lab report 4 (chemistry), Lab Reports of Chemistry

Download lab report 4 (chemistry) and more Lab Reports Chemistry in PDF only on Docsity! Lab Report Edric ong khai jieh (21040985) Experiment 4 - Determination on of a Solubility Product Constant Objective: To determine the dissolved iodate concentration in saturated solutions of calcium iodate hydrate. Procedure: Part A: Titration of Saturated Calcium Iodate 1. 35mL of 1.00M HCl is placed into a beaker and labelled as 1.00M HCl 2. 300mL of 0.24M KI solution is then filled into a 400mL beaker. 3. A burette is prepared and filled with the standard solution Na2S2O3. The initial reading of burette is observed and recorded. 4. 50mL of 0.024M KI solution, 10mL of Ca(IO3)2 solution and 10mL of 1.00M HCl is pipetted and added into a conical flask accordingly. The color is observed and recorded. 5. Na2S2O3 is then used to titrated the solution prepared in step4 until it turns pale yellow. Then, 15drops of starch indicator is added into the solution. 6. The titration is then undergone in drop wise until the solution turns into colorless. The final burette reading is observed and recorded. 7. Steps 4 to 6 is repeated three more times and the reading of burette is all observed and recorded. Part B: Titrations of Saturated Calcium Iodate in KIO3 1. 35mL of 1.00M HCl is placed into a beaker and labelled as 1.00M HCl 2. 300mL of 0.24M KI solution is then filled into a 400mL beaker. 3. A burette is prepared and filled with the standard solution Na2S2O3. The initial reading of burette is observed and recorded. 4. 50mL of 0.024M KI solution, 10mL of Ca(IO3)2 solution in KIO3 and 10mL of 1.00M HCl is pipetted and added into a conical flask accordingly. The color is observed and recorded. 5. Na2S2O3 is then used to titrated the solution prepared in step4 until it turns pale yellow. Then, 15drops of starch indicator is added into the solution. 6. The titration is then undergone in drop wise until the solution turns into colorless. The final burette reading is observed and recorded. 7. Steps 4 to 6 is repeated three more times and the reading of burette is all observed and recorded. Results 1 2 3 4 Final buret reading 7.4 7.3 7.1 7.6 Initial buret reading 0 0 0 0 Vol. Of Na2S2O3 7.4 7.3 7.1 7.6 [Na2S2O3] 0.12 0.12 0.12 0.12 Mol of S2O3 - reacted 8.88 x 10-4 8.76 x 10-4 8.52 x 10-4 9.12 x 10-4 Mol of IO3 - reacted 1.48 x 10-4 1.46 x 10-4 1.42 x 10-4 1.52 x 10-4 Vol. Of Ca(IO3)2 used 10 10 10 10 [IO3 -] 0.0148M 0.0146M 0.0142M 0.0152M [Ca2+] 0.0074M 0.0073M 0.0071M 0.0076M KHP 1.62 x 10-6 1.56 x 10-6 1.43 x 10-6 1.76 x 10-6 1) Calculate the moles of S2O3 - reacted M = n/v n = MV No of moles of S2O3 - = 0.12 M x 7.4/1000 = 8.88 x 10-4 2) Calculate the no moles of IO3 - reacted I2 + 2S2O3 2- -> 2I- S4O6 2- 2 moles of S2O3 2-react with 1 mole of I2 8.88 x 10-4 of S2O3 2-react with 4.44 x 10-4 mole of I2 IO3 - + 5I- + 6H2O ->3I2 + 9H2O 3 moles of I2 is produced from 5 moles of I- 4.44 x 10-4 of I2 is produced from 7.4 x 10-4 moles of I- 7.85 x 10-3 moles of I- react with 1.57 x 10-3 mole of IO3 - no moles of IO3 - reacted = 1.57 x 10-3 mole 8) Calculate [IO3 -] [IO3 -] x10/1000 = 1.57 x 10-3 [IO3 -] = 0.157M 9) Calculate [Ca2+] Ca(IO3)2 H2O ⇌Ca2+ +2IO3 - Initial (M) Change (M) -x x 2x Equilibrium (M) x 2x 2x = 0.157M x = 0.0785M [Ca2+] = 0.0785M 10) Calculate Ksp 2. Calculate Ksp Ksp = [Ca2+][IO3 -]2 = ( 0.0785 )( 0.157 )2 = 1.93 x 10-3 Discussion: 1. When 50mL of 0.024M KI solution, 10mL of Ca(IO3)2 solution, and 10mL of 1.00M HCl are pipetted into a conical flask, the colourless solution turns reddish brown. This occurs when the iodine ions in KI solution are oxidised by the iodate ions in Ca(IO3)2 to reddish-brown iodine. 3I2 (aq) + 9H2O = IO3- (aq) + 5I-(aq) + 6H3O+ (aq) is the formula for this reaction (l). HCl is employed in this process as a reagent to speed up the reaction. 2. When KI solution, Ca(IO3)2 solution, and HCl are added to the reddish- brown solution, sodium thiosulphate solution is used to titrate the mixture. This is due to the fact that the thiosulfate ions change the iodine ion back into the colourless iodide ion. The consequence is a lightening of the solution's colour. 2I-(aq) + 2S2O32-(aq), 2I-(aq) + S4O62-(aq), and 2I-(aq) + 2S2O32-(aq) are all examples of the same compound (aq). The solution is made blue black by adding starch solution when it starts to seem paler. This is so that it can be determined whether all of the iodine has been converted to iodide ions, making the solution colourless, before the colour fades. In order to acquire a more accurate result and to make the colour changes clearer and more noticeable, the starch solution is thereafter added. 3. Calcium iodate is less soluble in potassium iodate solution when compared to plain water. The iodate ion is known as the common ion because it may be found in both the potassium iodate solution and the calcium iodate solid. The common ion is an ion that exists in the actual ionic composition. Le Chatelier's principle states that the presence of a common ion can affect how an ionic material balances. The common ion will make a small change to the equilibrium, possibly to the right or even to the left. This is because the common ion would be seen as a product, leading to preference in products and reducing the solubility of the ionic substance.This decrease in solubility is known as the "common ion effect." We can see that when potassium iodide, calcium iodate in iodate solution, and HCl are combined, a little quantity of precipitate occurs in the solution because calcium iodate is less soluble in potassium iodate solution. 4. We found, however, that the Ksp value of calcium iodate in potassium iodate solution is actually higher after performing all the calculations. This is because Ksp, the solubility product constant, is unaffected by the solubility of calcium iodate solid. Only variations in temperature and dissolved ion concentration have an effect on the Ksp value. Due to the iodate ions from the potassium iodate solution, the calcium iodate in potassium iodate solution has about twice the concentration of iodate ions when compared to calcium iodate in pure water..As a result, the twofold concentration of iodate ions from potassium iodate raises the iodate ions concentration and hence raises the Ksp value, which can be computed using the formula Ksp = [Ca2+][IO3 - ] 2. 5. We may get started by figuring out how many moles of thiosulphate ions were used in the calculation. Next, calculate the concentration and number of moles of iodate ions used in the process using stoichiometry. Once we have approximated the concentration of iodate ions, we may use the ICE to determine the concentration of calcium ions. Calculate the Ksp using the calcium and iodate ion concentrations and the equation Ksp = [Ca2+][IO3 -] 2. Conclusion: 1. Investigating the dissolved iodate concentration in saturated solutions of calcium iodate hydrate and calcium iodate hydrate in KIO3 proves the experiment's goal and yields the solubility product constant.. 2. The amount of dissolved ions might have an impact on the Ksp value. The Ksp value increases with the amount of dissolved ions present.