Kelsey Joyce Nandini Pandya Determining the Ksp of Calcium Hydroxide 12/7/12 Abstract The goals of this experiment were to titrate a saturated Cs(OH)2 solution with an HCl solution. Then determine the [OH-­‐] for the solution, and calculated the Ksp of the Ca(OH)2 solution. The results of this experiment were found using the Ksp formula and the balanced chemical equation for this experiment. The Ksp of this experiment was determined to be 3.47x10-­‐5.The goals of this experiment were meet, by the value for the Ksp being able to be calculated. Introduction The purpose of this experiment was to determine the [OH-­‐] for the Ca(OH)2 solution, and calculate its Ksp. The method used to perform this experiment was to titrate the Ca(OH)2 solution, and then use the concentrations of OH-­‐ and Ca2+ in the Ksp formula for the reaction. Procedure To begin this experiment set up the materials that will be used in this experiment. Gather about 15 mL of the calcium hydroxide solution, and fill the buret with the 0.05 M HCl solution. Then pour the 15 mL of Ca(OH)2 solution in to a beaker with a magnetic stirring rod inside of the beaker. Next position the pH sensor in the Ca(OH)2 solution where it won’t get hit be the magnetic stirrer when turned on. The last step to set up this experiment is to open the file “23 Ksp ” in the Logger Pro program on the computer, and to turn on the magnetic stirrer. Now the titration of the solution can begin. Click collect on the computer and record the initial pH reading at 0 mL of 0.05 M HCl. Add the titrant in small amounts and record each volume after the pH stabilizes. Keep adding HCl in small amounts until the equivalence is reached and the pH readings become constant. When the titration is complete click stop on the computer and store the data. Use page number three to find the equivalence point of the data that was just recorded. Then clean the materials and perform another titration using the same method as the first trial. When both trials have been completed clean up all of the materials that were used and save the graphs for each trial. Results Using the procedure above we were able to determine the equivalences for the titration in trial one and two. The equivalence point for trial one was 12.21 mL, and the equivalence point for trial two was 12.46 mL. Data Table: Trial Equivalence Point (mL) 1 12.21 2 12.46 Graph: Analysis Questions: 1) Calculate the [OH-­‐] from the results of your titrations. Explain your calculations. 0.05 mol = x L 0.012335 L X = 6.1675x10-­‐4 6.1675x10-­‐4 = 0.0411 mol/L 0.015 L [OH-­‐] = 0.0411 mol/L I took the molarity of the HCl solution and multiplied it by the average volume for the equivalence points for each trial and found the moles of OH produced. I then took the number of moles and divided it by the volume of calcium hydroxide for this experiment to get the concentration of OH. 2) Calculate the [Ca2+]. Explain your calculations. [Ca2+] = ½[ OH-­‐] = 0.0206 mol/L To find the concentration of Ca I found half of the concentration of OH. 3) Calculate the Ksp for calcium hydroxide. Explain your calculations. Ksp = [Ca2+][OH-­‐]2 = 3.47x10-­‐5 To find the Ksp for this experiment I inserted the concentrations of Ca and OH in to the Ksp equation for this chemical reaction. 4) Find the accepted value of the Ksp for calcium hydroxide and compare it with your value. Discuss the discrepancy and suggest possible sources of experimental error. Accepted Ksp = 6.5x10-­‐6 Discussion The Ksp for this experiment was calculated to be 3.47x10-­‐5. The calculated value for the Ksp was significantly different than he accepted value of Ksp for Ca(OH)2. This can be caused by error occurring in our experiment. Some places where error could have occurred are during the recording of our data. The Measurements for the pH change during the titration in our experiment were very rough values and could have been more specific for each trial. This discrepancy could have greatly swayed the calculated Ksp for our experiment. Conclusion In conclusion our experiment was successful because we were able to calculate a Ksp for the experiment. However, there was a large degree of error present. Appendix MaXb(s) à aMb+(aq) + bXa-­‐(aq) Ksp = [Mb+]a[Xa-­‐]b Ca(OH)2à Ca2+ + 2OH-­‐ pOH = 14 – pH [Ca2+] = ½[OH-­‐] Ksp = [Ca2+][OH-­‐]2x