Molar Solubility

2063 words 9 pages
Mateo Castro
April 3, 2013
Lab Partner: Unur Abdul Kader
T.A: Katie

Experiment 22: Molar Solubility, Common-Ion Effect

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
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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


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