Le Chateliers Principle Lab Report

Experiment 11: Equilibrium Constant and Le Chatelier’s Principle

The purpose of this experiment is to expand previous knowledge about equilibrium and solubility of certain ionic compounds. This laboratory analyzes the behavior of an exceptional salt in response to normal “stressors” on the reaction, such as temperature increase or decrease, which would cause a shift in the direction of the reaction. Another main focus is to be able calculate and determine the relationship between thermodynamics and solubility based on the observations of those behaviors during the reaction. Last but not least important, this lab reemphasizes proper equipment and safety use of pipettes, ring stands, flasks and chemical solutions.

There were no changes to the procedure outlined in the online lab manual http://www.chem.fsu.edu/chemlab/chm1046lmanual/chemeq/index.html “Chemical Equilibrium and Le Chatelier’s Principle. Dillon, Stephanie. CHM1046L, Fall2011.

Data:

Ca²(OH¯)₂ (s) Ca² (aq) + 2 OH¯(aq)

Temp (C)|Change in volume (Vfinal-Vinitial)|[OH-]|Molar solubility|Ksp|DG|

17|9.59 mL|0.04795 M|.024|5.52x10^-5|23.65 KJ/mol|

52.2|13 mL|0.065 M|.032|1.40x10^-4|24.00 KJ/mol|

17|9.95 mL|0.04975 M|.025|6.19x10^-5|23.38 KJ/mol|

71|8.61 mL|0.04305 M|.022|4.07x10^-5|28.93 KJ/mol|

Sample calculations: [OH-], molar solubility, Ksp, DG, DH, and DS

[OH-]: mL HCl x 0.05 mol HCl/1 L x 1 mol OH-/1 mol HCl = mol OH-

mol OH-/0.010L = [OH-]

9.59 mL HCl x 0.05 mol HCl/1000L x 1 mol OH-/1 mol HCl = .0004795/0.010L = .04795 M OH-

Molar Solubility: [OH-]/2

.04795/2 = 0.024

Ksp: [OH-]m[Ca2+]n

[.04795]2[.024]1 = 5.52x10^-5

DG: -RTlnKsp

-(.0083145 KJ/mol )(290.15 K)ln(5.52x10^-5) = 23.65 KJ/mol*K

DG = DH-T*DS

DH: Y-int

0.0675

DS: - Slope

-(.082)

Conclusion

The OH- concentrations were calculated from mL of titrant used and the molarity of the samples and so were found to be .04795, .065, .04975, and .04305, respectively. The molar solubility of Ca(OH)2 were...