Answer Happy

= 13.3 ml 1- 1.0M of NH2. (use the stock solution of 25% of NH3, 0.91 g/mL) in 100.Oml volumetric flask. 25 ml 0.919 1 mol 1000 ml mol 100 ml 1700 g 2 13.3 x v1 = 10 x 100 + v1 = 7.5 ml will be diluted til 100 ml 2-0.10M of copper il)sulfate pentahydrate in 50.0 mL volumetric flask. 0.1 M X0.05 = 0.055 moel249.69 g/mol =1.248 g will be made a 50 ml solution 3-0.055 M of HCl Luse the dilution method to prepare it from the stock solution of 0.10M HCL) in 50.0 mL volumetric flask. 0.1 x v1 = 0.055 x 50 → Vi=27.5 ml of HCL will be diluted till 50 ml Part A: Determination of ammonia distribution coefficients between water and chloroform 1. Mix 10.0 mL of 1.0 M of NH3 solution with 10.0 mL of deionized water in a separatory funnel and shake the mixture for 5 minutes to achieve a homogenous mixture; then add 25.0 mL of chloroform to the same mixture and shake for 5-10 minutes. Caution upon shaking by opening regularly to avoid build up of gases. 2. Attach the separatory funnel to the metallic stand, and keep the mixture for 10 minutes, separate the two layers into two separate conical flasks (100.0 mL). 3. Transfer 10.0 mL of a chloroform solution into another conical flask and add 10.0 mL of deionized water with 10.0mL of methyl orange 4. Titrate the solution gradually with a 0.055 M HCl standard solution using a 5 ml plastic syringe. 5. The equivalent point is indicated by the change of color. Repeat the titration three times. 6. Calculate ammonia distribution coefficient by using the following equation: K- ammonia chlorofor mewater Trial 1 → Sml - 1.8 ml = 3.2 ml of HCl titrated Trial 2 → 5ml -2.3 ml = 2.7 ml of HC titrated 1- Determination of ammonia distribution coefficients between water and chloroform: A-Calculate the concentration of Ammonia in water. Remember to use the dilution equation : "MiVi= M2V2". B- Find the concentration of Ammonia in chloroform through the titration with acid. C. Calculate the numerical value of the Kd.