It is important to note that comparing the Ksp values of two or more compounds can only be done if the compounds share the same dissociation stoichiometry. In this experiment, 1 mol of Ca(OH)2 is used to produce 3 mol of products. In order to compare this constant to another compound, …show more content…
This could be due to a calculation, or measurement error during the experiment. In addition, the values between both runs for Part A, are also very different further supporting an error in measurement of data. In addition, the temperature of the room, and of the solution, were not measured. These temperatures may have been above 25℃.
Data - Part B (80℃):
Volume of HCl (mL) pH Volume of HCl (mL) pH Volume of HCl (mL) pH
0 11.22 4.5 9.16 6.6 2.85
2.5 11.06 4.7 8.35 7.1 2.77
3 10.94 4.8 6.64 7.6 2.72
3.5 10.72 5.6 3.37
4 10.35 6.1 …show more content…
In the solution heated to 80℃, the value for Ksp was nearly three times lower (6.9x10-6) than both the Ksp values for the 25℃ solutions (average 2.1x10-5). In the experiment, the solubility product constant observed, was significantly higher than the expected value. Le Châtelier’s principle states that a change in temperature on a system at equilibrium, will result a new equilibrium as the system adjusts, shifting in a direction that minimizes the disturbance (Tro, 636). In this case, higher temperature causes the system to shift left, resulting in less Ca2+ and OH- being produced, and reducing the solubility of Ca(OH)2. In this experiment, the effects of temperature on solubility and the rate of solubility were learned. Techniques for titration were used again, and expanded upon from the previous lab in which they were also used.
References:
Thorne, E. J. (2015). Laboratory Manual for General Chemistry. Wiley Custom Learning Solutions.