Geology 401 Problem set #2

Due: 7 October, 2003

1) Speciation of weak acids and bases.

a. Create a plot that describes the speciation of As(V) in solution as a function of pH.
** No need for full calculations here, just sketch the diagram and show me the critical points you used to constrain your sketch.

pK values for As(V) are 2.2, 7.0, and 11.5. These are the crossover points. At one pH unit away from a crossover point, the speciation should be about 90%/10%.

b. I would like to create a pH=9.0 buffer using ammonia (i.e., an ammonium hydroxide solution) and ammonium ions.

i. What will the proportions of ammonia and ammonium chloride be when I have made the buffer correctly? State in terms of activities.

Use the mass action expression for dissociation of ammonium hydroxide:

So a 2:1 ratio of ammonium activity to ammonium hydroxide activity would be needed. We might try a 2:1 concentration ratio and adjust as necessary to compensate for the fact that activity coefficient for the charged species is <1.

ii. If the buffer contains 0.1 molal NH₄OH, how much HCl can I add before the buffering capacity of the solution is exhausted completely?

The buffering capacity is used up when all the NH₄OH is converted to NH₄⁺, so the answer is 0.1 moles per kg H₂O.

iii. If the buffer contains 0.1 molal NH₄OH, how much would pH change if I added 0.01 moles HCl to one liter of the buffer (use approximation: one liter buffer contains exactly one kg H₂O).

This would convert 0.01 moles of the NH₄OH to NH₄⁺. The original amounts were 0.2 moles and 0.1 moles, so the final amounts would be 0.21 moles and 0.09 moles. Plug this into the mass action expression above and get [OH^-] = 8.55 x 10^-6 molal. pH is thus 8.93.

2) Solubility of speciation-sensitive solids. No need for full calculations here, just sketch the diagrams and show me the critical points and/or line slopes you used to constrain your sketch. (Approximation: Let activity coefficients be 1.0000 for 2a and 2b)

a. The pK values for the 1^st and 2^nd deprotonation reactions for Germanic acid (H₄GeO₄) are 9.0 and 12.3, respectively. If the solubility of GeO₂ in pure water at room temperature is 27 mg/L, sketch a diagram of the solubility of GeO₂ as a function of pH (pH = 0 to 14).

As long as the H4GeO4 is the only species with a substantial concentration, the solubility remains constant at 0.258 millimolal and is not a function of pH. You should have a horizontal line this species. However, at a pH of about 8.0, H3GeO4- is present in substantial proportions. With every 10-fold increase in [H+], the ratio of H3GeO4- to H4GeO4 increases ten fold, and thus the H3GeO4- concentration increases ten fold. So the line for H3GeO4- on a log-log plot has a slope of 1.000. It intersects the H4GeO4 line at pH=9.0.

Using similar reasoning, you should have a line with a slope of 2.000 for H2GeO42-. It should intersect the H3GeO4- line at pH = 12.3.

3) Solubility of Aluminum.
Use the data in Chapter 9 of Faure’s text to calculate the concentrations of the various Al(III) species as a function of pH.

· Do these calculations in a spreadsheet application like Excel and turn in your results- both the table and the plot.

· Show me how you set up your calculations for all parts.

· Hint: You might find it convenient to define and calculate pOH (similar to pH but it applies to the OH^- ion instead of H⁺).

a. Calculate the activity of the Al(OH)₂⁺ ion in equilibrium with Gibbsite, as a function of pH. pK₁ in Table 9.3 is needed for this.

The law of mass action yields: log[Al(OH)₂⁺] = logK – log[OH^-]

OR log[Al(OH)₂⁺] = logK + pOH

OR log[Al(OH)₂⁺] = logK + 14 – pH

= -0.8 – pH

b. Calculate the activity of the Al(OH)²⁺ ion in the same solution, as a function of pH. (I think Faure interchanged pK₃ and pK₂ by mistake in his table- tell me if you agree and why/why not).

The speciation of Al predicted by the table in Faure seems impossible. AlOH²⁺ is never the dominant species. Al(OH)₂⁺ would give way to Al³⁺ with decreasing pH, without ever forming much AlOH²⁺. This seems improbable.

(Note: I did some more research- the correct values are: pK₂ = 8.86, pK₃ = correct as written in Faure.)

Calculation: log[AlOH²⁺] = logK + log[Al(OH)₂⁺] + pOH I used log k = 9.0, it was okay if you used 10.3.

c. Calculate the activity of the Al³⁺ ion in the same solution, as a function of pH. (The error in the table affects this one also.)

Calculation: log[Al³⁺] = logK + log[AlOH²⁺] + pOH I used log k =10.3, it was okay if you used 9.0.

d. Calculate the activity of the Al(OH)₄^- ion in the same solution, as a function of pH. (Note: I don’t like the way Faure uses pK_A; for most people pK₁, pK₂, and pK₃ are all pK_a values- the subscript meaning “acid”).

Calculation: log[Al(OH)₄^-] = logK - pOH

e. Note that Al(OH)_{3 (aq)} is missing as an aqueous species. Write into Table 9.3 a value of –8.7 for pK_aq for the “Aluminum, Gibbsite” row. Calculate the activity of Al(OH)₃ as a function of pH.

Calculation: log[Al(OH)_3(aq)] = logK_sp (I don’t know why he calls it K_aq in the table, it is just a K_sp)

f. Calculate the total of all Al species.

You must convert from log values to normal values, then add, and maybe convert back to log values for convenience.

This table gives all the results if you used Faure’s data with pK₂ and pK₃ switched.

		Log molal Activities
pH	pOH	Al(OH)4	Al(OH)3	Al(OH)2	AlOH	Al	total
0	14	-15.4	-8.7	-0.8	4.2	7.9	7.9
1	13	-14.4	-8.7	-1.8	2.2	4.9	4.901
2	12	-13.4	-8.7	-2.8	0.2	1.9	1.909
3	11	-12.4	-8.7	-3.8	-1.8	-1.1	-1.02
4	10	-11.4	-8.7	-4.8	-3.8	-4.1	-3.6
5	9	-10.4	-8.7	-5.8	-5.8	-7.1	-5.49
6	8	-9.4	-8.7	-6.8	-7.8	-10.1	-6.75
7	7	-8.4	-8.7	-7.8	-9.8	-13.1	-7.66
8	6	-7.4	-8.7	-8.8	-11.8	-16.1	-7.36
9	5	-6.4	-8.7	-9.8	-13.8	-19.1	-6.4
10	4	-5.4	-8.7	-10.8	-15.8	-22.1	-5.4
11	3	-4.4	-8.7	-11.8	-17.8	-25.1	-4.4
12	2	-3.4	-8.7	-12.8	-19.8	-28.1	-3.4
13	1	-2.4	-8.7	-13.8	-21.8	-31.1	-2.4
14	0	-1.4	-8.7	-14.8	-23.8	-34.1	-1.4

If you used Faure’s data as given in the table, your Al(OH) values should be 1.3 log units less.

g. Plot the activities of all the species, and the total Al, on a full page plot, and turn this in.

h. Mark on the plot the crossover pH’s and make sure they are at the correct pH values. The two crossovers at higher pH are challenging, so you can skip those if you are pressed for time.

Note crossovers at pH = 3.7 (=14.0 - 10.3), 5.0 (= 14.0-9.0), 7.9, and 6.7. The last two can be calculated by using the equation along with the mass action expressions Faure gives.

i. If assume the Al concentration in a mountain watershed is limited by gibbsite solubility, how much would we expect the total Al concentration to increase if acid rain decreases the pH of the soil waters from 6.0 to 5.0? Ignore activity coefficient issues for this.

Al(OH)₂⁺ is the dominant species in this range. From the slope of the line on the plot, you can see that for a factor of 10 decrease in OH^- concentration, the concentration of Al(OH)₂⁺ increases by a factor of ten. It turns out the overall increase, including all of the species is a factor of 18.

j. How much would we expect the total Al concentration to increase if acid rain decreases the pH of the soil waters from 5.0 to 4.0? Ignore activity coefficient issues for this.

Same reasoning but this time the dominant species is Al(OH)²⁺ and the increase is close to a factor of 10²!!!! The point here is that speciation changes can cause some drastic changes in solubility as a function of pH.

k. If we mix some of this pH 4.0 water with a less acidic water, and the final pH of the mixture ends up at pH 5.0, what percentage of the Al originally dissolved in the pH 4.0 water would precipitate out (assuming equilibrium were reattained)? What is this in mg/L Al?

About 99%, because the solubility at pH 5 is about 100x less than at pH 4. So essentially, 0.2 millimolal, or 5.4 mg/L Al ends up precipitating somewhere, maybe on some fish’s gills.

1) Now let’s see what happens when we add some sulfate to the system. The log K for the formation of an AlSO₄⁺ ion pair is 3.01. Recalculate your plot for part g. above, but now include this ion pair. At what value of [SO₄²⁺] should we start considering this ion pair? Assume we can ignore it if it is less than 10% of the total Al in solution. Ignore activity coefficient issues for this.

The effect of the complex is greatest at low pH, where Al³⁺ is dominant. The minimum sulfate concentration for which we must consider the complex is found by setting the AlSO₄⁺ : Al³⁺ ratio equal to 0.1 in the mass action expression and solving for the sulfate concentration….. the answer is 0.1 millimolal.

4) Wastewater containing a high concentration of Pb²⁺ ions (e.g., 10^-3 mmol/L) is injected into a rock formation containing gypsum. Assume that the waters in the formation are saturated with respect to gypsum before injection. Consider what happens after the injection.

a. How would you determine if PbSO₄ is supersaturated initially? Assume you have a full chemical analysis of the formation waters.

Calculate , then compare to K. If Q>K, then it is supersaturated. The activities for the Q calculation are the product of the molalities and the activity coefficients. The activity coefficients are calculated using an activity model such as the Debye-Hückel model, which considers the effects of all the ions on activities.

b. Assuming PbSO₄ is supersaturated, what will happen to the gypsum as PbSO₄ precipitates? Why?

Sulfate concentration will be drawn down as PbSO₄ precipitates, and this will cause gypsum to dissolve.

5) Saturation index. Determine the saturation index (log(Q/K)) for Barite (BaSO₄) in the brackish water given in Table 1 at 5˚C.

Table 1. Composition of estuarine water.
Component	mg/L
Ba	0.66
Ca	41.2
Mg	129.2
Na	1076.8
K	39.9
Cl	1935
HCO3^-	14.2
SO4^2-	271.2
pH	8.22

Ionic strength is 0.070.

Activity coefficients for Ba²⁺ and SO₄^2-, according to the Debye-Hückel model, are 0.43 and 0.41, respectively.

Q = 2.43743 x 10^-09, K_sp = 10^-10

Log Q/K = 1.4

Barite is supersaturated.