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 ra
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 )(1
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.
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+ ] =
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 dissociate
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
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/
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
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 s
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 =
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
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 o
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
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. Th
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, 3.28x10
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 r
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
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