CEE 543 Autumn 2015 Midterm Exam #2 Solutions
1. A solution is made by adding 3x10 4 M Ca(OCl)2, 4 104 M CaAc2, 7 104 M NH4Ac, and
2 104 M H2SO4 to water.
(a) (13) The solution is titrated over the range from pH 4 to 10. On the graph provided below,
sketc
CEE 543 Aut 2015 Example Midterm #2 Questions and Answers
Note: The first four questions are the actual exam from Aut 2014; the remaining questions
are either HW or exam questions from prior years
1. A triprotic acid has pKa1 = pKa2 = 5.8, and pKa3 = 8.8.
ii
CHAPTER 1 HOMEWORK PROBLEMS
1. An aqueous solution is prepared by adding 110 mg of calcium chloride (CaCl2 )
and 50 mg of calcium sulfate (CaSO4 ) to 500 mL of water. The solution pH is
8.0 [H+ ] = 108 M, [OH ] = 106 M .
(a) Assuming that the salts dis
CEE 543 Aut 2014 Final Exam
Name_
Answer all questions. Assume ideal solution behavior in all cases.
For H2CO3: Ka1 = 106.35, Ka2 = 1010.33
For CaCO3(s): Ks0 = 108.48
For Zn2+/Zn(s), peo = 12.895; for Fe2+/Fe(s), peo = 7.45
1. (15) Write the oxidation and
CEE 543 Final Exam, Aut 2013
Atomic wts: H = 1, C = 12, O = 16, Fe = 55.8, Hg = 200.6
Nernst equation with H+ term separated: pe peo
Red nH pH
1
log
ne
Ox ne
1. A well water at pH 8.40 contains 4 mg/L TOTFe(II) and is thought to be in equilibrium with
si
9.12.The raw water supply for a community contains 18 mg/L total
particulate matter and is to be treated by addition of 60 mg alum (Al 2(SO4)3
18H2O) per liter of water.
(a) For a total flow of 8000m3/d, compute the daily alum requirement and
the concentr
Errata for 1ST printing of Water Chemistry 2/e, by Mark Benjamin
page 1
Chapter 1
pp.3134
Complete updated problems provided in separate file.
Chapter 2
pp.7375
Complete updated problems provided in separate file.
Chapter 3
pp.122130
Complete updated prob
CEE 543 Aut 2015 HW#8 Solutions
8.37. (a) First, we compute the molar concentrations of total acetate and total sulfite in the
system:
TOTAc = 5000 mg/L (60,000 mg/mol) = 0.0833 M = 101.08 M
TOTSO3 = 300 mg S/L/(32,000 mg/mol) = 0.00938 M = 102.03 M
Since
CEE 543 Aut 2014 HW#5 Solutions
5.20. If the solution behaves ideally, the equilibrium composition can be determined by solving
the mass balance on TOTOCl, the equilibrium constants for dissociation of HOCl and H2O, and
the charge balance equation. These
CEE 543 Aut 2014 HW#6 Solutions
5.20. To solve this problem using Visual Minteq, we simply input 0.01 M Ca2+ and 0.02 M OCl
as components and use the default settings for pH (calculated based on mass balance). We can
run the simulation twice, once with th
CEE 543 Aut 2014 HW#9 Solutions
9.20. Two approaches are shown below for each part of this question. The first approach uses
manual calculations, the second uses Visual Minteq.
(a) Manual approach. The initial ALK in meq/L is:
40 mg/L as CaCO3
= 0.8 meq/L
CEE 543 Aut 2014 HW#4 Solutions
5.3. The carbonic acid and carbonate ion concentrations can be computed from the concentration
of bicarbonate and the pH:
( H 2CO3 )
( CO ) =
2
3
( HCO )( H ) = (10 )(10 ) = 4.47 10
=
3
+
3.2
7.50
5
106.35
K a1
K a 2 ( HCO3
CEE 543 Aut 2014 HW#1 Solutions
1.2. The atomic weight of Ba2+ is 137.3, so its molar concentration in solution is:
2+
mg Ba 2+ 1 mol Ba 2+
4 mol Ba
45
3.28x10
=
2+
L
L
137,300 mg Ba
Thus, 3.28x104 mol/L SO42 would be needed to precipitate all the Ba2
CEE 543 Aut 2014 HW#3 Solutions
3.11. Because the reaction is occurring in a batch reactor, we can equate the rate of change of the
dioxin concentration with the rate at which it is generated by the reaction. The given reaction rate
expression describes t
CEE 543 Aut 2014 HW#7 Solutions
8.1. (a) We know that base must have been added to adjust Solution 1 to pH 8. Also, because
adding equal amounts of an acid and its conjugate base generally leads to a solution at pH
near the pKa, and pKa for HOCl is 7.53,
CEE 543 Aut 2014 HW#8 Solutions
8.9. (a) Since alkalinity is conservative, any alkalinity of the material added contributes that
amount of alkalinity to solution.
(i)
(ii)
HCl is completely dissociated at pH 4.5, so each molecule of HCl added corresponds
CEE 543 Aut 2014 HW#2 Solutions
1.14. When the organic matter is oxidized, each mole of organic carbon is converted into one
mole of bicarbonate ion, so 10 mg/L of organic carbon would be converted to 10 mg/L of C in
the form of HCO3, i.e., 10 mg/L DIC wo
CEE 543 Aut 2014 HW#10 Solutions
11.6. One milligram per liter of NiCO3(s) corresponds to a molar concentration of:
1 mg NiCO3 ( s ) / L
mol
= 8.42 106
118, 700 mg/mol
L
This question can be answered using Visual Minteq by specifying the given concentrati