I've searched the web, and I can't seem to find an agreement for what the solubility product is for calcium hydroxide.

My Chemistry book from 2007 says it is $\pu{5.5*10^{-6}M3}$ at $\pu{25 ^\circ C}$. On the fact sheet of the National Lime Association (pdf) it says that the solubility in water is $\pu{0.159 g//100ml}$ at $\pu{25 ^\circ C}$.
When you calculate the solubility product from the solubility in the water you get:

\begin{align} \frac{\pu{0.159 g//100 mL}}{\pu{74.093 g//mol}}\cdot \pu{10 mL} &=\pu{0.0215 mol//L}\\ \ce{CaOH2(s) &<=> Ca^2+(aq) +2OH-(aq)}\\ K_\mathrm{sp} &=[\ce{Ca^2+}][\ce{OH-}]^2\\ \pu{0.0215 mol//L}\times\left(2 \times \pu{0.0215 mol//L}\right)^2 &= \pu{3.98\times10^{-5}M^3} \end{align}

I can also try to calculate the solubility in water when the solubility product is $\pu{5.5*10^{-6}M^3}$:

\begin{align} x\cdot(2\cdot x)^2 &= \pu{5.5\cdot10^{-6}M^3}\\ &= \pu{0.0111M} \end{align}

Converting to $\pu{g//100 mL}$:

$$\pu{0.0111 M}\times \pu{74.093 g//mol}\times \pu{0.100 L} =\pu{0.0822 g//100mL} $$

This doesn't make any sense. According to the website I linked to earlier this is at about $\pu{90 ^\circ C}$:

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For the tabel above I found the sobulity on the fact sheet of the National Lime Association (pdf), and then I just calculated $K_\mathrm{sp}$/$K_\mathrm{o}$

  • On Wiki it also says the solubility product is 5.5*10^-6 – Hilmar Nov 8 at 15:29
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    One of your equations should be $$\dfrac{0.159 \pu{g//100 mL}}{\pu{74.093 g//mol}}\cdot \dfrac{1000\pu{mL}}{\pu{L}} =\pu{0.0215 mol//L}$$ – MaxW Nov 8 at 16:31
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    On the fact sheet it says 'Grams per 100 gms. sat. sol.'. So you need the density of the saturated solution to convert to amount concentration. I'd also be wary of a fact sheet that doesn't use proper units. – Martin - マーチン Nov 8 at 16:34
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    @MaxW, It is linked now. – Hilmar Nov 8 at 21:29

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