Common Ion

Question

At some other temperature, a solid sample of Ca(OH)2 is shaken with 0.0100 M CaCl2.
Once equilibrated, some solid Ca(OH)2 remains undissolved.
The solution is filtered and a 10.00 mL sample requires 12.96 mL of 0.0150 M HCl to neutralize it.
1. Calculate the value for Ksp of Ca(OH)2 from this data.
2.What is the molarity of Ca(OH)2 which is dissolved in this solution?
3.What is the solubility of Ca(OH)2 in this solution, in g/100mL?

Answer 1

You have two equilibria occurring in the solution.
Ca(OH)2 --> Ca^2+ + 2OH^- and
CaCl2 ==> Ca^2+ + 2Cl^-
The first one is limited solubility. The second one is 100% ionized. So
...................Ca(OH)2 ==Ca^2+ + 2OH^-
I......................solid............0...........0
C......................-x................x...........2x
E......................solid.............x..........2x where
Ksp for Ca(OH)2 = (Ca^2+)(OH^-)^2. For the CaCl2, we have
.................CaCl2 ==> Ca^2+ + 2Cl^-
I.............0.0100M...........0.............0
C...........-0.0100..........0.0100.....0.0200
E................0.0...........0.0100......0.0200
Therefore, Ksp for Ca(OH)2 = (Ca^2+)(OH^-)^2
You know (Ca^2+) = 0.0100 from CaCl2 and x from CaCl2 to make 0.0100 + x so Ksp = (0.0100 + x)(2x)^2. The titration with HCl determines the (OH-)
HCl + OH^- ==> H2O + Cl^-
mols HCl used = M x L = 0.1500 M x 0.01296 L = 0.001944 moles for a 10.00 mL sample. (OH^-) in mols/L = 0.001944 x (1000 mL/10.00 mL) = 0.1944 M = 2x
Plug that in for Ksp expression and solve for Ksp. This is answer to 1.
You shouldn't have any trouble with question 2 and 3.

Answer 2

By the way, in question 1 you have a grammar error. That should be
"1. Calculate the value for Ksp of Ca(OH)2 from THESE data."
Data is plural but this is singular. These is plural.