Eric Brandon Ngaba

INTRODUCTION. In this experiment, we take an advanced approach to solubility in water. This approach abandons looking at solubility as two distinct states - soluble and insoluble - and instead relates it as a continuous state where the solute and solvent are always in equilibrium. This is because dissolving a compound causes energy to be released as the constituent ions separate from their lattice and scatter in the solvent. Thus, the ratio of hydration energy to lattice energy holding the ions together as molecules determines the degree of solubility. The purpose of this experiment is thus to apply Le Chateliers principles of equilibrium in investigating the effect of temperature on the solubility of Ca(OH)₂. Temperature is one of the four conditions affecting solubility, the rest being; the aforementioned intermolecular between solute and solvent, the concentration of products and reactants, and the change in entropy that accompanies the process. MATERIALS. Instrumentation; Clamp and Stand 50mL Beakers 100mL Beakers 250mL Beakers 125mL Erlenmeyer Flask Filter papers Filter funnel Magnetic Stirrer Thermometer Reagents; Calcium Hydroxide 0.05M Hydrochloric acid Distilled water Bromothymol blue indicator PROCEDURE. 1. A buret was rinsed and attached to a Clamp and Stand. 30mL of HCl was then measured into a dry 100mL beaker and its concentration noted down. 2. The buret was rinsed and filled with HCl, with waste HCl from cleaning and removing air bubbles collected in a beaker labeled waste. 3. 100mL of distilled water at room temperature was used to dissolve 2g of Ca(OH)₂ in a 250mL beaker. It was filtered to clarify any remaining solute after reaching saturation. 10mL of this solute was then pipetted into a 100mL beaker followed by adding 25mL distilled water and 10 drops of Bromothymol blue indicator. 4. While using a magnetic stirrer bar, the Ca(OH)₂ sample was titrated with HCl until the dissolved indicator's color changed to yellow at which point the amount of HCl used was calculated. 5. Repeat steps 3 and 4 using 100mL of distilled water heated to a temperature ranging between 50°C - 70°C. 6. Repeat steps 3 and 4 using 100mL of distilled water at temperatures of less than 15°C which can be achieved by immersing the beaker in an ice bath for 5 minutes. 7. Repeat steps 3 and 4 using 100mL of distilled water at a selected temperature based on molar solubility calculations. RESULTS. Ionic equation for dissolving calcium hydroxide; Ca(OH)₂(s) ⇔ Ca²⁺(aq) + 2OH⁻₍ₐq) Ionic equations for neutralization with HCl; H⁺(aq) + OH⁻(aq) ⇒ H₂O(l) Data Table 1; Titration volumes Iteration Initial Volume (mL) Final volume (mL) Titration volume (mL) Low temperature - Room temperature - High temperature - Selected temperature- Data Table 2; Experimental Data Temperature (°C) Temperature (K) Moles of H⁺ (moles) Moles of OH⁻ (moles) Molarity of OH⁻ (M) Molarity of Ca²⁺ (M) Ksp (nx10^-7) ∆G (kJ/mol) ∆S (J/K) ∆H (kJ/mol- - - - - - - - -5.706 CALCULATIONS. Graph 1; ∆G against T(K). Graph 2; Molar solubility against T(K) Examples based on low temperature reaction Amount of HCl used = 7.4mL Concentration of HCl = 0.05M Moles of HCl = (7.4mL/1000) * 0.05M = 0.00037 moles According to molar ratios: Since moles of HCl = moles of H+; 1. And moles of H+ = moles of OH⁻ 1. Moles of OH⁻ = 0.00037 moles 2. Volume of OH⁻ = 35ml Molarity = 0.00037moles/(35ml/1000) = 0.01057M 2. Molar solubility Molar solubility = [Ca²⁺] [Ca²⁺] = [OH⁻]/2 = 0.01057M/2 =0.00529M 3. Ksp Ksp = [Ca²⁺] * [OH⁻]² = 0.00529M * (0.01057)² = 5.91 * 10⁻⁷ 4. ∆G ∆G = -RTln(Ksp) = -(8.3145 * 287.1 *ln(5.91 * 10⁻⁷)) = - J/mole = -34.234kJ/mole 5. ∆H and ∆S Trendline equation for linear regression of graph for ∆G against T(K) is; Y = - + 140.584X Recalling; ∆G = ∆H - T∆S Therefore ∆H = - J/mole = -5.706kJ/mole And ∆S = -(140.584) = -140.584 J/K CONCLUSION. The molarity of hydroxide ions and the molar solubility of calcium hydroxide per trial were as follows: At low temperatures of 14.1°C molarity was 0.01057M while molar solubility was 0.00529M. At room temperature, it was 0.00786M and 0.00393M. At 61.1°C it was 0.00914M and 0.00457M. Finally, at the selected temperature of 16.8°C it was 0.01129M and 0.00565M. The graph for molar solubility against temperature did not have a definitive curve because it rose or fell at several points. Generally, however, it revealed that molar solubility was greater at lower temperatures because this is where the value graphed curve peaked. Between room temperature and the high temperature existed a slight uptick possibly as a result of increased reaction rates at higher temperatures. Since this uptick did not exceed the points at lower temperatures it can be Interpreted that indeed for calcium hydroxide solubility does increase with a drop in temperature. The calcium hydroxide was filtered before reacting so that only the amount that had dissolved could take part in the reaction with HCl. Leaving the undissolved amount in solution would have resulted in longer reactions because as the dissolved calcium hydroxide gets depleted by the reaction, the negative pressure of the dilution equilibrium would cause more of the solute to dissolve thereby skewing results. We did not need to accurately record the precise amount of calcium hydroxide that dissolved because titration would offer up a way for us to calculate it. Overall, the values for the solubility product constant, and Gibbs free energy retained similar trends to the molar solubility graph. Entropy was a constant extracted from the gradient for the Gibbs free energy graph, the same is true for enthalpy change since it was the y-intercept in that graph. The experiment was therefore successful at showing how solubility actually works, although it suffered from slight inconsistencies possibly as a result of the temperature fluctuations as the reaction progressed.