Purpose
To examine the equilibrium system (represented by the net ionic equation Co(H2O)6^2+ + 4Cl- ---> CoCl4^-2 + 6H2O) after subjecting various stresses to the system, and observing Le Chatelier's Principle work.
Procedure
1. Place a beaker containing nearly 100mL of water on a hot plate. Do not let the water boil. Go on to the next steps while waiting for the water to heat up.
2. Label a well plate from left to right at the top of the columns with numbers 1-6. Down the left side, label the rows with A-D.
3. Put on latex gloves for safety precautions. Obtain solutions of CoCl2, HCL, distilled water and AgNO3.
4. Place five drops of CoCl2 solution in all 24 wells.
5. Add two drops of HCL in the wells of column one (A1, B1, C1, D1). Add four drops of HCL in the wells of column 2, six drops in the wells of column 3, eight drops in the wells of column 4, ten drops in the wells of column 5, and twelve drops in the wells of column 6. Mix the contents with a toothpick and rinse the toothpick. Record the color on the data table. Row A should be your control.
6. In row B, add one drop of HCL in each well , stir it with the toothpick (make sure to rinse after each use), and record observations.
7. In row C, add five drops of distilled water, stir it with the toothpick, and record observations
8. In row D, add five drops of AgNO3, stir it with the toothpick, and record observations.
9. In a test tube, place 5mL of cobalt solution. Add HCL just enough to get to a purple color (between blue and pink).
10. Place the test tube in the beaker of hot water from step 1 until a color change occurs. Record observations.
11. Obtain a 250mL beaker and place ice cubes and water to prepare an ice bath. Place the test tube in the ice bath until a color change occurs. Record observations.
12. Let the teacher dispose the chemicals. Clean up work area and wash hands before leaving.
2. Label a well plate from left to right at the top of the columns with numbers 1-6. Down the left side, label the rows with A-D.
3. Put on latex gloves for safety precautions. Obtain solutions of CoCl2, HCL, distilled water and AgNO3.
4. Place five drops of CoCl2 solution in all 24 wells.
5. Add two drops of HCL in the wells of column one (A1, B1, C1, D1). Add four drops of HCL in the wells of column 2, six drops in the wells of column 3, eight drops in the wells of column 4, ten drops in the wells of column 5, and twelve drops in the wells of column 6. Mix the contents with a toothpick and rinse the toothpick. Record the color on the data table. Row A should be your control.
6. In row B, add one drop of HCL in each well , stir it with the toothpick (make sure to rinse after each use), and record observations.
7. In row C, add five drops of distilled water, stir it with the toothpick, and record observations
8. In row D, add five drops of AgNO3, stir it with the toothpick, and record observations.
9. In a test tube, place 5mL of cobalt solution. Add HCL just enough to get to a purple color (between blue and pink).
10. Place the test tube in the beaker of hot water from step 1 until a color change occurs. Record observations.
11. Obtain a 250mL beaker and place ice cubes and water to prepare an ice bath. Place the test tube in the ice bath until a color change occurs. Record observations.
12. Let the teacher dispose the chemicals. Clean up work area and wash hands before leaving.
Data Table
CoCl2 at room temperature: 27 degrees Celcius - purple
CoCl2 in hot water: 35 degrees Celcius - blue
CoCl2 in cold water: 4 degrees Celcius - pink
CoCl2 in hot water: 35 degrees Celcius - blue
CoCl2 in cold water: 4 degrees Celcius - pink
Conclusion
This lab, in conclusion, shows the effects of adding more concentrations of reactants or products and the changing of temperature that would prove Le Chatelier’s principle. As stress was placed on the system, the equilibrium shifted and changed the colors on each well. Both times when temperature was increased and when HCl was added, the equilibrium shifted right (it turned blue). When water, AgNO3, and temperature decreased, the equilibrium shifted left (it turned pink).
Discussion of Theory
Le Chatelier’s principle states that if we alter something that is on the products or reactants side of an equilibrium, then the system will respond in order to re-establish equilibrium. The stresses are either changes in concentration of concentration (such as adding more of a product/reactant), changes in temperature, or changes in pressure. In this lab, only the changes in temperature and changes in concentration were studied.
Adding a solid or a liquid does not effect the concentration, however, thus, it does not affect the equilibrium. Gases and aqueous solutions would be the only ones to affect equilibrium. If one were to increase concentrations of the reactants, the equilibrium will shift right. If one were to increase concentrations of the products, the equilibrium will shift left. If one were to decrease concentrations of the reactants, the equilibrium will shift left. If one were to decrease concentrations of the products, the equilibrium will shift right.
To increase temperature is to add heat, which would be an endothermic reaction. In endothermic reactions, heat is a “reactant”. Therefore, if more heat is added, the equilibrium shifts right. Removing this heat would shift left. In exothermic reactions, heat is a “product”. This means that if heat were added in an exothermic reaction, the equilibrium would shift left. Removing this heat would shift right.
Only changes in temperature would cause changes in the Kc or Kp value. If the equilibrium shifts to the right, the K value increases. If the equilibrium shifts to the left, the K value decreases.
The amount of volume determines the pressure which also shifts the equilibriums. Decreasing the volume increases the pressure. When this happens, the equilibrium shifts towards the side with a fewer sum of moles. Increasing the volume would decrease the pressure which shifts the equilibrium towards the side with more moles.
Inert gases are gases not involved in the reaction. If they are added to the system at a constant pressure, the volume increases. When the volume increases, pressure decreases and equilibrium shifts towards the side with more moles. It’s basically already explained in the paragraph above. If the inert gas is added at a constant volume, the concentrations and partial pressure of the reactants and products don’t change; thus, the equilibrium doesn’t change either.
Catalysts speed up the process of reaching equilibrium. However, catalysts do not change the equilibrium point. They lower activation energy, provide alternate reaction pathways, and make a higher percentage of collisions.
Adding a solid or a liquid does not effect the concentration, however, thus, it does not affect the equilibrium. Gases and aqueous solutions would be the only ones to affect equilibrium. If one were to increase concentrations of the reactants, the equilibrium will shift right. If one were to increase concentrations of the products, the equilibrium will shift left. If one were to decrease concentrations of the reactants, the equilibrium will shift left. If one were to decrease concentrations of the products, the equilibrium will shift right.
To increase temperature is to add heat, which would be an endothermic reaction. In endothermic reactions, heat is a “reactant”. Therefore, if more heat is added, the equilibrium shifts right. Removing this heat would shift left. In exothermic reactions, heat is a “product”. This means that if heat were added in an exothermic reaction, the equilibrium would shift left. Removing this heat would shift right.
Only changes in temperature would cause changes in the Kc or Kp value. If the equilibrium shifts to the right, the K value increases. If the equilibrium shifts to the left, the K value decreases.
The amount of volume determines the pressure which also shifts the equilibriums. Decreasing the volume increases the pressure. When this happens, the equilibrium shifts towards the side with a fewer sum of moles. Increasing the volume would decrease the pressure which shifts the equilibrium towards the side with more moles.
Inert gases are gases not involved in the reaction. If they are added to the system at a constant pressure, the volume increases. When the volume increases, pressure decreases and equilibrium shifts towards the side with more moles. It’s basically already explained in the paragraph above. If the inert gas is added at a constant volume, the concentrations and partial pressure of the reactants and products don’t change; thus, the equilibrium doesn’t change either.
Catalysts speed up the process of reaching equilibrium. However, catalysts do not change the equilibrium point. They lower activation energy, provide alternate reaction pathways, and make a higher percentage of collisions.
Sources of Error
In the lab, distilled water was suppose to be used, but we used tap water instead. Tap water has impurities that could have changed the observation of the colors slightly. Not to mention, the beaker used to hold the tap water was unclean itself. There were multiple rinses to get the dirty particles out, but not all of them were gone. They were most definitely not tiny air bubbles. With distilled water, the color might have been a much more clear pink that is closely similar to well A1. Another mechanical error were the size of the dropper of the AgNO3 solution. The dropper for that was fairly larger than that of the other droppers which means the amount of AgNO3 dropped could have been an unequal ratio to what was in row D.
Pre-lab Questions
1. Le Chatelier’s Principle states that if a system in equilibrium is put under a stress, the system will respond by shifting to reduce the stress.
2. When the equilibrium is reached, there is no net change in concentrations of reactants. The rate of the forward and reverse reactions are equal.
3. The stresses studied in the experiment are changes in concentrations and the changes of temperature on the system.
4. Cobalt (II) chloride hexhydrate
5. A few safety precautions are to wear goggles and aprons at all times. Make sure to wear gloves when handling with the hydrochloric acid and silver nitrate. If hydrochloric acid is touched, it will irritate the skin and it will burn. Silver nitrate will turn fingers into a silvery color and will stain skin and clothes.
6. A. If HCl is added, the equilibrium will shift to the right because H+ would be increased.
B. If H2O is added, the equilibrium will not shift because water is a liquid. Solids and liquids do not affect concentration.
C. If NaOH is added, the equilibrium will shift to the left because it would reduce and neutralize the amount of H+ ions.
2. When the equilibrium is reached, there is no net change in concentrations of reactants. The rate of the forward and reverse reactions are equal.
3. The stresses studied in the experiment are changes in concentrations and the changes of temperature on the system.
4. Cobalt (II) chloride hexhydrate
5. A few safety precautions are to wear goggles and aprons at all times. Make sure to wear gloves when handling with the hydrochloric acid and silver nitrate. If hydrochloric acid is touched, it will irritate the skin and it will burn. Silver nitrate will turn fingers into a silvery color and will stain skin and clothes.
6. A. If HCl is added, the equilibrium will shift to the right because H+ would be increased.
B. If H2O is added, the equilibrium will not shift because water is a liquid. Solids and liquids do not affect concentration.
C. If NaOH is added, the equilibrium will shift to the left because it would reduce and neutralize the amount of H+ ions.
Post-lab Questions
1. A. with the addition of HCl, the system shifted right.
B. with the addition of water, the system shifted left.
C. with the addition of AgNO3, the system shifted left.
D. With the increase in temperature, the system shifted right.
E. With the decrease in temperature, the system shifted left.
2. When the HCl was added, it split itself into ions of H+ and Cl-. In that case, ions of Cl- increased on the reactant side, so the equilibrium shifted right to maintain balance. When water was added, it increased the amount of water already on the product side, so it shifted back to the left to reach equilibrium once again.
3. With the addition of AgNO3, a small precipitate was formed in the wells of row D. This is because AgNO3 mixed with Cl- and resulted in a reaction of AgCl(s) + NO3(aq). A removal of Cl- ions caused the equilibrium to shift left.
4. The reaction shown in the introduction was endothermic. The equilibrium shifted right when temperature increased and shifted left when temperature decreased. This tells that heat must have started off on the reactant side.
5. Ka = [CoCl4^2-] / ([Co(H2O)6^2+][Cl-]^4)
B. with the addition of water, the system shifted left.
C. with the addition of AgNO3, the system shifted left.
D. With the increase in temperature, the system shifted right.
E. With the decrease in temperature, the system shifted left.
2. When the HCl was added, it split itself into ions of H+ and Cl-. In that case, ions of Cl- increased on the reactant side, so the equilibrium shifted right to maintain balance. When water was added, it increased the amount of water already on the product side, so it shifted back to the left to reach equilibrium once again.
3. With the addition of AgNO3, a small precipitate was formed in the wells of row D. This is because AgNO3 mixed with Cl- and resulted in a reaction of AgCl(s) + NO3(aq). A removal of Cl- ions caused the equilibrium to shift left.
4. The reaction shown in the introduction was endothermic. The equilibrium shifted right when temperature increased and shifted left when temperature decreased. This tells that heat must have started off on the reactant side.
5. Ka = [CoCl4^2-] / ([Co(H2O)6^2+][Cl-]^4)
Critical Thinking
1. Adding NaCl would most likely shift the equilibrium right. NaCl would dissociate into ions of Na+ and Cl-. The increase Cl- on the reactant side would cause a stress in the equilibrium, so it would shift right.
2. +50 kj/mol + Co(H2O)6^2+ + 4Cl <--> CoCl4^2- + 6H2O
3. Keq = [products] / [reactants] If Keq is 6 x 10^9 (a large number), that means the product’s concentration must be very high while the reactant’s concentration is fairly low. An assumption would be that AgCl(s) would be more than silver and chloride ions, since AgCl(s) is a product.
2. +50 kj/mol + Co(H2O)6^2+ + 4Cl <--> CoCl4^2- + 6H2O
3. Keq = [products] / [reactants] If Keq is 6 x 10^9 (a large number), that means the product’s concentration must be very high while the reactant’s concentration is fairly low. An assumption would be that AgCl(s) would be more than silver and chloride ions, since AgCl(s) is a product.