Part B: Calculate the molar solubility of Ca(OH)2 in the presence of the dissolved CaCl2 for each trial. Thank you!! **Part B**| | Trial 1 | Trial 2 ||--------------------------|---------|---------|| Volume of sample analyzed (mL) | 25.00 | 25.00 || Initial buret reading (mL) | 0.03 | 19.85 || Final buret reading (mL) | 19.85 | 38.90 |This table displays the results of two trials in a titration experiment. The volume of the sample analyzed is consistent at 25.00 mL for both trials. The initial and final buret readings for Trial 1 are 0.03 mL and 19.85 mL, respectively. For Trial 2, the initial reading is 19.85 mL, and the final reading is 38.90 mL. This data is typically used to calculate the volume of titrant used in the experiment. ### Part A: Molar Solubility of Calcium Hydroxide1. **Obtain a Sample**: Obtain a sample of saturated Ca(OH)₂ solution (without added CaCl₂) from your instructor. Transfer 25 mL of the saturated Ca(OH)₂ solution into a 125 mL flask and add 2 drops of methyl orange indicator.2. **Pipetting**: Pipet 25 mL of the saturated Ca(OH)₂ solution.3. **Titration**: Titrate the Ca(OH)₂ solution to the methyl orange endpoint, where the color changes from yellow to red. Remember, the addition of HCl should stop within one-half drop of the end point. Read and record the final volume of standard HCl in the buret.4. **Data Analysis**: Titrate an additional 25 mL aliquot of the saturated Ca(OH)₂ solution. Calculate the molar solubility and solubility product of Ca(OH)₂ for each trial and report their average values.### Part B: Molar Solubility of Calcium Hydroxide in the Presence of a Common Ion1. **Repeat Procedure**: Repeat steps 2–5 using a sample of saturated Ca(OH)₂ solution containing added CaCl₂.2. **Discard Solutions**: Discard all of the reaction mixtures and HCl solution in the buret into the sink, followed by a generous supply of tap water. Rinse the buret twice with tap water and twice with deionized water.3. **Data Analysis**: Calculate the molar solubility of Ca(OH)₂ in the presence of the dissolved CaCl₂ for each trial and report its average value.

molar solubility is used to measure the no of moles of solute that dissolves in one liter of solution. molar solubility and solubility product are directly related to each other. the chemical reaction ,CA(OH)2⇔Ca2+ + 2OH+ Ksp=[Ca2+] [OH-]2 by using titration data calculate the no of moles of OH- in the 25.00 ml of sample every one H+ neutralizes one OH- by considering 0.00250 M Hcl for neutralization and volume of Hcl reqired for first trial isno mof moles=0.00250m×0.01982l=0.00049 mole =5×10-4 molesby using moles of OH- & sample volume determine the [OH-]=5×10-4 mole 0.002500 l=0.2 mole/ldetermine [Ca2+] it is exactly half of [OH-] because [Ca2+] [OH-] is in 1:2 molar ratio from balanced equation. therefor the [Ca2+]= 0.1 mole/lKsp for Ca(OH)2=(0.1) (0.2)2= 0.004molar solubility=0.0044(1+2)=0.0013 =0.1 mole/las there is not much difference in two trials , that is the volume of HCl is required for neutralization is very close the molar solubility of second trial is nearly equal to 0.1 mole /lit