Mateo Castro
April 3, 2013
Lab Partner: Unur Abdul Kader
T.A: Katie

Experiment 22: Molar Solubility, Common-Ion Effect

Abstract
The purpose of this experiment was to determine the molar solubility, the solubility constant, and the effect of a common ion on the molar solubility of calcium hydroxide. To accomplish this the experiment was split into two parts; part A and Part B. in Part A of the experiment a standardized 0.05 M solution of HCl was titrated into a 25 mL solution of saturated Ca(OH)2 which contained 2 drops of orange methyl identifier. Once the titration began, the HCl was added until the methyl orange endpoint was reached, and as a result the volume of the HCl needed for the endpoint to be reached could be used in …show more content…

Molar Solubility and solubility product of calcium hydroxide 1. Volume of HCl added= final reading – Buret reading initial
25.8-0= 25.8 mL 2. Moles of HCl added= molar concentration HCl *(Vol. HCl added/1000)
0.05 M HCl X (25.8 mL/1000)= 0.00129 mol 3. Moles of OH- in saturated solution= moles of HCl added
0.00129 mol 4. [OH-], equilibrium (mol/L)= moles of HCl added/ 0.025L
0.00129/0.025= 0.0516 M 5. [Ca2+], equilibrium (mol/L)= [OH-]/2
0.0516/2= 0.0258 M 6. Molar solubility of Ca(OH)2 = [Ca2+]
0.0258 M 7. Ksp of Ca(OH)2= [Ca2+][OH-]2
0.0516 X (0.0258) 2= 0.06.87E-05
Part B. Molar Solubility of Calcium Hydroxide in the Presence of a Common Ion 8. Volume of HCl added= buret reading initial- reading final
21.1-0= 21.1 mL 9. Moles of HCl added= molar concentration of HCl X (Vol. HCl /1000)
0.05 X (21.1/1000)= 0.001105 mol 10. Moles of OH- in saturated solution= moles of HCl
0.001105 mol 11. [OH-], equilibrium = moles of HCl/0.025 L
0.0442 mol/L 12. Molar solubility = [OH-]/2
0.0442/2= 0.0221 mol/L

Discussion The purpose of this