How do I find 7, 8, 9? Molar solubility of Calcium Hydroxide in the Presence of

Question

How do I find 7, 8, 9? Molar solubility of Calcium Hydroxide in the Presence of a common Ion Trial I Trial 2 Trial 3 1. Volume of saturated Ca(oH, with added 25.0 25.00 Caci2 solution (ml) 25.0 2. Concentration of standardized HCI solution (moun) ae Buret reading, initial (ml) IO. lo su 24.29 4. Buret reading, final (ml) 13 85 5. volume of HCI added (ml) 6. Moles of HCI added (mo) 7. Moles of OH in saturated solution (mo) 8. LOH equilibrium (moVL) 9. Molar solubility ofCa(oH) with added Caci (molL) 10. Average molar solubility of Ca(oH), with added CaCl2 (molL) Account for the different molar solubilities in Part A.11 and Part B.10 (Report Sheet). Laboratory Questions Circle the questions that are to be answered. 1. Part A2. suppose some of the solid calcium hydroxide is inadvertently transferred along with the supermatant liqu a. Will more, less, the same amount of hydrochl acid used for the analysis in Part A6?Explain. b. this inadvertent transfer titrant be increase, decrease, or have no effect on calculated solubility product for calcium the 2. Will this inadvertent transfer increase, decrease, or have effect on the calculated molar solubility of calcium hydroxide? Explain. no 3 Part A6 Does adding deionized water to the titrating flask to wash wall of the Erlenmeyer flask and the buret tip increase, boiled, Pan A6. decrease, or have no effect on the Ke value of the the dark While turting the Ca(oH)? Explain. kow red-orange. As a result of soluti was distracted, and the endpoint was ?Explain. this technique crror, will the reported molar solubility high or too of be too .4. Part B.1. How ill using tap water instead of boiled, deio water affect the s. decrease or have no effect? Hint: How will the minerals in the water affect the solubility of Ca(oHi Part Jerry forgot to record the actual molar concer As. 's. Caloi complete the quickly, -005 standard HCI which was actually oous). the of too high or low? M concentra The ethylenedi Explain.

Explanation / Answer

mole of OH- used = moles of H+

so

mol of HCl = mol of H+´= mol of OH- = 6.63*10^-4 mol of OH-

[OH-] = mol /V = (6.63*10^-4) / (25*10^-3) = 0.02652 M of OH-

so...

Molar solubility

1 mol of OH- = 0.5 mol of Ca(OH)2

then

0.02652 mol of OH- --> 0.02652/2 = 0.01326 mol of Ca(OH)2

per liter so:

[Ca(OH)2] = 0.01326 mol /L

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