I've been struggling with this question... in vain. Some hints?

Saturated solutions of calcium hydroxide were made up in (a) water, and in sodium hydroxide solutions of concentrations, (b) 0.025 moldm-3, (c) 0.05 moldm-3, (d) 0.1 moldm-3. 20.0 cm3 of each solution were titrated with 0.05 moldm-3 hydrochloric acid, and required (a) 19.0 cm3, (b) 23.2 cm3, (c) 28.4 cm3, (d) 43.6 cm3 of the acid. Show that an almost constant value of about 5.5 x 10‑6 can be obtained, from these observations, for the solubility product of calcium hydroxide.

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I worked out the number of moles of acid used;
(a) 9.5x10^-4 mol
(b) 1.16x10^-3 mol
(c) 1.42x10^-3 mol
(d) 2.18x10^-3 mol

Then, I found the concentration of OH^- and Ca(OH)2 from the 20 cm^3 sample:
(a) 0.0475M, 0.0237 M
(b) 0.058 M, 0.029 M
(c) 0.071 M, 0.035 M
(d) 0.108 M, 0.054 M

Now, I know

Assuming the concentration of OH^- be as I obtained before, the Ksp for the first one should be:

Let x be the number of moles of dissociated ions.
There is x moles of Ca^2+ and 2x^2 moles of OH^-, or a product of 4x^3 mol^3dm^-9.

Ksp = 4 * (0.0475)^3 = 0.00043 = 4.3x10^-4

Which is way too far from 5.5x10^-6

:(