygen to manganese ratio of 3:1. Removal of every sixth manganese and its two axial oxygens would give the sheet an overall formula of Mn₅O₁₆. If all the sheet manganese atoms are in the +4 oxidation state, the empirical formula of the sheet is MnO_3.2 ^2.4–. The remaining manganese, assumed to be Mn³⁺ cations, would be in the interlayers above and below the vacant positions, octahedrally coordinated by water molecules and the axial oxygen atoms. The remaining cations (Ca, Na, ¹ ) would be held somewhere in the layers between the manganese oxide sheets and the interlayer water sheets. The interlayer thickness d (001) would be about 7.2 �.

Giovanoli's work yielded two slightly different formulae for birnessite than Jones and Milne's work: Na₄Mn₁₄^3.3+O₂₅ á 11 H₂O and Na₄Mn₁₄^3.7+O₂₈ á 8 H₂O, depending on the dehydration temperature used. Giovanoli settled on an intermediate formula, Na₄Mn₁₄^3.6+O₂₈ á 9 H₂O. There are other formulae that might make sense, but they are all in the same family.

Charge balance in birnessites of different compositions is probably achieved by substitution of some Mn(II) or Mn (III) in place of some of the Mn (IV) atoms in the sheets and by varying the number and type of M⁺ and M²⁺ ions in the interlayers. Interlayer cations may include K⁺, Na⁺, Ca²⁺, Mn^x+, Ba²⁺, and/or Cs⁺. Natural birnessite has Na, Ca, K, and Mn ions.

The high layer charge density of birnessite gives it a high cation exchange capacity: 240 mEq/100 g for sodium birnessite.

Investigations previous to the Moore, Walker, and Hayes paper, of As (III) oxidation to As (V), had identified birnessite as an electron-acceptor but had not addressed the role played by its crystal structure. For example, Oscarson et al. showed it as an MnO₂ ¨ Mn²⁺ reaction. They saw the scant release of Mn²⁺ as evidence of complexation or possibly a precipitate of Mn₃(AsO₄)₂.

To previous lecture: Lecture No 18. Determining Geochemical Background

To next lecture: Lecture No. 20. Pit Lakes

Back to list of lecture notes