Instructor’s Guide
Inquiry Material and Pre-/Post-Test
To Supplement “Teaching chemical equilibrium using field-lab experiences”
S.K. Lunsford1, K.V. Nedunuri,2 and Michael Sandy3
1
Department of Chemistry, Wright State University, Dayton, OH 45435
2
Department of Water Resources Management, Central State University, Dayton, Ohio 45384
3
Michael Sandy, Department of Geological Sciences, University of Dayton, Dayton, Ohio 45435
Introduction
This document provides instructional materials for pre-service teachers to teach the concepts of
chemical equilibrium using a combination field and laboratory experiences. The field visit
consists of a trip to a nearby stream or a river and/or a geologic outcrop for the collection of
water samples. Samples will be analyzed for some basic water quality parameters using online
sensors. Later these samples will be taken to the laboratory for detailed analysis using wet
chemistry. Teachers then let students go through a guided inquiry unit to learn the concepts of
equilibrium by integrating their field and lab experiences.
Theory of Carbonate Equilibrium
The chemical equilibrium concept will be illustrated using carbonate chemistry. The chemistry
of natural waters is dominated by carbonate species. Carbon dioxide gas is dissolved in natural
water forming various hydrolysis species. The following equilibrium reactions are important.
The equilibrium constant provided for each reaction was evaluated at 250 C. The aqueous phase
concentrations are expressed in mol/L and the partial pressure of gases are expressed in
atmospheres.
CO2 (g) + H2O

H2CO3
KH=10-1.47
H2CO3

H+ + HCO3-
Ka1=10-6.4
HCO3-

H+ + CO3 2-
Ka2 = 10-10.3
Where calcite is predominant in the bedrock, the following equilibrium reaction will be
controlling the chemistry of natural waters.
CaCO3 (s) + H+

Ca2+ + HCO3- Ksp = 101.92
The dissociation of water is always included in the chemistry of natural waters as
H20

H+ + OH-
K= 10-14
Oxygen is also dissolved in water according to the following gas-liquid equilibrium
O2 (g)

O2 (l)
KH=10-2.9
1
Field visit
Students go through a primer on stream quality assessment before the field trip. Then, they
collect water samples from the river both upstream and downstream of a dam. Samples will be
collected from the river at two locations: one at the inflow to the dam and another at the effluent
of the dam. Online water quality monitoring instruments (Hach Inc) will be used to measure the
parameters. At least three samples along the river depth will be collected from middle section of
the stream using samplers: one from the surface, one from the middle and another from the
bottom. Flow data will be collected from the USGS stream gage. Students will then tabulate the
data as follows.
Flow Temp.(0C) pH Dissolved Calcium
Oxygen
(mg/L)
(mg/L)
Upstream
Slow
22.3
8.7
9.4
180
Downstream Fast
23.1
8.4
9.2
140
Location/
Parameter
Total
Phosphate
Carbonate (mg/L)
(mg/L)
0.0024
4
0.0020
2
Water samples can also be obtained from the stream closer to the geologic outcrop to illustrate
the process of erosion and its effect on chemical quality of water (this can also be investigated
from United States Geological Survey Gage Station information available online depending on
weather. A typical table for collecting data is given below.
Flow Temp. pH Dissolved Total
Oxygen
carbonate
Nitrite/ Phosphate
Nitrate
Calcium
From this point onwards teachers will use the field and lab experiences in a guided inquiry unit
to teach chemical equilibrium. The solutions to the guided inquiry questions are provided in red
italicized fonts.
Observations
Students will observe changes in parameters, and explore the cause and effect relationships.
What would be the ideal positions both upstream and downstream to measure water quality
parameters?
The equilibrium was attained quickly under well mixed conditions. A location along the river
where the water current is rapid and turbulent enhances transfer of gases such as oxygen and
carbon dioxide between air and water phases. These locations are normally found when water
flows at moderate to high velocities across riffles and meanders. Since the measured parameters
will be used in equilibrium calculations, these are ideal locations for measurements.
What are independent parameters (variables)? (Causes)
The equilibrium concentrations of inorganic ions in water depend on temperature and pH. So,
these are independent parameters.
2
What are dependent parameters (variables)? (Effects)
The equilibrium concentrations of inorganic ions such as dissolved oxygen, carbon dioxide,
carbonates, phosphates, and calcium are dependent parameters, since these can be calculated
using law of mass action equations as outlined in earlier section on the “Theory of carbonate
chemistry”.
Use graphs to derive the cause-effect relationships
Using Henry’s law, students calculate dissolved oxygen and carbon dioxide at several
temperatures. Then they would plot corresponding graphs. They should mark on the graph the
temperature and dissolved oxygen measured on the stream. Students should observe that
concentrations of dissolved oxygen and carbon dioxide decrease with increasing temperatures.
Thus gases dissolve less as water gets warmer.
Guided Inquiry
Define Henry’s law.
The concentration of dissolved gas in a dilute aqueous solution is proportional to its partial
pressure.
Describe how it can be used to predict dissolved oxygen concentration when partial pressure of
oxygen in the air is known.
xO 2  K H ,O 2 (T ). pO 2
where xO2 is molar concentration of the dissolved gas in aqueous phase and KH,O2(T) is Henry’s
law constant for oxygen at a given temperature, and pO2 is partial pressure of O2 in atm.
The temperature dependence of Henry’s Law is illustrated for oxygen below using a graph.
Henry's Law Constant K
H
(mg/L/atm)
43
42
41
40
39
38
37
36
35
34
33
20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35
Tem perature (deg. C)
3
Apply Henry’s Law to estimate the concentration of dissolved oxygen in the Mad River at the
time of collection of samples assuming equilibrium between gaseous O2 and dissolved oxygen?
The temperature of the stream is 220 C. Henry’s law constant KH,O2(220 C) is 42 mg/L/atm.
Partial pressure of oxygen is 0.21 atm. Dissolved oxygen concentration is calculated as
mg
xO 2  42 L .0.21 atm
atm
mg
 8.82
L
How well does the estimates match with dissolved oxygen in the upstream and downstream of
the river? What could be the reasons for the mismatch?
The dissolved oxygen calculated for upstream conditions was 8.82 mg/L.
The temperature of the river downstream was 230 C. The dissolved oxygen for downstream
conditions is calculated below.
mg
xO 2  41 L .0.21 atm
atm
mg
 8.61
L
The values estimated assume sea level conditions. Altitude correction factor for Dayton, Ohio is
1.09. When this correction is applied, the upstream and downstream values will be 9.6 mg/L and
9.38 mg/L respectively.
The estimated dissolved oxygen values matches reasonably well with measured values both
upstream and downstream. In case the values do not match, the possible reasons could be under
saturation or super saturation, and instrumentation errors.
Using the Dissolved Oxygen Nomogram, estimate increase/decrease in oxygen saturation as
water moves downstream.
The dissolved oxygen nomograph of dissolved oxygen and temperature at different percent
saturation values can be obtained. This nomograph has several locations online or from
standard textbooks. The DO nomograph used in this lab inquiry was obtained from the chapter
on Dissolved Oxygen and Carbon dioxide, Limnology, Second Edition, Alexander J. Horne and
Charles R. Goldman, Mc Graw Hill Inc. Another location for DO is “A Beginner’s guide to
water management – oxygen and temperature” available online at the following website
http://lakewatch.ifas.ufl.edu/circpdffolder/Circ109_OxygenTemp.pdf.
From the graph, DO upstream is 8.5 mg/L and corresponding value downstream is 8.4 mg/L.
Since Dayton is at an elevation of 700 feet from the sea level, the DO values read from the
4
nomograph need to be multiplied by a factor of 1.09. A table insert in the nomograph provides
correction factors for oxygen saturation at various altitudes (Limnology, Second Edition,
Alexander J. Horne and Charles R. Goldman, Mc Graw Hill Inc). These correction factors
allowed students to regress DO results back to sea level conditions. The percentage decrease in
dissolved oxygen is 1.2%.
Assuming urea (NH2CONH2) is the source of ammonia, nitrate in water, can you write
equilibrium reactions between these species?
Urea Hydrolyis
NH2CONH2 + H2O 
NH2CO2NH4 + H2O 
NH2CO2NH4
2 NH4+ +OH- + HCO3-
Nitrification
NH4+ + 5/2 O2 + 2e- 
NO3- + 2H2O
Assuming farmers applied calcium phosphate (Ca3(PO4)2) as fertilizer in this region, can you
estimate concentration of phosphorus (H2PO4-) in the river, and how well does it match with the
value observed on the stream and also near the outcrop?

Ca3(PO4)2 (s) +4H+
Ca  H PO 
H 
Ksp = 1013.61
 2
2 3
2
4
 4
 1013.61
 
pH = 8.7 which implies H   10 8.7
H PO   10 
 2
2
Ca2+ + 2 H2PO4-
4
8.7 4
10 
 2.75 3
H PO   10
mol
L
X 1013.61  1013.6134.88.25  10 12.94
mol
L
Experimentally found value of [H2PO4-] is 10-4.67 mol/L. The equilibrium value does not match
with the observed value. This means that the phosphorus solubility is not controlled by the
calcium phosphate fertilizer.

2
4
12.94 / 2
 10 6.47
Prediction of some chemical water quality parameters on the stream
The first module will attempt to explain observed chemistry on the Mad River.
5
Huffman Dam
Upstream of the Huffman Dam
Downstream of the Huffman Dam
The pH of an aqueous solution is defined as a negative logarithm (to the base 10) of hydrogen
ion concentration. In an open well-mixed stream, the pH of water is controlled by the carbonate
equilibrium.
Observe both upstream and downstream sections of the dam on the Mad River. Knowing
equilibrium chemistry, can we explain changes occurring as water flows across the dam?
One can predict concentrations of species assuming carbonates as controlling the equilibrium
chemistry. One can predict pH and concentrations of other components.
Can we assume equilibrium between CO2 (g) and dissolved CO2 (aq.)?
Yes under turbulent well mixed conditions.
If yes, what would be those sampling locations? If we cannot assume equilibrium, why cannot
we?
On the surface where water is essentially turbulent and well mixed.
Can we still use equilibrium chemistry to learn what is happening across the dam?
Yes.
From your earlier geology field trip, have we observed different minerals on the dam?
Calcite, dolomite, and silica.
6
Could you identify primary minerals and secondary minerals?
Silicates are primary minerals. Calcite and dolomite are secondary minerals.
As chemical weathering had taken place over geologic times, what element was lost from these
minerals, and what element had been added?
Silicon has been lost. Calcium and magnesium have been added.
Okay, it is time to say out loud some secondary minerals that were controlling the solubility of
some ions in water.
What are these minerals? What are these ions?
Calcite and dolomite.
.
Can we assume water in the flowing stream is in equilibrium with these minerals?
Yes.
We all know pH determines the overall chemical condition of the lake. What species are likely to
control pH of water on the river?
CO2, H2CO3, HCO3-, CO32-, CaCO3
If we determine these species, then can we predict pH?
If we write equations that relate species concentrations to pH or hydrogen ion concentrations,
then we can predict pH.
What could be those equations?
The equations come from writing law of mass action equations for carbonate chemistry.
What we know?
The total carbonate (CO3,T) can be obtained by titrating a known volume of sample against
hydrochloric acid to pH 4.3. Obtain CO3,T.
[H2CO3]+ [HCO3-] + [CO3 2-]
[H2CO3]+ [HCO3-] + [CO3 2-]
Partial pressure of CO2 (g)
=
=
=
[CO3,T]
= 2.4X10-3 mol/L
10-3.5 atm.
Now, we need to solve these equations to obtain pH given [CO3,T] and pCO2 = 10-3.5 atm.
7
We can also write down the following law of mass action equations corresponding to the
carbonate equilibrium. (The instructor must allow students to write these equations and facilitate
finding the unknown species in a sequential manner).
What species concentration can you find from the equation below?
[ H 2 CO3 ]
 10 1.47
PCO 2
You can find [H2CO3] given pCO2
What species concentration can you find from the equation below?
[HCO 3 ][ H  ]
 10 6.4
[H 2 CO 3 ]
You can find [HCO3 ] provided you know H+.
What species concentration can you find from the equation below?
[CO 3 2- ][H  ]
 10 10.3
[HCO 3 ]
2Also, you can find [CO3 ] provided you know H+.
You can find the individual species concentration [HCO3-] and [CO3 2-] provided you know
[H+]. However, [H+] is not given, it is what we should find!
Looks like, we have reached a deadlock here. How can we break it? One clue, if somehow we
can find H+, we can find concentration of all species, and add them up.
Then, we can check whether the estimated [CO3,T] matches with measured [CO3,T] titration.
And what if does not match? Where possibly could we have gone wrong? Well, the only variable
we guessed in the process is the value of pH. We can try another value of pH and see if we can
match the estimated and predicted values of [CO3,T].
This looks like a repetitive procedure. Can we automate the process?
Well, we can use an EXCEL table.
[H+] and all [ ] represent units in moles/liter
pH
[H+]
[H2CO3]
[HCO3-]
[CO32-]
4
5
6
7
8
9
10
10-4
10-5
10-6
10-7
10-8
10-9
10-10
10-4.96
10-4.96
10-4.96
10-4.96
10-4.96
10-4.96
10-4.96
10-7.36
10-6.36
10-5.36
10-4.36
10-3.36
10-2.36
10-1.36
10-13.69
10-11.69
10-9.69
10-7.69
10-5.69
10-3.69
10-1.69
[CO3,T]
estimated
10-4.95
10-4.94
10-4.81
10-4.26
10-3.35
10-2.34
10-1.19
[CO3,T]
measured
10-2.62
10-2.62
10-2.62
10-2.62
10-2.62
10-2.62
10-2.62
8
From the table, estimated and measured [CO3,T] do not exactly match, however, one can observe
that pH 8 and 9, these values can be matched. The corresponding pH will be the estimated pH.
So, what if CO3,T estimated matched closely with CO3,T measured, but not exactly!
What is the next step?
Well, we can graph these values and see the at what pH the line of CO3,T against pH crosses the
horizontal CO3,T measured value. This will be the estimated pH.
While graphing, all concentrations are transformed into log transformed values as given below
p[X]=-log[X]
where X is the species and [] denotes concentration in mol/L The log transformed values of
concentrations of the species in given below.
pH
pH2CO3
pHCO3
pCO3
4
5
6
7
8
9
10
-4.96
-4.96
-4.96
-4.96
-4.96
-4.96
-4.96
-7.36
-6.36
-5.36
-4.36
-3.36
-2.36
-1.36
-13.69
-11.69
-9.69
-7.69
-5.69
-3.69
-1.69
pCO3,T
estimated
-4.95
-4.94
-4.81
-4.26
-3.35
-2.34
-1.19
pCO3,T
measured
-2.62
-2.62
-2.62
-2.62
-2.62
-2.62
-2.62
Log transformed concentrations of carbonate species
16
6
14
12
4
p[X]=-log[X]
10
8
3
6
2
pCO3, T estimated and measured
5
4
1
2
0
0
4
4.5
5
5.5
6
6.5
7
7.5
8
8.5
9
9.5
10
pH
H2CO3
HCO3
CO3
CO3, T estimated
CO3,T measured
How well the estimated pH matches with the measured pH on the stream.
9
From the graph shown above, the measured and estimated values of total carbonate intersect at
a pH 8.65. This value is in close agreement with the measured value of 8.7 upstream the river.
At what pH ranges, does each of the species [H2CO3], [HCO3-], [CO3 2-] is dominant?
From the graph, H2CO3 is dominant until pH 6.3. HCO3- is a dominant species in the pH range
6.3 to 10. CO32- is dominant beyond pH 10.
In the pH range seen on the river, does limestone dissolve or precipitate?
Depends on the ion activity product given as IAP = [Ca2+][HCO3-]/[H+]. If the ion activity
product is less than Ksp, limestone dissolves, otherwise, it would precipitate.
Comment on whether you can generalize this procedure to estimate pH of any chemical system.
We can generalize this method for other systems such as phosphate provided the controlling
solid phases and gases are known.
Prediction of pH of natural waters on the geologic outcrop
Water close to the outcrop
Crystal found near the outcrop
In this module, we will collect samples from water in intimate contact with the outcrop.
What is the dominant mineral of calcium found in this geologic outcrop from prior visit?
Calcite and dolomite.
This pool of water exists despite high temperatures around this time of summer. What is the
source of this water?
Groundwater which is also the water seeping from the outcrop.
Do you believe water here is in equilibrium with a secondary mineral? If yes, what could that
mineral be?
Calcite
10
Can you predict the pH of samples collected from this region if these were supposed to be in
intimate contact with the mineral?
Assume that the total carbonate is now obtained from the dissolution of CaCO3. Assume closed
system and solve for pH just as in the previous example.
The students now calculate total carbonate assuming that it comes from the solubility of
limestone according to the following solubility product equation.
[Ca 2 ][HCO 3 ]
 101.92
[H ]
Students will now adopt the same method as developed to estimate pH on the stream.
-
101.92 X 10 8.1
[HCO 3 ] 
10 2.75
 10 3.43
mol
L
10 10.3 X 10 3.43
10 8.1
 10 5.6
mol
L

2
[CO 3 ] 
mol
10 8.1 X 10 3.43
 10 5.13
6.4
L
10
2Total carbonate in the system is [H2CO3]+ [HCO3 ] + [CO3 ]
=
In this example, it is given as 3.8146 X 10-4 mol/L.
[H 2 CO 3 ] 
[CO3,T]
11
Appendix C: Post test
Solutions are provided in red italicized font.
Item No. 1. Define Henry’s law.
Solution:
Concentration of dissolved gas is directly proportional to its partial pressure.
Item No. 2. Describe how it can be used to predict dissolved carbon dioxide concentration when
partial pressure of carbon dioxide in the air is known.
XCO2 = KH(CO2, T).pCO2
Where x is the concentration of CO2 in mol/L and pCO2 is partial pressure in atm.
Item No. 3. Estimate dissolved carbon dioxide concentration when the partial pressure of carbon
dioxide in the air is 0.01 atm. Use Henry’s Law constant provided in the equation above.
mol
xCO2  10 3.5 L X 0.01atm
atm
xCO2  3.162 X 10 6 mol
L
Item No. 4. Water temperature at a certain location on the Little Miami River was found to be
250C. The Henry’s law constant for oxygen in water at this temperature was 39.71 (mg/L/atm).
Apply Henry’s Law to estimate the concentration of dissolved oxygen in the river in mol/L?
Take partial pressure f oxygen in the air to be 0.2019 atm.
mg
mg 
mol
mg  1gm  1mol 

 = 2.505 X 10  4
xCO2  39.7 L X 0.2atm = 8.02
=  8.02

L 
L
atm
L  1000mg  32 gm 
Item No. 5. Using the Dissolved oxygen nomograph, estimate the concentration of dissolved
oxygen at 100% saturation, 80% saturation, and 120% saturation.
Item No 6. Assuming urea (NH2CONH2) is the source of ammonia, nitrite/nitrate in water, can
you write equilibrium reactions between these species?
Urea Hydrolyis
NH2CONH2 + H2O 
NH2CO2NH4 + H2O 
NH2CO2NH4
2 NH4+ +OH- + HCO3-
Nitrification
NH4+ + 5/2 O2 + 2e- 
NO3- + 2H2O
12
Item No. 7. Oxygen level in certain parts of Lake Erie during the period of algal bloom was
found to be 5 mg/L at 200 C. Find the oxygen depletion. Assume fresh water in the lake will be
100% saturated with dissolved oxygen. (Application)
Dissolved oxygen at sea level = 9 mg/L (1/2 point)
Lake Erie is 570 ft from the sea level (1/2 point). DO at 570 ft must be calculated using linear
interpolation. The correction factor is calculated as 1.025 (1/2 point).
DO in lake Erie is 9 X 1.025 – 9.225 mg/L. (1/2 point)
Oxygen depletion is 9.225 – 5 = 4.225 mg/L (1).
Item No. 8. During our field trip to the Huffman dam, you have observed rapids downstream of
the dam. Can we assume equilibrium between oxygen in the air and dissolved oxygen at this
location? If yes, explain the reasons. (Application and Analysis)
Yes. Rapids ensure intimate contact between air and water phases. Equilibrium between the two
phases is more likely.
Item No. 9. Assuming farmers applied calcium phosphate (Ca3(PO4)2) as fertilizer in this region,
estimate concentration of phosphorus as orthophosphate (H2PO4-) when the pH of water is 5 and
calcium ion concentration is 10-2.5 M. Also find the concentrations of HPO42-, PO43-.

Ca3(PO4)2 (s) +4H+
Ca  H PO 
H 
Ca2+ + 2 H2PO4-
 2
2 3
2
4
 4
 1013.61
 
pH = 5 which implies H   10 5
H PO
2
  10 
10 
5 4
 2
4
 2.5 3
H PO   10

2
Ksp = 1013.61
1.11 / 2
4
mol
L
X 1013.61  1013.61 20 7.5  101.11
 3.55
mol
L
In a similar manner one can use law of mass action equations to calculate HPO42- and PO43-. (5
points were awarded).
mol
mol
2
3
HPO4  10 1.65
and PO4  10 9
L
L




Item No. 10. What happens to the concentration of orthophosphate when the pH is increased to
8? Find the concentration of HPO42-? Do you notice as pH increases, orthophosphate and HPO42decreases while PO43- increases? Explain the reason for such a trend.
H PO
2
4
  10 
10 
 2
8 4
 2.5 3
X 1013.61  1013.6132 7.5  10 10.89
13
H PO   10

2
10.89 / 2
4
 10 5.45

H2PO4-
mol
L
HPO   10 1010   10
2
7.2
5.45
HPO42- + H+
7.25.458
8
4
 10 4.65
Ka2=10-7.20
mol
L
As pH was increased to 8 (3 units), concentration of H2PO4- decreased 1000,000 fold.
Concentration of HPO42- decreased 1000 fold.
In the alkaline range (pH>7), lowering of acid concentrations is more pronounced. H2PO4- is
more acidic than HPO42-. Hence lowering of H2PO4- is much more than HPO42-

HPO42-

. PO4
3
  10 1010   10
12.35
4.65
8
PO43- + H+
12.35 4.658
 10 9
Ka3=10-12.35
mol
L
PO43- concentration is directly proportional to HPO42-.and inversely proportional to H+. As pH
is lowered from 5 to 8 (1000 fold reduction in H+), concentration of HPO42- is also lowered 1000
fold. Thus there is no change in PO43-. Change on pH does not influence phosphate
concentration in equilibrium with calcium phosphate mineral.
Item No. 11. If you start with 0.01 mole/L of Ca2+ and 0.01 mole/L of HCO32-, find the
concentrations of Ca2+ and HCO32 in equilibrium with pure solid CaCO3 (s) when the solution
pH is maintained at 6.0.
Ca2+ + HCO3
CaCO3 (s) + H+
Ksp = 4.8X10-9
Initial
Reacted
Equilibrium
0.01 0.01
0
10-6 (1 point)
x
x
x
10-6 (1 points)
_______________________________________
0.01-x 0.01-x
x
10-6 ( 2points)
10 
6
10 9

(2 points)
0.01  x2 4.8
Solving for x yields 0.01 – x equals 6.928X10-8 mol/L ( 1point).
Hence the equilibrium concentration is 6.928X10-8 mol/L.
14
Useful chemical reactions for the Pre-test and Post-test
Note: Not all reactions are balanced. You need to balance the reactions before you use
them. The equilibrium reactions are assumed to be carried out at 250 C.
CO2 (g)

CO2 (l)
Where KH is Henry’s law constant.
5KH=10-1.47
H3PO4

H2PO4- + H+
Ka1=10-2.15
H2PO4-

HPO42- + H+
Ka2=10-7.20
HPO42-

PO43- + H+
Ka3=10-12.35
Ca3(PO4)2 (s) + H+

Ca2+ + 2 H2PO4-
Ksp = 1013.61
CaCO3 (s) + H+

Ca2+ + HCO3-
Ksp = 101.92
15