Lab 9 Cobalt Chloride and LeChatlier's Principle (1)
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Lab 9:
Cobalt Lab
: Cobalt(II) Complexes & LeChatlier's Principle
This lab assignment uses the virtual lab. Using a computer that is running Microsoft windows or Macintosh OS 10.1
or higher, go to
https://chemcollective.oli.cmu.edu/vlab/85
Objective:
To gain an understanding of LeChatlier’s principle through the observation of the effect of
induced stresses on the equilibrium distribution of various cobalt(II) complexes.
Background:
Cobalt(II) does not exist in aqueous solution as a free ion, but forms a complex ion where 6
water molecules, acting as a Lewis bases (electron pair donors), donate electrons into the vacant orbitals
of the cobalt (II). This results in the pink complex ion Co(H
2
O)
6
+2
. In the presence of chloride ions, a
different complex forms, the blue CoCl
4
-2
complex ion. We use their different colors to indicate the
equilibrium concentrations for the following reaction.
Co(H
2
O)
6
+2
+ 4Cl
-
<==> CoCl
4
-2
+ 6H
2
O
In this lab activity you not only observe the equilibrium concentrations through their colors, but also
directly read their concentrations. Note how the stock solutions exist as their ions (Co(No
3
)
2
(aq) exist as
Co(H
2
O)
6
+2
and NO
3
-2
.
Assignment:
Use the equilibrium concentrations after each step to determine the K for the above
equation. Be sure to include appropriate dilution factors (as they will not cancel).
1.)
Add 25 mL of [Co(H
2
O)
6
]
+2
to an empty Erlenmeyer flask (this will be the 1 M CoCl
2
in the stockroom.
You will see the species and molarity in the data table when you place the flask of CoCl2 on the
workbench. See figure to the right. )
2. Now add 12 M HCl in 1mL increments until the equilibrium color has changed. (
Hint: Type in 1 for
the volume to be transferred, and then keep clicking “pour” until you see a change, counting clicks to
determine total volume added.)
Record your data in the given data table:
Amount of HCl added
Color
1 mL
Color looks similar to original
2 mL
Still looks like original
3 mL
Color is starting to change but
looks pink
4 mL
Has a slightly different pink
5 mL
Has a gray undertone to it
6 mL
Same as before has a slight gray
undertone
3.)
Predict the effect of removing chloride ions. Now remove some of the free chloride ions by adding
some silver nitrate (Hint: add 1 mL amounts of the silver nitrate successively until the equilibrium has
been shifted instead of a whole bunch at once).
AgNO
3
(aq) <==> Ag
+
+ NO
3
-
Ag
+
+ Cl
-
<==> AgCl(s)
AgNO
3
(aq) + Cl
-
<==> AgCl(s) + NO
3
-
Note by the above equation how Ag
+
scavenges free chloride ions by tying them up in a precipitate, and
thus removes them from the solution.
Record your observations in the table below:
Equilibrium shift.
Favors reactants or products:
Addition of silver nitrate
Favors product
Observations: Made a little bubble at the
bottom of the flask.
4.)
Predict the effect of adding HCl to the reaction. Now do so in very small incremental steps until the
equilibrium has shifted
.
Record your data in the table below:
Addition of HCl (mL)
Observations
1ml (7ml total)
Took bubble away, still pink
2ml
Looks like the same color as
before
3ml
Starting to look a bit more gray
4ml
Bubble came back but color has
changed
5ml
Looks gray with pink undertone
6ml
Looks gray with pink undertone
Equilibrium shift? (favors
products or reactants?)
favors reactants
5.)
Is the reaction as written endo or exothermic? Right click on the flask and choose “thermal
properties”. You can now change the temperature between 0 and 99 deg C. Heat or cool the system until
you have perturbed the equilibrium. Then apply LeChatlier’s principal to determine if it is exothermic or
endothermic. Experiment with 4 different temperatures.
Record your data in the given table:
Temperature
Observations
1 degree
The color has changed to bright pink
89 degrees
The color was gray
40 degrees
The color was gray and had a slightest shade of pink
10 degrees
The color was pretty bright in pink
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Endothermic or Exothermic
Reaction? :
Exothermic
6.)
Allow the system to reach thermal equilibrium (constant temperature). Use the concentration values
to determine K. Now go to the thermal properties, change the temperature and click on the thermally
isolated system option. Determine the new K at the new temperature. From the new K at the new
temperature, determine if the system is endothermic or exothermic.
Record your data in the table
below:
Temperature
Observations
[Co(H
2
O)
6
+2
]
[CoCl
4
-2
]
K
10 degrees
pink
.841595
0.0279699
30.0893103
52 degrees
Gray with bubble
.527862
.341704
1.54479315
79 degrees
Dark gray
.361971
.507594
.71311127
0 degrees
Bright pink
0.862062
0.00750365
114.885689
Endothermic or
Exothermic
Reaction? :
Exothermic
Show all your calculations for K below.
25°C
[Co(H2O)+2] = 0.02M
[CoCl2] = 0.08M
K = ([CoCl2]/[Co(H2O)+2]) = 4
Exothermic
[Co(H2O)+2] + 2Cl-
⇄
[CoCl2] + 2H2O
50°C
[Co(H2O)+2] = 0.02M
[CoCl2] = 0.07M
K = ([CoCl2]/[Co(H2O)+2]) = 3.5
Endothermic
[Co(H2O)+2] + 2Cl-
⇄
[CoCl2] + 2H2O
7.)
Write a short conclusion of your findings using your collected data. Submit this lab report directly to
D2L in the Assignment Folder: Lab 9 Cobalt Lab
The more Hcl was added the more the color would change would make the reaction a exothermic. Changing
the temperature also made a difference in the lab which gave the conclusion of an exothermic.
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