Lecture No. 18. Determining Geochemical Background

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1. Assigned Reading: Poster: Steve Helgen, "Determining natural background conditions in a contaminated river system by modeling the downstream dilution of trace metals in fine-grained sediments" on the wall outside Room SC304.

2. Assigned Reading: Moore, Johnnie N., Jeffrey R. Walker, and Thomas H. Hayes. 1990. "Reaction scheme for the oxidation of As (III) to As (V) by birnessite," Clays and Clay Minerals 38(5), 549-555. On reserve, Mansfield Library.

3. Recommended Reading: Hawkes, H.E. 1976. "The downstream dilution of stream sediment anomalies," Journal of Geochemical Exploration 6(3), 545-558.

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Consider now the determination of background concentrations in various media.

A. Water

Consider the water contacting a sulfide orebody or a sulfidic waste. Here are some important reactions. Some occur abiotically, and some are promoted by Thiobacillus species.

4 FeS₂ + 14 O₂ + 4 H₂O ¨ 8H⁺ + 8 SO₄^2– + 4 Fe²⁺ (abiotic)

4 Fe²⁺ + O₂ + 10 H₂O ¨ 4 Fe(OH)₃ + 8 H⁺ (abiotic or T. ferrooxidans)

4 Fe²⁺ +4 H⁺ + O₂ ¨ 4 Fe³⁺ + 2 H₂O (T. thiooxidans)

14 Fe³⁺ + FeS₂ + 8 H₂O ¨ 2 SO₄^2– + 15 Fe²⁺ + 16 H⁺ (abiotic)

The first 2 reactions occur at higher pH (> 2.5). The third and fourth occur after enough acid has built up to keep iron (III) in solution.

The net yield of acid, as we have seen before, is 4 mols of acid per mol of pyrite. The pH controls the phase the metal is in — solute or particulate phase.

Metals, sulfate, and pH of water tell us a lot.

Natural erosion on an unmined site will produce acid, sulfate, and probably elevated levels of metals. The water in naturally mineralized zones is commonly between pH 2.6 and pH 4.0. Common metal concentrations might be

The Question of Dispersion

The real question — and the tough one — is how far away the metal anomaly in water gets. If we look again at the paper on the Blackfoot River by Moore et al., we see that, in a natural system (or a system not too badly disturbed), metal concentrations should drop fairly rapidly. Dissolved concentrations drop faster than sediment concentrations, but both drop off reasonably quickly.

Example of a Small Unmined Ore Deposit: Limerick, Ontario

The main ore body has about 2.9 tons of 0.69% Ni and 0.2% copper ore, or ~ 4.2 tons of lower-grade ore (0.57% Ni, 0.17% Cu). In water, the following concentrations of metals are found in solution (ppb; BD = Below Detection).

Another Example of an Unmined Deposit: Agricola Lake System, NWT

Mean tributary values for zinc (for tributaries with no contact with orebody) were ~ 14 ppb. The mainstem stream reaches background in about 2.6 km.

Counterexample: Butte

When the mines were operating, the water 100 km downstream had pH 5.8 and high metals. Here are some more recent data:

B. Soil

Now consider soil around or above a weathering orebody. In an unmined site, the anomaly is usually from weathering. Physical transport and chemical/hydrological transport both come into action. Physical weathering extends just a little bit downslope. Groundwater transport can extend the anomaly a bit farther. The result of groundwater transport is a separate anomaly where the metal-bearing groundwater intersects the surface.

A useful measure is the area of soil dispersion. Here are some pre-mining data from orebodies in Canada.

Soil dispersion from unmined ore is seldom more than 300%.

In Butte, the Cu-Ag vein system was about 4 km². The total deposit (including the porphyry (disseminated low-grade) deposit, was about 60 km². Maybe 100 km² of elevated soil concentrations existed. We cannot study that now, because of the mining and (especially) smelting that have gone on.

Anaconda now has about 1260 km² contaminated above background. Butte proper has about 200 km² of contamination (of which some is from Anaconda). Nearly all the local dispersion is by airfall. The airfall distribution from Anaconda was generally SW to NE, but with a 6-20 mile radius crosswind. Butte got some of it.

C. Sediments

Consider a drainage system, with an orebody somewhere in it.

Particulates (mostly oxidized gossan) are eroded and transported. Because of downstream dilution by clean tributaries, sediment concentration shows a fall-off with downstream distance. For background, one could look at the mainstem upstream of the orebody or at clean tributaries. However,

1. If mining activity is going on, airfall will affect upstream values, and
2. In any case, a clean tributary does not represent an undisturbed orebody.

See handouts on dispersion trains. Note that in most cases the enrichment in sediment drops off very rapidly in unmined areas. Also note the plot showing the Clark Fork and three other mineralized drainages. The sediment concentrations in the Clark Fork are high and stay high for over 100 miles. The next overhead lists concentrations of As, Cd, Cu, Pb, and Zn in various sediments in the western US, and the next one lists global abundances for various elements in soil, sediment, shale, granite, and the overall crust. The crustal abundance for aluminum is somewhat higher than the figure usually accepted, and the one for arsenic is much lower than the commonly accepted 5 ppm.

Determining Dilution

This method was developed by Hawkes (1976). (See reading assignment.) It applies to concentrations of metals in sediments.

Look again at the drainage basin containing an orebody.

A_o ¼ area of orebody

C_o ¼ concentration of orebody

A_t ¼ area of drainage basin

C_b ¼ background concentration (rest of drainage basin)

and

Me ¼ sediment concentration at sampling point.

The concentration of the mixed sediment is a linear combination of the enriched sediment at the orebody and the rest of the sediment with background concentrations. The contribution of each to the total sediment is the sediment concentration times its surface area.

Me A_t = A_oC_o + C_b(A_t – A_o)

Therefore

A_oC_o = Me A_t – C_b (A_t–A_o)

or

A_oC_o = A_t (Me – C_b) + A_oC_b

Assumptions

1. Erosion and similar processes do not significantly affect the chemistry of sediments.
2. Only physical mixing is going on.
3. The erosion rate is constant throughout the basin.
4. There is a single source (the ore body).
5. The area is not mined. (This is crucial!)
6. C_b, the background concentration, must be well-established.

We can rewrite the equation to get at the ore grade:

Example:

C_b=10 ppm A_o=1 km² (small orebody) A_t=100 km² Me=109 ppm

C_o = (100/1)(109–10) + 10 = 9910 ppm

Now we will go about it the other way: given an orebody, what should Me be?

Rearranging the equation for C_o,

Moving C_b to the right side,

Rearranging,

The first term represents the fractional area that represents background, and therefore the second term represents the fractional area that is orebody.

If we go back to our example and suppose we know the orebody grade, the background concentration, and the surface areas, then

Me = 10 [(100-1)/100] + 9910 [1-(100-1)/100]

Me = 109

If we know the area and concentration of an unmined ore body, we can predict the dispersion train (Me at various distances downstream).

Examples from Hawkes’ Paper

The method is tested on four porphyry copper deposits. (In a mining context, a "porphyry deposit" is a large disseminated deposit — low grade, high extent.)

• Cerro Colorado, Panam‡

From Table I we can see that the tenor (grade) of the ore deposit was calculated pretty closely from sediment samples as far as about 4 km downstream. The 6 km sample was markedly low, but the sample from 47 km downstream was not too bad. Four out of the five sampling locations yielded estimates within ± 50% of the value measured at the ore deposit itself.

• Chaucha, Ecuador

The Chaucha deposit is unmined. One of the samples was taken in an area where the copper is apparently not behaving conservatively. Neither one of the sample sites yielded a good estimate of the ore’s grade, but both did indicate mineralization; this makes them at least useful from a prospecting standpoint.

• Rosemont, Arizona

In the case of the Rosemont deposit, sediment samples yielded consistent estimates of ore grade as far as 26 km downstream. Accuracy is hard to judge because the deposit has not been mined.

• Casino, Yukon Territory

This is a porphyry copper-molybdenum deposit in an unglaciated permafrost terrain. For copper, the estimates at 10 and 13 km may be good. The one at 5 km was in an area where the pH was low and copper was not acting conservatively. This is also probably true of the site closest to the orebody. The molybdenum numbers are of limited value because of the poor precision of the analytical method used.

• Example (Clark Fork Basin)

We will try to predict Me at a point where A_t = 2025 km²;

A_o = 4 km², C_o = 5000 ppm Cu (in weathered zone). C_b = 20 ppm.

Me = 20 [(2025-4)/2025] + 5000 [1–(2025-4)/2025]

Me = 30 ppm Cu

But when we measure the sediment concentration, we get 1500 ppm Cu! Why the 50« enrichment?

What happened was that Butte was mined. With the ground dug up and tailings scattered all over, the effective area is much larger.

To get the effective area (or virtual area) A_x, plug it into the equation for A_o.

This can be used as a way to measure site contamination.

Partitioning the Blame

Now let’s look at how we may be able to calculate where a contaminant came from, or how much of a contaminant came from a given source. Consider the case where the contaminants in sediment may have come from either of two sources, A and B:

Let f_A be the fraction of sediment that comes from A, and f_B = (1-f_A) be the fraction of sediment that comes from B. Let X be the concentration of element X in the sediment.

X_M = X_Af_A + X_B (1-f_A)

We can rewrite this as

X_M = f_A (X_A – X_B) + X_B

This is the equation of a linear plot of X_M vs. f_A, with slope (X_A–X_B) and y-intercept X_B.

In general this will be true for any conservative species. If X and Y are two conservative species (say Cl^– and Na⁺), then we can write a similar equation for Y:

Y_M = f_A (Y_A – Y_B) + Y_B

We can solve both equations for f_A:

And we can then solve for Y_M:

Finally simplifying, we get

This is again the equation of a line, 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. 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.

To previous lecture: Lecture No 17. Biomethylation

To next lecture: Lecture No 19. Determining Geochemical Background II

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