Lab 9 Cobalt Chloride and LeChatlier's Principle (1)

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Pima Community College *

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Chemistry

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Jan 9, 2024

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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.