Question

Problem #1 Use this data to calculate Ksp of Ca(OH)2. The Ca(OH) used was 15.0 ml, titrated with 0.05M HCI

yes, which the data for the experiment is given in the pictures

Answer 1

The neutralization reaction taking place is:
$Ca(OH)_{2}(aq)\, +\, 2HCl(aq)\, \rightarrow \, CaCl_{2}(aq)\, +\, 2H_{2}O(l)$
Ca(OH)2 has low solubility and only the amount dissolved will participate in the reaction.
Vol. of HCl used = 8.5 mL (From the data, sharp dip in the pH at this vol.)
Molarity of HCl = 0.05 M
Number of moles of HCl used = molarity x Vol. in L = 0.05 x 8.5 x 10-3 = 4.25 x 10-4
From the balanced equation, the number of moles of Ca(OH)2 reacted = (4.25 x 10-4)/2 = 2.125 x 10-4
Vol. of Ca(OH)2 taken = 15.0 mL
Molarity of Ca(OH)2 in the solution is also its solubility and is refered to as 'S'.
S = number of mole)/ Vol. in L = 2.125 x 10-4/(15.0 x 10-3) = 0.014
S = 0.014 M

Ksp = [Ca+2][OH-]2
[Ca+2] = S
[OH-] = 2S
Therefore, Ksp = S x (2S)2 = 4S3 = 4 x (0.014)3 = 1.0976 x 10-5