Cobalt Chloride and LeChatlier’s Principle

Objective: To gain an understanding of LeChatlier’s principle through the observation of the effect of induced perturbations 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(H2O)6+2. In the presence of chloride ions, a different complex forms, the blue CoCl4-2 complex ion. We use their different colors to indicate the equilibrium concentrations for the following reaction. Co(H2O)6+2 + 4Cl – <==> CoCl4-2 + 6H2O In this simulation 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(No3)2(aq) exist as Co(H2O)6+2 and NO3-2. 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(H2O)6]+2 to an empty Erlenmeyer flask. 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.) 2.) 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). 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. AgNO3(aq) <==> Ag+ + NO3-Ag++ Cl – <==> AgCl(s) AgNO3(aq) + Cl – <==> AgCl(s) + NO3- 3.) Predict the effect of adding HCl to the reaction. Now do so in very small incremental steps until the equilibrium has shifted. 4.) 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. 5.) 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.

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