Ksp=5.5×10−6at298K The equilibrium in a saturated solution of Ca(OH)2 is represented above. In an experiment, a student places 5.0g of Ca(OH)2(s) into 100.0mL of distilled water and stirs the mixture. How would the results be affected if the student repeats the experiment but this time places 5.0g of Ca(OH)2(s) into 100.0mL of 0.0010MNaOH(aq) instead of distilled water?

Explanation:

The result will be different since the mixture is out inside a Base giving a different reading to the reaction as the previous

Less solid dissolve equilibrium lies to left.

Explanation:

We have the reaction:

Ca(OH)2>Ca2+ + 2OH-

NaOH>Na+ + OH-

This has to do with the ionic effect of the hydroxide ions that are produced by the base, in this case, sodium hydroxide, therefore, the solubility of calcium hydroxide is decreased. We conclude that the balance of calcium hydroxide will be further to the left so that the equilibrium concentration of hydroxide is maintained.