116095170
Name
Le Chatelier’s Principle (Chemical Equilibria)
Introduction
We will investigate Le Chatelier’s Principle in several systems by shifting the equilibrium concentrations of reactants and
products by applying a “stress” to the equilibrium. This experiment will demonstrate some observable equilibrium
concentration shifts and give you a chance to review the writing of net-ionic equations.
Le Chatelier’s Principle states: “When a stress is applied to a chemical system at equilibrium, the equilibrium concentrations
will shift in a direction that reduces the effect of the stress.”
The stress that will be applied in our investigation will be either:



the addition of more reactant or product to the equilibrium mixture, increasing their concentration.
the addition of a species that chemically reacts with either a reactant or product in the equilibrium mixture, lowering their
concentration.
a change in temperature.
The systems will involve equilibriums that you may have already studied: weak acid and base, (K a and Kb), solubility, (Ksp),
and complex ion formation, (Kf). Although no calculations will be done, you should be able to recognize the type of
equilibrium under investigation.
The following is an example of an equilibrium system similar to the five you will investigate. The first step in the process is to
establish the equilibrium given the allowed reagents. The equilibrium concentrations are shifted (to the left or right) by
applying a stress to the equilibrium system by adding one of the given reagents. Changes in the appearance of the system are
recorded. Finally, a net-ionic equation is written to understand the chemistry behind the change in equilibrium concentrations.
An example equilibrium
Consider the following equilibrium between aluminum ions and ammonia:
Al3+(aq) + 3 NH3(aq) + 3 H2O(aq) ⇌Al(OH)3(s) + 3 NH4+(aq)
Reagents:
To establish equilibrium:
0.1 M aluminum nitrate
To shift equilibrium concentrations: 1M hydrochloric acid
1.
2.
6 M ammonia
Al(OH)3(s) might be produced by:
“Add 6 M ammonia solution, drop-wise, to 1 mL of 0.1 M aluminum nitrate solution.”
a)
Observation:
“When the two clear and colorless solutions were mixed, a white precipitate formed.”
b)
Net-ionic equation:
“Al3+(aq) + 3 NH3(aq) + 3 H2O(l)  Al(OH)3(s) + 3 NH4+(aq)”
Note: the arrow is shown to the right since Al(OH)3(s) is produced. However, equilibrium is quickly
established within this system.
The equilibrium concentrations can be shifted left by adding: 1M HCl; Add drop-wise, mix well after each
addition, and carefully observe.”
a)
Observation:
“The precipitate dissolved leaving a clear, colorless solution.”
b)
Net-ionic equation:
“Al3+(aq) + 3 H2O(l)  Al(OH)3(s) + 3 H+(aq)”
Or, you can reverse the chemical equation to show it in the more usual form:
“Al(OH)3(s) + 3 H+(aq)  Al3+(aq) + 3 H2O(l)”
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Writing net-ionic equations
This will be the most troublesome part for many students. There are two different methods to use when writing the net-ionic
equations depending on the type of stress applied to the equilibrium.
Method 1 for writing the net-ionic equations.
If the stress applied to the equilibrium system is simply an increase in concentration of a reactant or product, without any
new chemistry, then the net-ionic equation is given by the original equilibrium net-ionic equation, with the arrow pointing in
the direction of the reaction (the direction of the shift in equilibrium), left or right.
In the aluminum example, the original equilibrium equation indicated how the precipitate would form based on the reagents,
Al3+(aq) and NH3(aq), needed to establish the equilibrium.
Method 2. This is used when NEW chemistry takes place.
This method is somewhat more complicated and requires some thought about chemical reactions. If the addition of a species
to the equilibrium system results in a chemical reaction with either a reactant or product then the stress applied to the
equilibrium lowers the concentration of a reactant or product. The equilibrium concentrations will shift accordingly. To write
the overall net-ionic equation several steps are involved. This method will work with all the systems you will study today
when NEW chemistry takes place.
In the previous example H+ was added to shift the equilibrium concentrations to the left. To write the final net-ionic equation,
follow these steps (in order!).
1.
Start with the original equilibrium net-ionic equation:
Al3+(aq) + 3 NH3(aq) + 3 H2O(l) ⇌ Al(OH)3(s) + 3 NH4+(aq)
2.
Since the equilibrium is shifting to the left, a reactant in the original equilibrium must be reacting with the added
acid, THIS is LE CHATELIER’S PRINCIPLE! The two possible reactants are Al 3+(aq) and 3 NH3(aq). Your chemical
knowledge should tell you H+ ions are reacting with the ammonia in an acid-base reaction. We now add enough H+ ions
to both sides of the original equilibrium equation to react all the ammonia:
+
Al3+ + 3 NH3 + 3 H2O ⇌Al(OH)3(s) + 3 NH4+
3 H + ⇌ 3 H+
3+
Al + 3 NH4+ + 3 H2O  Al(OH)3(s) + 3 NH4+ + 3 H+
(original equilibrium)
(acid added)
(result of chemical rxn)
3.
We now cancel any species that appear on both sides of the new equation, NH 4+ in this example:
Al3+(aq) + 3 H2O(l) ← Al(OH)3(s) + 3 H+(aq)
4.
Since the original equilibrium concentrations were being shifted to the left it is good practice to write the new equation
with the equilibrium double arrow replaced by a single arrow pointing in the direction of the shift, left in this case.
Alternatively, you can reverse the equation and use a single arrow to the right to show it in the more usual form. Also
include all phase symbols.
Ammonia as a reagent – more than one use.
Ammonia will be used in this lab to accomplish three things for us: 1) form hydroxide precipitates, 2) neutralize acids, and 3)
form complex ions. You should already be comfortably familiar with these three types of chemical reactions. As a review, the
chemistry of each is discussed in more detail below.
1.
Using Ammonia as a source of hydroxide ions.
Remember ammonia is a weak base and maintains an equilibrium with water that produces hydroxide ions:
NH3(aq) + H2O(l) ⇌NH4+(aq) + OH–(aq)
Kb = 1.8x10–5
Thus, in an aqueous solution of ammonia about 1% of the ammonia molecules have undergone hydrolysis producing
hydroxide and ammonium ions that are in equilibrium with ammonia molecules. Whenever you add aqueous ammonia to an
equilibrium system, you are adding a source of hydroxide ions.
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When 6 M ammonia is added drop wise to a 0.10 M solution of aluminum nitrate a precipitate forms. The precipitate is
aluminum hydroxide. The small concentration of hydroxide ions in the ammonia solution is enough to precipitate some Al3+
ions. Since there is already a Kb equilibrium present in the ammonia solution, the net-ionic equation for the precipitation is
written according to method 2 with the chemistry being the reaction of Al 3+ with the hydroxide ions. In other words, we use
the Kb equilibrium of ammonia to obtain the hydroxide ions needed for the chemical reaction; Al 3+ and NH3 are
written on the reactant side.
First the coefficients in the ammonia equilibrium are tripled to provide 3 hydroxide ions for reaction with the Al3+.
3 NH3 + 3 H2O ⇌3 NH4+ + 3 OH–
Al3+ ⇌ Al3+
3+
Al (aq) + 3 NH3(aq)+ 3 H2O(l)  Al(OH)3(s) + 3 NH4+(aq)
(equilibrium already present)
(species added)
(result of chemical rxn on product side)
2.
Using ammonia to neutralize acids.
Often ammonia is used as a base to consume extra acid in solution and therefore shift an equilibrium. In this
case it is the ammonia directly reacting with the hydronium ions, not the hydroxide ions present in the
ammonium solution. As an example, consider the following equilibrium system:
Ba3(PO4)2(s) + 2H3O+(aq) ⇌2 HPO42–(aq) + Ba2+(aq) + 2 H2O(l)
Adding aqueous ammonia to this system reduces the hydronium ion concentration. This will result in a shift of
the equilibrium to the left. Use the following procedure to write the net-ionic equation:
Ba3(PO4)2(s) + 2H3O+ ⇌2 HPO42– + Ba2+ + 2 H2O
2 NH3 ⇌
2 NH3
(original equilibrium)
(ammonia added)
Ba3(PO4)2(s) + 2 NH4+(aq)  2 HPO42–(aq) + Ba2+(aq)+ 2 NH3(aq)
(result of chemical rxn on reactant side,
waters have been canceled)
Note that, in the above resulting reaction, we can see that the NH3 acts to deprotonate the hydrogen phosphate ion, thus
providing phosphate ions to precipitate as insoluble barium phosphate.
3.
Using Ammonia to form complex ions. (Recall that this is categorized as strictly a Lewis acid-base reaction.)
For many metal cations, adding excess ammonia results in the formation of a soluble complex ion. Remember too that the
formation constants, Kf, for complex ions tend to be high, thus these provide a method for dissolution of insoluble ionic
compounds. Cd(OH)2(s) is one example that can be dissolved by adding excess ammonia. When ammonia is in high enough
concentration, it will act a a Lewis acid and replace the OH– ions on the cadmium forming the complex ion Cd(NH3)42+. Use
method 2 to write the net-ionic equation realizing Cd2+ is reacting with excess NH3, thus shifting the original equilibrium to
the left:
Cd(OH)2(s) ⇌
4 NH3
2 OH– + Cd2+
⇌
4 NH3
(original equilibrium)
(ammonia added)
Cd(OH)2(s) + 4 NH3(aq)  2 OH–(aq) + Cd(NH3)42+(aq)
(result of chemical rxn on product side)
Overview of the Procedure
Prepare your notebook prior to coming to lab by writing in your notebook the title, purpose, background information
and a summary of the procedural notes given below. You can then either hand write each equilibrium system,
numbers 1 to 5, onto separate pages in your notebook, making sure to list all reagents with concentrations, OR you
can cut the printed pages to fit and tape/paste them directly onto SEPARATE pages in your notebook.
For each equilibria investigated, use only the reagents listed to set-up the equilibrium. Use small test tubes. After the
equilibrium is established, again use only the reagents listed and/or a temperature change to shift the equilibrium
concentrations to the left or right.
Record all observations that indicate the equilibrium concentrations have shifted and then write the net-ionic equation that
produced the shift in equilibrium concentrations.
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Other Procedural notes:
1. Except for equilibrium system number 1 where a 3 mL quantity is used, use one mL quantities where indicated to
establish the initial equilibrium in the test tube.
2.
Reagents are added slowly, drop-wise to shift the equilibrium.
3.
Do not add more reagent than the minimum needed to produce the desired effect. If the concentration of a reagent
becomes too high, you may not be able to shift the equilibrium concentrations back again or an unexpected side-reaction
may occur. If you carefully add reagents drop-wise, and mix thoroughly after each addition, you should be able to shift
the equilibrium concentrations back and forth several times.
4.
When needed, heat test tubes in a hot water bath. Use an ice bath for cooling.
5.
Rinse your test tubes thoroughly with de-ionized water between investigations to avoid contamination.
6.
All reagents should be disposed of in a proper Hazardous Waste container, not in the sink.
7.
Reagent solutions are aqueous, unless otherwise noted.
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Name:
Lab section: MW or TTh
Partner’s Name:
1. CoCl42–(aq) (blue) + 6 H2O(l) ⇌Co(H2O)62+(aq) (pink) + 4 Cl–(aq)
Reagents:
0.1 M CoCl2 in 95% ethanol (Measure 3 mL into a clean, dry test tube)
Caution: CoCl2 in ethanol is flammable, toxic and irritating!
12 M hydrochloric acid, use drop wise (Caution: Use in the hood with gloves!)
deionized water, use drop-wise
Prepare the equilibrium by adding deionized water, 1 drop at a time and mixing well after each drop, to 3 mL of the
CoCl2 solution until the solution turns a light purple color.
a.
b.
c.
Equilibrium concentrations in equation (1) are shifted left by adding:
i)
Observation:
ii)
Net-ionic equation:
Equilibrium concentrations in equation (1) are shifted back to the right by adding:
i)
Observation:
ii)
Net-ionic equation:
Prepare a fresh 3 mL sample as instructed above. Separate this sample into two different test tubes. Place one test
tube in hot water and the other in ice water. Wait five minutes.
i)
Observations:
ii)
Is the reaction in equation (1) exothermic or endothermic as it is written in equation (1)? How do you
know?
iii)
Rewrite the equilibrium reaction by adding heat to the appropriate side of the equation.
iv)
Does the value of the equilibrium constant, K, increase, decrease or stay the same as temperature increases
for this reaction as it is written in equation (1)?
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Name:
Lab section: MW or TTh
Partner’s Name:
2. BaC2O4(s) + H3O+(aq) ⇌HC2O4–(aq) + Ba2+(aq) + H2O(l)
Reagents
a.
c.
0.1 M ammonium oxalate, use 1 mL
6 M ammonia solution, use drop-wise
BaC2O4(s) is produced by mixing:
i)
b.
0.1 M barium chloride, use 1 mL
1M hydrochloric acid, use drop-wise
Net-ionic equation:
The equilibrium in equation (2) is shifted right by adding:
i)
Observation:
ii)
Net-ionic equation:
The equilibrium in equation (2) is shifted back left by adding:
i)
Observation:
ii)
Net-ionic equation:
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Name:
Lab section: MW or TTh
Partner’s Name:
3. AgCl(s) (white) + 2 NH3(aq) ⇌[Ag(NH3)2]+(aq) (colorless) + Cl–(aq)
Reagents:
a.
c.
1 M hydrochloric acid, use drop-wise
AgCl(s) is produced by mixing:
i)
b.
0.01 M silver nitrate, use 1 mL
6 M ammonia solution, use drop-wise
Net-ionic equation:
The equilibrium in equation (3) is shifted right by adding:
i)
Observation:
ii)
Net-ionic equation:
The equilibrium in equation (3) is shifted back left by adding:
i)
Observation:
ii)
Net-ionic equation:
Note: You should get a new net-ionic here (The equilibrium is not shifted simply due to an increase in the
Cl– ion concentration!).
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Name:
Lab section: MW or TTh
Partner’s Name:
4. Copper ions and ammonia produce multiple equilibria.
4a.
4b.
2 H2O(l) + Cu2+(aq) (pale blue) + 2 NH3(aq) ⇌Cu(OH)2(s) (light blue solid) + 2 NH4+(aq)
Cu(OH)2(s) + 4 NH3(aq) ⇌[Cu(NH3)4]2+(aq) (deep blue solution) + 2 OH–(aq)
Reagents:
a.
1 M hydrochloric acid, use drop-wise
Cu(OH)2 (s) is produced by mixing:
i)
b.
0.1 M copper (II) nitrate, use 1 mL
6 M ammonia solution, use drop-wise
Net-ionic equation:
The equilibrium in equation (4a) is shifted left by adding:
i)
Observation:
ii)
Net-ionic equation:
c.
Using the sample from step (b), shift the equilibrium back right so that you have solid Cu(OH)2 present in the test
tube. What reagent did you add to accomplish this shift?
d.
Use the sample prepared in step (c), that should contain some solid Cu(OH)2, and now consider the equilibrium
in equation (4b). The equilibrium in equation (4b) is shifted right by adding:
i)
Observation:
ii)
Net-ionic equation:
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Lab section: MW or TTh
5. Zinc ions and bases produce multiple equilibria.
5a.
5b.
5c.
Zn2+(aq) + 2 OH–(aq) ⇌Zn(OH)2(s)
Zn2+(aq) + 4 OH–(aq) ⇌[Zn(OH)4]2–(aq)
Zn2+(aq) + 4 NH3(aq) ⇌[Zn(NH3)4]2+(aq)
Reagents:
a.
Name:
Partner’s Name:
0.1 M zinc nitrate, use 1 ml
1 M sodium hydroxide, use drop-wise
1M hydrochloric, use drop-wise
6 M ammonia solution, use drop-wise
Produce Zn(OH)2(s) two DIFFERENT ways:
i)
Reactants:
Observation:
Net-ionic equation:
ii)
Reactants:
Observation:
Net-ionic equation:
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Name:
Lab section: MW or TTh
Partner’s Name:
b.
Using one of the previous methods to produce Zn(OH) 2(s), dissolve the Zn(OH)2(s) three DIFFERENT ways.
Use the Zn(OH)2(s)  Zn2+ (aq)+ 2 OH– (aq) equilibrium as the starting point for your net ionic equations.
Each net ionic equation MUST CONTAIN Zn(OH)2(s) when complete!
i)
Reactant added to dissolve Zn(OH)2(s):
Observation:
Net-ionic equation:
ii)
Reactant added to dissolve Zn(OH)2(s):
Observation:
Net-ionic equation:
iii)
Reactant added to dissolve Zn(OH)2(s):
Observation:
Net-ionic equation:
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