Overview 

In this experiment you investigated Le Châtelier's Principle, as applied to six equilibrium systems. Le Châtelier's principle states that, when a change is made to a system that is already in equilibrium, the system reacts in such a way as to return to equilibrium. For a chemical reaction system, when the system "reacts so as to return to equilibrium", this means that the system will either react in the forward direction (so as to produce additional product) or in the reverse direction (so as to consume some of the product).

Le Le Châtelier's principle is very important in the chemical and chemical engineering industries. The goal in these industries is to get as much product as possible (which can be then sold) for the given amounts of starting materials. Chemists and chemical engineers designing new processes make great use of Le Le Châtelier's Principle in designing their systems

Results, Explanations, and Discussion 

Since this was a "qualitative" experiment, there are obviously no calculations to go through for your report..

Instead, let's try to understand each step of the procedure (what you did and observed in lab) and try to relate this to Le Châtelier's Principle. Sometimes, given a long procedure, it has been hard for students to appreciate exactly where Le Le Châtelier's Principle enters into the discussion in some of the reactions. You should have ready both your data pages (Pages 157-161) and also the procedure pages (Pages 149-155) for this.

Be careful!  In your report, you are asked to tell what the stress was that was applied to the equilibrium reaction.  The stress is not the chemical you added that caused a change.  The stress can only be a change in the concentration of one of the species that is written in the equilibrium reaction equation. So, for example, if you add HCl to the system, and it causes a change, the stress is not "adding HCl": the stress must be interpreted in terms of the component species of the equilibrium only. 

A. The Chromate-Dichromate Equilibrium

Pages 149-150 and Page 157.

The equilibrium between chromate ion (CrO₄^2-) and dichromate ion (Cr₂O₇^2-) was studied. The indication for which way the reaction shifts when changes were made in the system is based on the color of these two ions in solution: chromate ion is bright yellow in solution, whereas dichromate ion is bright orange. So, if the solution turned orange when you added something to it, the equilibrium was shifted to the right; if the solution turned yellow, the equilibrium was shifted to the left.

CrO₄^2- (yellow) + 2H⁺ = Cr₂O₇^2- (orange) + H₂O

1. In Step A1 on Page 150, you added sulfuric acid (H₂SO₄) to a sample of potassium chromate, K₂CrO₄. What happened to the color of the K₂CrO₄ when you did this? (see your observation on Page 157). Based on the color change you saw, which way did the equilibrium shift, right or left? Sulfuric acid is a source of H⁺: any source of H⁺ would have caused the same shift in this equilibrium. The "stress" applied to the system is not the addition of sulfuric acid!  The "stress" must be explained only in terms of the species present in the equilibrium reaction equation: sulfuric acid is a source of hydrogen ions and it is an increase in the hydrogen ion concentration that causes the equilibrium to shift.
2. This is an example of where students have trouble interpreting what they have done, and how the equilibrium is affected by what has been done to the system. First of all, NaOH is clearly not one of the species involved in the equilibrium written above. Yet, when you added a few drops of NaOH to the mixture from Step A1, you should have seen a change. Although NaOH is not part of the equilibrium above itself, NaOH reacts with one of the components of the equilibrium, effectively removing that species from the equilibrium. NaOH is a strong base and reacts with acids: NaOH reacts with the hydrogen ion in the equilibrium system and removes it. If the H⁺ is removed from the equilibrium system, the system will have to react in the direction that replaces some of the H⁺ to restore equilibrium. Is this consistent with the color change you observed?

H⁺ (from the equilibrium) + OH^- (from the NaOH) H₂O

The change in the system occurred when you added the NaOH, but the "stress" to the system must be described in terms of the equation for the equilibrium

CrO₄^2- (yellow) + 2H⁺ = Cr₂O₇^2- (orange) + H₂O

So, since NaOH effectively removes hydrogen ion from the system by converting it to water, the "stress" from the equilibrium's point of view would be a decrease in the hydrogen ion concentration.

B. The Iron(III) - Thiocyanate Equilibrium

Pages 150-151 and Page 158

As you saw in the previous experiment, Fe³⁺ ion and SCN^- react with each other to form a red complex ion, [FeSCN]²⁺. In the previous experiment, you used very dilute (0.00200 M) solutions of Fe³⁺ and SCN^- and so the color of the product was a faint reddish-orange. In this experiment, you use much more concentrated solutions (0.1 M), and so the product solution is a deep red (almost the color of blood).

Fe³⁺(colorless) + SCN^- (colorless) = [FeSCN]²⁺ (blood red)

In this system, we can tell in which direction the equilibrium shifts when a change is made by monitoring the intensity of color of the system. If the system gets darker red in color, then the equilibrium must be shifting to the right (toward producing more of the colored product). If the color of the system gets fainter (or disappears altogether), the equilibrium must be shifting toward the left (toward the colorless components). You had to be very observant on this part to see the change in intensity of the red color when a drop of reagent was added: if you waited too long, the localized change in color as the droplet enters the equilibrium reaction would fade out.

1. This step of the procedure is just setting up the equilibrium system: you mixed some Fe(NO₃)₃ and some KSCN to generate the system in the equation above. Ionic reactions are very fast, and the system came to equilibrium within a few seconds, which was indicated by the appearance of the deep red color. Since this color is so deep, you were told to divide the mixture into several wells, and to add water to it until it had been diluted enough that you would be able to see color changes in the system.
2. In this step, you added additional Fe(NO₃)₃ dropwise to one of the wells containing the colored equilibrium mixture. If additional Fe³⁺ was added to the system which was already in equilibrium, this would have been too much Fe³⁺ present in the system. The system would have to react (either to the left or to the right) to remove some of the additional Fe³⁺ from the mixture. In which direction would the system react to remove Fe³⁺? If the system reacted to remove the excess Fe³⁺, would the net result be more or less of the colored product? Based on your observation (Page 158) as to whether the mixture turned darker or lighter when the additional Fe³⁺ was added, which way did the equilibrium shift? Was your color observation consistent with the idea that the system would have to remove some of the additional Fe³⁺? You should realize that the change in the system would have been taken with the first drop of additional Fe³⁺ added: if you were not observant, you might have missed the color change!
3. In this step, you added additional KSCN dropwise to one of the wells containing the colored equilibrium mixture. If additional SCN^- were added to the system which was already in equilibrium, this would have been too much SCN^- present in the system. The system would have to react to remove some of the additional SCN^- from the system. Based on this discussion, and your observations of the system from Page 158, which way did the equilibrium shift: toward the right (more product, more color) or to the left (less product, less color).
4. Here's a situation which is similar to that raised in section A2 above: NaOH is not part of the equilibrium reaction given. However, the procedure (Page 151) has a special note, reminding you that Fe(OH)₃ is insoluble in water. If Fe(OH)₃ is insoluble in water, then adding NaOH to the equilibrium system should cause the reaction below to take place: adding NaOH precipitates Fe³⁺ as Fe(OH)₃ from the equilibrium system. If Fe³⁺ is removed from the equilibrium system, then the system will no longer be in equilibrium and will have to react in the direction that restores some Fe³⁺. Based on your observations (Page 158) and this discussion, which way would the equilibrium shift if Fe³⁺ is removed from the system? Would the amount of colored complex increase or decrease?

Fe³⁺ (from the equilibrium) + 3OH^- (from the NaOH) Fe(OH)₃ (solid precipitate)

C. The Cobalt(II)-Chloride Ion Equilibrium

We omitted this part of the experiment!

D. The Equilibrium in a Saturated NaCl Solution

Page 152 and Page 159

In Parts A and B above, we considered equilibrium systems that were based on true chemical reactions. In this Part, we consider a physical equilibrium based on solubility.

When a crystalline ionic solid is dissolved in water, particles of solid enter the solvent to form the solution. But this only happens up to a point. As the concentration of ions in solution increases, the likelihood that ions will be attracted by the remaining ions in the crystal increases, and ions which were dissolved begin to re-enter the crystals of solid. Eventually, the rate at which ions re-enter the crystals becomes equal to the rate at which ions leave the crystals, and a "steady state" is established. Once this steady state is reached, the net concentrations of the ions in solution no longer increases and the solution is said to be "saturated". Since we have two opposing processes going on at the same rate, an equilibrium exists between undissolved solute and the saturated solution.

Na⁺ (in solution) + Cl^- (in solution) = NaCl (undissolved solid)

1. This is just setting up the 24-well plate for the tests that follow. Perhaps you noticed that there were some crystals at the bottom of the bottle of saturated NaCl solution as you took your samples. These were to ensure that the solution would be saturated with NaCl when you took some. Saturated NaCl contains Na⁺ and Cl^- ions each at a concentration of 5.4 M (moles/L).
2. In this step, you added a few drops of concentrated hydrochloric acid (HCl) to one of your saturated NaCl samples. Concentrated HCl contains H⁺ and Cl^- ions each at a concentration of 12.0 moles/L. In other words, when you added concentrated HCl to the saturated NaCl solution you were adding a large amount of chloride ion, Cl^-, to the solution. The solubility equilibrium should have responded by shifting in the direction which would remove some of the excess chloride ion from solution. What did you observe (Page 159) in your solution that indicated that excess chloride ion was being removed from the solution?
3. In this part of the experiment, you added 3 M HCl to a second sample of saturated NaCl. Notice that 3 M HCl contains Cl^- ion at a lower concentration than in the saturated NaCl solution (5.4 M) you added it to. When you added 3 M HCl to the saturated NaCl solution, you probably did not observe any change (Page 159). Adding the 3 M HCl to the saturated sodium chloride solution actually lowers the concentration of chloride ion in the solution.

E. The Equilibrium of Saturated Barium Chromate

This is another example of a solubility equilibrium, with a little "twist" coming up in Step 2.

Ba²⁺ (in solution) + CrO₄^2- (in solution; yellow) = BaCrO₄ (yellow solid)

1. In this step, you just generate the equilibrium by mixing BaCl₂ and K₂CrO₄ solutions. A yellow solid should have precipitated, with a clear yellow solution above it. The clear yellow solution is a saturated BaCrO₄ solution (see your observations on Page 159).
2. In this step, you added a few drops of concentrated hydrochloric acid, HCl, to the mixture prepared in Step 1. What happened to the precipitate of BaCrO₄ (see your observations on Page 159) produced in Step 1 when you did this? On Page 159, you are asked a question about why HCl affects the BaCrO₄ equilibrium, even though HCl does not appear as part of the equilibrium reaction above for BaCrO₄. Look back at Part A of this experiment and see if you can figure why an acid like HCl affects the BaCrO₄ equilibrium. What does addition of an acid do to CrO₄^2- ion in solution, and how would this affect the BaCrO₄ equilibrium?

F. Equilibrium of Saturated Ferric Hydroxide

In this part you study really two equilibria involving Fe³⁺ ion and hydroxide ion, OH^-.

Fe³⁺ (in solution) + 3OH^- (in solution) = Fe(OH)₃ (reddish precipitate)

and

Fe³⁺ (in solution) + 4OH^- (in solution) = Fe(OH)₄^- (in solution)

which can also be written as

Fe(OH)₃ (reddish solid) + OH^- (in solution) = Fe(OH)₄^- (in solution)

1. In this step, you are merely generating the first equilibrium written above. When you add sodium hydroxide solution to the iron(III) nitrate solution, a reddish precipitate of Fe(OH)₃ forms (see your observations on Page 160).
2. When you add HCl (a strong acid) to the mixture from Step 1, the acid reacts with hydroxide ion according to the reaction

H⁺ + OH^- H₂O.

• As the introduction to Part F states, an acid effectively removes OH^- from the equilibrium system. What happened to the precipitate from Step 1 when you added the HCl (see your observations on Page 160) and how is this consistent with the equilibrium?

3. In this part you add additional NaOH to the second well containing Fe(OH)₃ you generated in Step 1. Adding additional NaOH brings into play the second equilibrium. How are your observations about what happened to the precipitate of Fe(OH)₃ when additional NaOH added consistent with the second (and third) reactions above?

G. Equilibrium of Ammonia in Water

We omitted this part of the experiment!

H. Equiliibrium of Saturated AgCl

In this Part, you study two related equilibria involving silver chloride. You'll have to pay close attention in the following discussion to keep straight which equilibrium we're talking about, since the procedure is quite involved.

When silver ion (Ag⁺) and chloride ion (Cl^-) are combined, a precipitate of silver chloride forms, leaving a saturated solution of AgCl above the solid. The silver ions and chloride ions remaining in solution are in equilibrium with the precipitate

AgCl ( solid) = Ag⁺ (in solution) + Cl^- (in solution)

Note that this solubility equilibrium has been written in the opposite sense to the solubility equilibria in Parts D and E above: this doesn't really matter, since an equilibrium reaction "goes in both directions". Just keep straight that your explanations should be in terms of the way the reaction equation is given in this Part.

Silver ion also forms an equilibrium with ammonia

Ag⁺ (in solution) + 2NH₃(in solution) = Ag(NH₃)₂⁺ (in solution)

1. In this step, you are just generating the first equilibrium above (the solubility equilibrium for AgCl). When you added HCl (a concentrated Cl^- ion solution) to AgNO₃, a white precipitate of AgCl formed. The solution above the precipitate was a saturated solution of Ag⁺ and Cl^- ions.
2. All you are doing in this step is removing most of the saturated solution from the precipitate in Step 1. Realize though, that the precipitate in Step 1 is still wet with saturated AgCl solution after you remove the bulk of the liquid from it. The solution you remove is a saturated AgCl solution. The mixture left in the well from Step 1 is solid AgCl still in equilibrium with the remaining coating of saturated AgCl solution.
3. In this step, you play with both of the equilibria above. When you add additional AgNO₃ solution to the liquid removed in Step 2, this is an increase in the concentration of Ag⁺ ion. What direction would the first equilibrium above shift if additional Ag⁺ ion were added? How is this reflected in your observation on Page 161? What is the precipitate that formed when you added AgNO₃ to the liquid from Step 2? Then, you added concentrated ammonia to the mixture. What happened to the precipitate that had just formed? This is the second equilibrium above: when ammonia is added to a solution containing Ag⁺ ion, the ammonia forms a complex with the Ag⁺ ion and removes it from solution. What happens to the precipitate if Ag⁺ ion is removed from solution by ammonia?
4. When you add ammonia to the mixture of solid AgCl and saturated AgCl solution, what happens to the precipitate? As we saw in Step 3 above, ammonia removes Ag⁺ ion. What would have to happen to the precipitate if the Ag⁺ ion were removed from the solution coating it? Then you added some nitric acid (HNO₃). Nitric acid reacts with ammonia (ammonia is a base): HNO₃ + NH₃ NH₄NO₃. The ammonia is thus removed from the system. If ammonia is removed from the second equilibrium above, how does that equilibrium shift? If the second equilibrium above shifts, how does that affect the first equilibrium? This is getting really tricky, huh!!