Lecture No. 12: Wrap-up of Phase Diagrams. Case Studies

In response to some questions I have gotten on the midterm exam: In question 6, assume that the E_h-pH diagrams you have are all valid for 10¡ C and 1« 10^–3 M concentration. Remember that the conversion from E_h to pe is a function of temperature. (This was explained in the second lecture.)

For Question No. 6, you may assume that N₂ is not redox-active.

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Now for a more complex system: Consider carbonic acid in a closed system, i.e., one in which

[CO₂] + [HCO₃^–] + [CO₃^2–] = C_T.

Let C_T=1« 10^–5 M.

The dissociation equilibria are

CO₂ + H₂O ó H⁺ + HCO₃^–

HCO₃^– ó H⁺ + CO₃^2–

log K₁ = –6.4 = log [H⁺] + log [HCO₃^–] – log [CO₂]

log K₂ = – 10.3 = log [H⁺] + log [CO₃^2–] – log [HCO₃^–]

At low pH, CO₂ predominates. As the pH increases, the HCO₃^– concentration increases. Eventually [HCO₃^–] passes up [CO₂], and bicarbonate becomes the predominant species. At even higher pH, the second ionization becomes significant, and eventually carbonate predominates. The form of the log c curves, then, has log [CO₂] decreasing after an initial plateau at low pH; log [HCO₃^–] increasing, hitting a plateau, then decreasing again; and log [CO₃^2–] increasing and finally hitting a plateau at high pH. Now to put some quantitative values on these log c curves.

Rearranging the equations,

log [HCO₃^–] = log [CO₂] + pH – 6.4

log [CO₃^2–] = log [HCO₃^–] + pH – 10.3

[CO₂] = [HCO₃^–] at pH 6.4

[HCO₃^–] = [CO₃^2–] at pH 10.3

[H⁺] = [OH^–] at pH 7

At low pH, log [CO₂] is constant. After the crossover point, as long as log [HCO₃^–] is constant, log [CO₂] has a slope of –1 vs. pH. After the bicarbonate-carbonate crossover, when bicarbonate starts decreasing, we need to plug in a value for [HCO₃^–] in terms of pH:

log [CO₂] = log [HCO₃^–] – pH +6.4

log [CO₂] = log [CO₃^2–] – pH +10.3 – pH +6.4

log [CO₂] = log [CO₃^2–] – 2 pH + 16.7

In this pH range, carbonate concentration is a constant, so log [CO₂] is linear with a slope of –2 vs. pH.

In the pH range where [HCO₃^–] is constant, [CO₃^2–] has a slope of +1 vs. pH. However, in ranges where [HCO₃^–] is variable, it is necessary to plug in the equation for [HCO₃^–] in terms of pH.

log [CO₃^2–] = log [HCO₃^–] + pH –10.3

log [CO₃^2–] = log [CO₂] +pH –6.4 + pH – 10.3

log [CO₃^2–] = log [CO₂] +2 pH –16.7

Thus in the range where [CO₂] is constant, [CO₃^2–] has a slope of +2 vs. pH. In the range where [HCO₃^–] is constant, the log [CO₃^2–] slope vs. pH is +1. In the high pH region where bicarbonate is decreasing, carbonate is the predominant species and has a constant concentration at 1« 10^–5 M.

The slope of log [HCO₃^–] varies, depending on the pH. At low pH its slope is +1; at high pH it is –1. In between, there is a range where [HCO₃^–] is constant.

Now to put all this together: We can plot the three equal-concentration points (system points) at (pH 7, –7) for H⁺ and OH^–, (pH 6.4, –5.301) for CO₂ and HCO₃^– and (pH 10.3, –5.301) for HCO₃^– and CO₃^2–. Connect the system points to the log C_T line with smooth curves. The log [CO₃^2–] line has a slope of +1 to the left of the carbonate-bicarbonate system point until it is at the pH of the bicarbonate-carbon dioxide system point. From there leftward its slope is +2. The log [CO₂] line has a slope of –1 from the CO₂-bicarbonate system point until it is directly below the bicarbonate-carbonate system point; to the right of that point, its slope is –2. First we construct all these lines. Then we can draw smooth curves connecting the system-point concentrations with the log C_T line, and connecting the segments of the log [CO₂] and log [CO₃^2–] curves with different slopes. Put in the water lines (log [H⁺] and log [OH^–]) and we’re done.

This illustrates the general principles for making a log c-pH diagram for any diprotic acid. These principles can be extended to acids with three or more ionizable protons, but we will not go into that in this course.

See the first overhead for a look at the finished product.

Note: You may encounter acids in which successive dissociation constants are close together, and the log c line for an intermediate species may not reach the log C_T line. Cases like that are rather obvious, because the increasing and decreasing legs of a log c line intersect below log C_T. A good approximate solution to this situation is to draw a smooth curve just below the intersection. It is always possible to get an exact solution by solving n equations in n unknowns, but it is usually not necessary.

Question: How would an open system have been different?

In an open system, carbon dioxide is available from the air. The dissolved CO₂ concentration is governed by the partial pressure of CO₂ and the Henry’s Law constant for CO₂ at the prevailing temperature. Therefore log [CO₂] is a constant. Since the supply of CO₂ is unlimited, bicarbonate and carbonate are not limited by C_T, but only by [CO₂]. The calculations are not really any more difficult, but the results are very different. (See the second overhead.) If anyone wants to learn more about log c-pH diagrams, see Snoeyink and Jenkins, Chapter 4. (The book is on reserve in the library.)

Case Studies

I. Arsenic Contamination at Milltown

Overview

The Milltown Reservoir is a small impoundment east of Missoula that dams the confluence of the Clark Fork and the Blackfoot Rivers. It was built in 1907. Since then it has trapped at least 2« 10⁶ m³ of sediment.

About 1976, Bailey looked at metal concentrations in the lake sediments and found elevated levels of Cu, Zn, Pb, and other metals.

In 1981 the Montana Department of Health and Environmental Services found As levels in 4 wells near Milltown to be over the USEPA maximum contaminant level (MCL) for drinking water (50 m g/L). The wells in question had to be closed down. Fortunately, these 4 wells and the ones subsequently found to be contaminated were between the railroad tracks and the lake, in a strip about ¸ mile wide.

In 1984, Woessner, Moore, et al. established that Milltown Reservoir was the source of the arsenic.

See Figure 4 in the Moore paper. The surface water recharging the aquifer (and carrying arsenic) mostly flows out of the northeast side of the lake. As can be seen in the map, this water runs smack dab into clean recharge water from the Blackfoot River. This is probably why the arsenic only travels about ¸ mile.

The Question: What are the processes of arsenic release and transport?

Geochemistry of the Sediment

There are millions of tons of sediment with significant enrichment above background. Examples: As ~ 32« background, Cd ~ 37« , Cu ~ 62« , Zn ~ 67« , Pb ~ 11« , Mn ~ 7« .

The source of all this is the mining of sulfide ores, mostly at Butte. The closed lead and silver mines in the Flint Creek drainage near Phillipsburg also make a contribution, especially of lead.

The contaminated wells contained high dissolved As — and also elevated Mn. The presence of dissolved manganese tells us that the water is anoxic.

Dissolved arsenic can be in either the +3 or the +5 oxidation state. In the well-water samples the arsenic was > 50% As (III). In some samples the As (III) approached 100%. However, a log-log plot of As (III) vs. total As was linear with little scatter, indicating that the percentage As (III) was fairly constant.

Among species controlling As, sulfur is an important one. Arsenic sulfide minerals can form. On the other hand, if sulfide levels get too high, soluble thioarsenites can also form.

Consider the diagram on the overhead showing geochemical profiles at low reservoir stage.

So what is the E_h in the zone above the water table (loosely referred to as the "oxic zone")? Between –0.2 V and + 0.7 V. (Between As reduction and H₂O oxidation.)

What should be the stable Fe phase in this E_h range? It could be Fe²⁺ (at the bottom) or FeOOH.

Is there some way to zero in? Yes. The dissolved iron concentration is low. Therefore E_h > –0.02 V.

In the saturated zone (anoxic), with dissolved As present, E_h is in a very narrow range (–0.12 V to –0.08 V). Fe will be Fe²⁺. (Or FeS?)

The stable S species should be SO₄^2– in both zones (until you get deep into the anoxic zone, where sulfur reduction starts). Therefore FeS will not be present in the upper anoxic zone. Nitrogen could be NO₃^–, NO₂^–, or NH₄⁺. (We don’t have analytical data on N. If we had, we could zero in more closely.)

In the saturated (and anoxic) zone, with dissolved arsenic present, E_h is in a very narrow range (–0.12 to –0.08 V). At any higher E_h, As would be As (V); any lower and sulfur would be reduced. Fe is present as Fe²⁺.

The stable S species in both zones is sulfate. (Important!)

Arsenic in the upper zone ought to be in solution as arsenate — but there is no measurable dissolved As in the oxic zone! Why? What is happening to the As?

It is well established that As (V) sorbs onto sediments, and co-precipitation with iron and manganese oxyhydroxides is known to happen. The surface reaction

¼ Fe–OH + H⁺ + H₂AsO₄^– ¨ ¼ Fe–OH_(x+1)AsO₃^(x–1)–

Depending on pH, you get

Lower pH
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Higher pH

Question: How do we know what is going on? Answer: look at the solid phases. A sequential extraction method was used.

1. For oxyhydroxides, hydroxylamine hydrochloride and acetic acid (NH₂OH•HCl + CH₃COOH) were used. This combination dissolves Fe and Mn oxyhydroxides and frees up sorbed matter. However, it also dissolves carbonates.
2. For organics, hydrogen peroxide + tetrasodium pyrophosphate (H₂O₂ + Na₄P₂O₇) were used. However, this method also dissolves some carbonates and some poorly crystallized sulfides.
3. Fur sulfides, potassium perchlorate + hydrochloric acid (KClO₄ + HCl) were used.
4. Residual silicates were extracted with a combination of strong acids in an extraction similar to the USGS total-in-sediment method (aqua regia, HF, HClO₄).

If a bunch of metals are in sulfides or in organics, they will be stable in an anoxic environment. On the other hand, if they are in oxyhydroxides, they will be stable in oxic environments, but unstable in anoxic ones.

See overheads.

The Milltown data confirm this for arsenic. Why do we not see As in the sediment from the vadose zone? We are up against the operational definition of dissolved metals again! Much of the Fe oxyhydroxide is colloidal.

The iron data look screwy, though, and the manganese data are even more so. Explanation: The system is probably never at equilibrium because the reservoir is continually being drawn down and then refilled as power generation demands change.

Sulfate is much more concentrated in the saturated zone. Why is organic carbon so much higher in the vadose zone? Probably because that is where is comes from — algae and other plants engaged in primary productivity. As you move downward, the organics get consumed.

Copper and zinc in the pore water hit maxima around 25 cm deep — still in an oxic zone. Or maybe the dissolved oxygen data were from a recent filling of the reservoir, and the Cu and Zn were holdovers from a previous drawdown.

To previous lecture: Lecture No 11. More Handy Computational Tools

To next lecture: Lecture No 13. Case Studies II. Milltown and the Upper Clark Fork

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