Experiment 10: Solubility Product for Calcium Hydroxide
GOAL AND OVERVIEW A saturated solution of Ca(OH)2 will be made by reacting calcium metal with water, then filtering off the solids: Ca(s) + H2O → Ca(OH)2(s) Ca2+(aq) + 2OH-(aq) The concentration of dissolved hydroxide will be determined by acid-base titration with standardized HCl solution. The Ksp for Ca(OH)2 will be calculated from the experimentally determined saturation concentration of hydroxide. Objectives of the data analysis understand solubility equilibria, acid-base neutralization, and the chemistry of lime (Ca(OH)2), limewater, and calcium carbonate (called limestone when occurring naturally as a mineral) manipulate equations for saturation equilibria, acid-base titrations, and volumetric dilutions understand how a large process is used to determine a single quantity SUGGESTED REVIEW AND EXTERNAL READING reference information on equilibrium and solubility; relevant reference and textbook information on solubility, acids, & bases BACKGROUND Whenever solid calcium hydroxide, Ca(OH)2 (commonly known as lime), is present in water, it dissolves according to this equation: Ca(OH)2(s) Ca2+(aq) + 2 OH-(aq) (1) until the rate of the backward reaction equals the rate of the forward reaction and the solution is saturated. The equilibrium constant for the reaction is the solubility product constant, Ksp, given by Ksp = [Ca2+][OH-]2 (2) The concentration of Ca(OH)2(s) does not appear the equilibrium constant expression because it is always present as the pure solid (constant concentration, or amount per volume), no matter how much or how little of it is present. PRELAB HOMEWORK (to be filled out in your bound lab notebook before you perform the experiment) Title and date Define: (1) saturated, (2) solubility product constant Answer: 1. When you filter the saturated Ca(OH)2 solution in this experiment, you are cautioned not to wash the precipitate with water, though washing is customary when you filter in order to obtain a pure sample of a solid material. What is the logic behind this caution? 2. If you have a mixture of a solid and its saturated solution and filter the mixture to get rid of the solid, how have you changed the concentrations of the quantities dissolved in the saturated solution? 3. Why does it not matter how much water is in the beaker before adding the Ca metal, as long as you see a white precipitate in the end? 4. Determine the extent to which you will dilute 2.0 M HCl to use as a titrant in the experiment (to use 10 mL per titration). Ksp (theoretical) = 6.5×10-6 = [Ca2+][OH-]2 = ½ [OH-]3 Procedure (Experimental Plan) You will not have time to do the entire experiment without preparation. Do what it takes to understand this experiment thoroughly. Data Tables
PROCEDURE 1) Check out the required equipment from the stockroom. Clean your equipment well before using it. Calcium metal is oxidized by the water, yielding calcium hydroxide and hydrogen gas: Ca + 2 H2O → Ca(OH)2(s) + H2(g) (3) The calcium hydroxide that is formed is sufficiently insoluble that it forms a saturated Ca(OH)2 solution, as in Eq. 1. **If your TA prepares the saturated Ca(OH)2 solution, carefully dispense approximately 30 mL of the liquid above the precipitate into an Erlenmeyer flask and stopper it immediately. PLEASE be sure to replace the lid on the reagent container (the TA-prepared solution). Proceed to step 5. 2) To prepare the saturated Ca(OH)2 solution, use your forceps to safely add two small pieces of calcium metal to 150 mL of distilled water in a beaker. Note: If you add too much metal, a large excess of calcium hydroxide solid can form and be difficult to remove by filtration. Note: If you add too little metal, you will not form enough calcium hydroxide to saturate the solution, which is the goal of adding the metal to the water. Note: Calcium metal is a strong reducing agent and should not touch skin. Note: You may have to score the metal to rub off the tarnish...
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