When the reaction equation is:
Ca(OH)2 → Ca2+ + 2OH-
and when we assume that the final concentration of the calcium ion is X
and we assume [OH-] = 2 X + 0.1
and X (the molarity) = moles of Ca(OH)2 / volume L
also, we have the solubility product constant Ksp = 6.5 x 10^-6
when the Ksp expression = [Ca2+][OH]^2
by substitution:
6.5 x 10^-6 = X*(2X+0.1)^2
∴ X = 6.3 x 10^-4 M
∴ [Ca2+] = X = 6.3 x 10^-4 M
