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Determining the Solubility Product Constant for Calcium Sulfate:The Effect of Ionic Strengths of Electrolyte SolutionsDavid LiuChemistry 102May 19, 2015
AbstractThe method of this experiment was done by filtering the solids of calcium sulfate out from its diluted solution. Once the solid calcium sulfate has been filtered out of the solution, ammonia buffer at pH of 10 is added into the remaining solution, also adding approximately 2 drops of eriochrome black 1 indicator to determine the endpoint. The final part of the preparationconsists of preparing a burette filled with EDTA to titrate into the solution until it turns from red to blue. Once data has been collected for three trials, three analyses in different methodologies will be performed, to calculate the solubility product constant of calcium sulfate. The different methods used for calculation is using molarities at room temperature, activities at room temperature, and Gibbs Free Energy. By employing the method molarity yielded a result of 6.22x10-5M, 1.17x10-5M, and 4.48x10-5M respectively. Further by using the activities of the ions, the calculations resulted in 1.19x10-5M, 1.12x10-5 M, and 1.20x10-5M. Lastly the method of using Gibbs free energy were used and determined to be 4.48x10-5. These numbers are then compared to the literature value for the solubility product constant (Ksp) of calcium sulfate, determined to be at 7.10x10-5 M.IntroductionElectrical conductivity in solutions has played a major role in the modern day era, from any industry using electricity to every residence living on electricity. Understanding the method and strength of how electricity is conducted, also known as electrolytes in solutions is vital for these applications. When a solid dissolves into water, the solid dissociates into ions, which wouldthen allow electricity to be conducted through the solution, which are also known as electrolytes. The equation to be used for calculating equilibrium constant is known to chemists asK = aCcaBb/aAaaBb(1)for any generalized chemical reactionaA(aq)+ bB(aq)cC(aq)+ dD(aq)(2)
where ais the activity of a given chemical species involved in the equilibrium, and the activity ofthe species is used as one of the method to calculate the solubility product constants, with abeing calculated with the expressiona = fM(3)This all relates to one of the method of calculating solubility constants, and the calculations for obtaining f (0<f<1), the activity coefficient will be mentioned later in the section. However, if were to allow fto be 1, then a more familiar equilibrium constant equation surfaces,K = [C]c[D]d/[A]a[B]b(4)Another common method to calculate equilibrium constant is to include Gibbs free energy change, since it is a thermodynamic quantity, by using the expressionΔGo= -RTlnK(5)where ΔGois the change in Gibbs free energy, R being the constant 8.314J/mol·K, and T would