Sample Questions for Exam 1

1. Sketch a log activity vs. pH diagram for "arsenate speciation" resulting from the dissolution of 0.01 M of H_{3}AsO_{4} in water.

Show how you determined the concentration and pH of *one* of the adjacent species crossing points.

2. A wetland water has activities of 734 mg/L SO_{4}^{-2} and 427 mg/L Ca^{2+}.

a) What are these activities as expressed in mol/L?

b) Assuming activity coefficients of 0.8, what are the concentrations in mg/L?

3. Describe the components that make up these typical rock types:

a) shale b) sandstone c) carbonates

4. What are the activities of Cu^{2+} and SO_{4}^{-2}, and the pH of a pure water (closed to atmosphere) in equilibrium with bronchantite (Cu_{4}(OH)_{6}SO_{4})?

5. Calculate the solubility of barite (BaSO_{4}) in water at 25 deg. C, assuming ideal behavior, expressed as:

a. ppm Ba

b. ppm SO_{4}^{-2}

c. ppm SO_{4}^{-2} if 0.001 M SO_{4}^{-2} had already been in solution previously.

6. Cobalt hydroxide, Co(OH)_{2}, is also known as the mineral transvaalite. Write equations for the dissociation of Co(OH)_{2} using K_{1}, K_{2}, K_{A}, and K_{Aq} in Faure, for a CO_{2} -free system. Derive relationships for log activities vs. pH for each species. Calculate the total species activity at pH = 10.

**Answers:**

1. Data from Faure, Table 9.3, p. 120.: pK_{1}=2.2; pK_{2}=7.0; pK_{3}=11.5;

Calculation of crossing point of H_{3}AsO_{4} with H_{2}AsO_{4}^{-1}:

10^{-2.2} = [H^{+}]*[H_{2}AsO_{4}^{-1}] / [H_{3}AsO_{4}]

at crossing, [H_{2}AsO_{4}^{-1}] / [H_{3}AsO_{4}] = 1.0

For concentration, remember the total activity is 10^{-2.0}M = sum of all As species in solution.

At pH - 2.2, [HAsO_{4}^{-2}] and [AsO_{4}^{-3}] will be negligible, so we can estimate that [H_{3}AsO_{4}] + [H_{2}AsO_{4}^{-1}] = approx. 10^{-2.0}.

Also, at the crossover, [H_{3}AsO_{4}] = [H_{2}AsO_{4}^{-1}].

By substituting, we will find that [H_{3}AsO_{4}] = [H_{2}AsO_{4}^{-1}] = 0.5*(10^{-2.0}) = 10^{-2.3}.

2.a.

7.64 x 10^{-3} or 10^{-2.12}M SO_{4}^{-2}

1.07 x 10^{-2} or 10^{-1.97}M Ca^{+2}

2.b.

I = 3.67 x 10^{-2} = 0.037

2c.

Using the Davies equation (assume 25 deg. C), gamma for both ions = 0.486 (compare with Table 10.5)

Questions 3-6: Similar to homeworks or exercises done in class.

Here is a homework with a solution from Dr. Alba Torrents of University of Maryland.