Lab – 19A:
Investigating Chemical Equilibrium
Purpose/Objective:
1.	To recognize the macroscopic properties of five chemical systems at equilibrium.
2.	To observe shifts in equilibrium concentrations as stresses are applied to the systems.
3.	To observe a shift in equilibrium concentrations associated with changes in temperature.
4.	To explain the observations obtained by applying Le Chatelier’s principle.
Materials:
As on the handout.
Procedure:
As on the handout.
Results:
Part 1: Equilibrium Involving Thymol Blue

Thymol Blue (Blue Form) + H+    Thymol blue+ (yellow form)
REAGENT ADDED	STRESS	COLOR OBSERVATION	DIRECTION OF EQUILIBRIUM SHIFT
HCl (step 3)	H+	Green	Right
HCl (step 4)	H+	Yellow	Right
NaOH (step 5)	OH-	Green  Yellow	Left
NaOh (step 6)	OH-	Blue	Left

Part 2: Equilibrium Involving Thiocyanatoiron III Ion

Fe3+          +            SCN           FeSCN2+
                                               (Light Yellow)              (Clear)               (Dark Red)
++ Solution turns amber when diluted
REAGANT ADDED	STRESS (Ion Added)	SPECTATOR ion	COLOR OBSERVATION	DIRECTION OF EQUILIBRIUM SHIFT
KCl (Test Tube B)	None	K+, Cl-	No Change	-
Fe(NO3)3 (Test Tube C)	Fe3+	NO3	Darker	Forward
KSCN (Test Tube D)	SCN-	K+	Darker	Forward
NaOH (Test Tube E)	OH-	Na+	Lighter	Reverse


Part 3: Equilibrium Involving Cobalt (II) Complexes

                                         Co(H2O)62           +                2Cl-                             Co(H2O)4Cl2 + 2H2O
                                   (Light Red/Pink)      (+ Energy, endothermic)                       (Dark Purple)
++ Reaction is endothermic because solution turned purple
	STRESS	COLOR Observation	DIRECTION OF EQUILIBRIUM SHIFT
Step 3	H2O	Violet to Pink (Light Red)	Reverse
Step 4	Heat 	Pink to Violet (Purple)	Forward
Step 5	Cold	Violet to Pink (Light Read)	Reverse


Part 5: Equilibrium Involving Copper (II) Complexes

NH3 (aq.) + H2O (l)     NH4  (aq.) + OH (aq.)
Cu2+ (aq.) + 2OH- (aq.)     Cu (OH)2 (s)
Cu(H2O)2+ (aq.) + 4NH3   Cu(NH3)42+ (aq.) + 4H2O (l)
                                            (Light Blue)                              (Deep Blue)

	0.1M CuSO4	+3 Drops NH3	+ More NH3	1M HCl Added
APPEARANCE	Light Blue	Light Blue (precipitate)	Dark Blue (solution)	Light Blue
(solution)

Questions And Calculations:

1.	In part 1, if we increase the molarity of NaOH from 0.1M to 0.2M, it will actually take less number of drops in order to change the color of the solution.

2.	When you add 6.0M NaOH into the iron (III) thiocyanate ion equilibrium system, the concentration of Fe3+ ion decreases. This causes the equilibrium system to shift to the left (reactant) side. This is why the solution becomes lighter. Fe (OH)3 is also formed during the experiment.

3.	If the hydrated cobalt (II) ion complex was refrigerated, the temperature would decrease. Since this is an endothermic reaction a decrease in temperature will actually cause the equilibrium system to move to the left, favouring the reactants. When the reactants are favoured the color of the solution will get lighter and change to pink/light red.

4.	According to the following equilibrium equation:
Co (H2O)62+ (aq.)  +  2Cl- (aq.) + energy   Co(H2O)4Cl2 (aq.) + 2H2O (l)
We can be sure that in this equation energy is on the reaction side (endothermic reaction) because when we added more hear (increased the temperature) into this solution we found out that the solution changed the colour to purple from light pink. This showed us that the equilibrium moved to the right (product) side. This is why temperature has to be on the reactants side (endothermic) in order for this shift to be possible.

5.	If you add Sodium chloride (NaCl) into the hydrated Cobalt (II) ion equilibrium, the concentration of the chloride ions (Cl-) would increase and some sodium ions (NA+) would be formed. Because of this the concentration of reactants increases, which causes the equilibrium system to move to the right (products) side. Therefore, the solution turns purple.
10. When you add HCl into the complex ion Cu(NH3)43+, it reacts with ammonia and decreases the concentration of ammonia. This decreases the concentration of reactants, which cause the equilibrium system to move to the left (reactants) side. Since the system moves towards the left, the solution gets very light and turns into light blue.