[Kaleyrvt]

1??? Man either I am just completely losing it or I am thinking way too hard!! It shouyldnt be this difficult!! BLAH! I am so sorry!

[Borek]

1 - you got it. Now, what the concentration of OH^- in 0.233M NaOH is?

[Kaleyrvt]

0.233 as well?

[Borek]

Exactly. Now, what about [Ca²⁺]?

And just to check if you really understand what is going on - what will be concentration of Cl^-?

[Kaleyrvt]

well the ration of CaCl2 is Ca:Cl2 = 1:2
Therefore Ca=0.066 and Cl= 2 x 0.066=0.132      ?
I hope this is right!

[Borek]

Therefore Ca=0.066 and Cl= 2 x 0.066=0.132

Good

Now plug concentrations into Ksp formula - if the result will be greater than the Ksp given for the particular compound, solid will precipitate.

[Kaleyrvt]

so....
Ksp= [Ca2+][OH-]2
=[0.066][0.233]2
=3.58 x 10^-3

are these calcs right???
If theya re, then Ca(OH)2, is 7.9 x 10-6  which is < Ksp= [Ca2+][OH-]2, which is 3.58 x 10^-3 thus a precipitate will form?

cross my fingers?

[Borek]

Don't call result of your calculations Ksp - Ksp is constant that is characteristic for every substance. You do calculate a value of Ksp formula using  known concentrations - and then you compare it with known Ksp. In this case calculated value is 3.58*10^-3 (you are OK), it is larger then Ksp for Ca(OH)₂ - so the precipitate forms.

In short: you are right this time