Title: Experiment 9: LeChatelier’s Principle
Purpose:
The purpose of this lab is to delve into the concept of solubility and look at it as having more properties than just it’s soluble or no it’s not.

Procedure:
In this lab we will be using Le Chatelier’s principle to predict the effects of temperature, pressure, and concentration changes on a given system at equilibrium. I will account whether the process is endothermic or exothermic in the given direction when discussing temperature.

Data tables and observations:

Balanced Equations: Ca(OH)2 + 2HCl --> CaCl2 + 2H2O
|Temperature |Average change in Vol.|[OH-] conc. (M) |Molar solubility |DG |
|(Celsius) |(mL) | | | |
|Ice (avg.9.25) |11.83 |59.2 |29.6 |-27.1 |
|Room Temp. |10.54 |52.7 |26.4 |-27.6 |
|30 |8.33 |41.7 |20.9 |-26.4 |
|40 |9.64 |48.2 |24.1 |-28.5 |
|50 |8.21 |41.1 |20.6 |-28.1 |
|60 |7.15 |35.8 |17.9 |-27.8 |
|70 |8.9 |44.5 |22.3 |-30.5 |

Calculations:
• mL HCl x 0.05mol HCl x 1mol OH- = mol OH- L HCl 1mol HCl
11.83 x 0.05mol x 1mol OH- = 0.59mol OH- 1 L 1mol HCl

[OH-] = mol OH- 0.010L
0.592mol = 59.2M
0.010L

• Molar Solubility of Ca(OH)2= [OH-]/2 • 59.2/ 2= 29.6

• Ksp = [OH-]m[Ca2+]n m= 2, n= 1
([59.2]^2) x [29.6]= 1.04*10^5

• DG=-RT ln Ks • (8.314*10^-3)x(273 + 9.25)x(ln(1.04*10^5))= 27.1

Post-Lab Questions: A. The reaction Mg(OH)2 Mg+2 + 2OH- will turn to the right when hydrochloric acid is added. This is because the acid eats away the hydroxides, and due to LeChatelier's principle, the equilibrium must shift to the right to account for this loss. B. The equilibrium shifts to the right, because EDTA sequesters (binds) Mg2+, thus taking it "out" of the solution C. Endothermic D. Equation should shift to the left because Na2SO4 is a salt that will dissociate. E. Equation should shift to the right because NaHSO4 is also a salt, but it dissociates into Na+ and HSO4- F. Exothermic

Conclusions: