Answer: The solubility of this compound in pure water is 0.012 M
Explanation:
Solubility product is defined as the equilibrium constant in which a solid ionic compound is dissolved to produce its ions in solution. It is represented as [tex]K_{sp}[/tex]
The equation for the ionization of the is given as:
[tex]Ca(OH)_2\rightarrow Ca^{2+}+2OH^-[/tex]
By stoichiometry of the reaction:
1 mole of [tex]Ca(OH)_2[/tex] gives 1 mole of [tex]Ca^{2+}[/tex] and 2 mole of [tex]OH^-[/tex]
When the solubility of [tex]Ca(OH)_2[/tex] is S moles/liter, then the solubility of [tex]Ca^{2+}[/tex] will be S moles\liter and solubility of [tex]OH^-[/tex] will be 2S moles/liter.
[tex]K_{sp}=[Ca^{2+}][OH^{-}]^2[/tex]
[tex]6.5\times 10^{-6}=[S][2S]^2[/tex]
[tex]S=0.012M[/tex]
Thus solubility of this compound in pure water is 0.012 M
