Question

1. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved Ca(OH)2 solid on the bottom of the test tube. If you add an additional scope of Ca(OH)2 to this solution will the concentration of [OH-] in the solution increase, decrease, of be unchanged? (Briefly explain and assume we let the solution reach equilibrium again)

2. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved Ca(OH)2 solid on the bottom of the test tube. If you add some pure water to the test tube. Will the amount of solid Ca(OH)2 at the bottom increase, decrease, of be unchanged? (Briefly explain and assume we let the solution reach equilibrium again)

3. Imagine you have a saturated solution of Ca(OH)2, at equilibrium, that has some undissolved Ca(OH)2 solid on the bottom of the test tube. If you add some pure water to the test tube. Will the concentration of [OH-] in the solution increase, decrease, of be unchanged? (Briefly explain and assume we let the solution reach equilibrium again)

Answer 1

1 ) Ca(OH)2 <-------------------------------> Ca+2 + 2OH-

on addition of Ca(OH)2 to this mixture reactant has more amount than product. according to lechatliers principle the equilibiurm will shift towards forward direction. OH- ion conc increases

2) on addition of water more amount of  Ca(OH)2 will dissolve due to solvation . so  Ca(OH)2 concentration will decreases

3)

on addition of water more amount of  Ca(OH)2 will dissolve due to solvation . so  Ca(OH)2 concentration will decreases and OH- ion concnetration will increases