510
Second FCH Midterm Exam
April 8, 1999
Use the attached booklet to answer all questions. For each problem, make sure
that you show all work and clearly state all assumptions that you have made. Check to
make sure that your assumptions are reasonable. Assume that all solutions are ideal,
unless otherwise noted. You may use a calculator to aid you in solving the exam
problems. Refer to the attached sheet for a list of equations and constants. You have
until 10:00 to finish the exam.
Short Answer Questions
1. (10 pts) What is the evidence for and against the formation of organometallic
complexes in natural waters?
2. (5 pts) What are three important reasons why a chemical speciation model may fail?
3. (5 pts) For most natural waters, how can the [Alk] expression be simplified? Provide
a brief explanation for you answer.
Long Answer Questions (20 pts each)
1. (20 pts) Calculate the equilibrium concentration of Hg+2 in anoxic lake water with the
following conditions: pH = 5.0, logCT,S = -5, logCT,Hg = -11
Hg+2 + 2HS-
Hg(HS)2o
logK1 = 37.7
Hg+2 + 2HS-
HgHS2- + H+
logK2 = 31.5
Hg+2 + 2HS-
HgS2-2 + 2H+
logK3 = 23.2
H2S HS- + H+
logKa,1 = -7.0
HS- S-2 + H+
logKa,2 = -17.0
2. (20 pts) A lake water sample has an pH initial = 6.5, CT,CO3 = 1.7 mM and
[Alk] = 1.0 mM. (a) What is the pH, [Alk] and CT,CO3 of the lake water if the sample
loses 0.7 mM carbon dioxide due to its uptake by algae? (b) Is the lake water in
equilibrium with the atmosphere after the algae have taken up the carbon dioxide?
3. (20 pts) (a) What is the total inorganic carbon (CT,CO3) concentration in a water
sample that is pH 6.5 and is in equilibrium with the atmosphere (pCO2 = 3.5)? (b)
What is the total alkalinity of this water?
4. (20 pts) Derive a pC vs pH expression for the CaCO3o ion pair in a sample that is in
equilibrium with the atmosphere. You are given [Ca+2] = 0.1 mM and pCO2 = 3.5
Assume that no calcium carbonate solid phase will form at any pH and that:
CaCO3o Ca+2 + CO3-2
logK = -3.2
FCH 510
Equations
[Alk] = [HCO3-] + 2[CO3-2] + other bases + [OH-] - [H+]
[Alk] = CT,CO3(1 + 22) + [OH-] - [H+]
[Acy] = [H+] + 2[H2CO3*] + [HCO3-] - [OH-]
[Acy] = CT,CO3(1 + 2o) + [OH-] - [H+]
o = [H+]2/E
1 = Ka,1[H+]/E
2 = Ka,1Ka,2/E
where E = [H+]2 + Ka,1[H+] + Ka,1Ka,2
o = [H2CO3*]/CT,CO3
1 = [HCO3-]/CT,CO3
2 = [CO3-2]/CT,CO3
pH = pKa + log[base]/[acid]
Some useful constants for the carbonate system (at 25 oC):
pKH = 1.5
pKa,1 = 6.3
pKa,2 = 10.3
pCO2 = 3.5